3.671 \(\int (a+b \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=171 \[ -\frac{\left (4 a d \left (4 c^2+d^2\right )+3 b \left (c^3+4 c d^2\right )\right ) \cos (e+f x)}{6 f}-\frac{d \left (20 a c d+6 b c^2+9 b d^2\right ) \sin (e+f x) \cos (e+f x)}{24 f}+\frac{1}{8} x \left (8 a c^3+12 a c d^2+12 b c^2 d+3 b d^3\right )-\frac{(4 a d+3 b c) \cos (e+f x) (c+d \sin (e+f x))^2}{12 f}-\frac{b \cos (e+f x) (c+d \sin (e+f x))^3}{4 f} \]

[Out]

((8*a*c^3 + 12*b*c^2*d + 12*a*c*d^2 + 3*b*d^3)*x)/8 - ((4*a*d*(4*c^2 + d^2) + 3*b*(c^3 + 4*c*d^2))*Cos[e + f*x
])/(6*f) - (d*(6*b*c^2 + 20*a*c*d + 9*b*d^2)*Cos[e + f*x]*Sin[e + f*x])/(24*f) - ((3*b*c + 4*a*d)*Cos[e + f*x]
*(c + d*Sin[e + f*x])^2)/(12*f) - (b*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(4*f)

________________________________________________________________________________________

Rubi [A]  time = 0.211378, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2753, 2734} \[ -\frac{\left (4 a d \left (4 c^2+d^2\right )+3 b \left (c^3+4 c d^2\right )\right ) \cos (e+f x)}{6 f}-\frac{d \left (20 a c d+6 b c^2+9 b d^2\right ) \sin (e+f x) \cos (e+f x)}{24 f}+\frac{1}{8} x \left (8 a c^3+12 a c d^2+12 b c^2 d+3 b d^3\right )-\frac{(4 a d+3 b c) \cos (e+f x) (c+d \sin (e+f x))^2}{12 f}-\frac{b \cos (e+f x) (c+d \sin (e+f x))^3}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^3,x]

[Out]

((8*a*c^3 + 12*b*c^2*d + 12*a*c*d^2 + 3*b*d^3)*x)/8 - ((4*a*d*(4*c^2 + d^2) + 3*b*(c^3 + 4*c*d^2))*Cos[e + f*x
])/(6*f) - (d*(6*b*c^2 + 20*a*c*d + 9*b*d^2)*Cos[e + f*x]*Sin[e + f*x])/(24*f) - ((3*b*c + 4*a*d)*Cos[e + f*x]
*(c + d*Sin[e + f*x])^2)/(12*f) - (b*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(4*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))^3 \, dx &=-\frac{b \cos (e+f x) (c+d \sin (e+f x))^3}{4 f}+\frac{1}{4} \int (c+d \sin (e+f x))^2 (4 a c+3 b d+(3 b c+4 a d) \sin (e+f x)) \, dx\\ &=-\frac{(3 b c+4 a d) \cos (e+f x) (c+d \sin (e+f x))^2}{12 f}-\frac{b \cos (e+f x) (c+d \sin (e+f x))^3}{4 f}+\frac{1}{12} \int (c+d \sin (e+f x)) \left (12 a c^2+15 b c d+8 a d^2+\left (6 b c^2+20 a c d+9 b d^2\right ) \sin (e+f x)\right ) \, dx\\ &=\frac{1}{8} \left (8 a c^3+12 b c^2 d+12 a c d^2+3 b d^3\right ) x-\frac{\left (4 a d \left (4 c^2+d^2\right )+3 b \left (c^3+4 c d^2\right )\right ) \cos (e+f x)}{6 f}-\frac{d \left (6 b c^2+20 a c d+9 b d^2\right ) \cos (e+f x) \sin (e+f x)}{24 f}-\frac{(3 b c+4 a d) \cos (e+f x) (c+d \sin (e+f x))^2}{12 f}-\frac{b \cos (e+f x) (c+d \sin (e+f x))^3}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.669611, size = 143, normalized size = 0.84 \[ \frac{3 \left (4 (e+f x) \left (8 a c^3+12 a c d^2+12 b c^2 d+3 b d^3\right )-8 d \left (3 a c d+b \left (3 c^2+d^2\right )\right ) \sin (2 (e+f x))+b d^3 \sin (4 (e+f x))\right )-24 \left (3 a d \left (4 c^2+d^2\right )+b \left (4 c^3+9 c d^2\right )\right ) \cos (e+f x)+8 d^2 (a d+3 b c) \cos (3 (e+f x))}{96 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^3,x]

[Out]

(-24*(3*a*d*(4*c^2 + d^2) + b*(4*c^3 + 9*c*d^2))*Cos[e + f*x] + 8*d^2*(3*b*c + a*d)*Cos[3*(e + f*x)] + 3*(4*(8
*a*c^3 + 12*b*c^2*d + 12*a*c*d^2 + 3*b*d^3)*(e + f*x) - 8*d*(3*a*c*d + b*(3*c^2 + d^2))*Sin[2*(e + f*x)] + b*d
^3*Sin[4*(e + f*x)]))/(96*f)

________________________________________________________________________________________

Maple [A]  time = 0.029, size = 182, normalized size = 1.1 \begin{align*}{\frac{1}{f} \left ( a{c}^{3} \left ( fx+e \right ) -3\,{c}^{2}da\cos \left ( fx+e \right ) +3\,ac{d}^{2} \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -{\frac{{d}^{3}a \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}-{c}^{3}b\cos \left ( fx+e \right ) +3\,b{c}^{2}d \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -c{d}^{2}b \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) +{d}^{3}b \left ( -{\frac{\cos \left ( fx+e \right ) }{4} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\sin \left ( fx+e \right ) }{2}} \right ) }+{\frac{3\,fx}{8}}+{\frac{3\,e}{8}} \right ) \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))^3,x)

[Out]

1/f*(a*c^3*(f*x+e)-3*c^2*d*a*cos(f*x+e)+3*a*c*d^2*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-1/3*d^3*a*(2+sin(
f*x+e)^2)*cos(f*x+e)-c^3*b*cos(f*x+e)+3*b*c^2*d*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-c*d^2*b*(2+sin(f*x+
e)^2)*cos(f*x+e)+d^3*b*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e))

________________________________________________________________________________________

Maxima [A]  time = 1.11255, size = 236, normalized size = 1.38 \begin{align*} \frac{96 \,{\left (f x + e\right )} a c^{3} + 72 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} b c^{2} d + 72 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a c d^{2} + 96 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} b c d^{2} + 32 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a d^{3} + 3 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} b d^{3} - 96 \, b c^{3} \cos \left (f x + e\right ) - 288 \, a c^{2} d \cos \left (f x + e\right )}{96 \, 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))^3,x, algorithm="maxima")

[Out]

1/96*(96*(f*x + e)*a*c^3 + 72*(2*f*x + 2*e - sin(2*f*x + 2*e))*b*c^2*d + 72*(2*f*x + 2*e - sin(2*f*x + 2*e))*a
*c*d^2 + 96*(cos(f*x + e)^3 - 3*cos(f*x + e))*b*c*d^2 + 32*(cos(f*x + e)^3 - 3*cos(f*x + e))*a*d^3 + 3*(12*f*x
+ 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*b*d^3 - 96*b*c^3*cos(f*x + e) - 288*a*c^2*d*cos(f*x + e))/f

________________________________________________________________________________________

Fricas [A]  time = 1.65525, size = 340, normalized size = 1.99 \begin{align*} \frac{8 \,{\left (3 \, b c d^{2} + a d^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (8 \, a c^{3} + 12 \, b c^{2} d + 12 \, a c d^{2} + 3 \, b d^{3}\right )} f x - 24 \,{\left (b c^{3} + 3 \, a c^{2} d + 3 \, b c d^{2} + a d^{3}\right )} \cos \left (f x + e\right ) + 3 \,{\left (2 \, b d^{3} \cos \left (f x + e\right )^{3} -{\left (12 \, b c^{2} d + 12 \, a c d^{2} + 5 \, b d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, 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))^3,x, algorithm="fricas")

[Out]

1/24*(8*(3*b*c*d^2 + a*d^3)*cos(f*x + e)^3 + 3*(8*a*c^3 + 12*b*c^2*d + 12*a*c*d^2 + 3*b*d^3)*f*x - 24*(b*c^3 +
3*a*c^2*d + 3*b*c*d^2 + a*d^3)*cos(f*x + e) + 3*(2*b*d^3*cos(f*x + e)^3 - (12*b*c^2*d + 12*a*c*d^2 + 5*b*d^3)
*cos(f*x + e))*sin(f*x + e))/f

________________________________________________________________________________________

Sympy [A]  time = 1.7594, size = 386, normalized size = 2.26 \begin{align*} \begin{cases} a c^{3} x - \frac{3 a c^{2} d \cos{\left (e + f x \right )}}{f} + \frac{3 a c d^{2} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{3 a c d^{2} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{3 a c d^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{a d^{3} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{2 a d^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{b c^{3} \cos{\left (e + f x \right )}}{f} + \frac{3 b c^{2} d x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{3 b c^{2} d x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{3 b c^{2} d \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{3 b c d^{2} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{2 b c d^{2} \cos ^{3}{\left (e + f x \right )}}{f} + \frac{3 b d^{3} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac{3 b d^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{3 b d^{3} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac{5 b d^{3} \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} - \frac{3 b d^{3} \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} & \text{for}\: f \neq 0 \\x \left (a + b \sin{\left (e \right )}\right ) \left (c + d \sin{\left (e \right )}\right )^{3} & \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))**3,x)

[Out]

Piecewise((a*c**3*x - 3*a*c**2*d*cos(e + f*x)/f + 3*a*c*d**2*x*sin(e + f*x)**2/2 + 3*a*c*d**2*x*cos(e + f*x)**
2/2 - 3*a*c*d**2*sin(e + f*x)*cos(e + f*x)/(2*f) - a*d**3*sin(e + f*x)**2*cos(e + f*x)/f - 2*a*d**3*cos(e + f*
x)**3/(3*f) - b*c**3*cos(e + f*x)/f + 3*b*c**2*d*x*sin(e + f*x)**2/2 + 3*b*c**2*d*x*cos(e + f*x)**2/2 - 3*b*c*
*2*d*sin(e + f*x)*cos(e + f*x)/(2*f) - 3*b*c*d**2*sin(e + f*x)**2*cos(e + f*x)/f - 2*b*c*d**2*cos(e + f*x)**3/
f + 3*b*d**3*x*sin(e + f*x)**4/8 + 3*b*d**3*x*sin(e + f*x)**2*cos(e + f*x)**2/4 + 3*b*d**3*x*cos(e + f*x)**4/8
- 5*b*d**3*sin(e + f*x)**3*cos(e + f*x)/(8*f) - 3*b*d**3*sin(e + f*x)*cos(e + f*x)**3/(8*f), Ne(f, 0)), (x*(a
+ b*sin(e))*(c + d*sin(e))**3, True))

________________________________________________________________________________________

Giac [A]  time = 1.29152, size = 205, normalized size = 1.2 \begin{align*} \frac{b d^{3} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac{1}{8} \,{\left (8 \, a c^{3} + 12 \, b c^{2} d + 12 \, a c d^{2} + 3 \, b d^{3}\right )} x + \frac{{\left (3 \, b c d^{2} + a d^{3}\right )} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac{{\left (4 \, b c^{3} + 12 \, a c^{2} d + 9 \, b c d^{2} + 3 \, a d^{3}\right )} \cos \left (f x + e\right )}{4 \, f} - \frac{{\left (3 \, b c^{2} d + 3 \, a c d^{2} + b d^{3}\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))^3,x, algorithm="giac")

[Out]

1/32*b*d^3*sin(4*f*x + 4*e)/f + 1/8*(8*a*c^3 + 12*b*c^2*d + 12*a*c*d^2 + 3*b*d^3)*x + 1/12*(3*b*c*d^2 + a*d^3)
*cos(3*f*x + 3*e)/f - 1/4*(4*b*c^3 + 12*a*c^2*d + 9*b*c*d^2 + 3*a*d^3)*cos(f*x + e)/f - 1/4*(3*b*c^2*d + 3*a*c
*d^2 + b*d^3)*sin(2*f*x + 2*e)/f