Integrand size = 23, antiderivative size = 106 \[ \int (a+b \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\frac {1}{2} \left (2 b c d+a \left (2 c^2+d^2\right )\right ) x-\frac {2 \left (3 a c d+b \left (c^2+d^2\right )\right ) \cos (e+f x)}{3 f}-\frac {d (2 b c+3 a d) \cos (e+f x) \sin (e+f x)}{6 f}-\frac {b \cos (e+f x) (c+d \sin (e+f x))^2}{3 f} \] Output:
1/2*(2*b*c*d+a*(2*c^2+d^2))*x-2/3*(3*a*c*d+b*(c^2+d^2))*cos(f*x+e)/f-1/6*d *(3*a*d+2*b*c)*cos(f*x+e)*sin(f*x+e)/f-1/3*b*cos(f*x+e)*(c+d*sin(f*x+e))^2 /f
Time = 0.89 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.85 \[ \int (a+b \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\frac {6 \left (2 b c d+a \left (2 c^2+d^2\right )\right ) (e+f x)-3 \left (4 b c^2+8 a c d+3 b d^2\right ) \cos (e+f x)+b d^2 \cos (3 (e+f x))-3 d (2 b c+a d) \sin (2 (e+f x))}{12 f} \] Input:
Integrate[(a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^2,x]
Output:
(6*(2*b*c*d + a*(2*c^2 + d^2))*(e + f*x) - 3*(4*b*c^2 + 8*a*c*d + 3*b*d^2) *Cos[e + f*x] + b*d^2*Cos[3*(e + f*x)] - 3*d*(2*b*c + a*d)*Sin[2*(e + f*x) ])/(12*f)
Time = 0.36 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3232, 3042, 3213}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (a+b \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \sin (e+f x)) (c+d \sin (e+f x))^2dx\) |
\(\Big \downarrow \) 3232 |
\(\displaystyle \frac {1}{3} \int (c+d \sin (e+f x)) (3 a c+2 b d+(2 b c+3 a d) \sin (e+f x))dx-\frac {b \cos (e+f x) (c+d \sin (e+f x))^2}{3 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} \int (c+d \sin (e+f x)) (3 a c+2 b d+(2 b c+3 a d) \sin (e+f x))dx-\frac {b \cos (e+f x) (c+d \sin (e+f x))^2}{3 f}\) |
\(\Big \downarrow \) 3213 |
\(\displaystyle \frac {1}{3} \left (-\frac {2 \left (3 a c d+b \left (c^2+d^2\right )\right ) \cos (e+f x)}{f}+\frac {3}{2} x \left (a \left (2 c^2+d^2\right )+2 b c d\right )-\frac {d (3 a d+2 b c) \sin (e+f x) \cos (e+f x)}{2 f}\right )-\frac {b \cos (e+f x) (c+d \sin (e+f x))^2}{3 f}\) |
Input:
Int[(a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^2,x]
Output:
-1/3*(b*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/f + ((3*(2*b*c*d + a*(2*c^2 + d^2))*x)/2 - (2*(3*a*c*d + b*(c^2 + d^2))*Cos[e + f*x])/f - (d*(2*b*c + 3 *a*d)*Cos[e + f*x]*Sin[e + f*x])/(2*f))/3
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[1/(m + 1) Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[b* d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ [{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]
Time = 0.30 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.08
\[\frac {a \,c^{2} \left (f x +e \right )-2 a c d \cos \left (f x +e \right )+a \,d^{2} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-b \,c^{2} \cos \left (f x +e \right )+2 b c d \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {b \,d^{2} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}}{f}\]
Input:
int((a+b*sin(f*x+e))*(c+d*sin(f*x+e))^2,x)
Output:
1/f*(a*c^2*(f*x+e)-2*a*c*d*cos(f*x+e)+a*d^2*(-1/2*sin(f*x+e)*cos(f*x+e)+1/ 2*f*x+1/2*e)-b*c^2*cos(f*x+e)+2*b*c*d*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+ 1/2*e)-1/3*b*d^2*(2+sin(f*x+e)^2)*cos(f*x+e))
Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.85 \[ \int (a+b \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\frac {2 \, b d^{2} \cos \left (f x + e\right )^{3} + 3 \, {\left (2 \, a c^{2} + 2 \, b c d + a d^{2}\right )} f x - 3 \, {\left (2 \, b c d + a d^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 6 \, {\left (b c^{2} + 2 \, a c d + b d^{2}\right )} \cos \left (f x + e\right )}{6 \, f} \] Input:
integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e))^2,x, algorithm="fricas")
Output:
1/6*(2*b*d^2*cos(f*x + e)^3 + 3*(2*a*c^2 + 2*b*c*d + a*d^2)*f*x - 3*(2*b*c *d + a*d^2)*cos(f*x + e)*sin(f*x + e) - 6*(b*c^2 + 2*a*c*d + b*d^2)*cos(f* x + e))/f
Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (97) = 194\).
Time = 0.20 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.88 \[ \int (a+b \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\begin {cases} a c^{2} x - \frac {2 a c d \cos {\left (e + f x \right )}}{f} + \frac {a d^{2} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {a d^{2} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {a d^{2} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {b c^{2} \cos {\left (e + f x \right )}}{f} + b c d x \sin ^{2}{\left (e + f x \right )} + b c d x \cos ^{2}{\left (e + f x \right )} - \frac {b c d \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {b d^{2} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 b d^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin {\left (e \right )}\right ) \left (c + d \sin {\left (e \right )}\right )^{2} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e))**2,x)
Output:
Piecewise((a*c**2*x - 2*a*c*d*cos(e + f*x)/f + a*d**2*x*sin(e + f*x)**2/2 + a*d**2*x*cos(e + f*x)**2/2 - a*d**2*sin(e + f*x)*cos(e + f*x)/(2*f) - b* c**2*cos(e + f*x)/f + b*c*d*x*sin(e + f*x)**2 + b*c*d*x*cos(e + f*x)**2 - b*c*d*sin(e + f*x)*cos(e + f*x)/f - b*d**2*sin(e + f*x)**2*cos(e + f*x)/f - 2*b*d**2*cos(e + f*x)**3/(3*f), Ne(f, 0)), (x*(a + b*sin(e))*(c + d*sin( e))**2, True))
Time = 0.03 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.06 \[ \int (a+b \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\frac {12 \, {\left (f x + e\right )} a c^{2} + 6 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} b c d + 3 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a d^{2} + 4 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} b d^{2} - 12 \, b c^{2} \cos \left (f x + e\right ) - 24 \, a c d \cos \left (f x + e\right )}{12 \, f} \] Input:
integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e))^2,x, algorithm="maxima")
Output:
1/12*(12*(f*x + e)*a*c^2 + 6*(2*f*x + 2*e - sin(2*f*x + 2*e))*b*c*d + 3*(2 *f*x + 2*e - sin(2*f*x + 2*e))*a*d^2 + 4*(cos(f*x + e)^3 - 3*cos(f*x + e)) *b*d^2 - 12*b*c^2*cos(f*x + e) - 24*a*c*d*cos(f*x + e))/f
Time = 0.14 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.88 \[ \int (a+b \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\frac {b d^{2} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} + \frac {1}{2} \, {\left (2 \, a c^{2} + 2 \, b c d + a d^{2}\right )} x - \frac {{\left (4 \, b c^{2} + 8 \, a c d + 3 \, b d^{2}\right )} \cos \left (f x + e\right )}{4 \, f} - \frac {{\left (2 \, b c d + a d^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \] Input:
integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e))^2,x, algorithm="giac")
Output:
1/12*b*d^2*cos(3*f*x + 3*e)/f + 1/2*(2*a*c^2 + 2*b*c*d + a*d^2)*x - 1/4*(4 *b*c^2 + 8*a*c*d + 3*b*d^2)*cos(f*x + e)/f - 1/4*(2*b*c*d + a*d^2)*sin(2*f *x + 2*e)/f
Time = 17.69 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02 \[ \int (a+b \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=-\frac {\frac {3\,a\,d^2\,\sin \left (2\,e+2\,f\,x\right )}{2}-\frac {b\,d^2\,\cos \left (3\,e+3\,f\,x\right )}{2}+6\,b\,c^2\,\cos \left (e+f\,x\right )+\frac {9\,b\,d^2\,\cos \left (e+f\,x\right )}{2}+3\,b\,c\,d\,\sin \left (2\,e+2\,f\,x\right )-6\,a\,c^2\,f\,x-3\,a\,d^2\,f\,x+12\,a\,c\,d\,\cos \left (e+f\,x\right )-6\,b\,c\,d\,f\,x}{6\,f} \] Input:
int((a + b*sin(e + f*x))*(c + d*sin(e + f*x))^2,x)
Output:
-((3*a*d^2*sin(2*e + 2*f*x))/2 - (b*d^2*cos(3*e + 3*f*x))/2 + 6*b*c^2*cos( e + f*x) + (9*b*d^2*cos(e + f*x))/2 + 3*b*c*d*sin(2*e + 2*f*x) - 6*a*c^2*f *x - 3*a*d^2*f*x + 12*a*c*d*cos(e + f*x) - 6*b*c*d*f*x)/(6*f)
Time = 0.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.28 \[ \int (a+b \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx=\frac {-2 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b \,d^{2}-3 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,d^{2}-6 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b c d -12 \cos \left (f x +e \right ) a c d -6 \cos \left (f x +e \right ) b \,c^{2}-4 \cos \left (f x +e \right ) b \,d^{2}+6 a \,c^{2} f x +12 a c d +3 a \,d^{2} f x +6 b \,c^{2}+6 b c d f x +4 b \,d^{2}}{6 f} \] Input:
int((a+b*sin(f*x+e))*(c+d*sin(f*x+e))^2,x)
Output:
( - 2*cos(e + f*x)*sin(e + f*x)**2*b*d**2 - 3*cos(e + f*x)*sin(e + f*x)*a* d**2 - 6*cos(e + f*x)*sin(e + f*x)*b*c*d - 12*cos(e + f*x)*a*c*d - 6*cos(e + f*x)*b*c**2 - 4*cos(e + f*x)*b*d**2 + 6*a*c**2*f*x + 12*a*c*d + 3*a*d** 2*f*x + 6*b*c**2 + 6*b*c*d*f*x + 4*b*d**2)/(6*f)