This is a classical control problem. You want to create a feedback loop that takes the divergence from optimal position and applies the appropriate torque to nudge it back into position.
btQuaternion targetOrientation = // whatever you need
btQuaternion currentOrientation = myObject->getOrientation();
Getting the delta orientation is quite simple, you need to find the "difference" between two the quaternions target and current orientation.
btQuaternion deltaOrientation = targetOrientation * currentOrientation.inverse();
Unfortunately this quaterion does not say much about any divergence that we can use.
But if you convert this delta orientation into euler angles.
btVector3 deltaEuler = QuaternionToEulerXYZ(deltaOrientation );
You basically get the scaled inverse of the torque you want to apply. Now you "just" need to find an appropriate amount to ease it in.
btVector torqueToApply = control(-deltaEuler);
The simple solution is to multiply it with a factor. You may need to play around with the factor until you find a good value.
// simple proportional controller
btVector control(btVector in)
// play around with the factor until you find a matching one
static Kp = 0.5;
return in * Kp;
You may also want to look into a PID-Controller to get a smoother more responsive easing.