3.672 \(\int (a+b \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=106 \[ -\frac{2 \left (3 a c d+b \left (c^2+d^2\right )\right ) \cos (e+f x)}{3 f}+\frac{1}{2} x \left (a \left (2 c^2+d^2\right )+2 b c d\right )-\frac{d (3 a d+2 b c) \sin (e+f x) \cos (e+f x)}{6 f}-\frac{b \cos (e+f x) (c+d \sin (e+f x))^2}{3 f} \]

[Out]

((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])/(3*f) - (d*(2*b*c + 3*a*d)*Cos[
e + f*x]*Sin[e + f*x])/(6*f) - (b*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(3*f)

________________________________________________________________________________________

Rubi [A]  time = 0.100707, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2753, 2734} \[ -\frac{2 \left (3 a c d+b \left (c^2+d^2\right )\right ) \cos (e+f x)}{3 f}+\frac{1}{2} x \left (a \left (2 c^2+d^2\right )+2 b c d\right )-\frac{d (3 a d+2 b c) \sin (e+f x) \cos (e+f x)}{6 f}-\frac{b \cos (e+f x) (c+d \sin (e+f x))^2}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^2,x]

[Out]

((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])/(3*f) - (d*(2*b*c + 3*a*d)*Cos[
e + f*x]*Sin[e + f*x])/(6*f) - (b*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(3*f)

Rule 2753

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] + Dist[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]

Rule 2734

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)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

\begin{align*} \int (a+b \sin (e+f x)) (c+d \sin (e+f x))^2 \, dx &=-\frac{b \cos (e+f x) (c+d \sin (e+f x))^2}{3 f}+\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{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}\\ \end{align*}

Mathematica [A]  time = 0.297141, size = 90, normalized size = 0.85 \[ \frac{6 (e+f x) \left (a \left (2 c^2+d^2\right )+2 b c d\right )-3 \left (8 a c d+4 b c^2+3 b d^2\right ) \cos (e+f x)-3 d (a d+2 b c) \sin (2 (e+f x))+b d^2 \cos (3 (e+f x))}{12 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^2,x]

[Out]

(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)

________________________________________________________________________________________

Maple [A]  time = 0.026, size = 115, normalized size = 1.1 \begin{align*}{\frac{1}{f} \left ( a{c}^{2} \left ( fx+e \right ) -2\,acd\cos \left ( fx+e \right ) +a{d}^{2} \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) -{c}^{2}b\cos \left ( fx+e \right ) +2\,bcd \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -{\frac{{d}^{2}b \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e))*(c+d*sin(f*x+e))^2,x)

[Out]

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)-c^2*b*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*d^2*b*(2+sin(f*x+e)^2)*cos(f*x+e))

________________________________________________________________________________________

Maxima [A]  time = 1.14654, size = 151, normalized size = 1.42 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.62584, size = 215, normalized size = 2.03 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.790579, size = 199, normalized size = 1.88 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e))**2,x)

[Out]

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))

________________________________________________________________________________________

Giac [A]  time = 1.34707, size = 130, normalized size = 1.23 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

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