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

Optimal. Leaf size=162 \[ -\frac{a \left (16 c^2 d+3 c^3+12 c d^2+4 d^3\right ) \cos (e+f x)}{6 f}-\frac{a d \left (6 c^2+20 c d+9 d^2\right ) \sin (e+f x) \cos (e+f x)}{24 f}+\frac{1}{8} a x \left (12 c^2 d+8 c^3+12 c d^2+3 d^3\right )-\frac{a \cos (e+f x) (c+d \sin (e+f x))^3}{4 f}-\frac{a (3 c+4 d) \cos (e+f x) (c+d \sin (e+f x))^2}{12 f} \]

[Out]

(a*(8*c^3 + 12*c^2*d + 12*c*d^2 + 3*d^3)*x)/8 - (a*(3*c^3 + 16*c^2*d + 12*c*d^2 + 4*d^3)*Cos[e + f*x])/(6*f) -
(a*d*(6*c^2 + 20*c*d + 9*d^2)*Cos[e + f*x]*Sin[e + f*x])/(24*f) - (a*(3*c + 4*d)*Cos[e + f*x]*(c + d*Sin[e +
f*x])^2)/(12*f) - (a*Cos[e + f*x]*(c + d*Sin[e + f*x])^3)/(4*f)

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Rubi [A]  time = 0.189431, antiderivative size = 162, 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} \[ -\frac{a \left (16 c^2 d+3 c^3+12 c d^2+4 d^3\right ) \cos (e+f x)}{6 f}-\frac{a d \left (6 c^2+20 c d+9 d^2\right ) \sin (e+f x) \cos (e+f x)}{24 f}+\frac{1}{8} a x \left (12 c^2 d+8 c^3+12 c d^2+3 d^3\right )-\frac{a \cos (e+f x) (c+d \sin (e+f x))^3}{4 f}-\frac{a (3 c+4 d) \cos (e+f x) (c+d \sin (e+f x))^2}{12 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(8*c^3 + 12*c^2*d + 12*c*d^2 + 3*d^3)*x)/8 - (a*(3*c^3 + 16*c^2*d + 12*c*d^2 + 4*d^3)*Cos[e + f*x])/(6*f) -
(a*d*(6*c^2 + 20*c*d + 9*d^2)*Cos[e + f*x]*Sin[e + f*x])/(24*f) - (a*(3*c + 4*d)*Cos[e + f*x]*(c + d*Sin[e +
f*x])^2)/(12*f) - (a*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+a \sin (e+f x)) (c+d \sin (e+f x))^3 \, dx &=-\frac{a \cos (e+f x) (c+d \sin (e+f x))^3}{4 f}+\frac{1}{4} \int (c+d \sin (e+f x))^2 (a (4 c+3 d)+a (3 c+4 d) \sin (e+f x)) \, dx\\ &=-\frac{a (3 c+4 d) \cos (e+f x) (c+d \sin (e+f x))^2}{12 f}-\frac{a \cos (e+f x) (c+d \sin (e+f x))^3}{4 f}+\frac{1}{12} \int (c+d \sin (e+f x)) \left (a \left (12 c^2+15 c d+8 d^2\right )+a \left (6 c^2+20 c d+9 d^2\right ) \sin (e+f x)\right ) \, dx\\ &=\frac{1}{8} a \left (8 c^3+12 c^2 d+12 c d^2+3 d^3\right ) x-\frac{a \left (3 c^3+16 c^2 d+12 c d^2+4 d^3\right ) \cos (e+f x)}{6 f}-\frac{a d \left (6 c^2+20 c d+9 d^2\right ) \cos (e+f x) \sin (e+f x)}{24 f}-\frac{a (3 c+4 d) \cos (e+f x) (c+d \sin (e+f x))^2}{12 f}-\frac{a \cos (e+f x) (c+d \sin (e+f x))^3}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.805007, size = 124, normalized size = 0.77 \[ \frac{a \left (3 \left (-8 d \left (3 c^2+3 c d+d^2\right ) \sin (2 (e+f x))+4 f x \left (12 c^2 d+8 c^3+12 c d^2+3 d^3\right )+d^3 \sin (4 (e+f x))\right )-24 \left (12 c^2 d+4 c^3+9 c d^2+3 d^3\right ) \cos (e+f x)+8 d^2 (3 c+d) \cos (3 (e+f x))\right )}{96 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-24*(4*c^3 + 12*c^2*d + 9*c*d^2 + 3*d^3)*Cos[e + f*x] + 8*d^2*(3*c + d)*Cos[3*(e + f*x)] + 3*(4*(8*c^3 + 1
2*c^2*d + 12*c*d^2 + 3*d^3)*f*x - 8*d*(3*c^2 + 3*c*d + d^2)*Sin[2*(e + f*x)] + d^3*Sin[4*(e + f*x)])))/(96*f)

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Maple [A]  time = 0.039, size = 182, normalized size = 1.1 \begin{align*}{\frac{1}{f} \left ( -a{c}^{3}\cos \left ( fx+e \right ) +3\,a{c}^{2}d \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -ac{d}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) +a{d}^{3} \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 ) +a{c}^{3} \left ( fx+e \right ) -3\,a{c}^{2}d\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{a{d}^{3} \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+a*sin(f*x+e))*(c+d*sin(f*x+e))^3,x)

[Out]

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

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Maxima [A]  time = 1.17298, size = 236, normalized size = 1.46 \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 )} a c^{2} d + 96 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a c d^{2} + 72 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a 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 )} a d^{3} - 96 \, a 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+a*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))*a*c^2*d + 96*(cos(f*x + e)^3 - 3*cos(f*x + e))*
a*c*d^2 + 72*(2*f*x + 2*e - sin(2*f*x + 2*e))*a*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))*a*d^3 - 96*a*c^3*cos(f*x + e) - 288*a*c^2*d*cos(f*x + e))/f

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Fricas [A]  time = 1.13413, size = 340, normalized size = 2.1 \begin{align*} \frac{8 \,{\left (3 \, a c d^{2} + a d^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (8 \, a c^{3} + 12 \, a c^{2} d + 12 \, a c d^{2} + 3 \, a d^{3}\right )} f x - 24 \,{\left (a c^{3} + 3 \, a c^{2} d + 3 \, a c d^{2} + a d^{3}\right )} \cos \left (f x + e\right ) + 3 \,{\left (2 \, a d^{3} \cos \left (f x + e\right )^{3} -{\left (12 \, a c^{2} d + 12 \, a c d^{2} + 5 \, a 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+a*sin(f*x+e))*(c+d*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.81782, size = 386, normalized size = 2.38 \begin{align*} \begin{cases} a c^{3} x - \frac{a c^{3} \cos{\left (e + f x \right )}}{f} + \frac{3 a c^{2} d x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{3 a c^{2} d x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{3 a c^{2} d \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \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 ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{3 a c d^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{2 a c d^{2} \cos ^{3}{\left (e + f x \right )}}{f} + \frac{3 a d^{3} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac{3 a d^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{3 a d^{3} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac{5 a d^{3} \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} - \frac{a d^{3} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{3 a d^{3} \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac{2 a d^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} & \text{for}\: f \neq 0 \\x \left (c + d \sin{\left (e \right )}\right )^{3} \left (a \sin{\left (e \right )} + a\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(c+d*sin(f*x+e))**3,x)

[Out]

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

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Giac [A]  time = 1.23924, size = 258, normalized size = 1.59 \begin{align*} \frac{a c d^{2} \cos \left (3 \, f x + 3 \, e\right )}{4 \, f} + \frac{a d^{3} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} + \frac{a d^{3} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} - \frac{3 \, a c d^{2} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} + \frac{1}{2} \,{\left (2 \, a c^{3} + 3 \, a c d^{2}\right )} x + \frac{3}{8} \,{\left (4 \, a c^{2} d + a d^{3}\right )} x - \frac{{\left (4 \, a c^{3} + 9 \, a c d^{2}\right )} \cos \left (f x + e\right )}{4 \, f} - \frac{3 \,{\left (4 \, a c^{2} d + a d^{3}\right )} \cos \left (f x + e\right )}{4 \, f} - \frac{{\left (3 \, a c^{2} d + a 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+a*sin(f*x+e))*(c+d*sin(f*x+e))^3,x, algorithm="giac")

[Out]

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