3.687 \(\int (a+b \sin (e+f x))^3 (c+d \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=315 \[ -\frac{\left (4 a^2 b^2 \left (20 c^2+13 d^2\right )+30 a^3 b c d-3 a^4 d^2+120 a b^3 c d+4 b^4 \left (5 c^2+4 d^2\right )\right ) \cos (e+f x)}{30 b f}-\frac{\left (60 a^2 b c d-6 a^3 d^2+a b^2 \left (100 c^2+71 d^2\right )+90 b^3 c d\right ) \sin (e+f x) \cos (e+f x)}{120 f}+\frac{1}{8} x \left (24 a^2 b c d+4 a^3 \left (2 c^2+d^2\right )+3 a b^2 \left (4 c^2+3 d^2\right )+6 b^3 c d\right )-\frac{\left (3 a d (10 b c-a d)+4 b^2 \left (5 c^2+4 d^2\right )\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{60 b f}-\frac{d (10 b c-a d) \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac{d^2 \cos (e+f x) (a+b \sin (e+f x))^4}{5 b f} \]

[Out]

((24*a^2*b*c*d + 6*b^3*c*d + 4*a^3*(2*c^2 + d^2) + 3*a*b^2*(4*c^2 + 3*d^2))*x)/8 - ((30*a^3*b*c*d + 120*a*b^3*
c*d - 3*a^4*d^2 + 4*b^4*(5*c^2 + 4*d^2) + 4*a^2*b^2*(20*c^2 + 13*d^2))*Cos[e + f*x])/(30*b*f) - ((60*a^2*b*c*d
+ 90*b^3*c*d - 6*a^3*d^2 + a*b^2*(100*c^2 + 71*d^2))*Cos[e + f*x]*Sin[e + f*x])/(120*f) - ((3*a*d*(10*b*c - a
*d) + 4*b^2*(5*c^2 + 4*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^2)/(60*b*f) - (d*(10*b*c - a*d)*Cos[e + f*x]*(a
+ b*Sin[e + f*x])^3)/(20*b*f) - (d^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^4)/(5*b*f)

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Rubi [A]  time = 0.459742, antiderivative size = 315, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2791, 2753, 2734} \[ -\frac{\left (4 a^2 b^2 \left (20 c^2+13 d^2\right )+30 a^3 b c d-3 a^4 d^2+120 a b^3 c d+4 b^4 \left (5 c^2+4 d^2\right )\right ) \cos (e+f x)}{30 b f}-\frac{\left (60 a^2 b c d-6 a^3 d^2+a b^2 \left (100 c^2+71 d^2\right )+90 b^3 c d\right ) \sin (e+f x) \cos (e+f x)}{120 f}+\frac{1}{8} x \left (24 a^2 b c d+4 a^3 \left (2 c^2+d^2\right )+3 a b^2 \left (4 c^2+3 d^2\right )+6 b^3 c d\right )-\frac{\left (3 a d (10 b c-a d)+4 b^2 \left (5 c^2+4 d^2\right )\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{60 b f}-\frac{d (10 b c-a d) \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac{d^2 \cos (e+f x) (a+b \sin (e+f x))^4}{5 b f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^3*(c + d*Sin[e + f*x])^2,x]

[Out]

((24*a^2*b*c*d + 6*b^3*c*d + 4*a^3*(2*c^2 + d^2) + 3*a*b^2*(4*c^2 + 3*d^2))*x)/8 - ((30*a^3*b*c*d + 120*a*b^3*
c*d - 3*a^4*d^2 + 4*b^4*(5*c^2 + 4*d^2) + 4*a^2*b^2*(20*c^2 + 13*d^2))*Cos[e + f*x])/(30*b*f) - ((60*a^2*b*c*d
+ 90*b^3*c*d - 6*a^3*d^2 + a*b^2*(100*c^2 + 71*d^2))*Cos[e + f*x]*Sin[e + f*x])/(120*f) - ((3*a*d*(10*b*c - a
*d) + 4*b^2*(5*c^2 + 4*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^2)/(60*b*f) - (d*(10*b*c - a*d)*Cos[e + f*x]*(a
+ b*Sin[e + f*x])^3)/(20*b*f) - (d^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^4)/(5*b*f)

Rule 2791

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
(d^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x
])^m*Simp[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

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))^3 (c+d \sin (e+f x))^2 \, dx &=-\frac{d^2 \cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac{\int (a+b \sin (e+f x))^3 \left (b \left (5 c^2+4 d^2\right )+d (10 b c-a d) \sin (e+f x)\right ) \, dx}{5 b}\\ &=-\frac{d (10 b c-a d) \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac{d^2 \cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac{\int (a+b \sin (e+f x))^2 \left (b \left (20 a c^2+30 b c d+13 a d^2\right )+\left (3 a d (10 b c-a d)+4 b^2 \left (5 c^2+4 d^2\right )\right ) \sin (e+f x)\right ) \, dx}{20 b}\\ &=-\frac{\left (3 a d (10 b c-a d)+4 b^2 \left (5 c^2+4 d^2\right )\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{60 b f}-\frac{d (10 b c-a d) \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac{d^2 \cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac{\int (a+b \sin (e+f x)) \left (b \left (150 a b c d+8 b^2 \left (5 c^2+4 d^2\right )+a^2 \left (60 c^2+33 d^2\right )\right )+\left (60 a^2 b c d+90 b^3 c d-6 a^3 d^2+a b^2 \left (100 c^2+71 d^2\right )\right ) \sin (e+f x)\right ) \, dx}{60 b}\\ &=\frac{1}{8} \left (24 a^2 b c d+6 b^3 c d+4 a^3 \left (2 c^2+d^2\right )+3 a b^2 \left (4 c^2+3 d^2\right )\right ) x-\frac{\left (30 a^3 b c d+120 a b^3 c d-3 a^4 d^2+4 b^4 \left (5 c^2+4 d^2\right )+4 a^2 b^2 \left (20 c^2+13 d^2\right )\right ) \cos (e+f x)}{30 b f}-\frac{\left (60 a^2 b c d+90 b^3 c d-6 a^3 d^2+a b^2 \left (100 c^2+71 d^2\right )\right ) \cos (e+f x) \sin (e+f x)}{120 f}-\frac{\left (3 a d (10 b c-a d)+4 b^2 \left (5 c^2+4 d^2\right )\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{60 b f}-\frac{d (10 b c-a d) \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac{d^2 \cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}\\ \end{align*}

Mathematica [A]  time = 1.64838, size = 246, normalized size = 0.78 \[ \frac{15 \left (4 (e+f x) \left (24 a^2 b c d+4 a^3 \left (2 c^2+d^2\right )+3 a b^2 \left (4 c^2+3 d^2\right )+6 b^3 c d\right )-8 \left (6 a^2 b c d+a^3 d^2+3 a b^2 \left (c^2+d^2\right )+2 b^3 c d\right ) \sin (2 (e+f x))+b^2 d (3 a d+2 b c) \sin (4 (e+f x))\right )+10 b \left (12 a^2 d^2+24 a b c d+b^2 \left (4 c^2+5 d^2\right )\right ) \cos (3 (e+f x))-60 \left (6 a^2 b \left (4 c^2+3 d^2\right )+16 a^3 c d+36 a b^2 c d+b^3 \left (6 c^2+5 d^2\right )\right ) \cos (e+f x)-6 b^3 d^2 \cos (5 (e+f x))}{480 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x])^3*(c + d*Sin[e + f*x])^2,x]

[Out]

(-60*(16*a^3*c*d + 36*a*b^2*c*d + 6*a^2*b*(4*c^2 + 3*d^2) + b^3*(6*c^2 + 5*d^2))*Cos[e + f*x] + 10*b*(24*a*b*c
*d + 12*a^2*d^2 + b^2*(4*c^2 + 5*d^2))*Cos[3*(e + f*x)] - 6*b^3*d^2*Cos[5*(e + f*x)] + 15*(4*(24*a^2*b*c*d + 6
*b^3*c*d + 4*a^3*(2*c^2 + d^2) + 3*a*b^2*(4*c^2 + 3*d^2))*(e + f*x) - 8*(6*a^2*b*c*d + 2*b^3*c*d + a^3*d^2 + 3
*a*b^2*(c^2 + d^2))*Sin[2*(e + f*x)] + b^2*d*(2*b*c + 3*a*d)*Sin[4*(e + f*x)]))/(480*f)

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Maple [A]  time = 0.036, size = 325, normalized size = 1. \begin{align*}{\frac{1}{f} \left ({a}^{3}{c}^{2} \left ( fx+e \right ) -2\,{a}^{3}cd\cos \left ( fx+e \right ) +{a}^{3}{d}^{2} \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) -3\,{a}^{2}b{c}^{2}\cos \left ( fx+e \right ) +6\,{a}^{2}bcd \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -{a}^{2}b{d}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) +3\,a{b}^{2}{c}^{2} \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -2\,a{b}^{2}cd \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) +3\,a{b}^{2}{d}^{2} \left ( -1/4\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+3/2\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -{\frac{{b}^{3}{c}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+2\,{b}^{3}cd \left ( -1/4\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+3/2\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -{\frac{{b}^{3}{d}^{2}\cos \left ( fx+e \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e))^3*(c+d*sin(f*x+e))^2,x)

[Out]

1/f*(a^3*c^2*(f*x+e)-2*a^3*c*d*cos(f*x+e)+a^3*d^2*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-3*a^2*b*c^2*cos(f
*x+e)+6*a^2*b*c*d*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-a^2*b*d^2*(2+sin(f*x+e)^2)*cos(f*x+e)+3*a*b^2*c^2
*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-2*a*b^2*c*d*(2+sin(f*x+e)^2)*cos(f*x+e)+3*a*b^2*d^2*(-1/4*(sin(f*x
+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)-1/3*b^3*c^2*(2+sin(f*x+e)^2)*cos(f*x+e)+2*b^3*c*d*(-1/4*(sin(f
*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)-1/5*b^3*d^2*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e))

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Maxima [A]  time = 1.0595, size = 424, normalized size = 1.35 \begin{align*} \frac{480 \,{\left (f x + e\right )} a^{3} c^{2} + 360 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a b^{2} c^{2} + 160 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} b^{3} c^{2} + 720 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} b c d + 960 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a b^{2} c d + 30 \,{\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^{3} c d + 120 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} d^{2} + 480 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} b d^{2} + 45 \,{\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 b^{2} d^{2} - 32 \,{\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} b^{3} d^{2} - 1440 \, a^{2} b c^{2} \cos \left (f x + e\right ) - 960 \, a^{3} c d \cos \left (f x + e\right )}{480 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^3*(c+d*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

1/480*(480*(f*x + e)*a^3*c^2 + 360*(2*f*x + 2*e - sin(2*f*x + 2*e))*a*b^2*c^2 + 160*(cos(f*x + e)^3 - 3*cos(f*
x + e))*b^3*c^2 + 720*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^2*b*c*d + 960*(cos(f*x + e)^3 - 3*cos(f*x + e))*a*b^2
*c*d + 30*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*b^3*c*d + 120*(2*f*x + 2*e - sin(2*f*x + 2*e
))*a^3*d^2 + 480*(cos(f*x + e)^3 - 3*cos(f*x + e))*a^2*b*d^2 + 45*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*
f*x + 2*e))*a*b^2*d^2 - 32*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*b^3*d^2 - 1440*a^2*b*c^2*c
os(f*x + e) - 960*a^3*c*d*cos(f*x + e))/f

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Fricas [A]  time = 1.80173, size = 568, normalized size = 1.8 \begin{align*} -\frac{24 \, b^{3} d^{2} \cos \left (f x + e\right )^{5} - 40 \,{\left (b^{3} c^{2} + 6 \, a b^{2} c d +{\left (3 \, a^{2} b + 2 \, b^{3}\right )} d^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left (4 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} c^{2} + 6 \,{\left (4 \, a^{2} b + b^{3}\right )} c d +{\left (4 \, a^{3} + 9 \, a b^{2}\right )} d^{2}\right )} f x + 120 \,{\left ({\left (3 \, a^{2} b + b^{3}\right )} c^{2} + 2 \,{\left (a^{3} + 3 \, a b^{2}\right )} c d +{\left (3 \, a^{2} b + b^{3}\right )} d^{2}\right )} \cos \left (f x + e\right ) - 15 \,{\left (2 \,{\left (2 \, b^{3} c d + 3 \, a b^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} -{\left (12 \, a b^{2} c^{2} + 2 \,{\left (12 \, a^{2} b + 5 \, b^{3}\right )} c d +{\left (4 \, a^{3} + 15 \, a b^{2}\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{120 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^3*(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-1/120*(24*b^3*d^2*cos(f*x + e)^5 - 40*(b^3*c^2 + 6*a*b^2*c*d + (3*a^2*b + 2*b^3)*d^2)*cos(f*x + e)^3 - 15*(4*
(2*a^3 + 3*a*b^2)*c^2 + 6*(4*a^2*b + b^3)*c*d + (4*a^3 + 9*a*b^2)*d^2)*f*x + 120*((3*a^2*b + b^3)*c^2 + 2*(a^3
+ 3*a*b^2)*c*d + (3*a^2*b + b^3)*d^2)*cos(f*x + e) - 15*(2*(2*b^3*c*d + 3*a*b^2*d^2)*cos(f*x + e)^3 - (12*a*b
^2*c^2 + 2*(12*a^2*b + 5*b^3)*c*d + (4*a^3 + 15*a*b^2)*d^2)*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 4.12948, size = 729, normalized size = 2.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))**3*(c+d*sin(f*x+e))**2,x)

[Out]

Piecewise((a**3*c**2*x - 2*a**3*c*d*cos(e + f*x)/f + a**3*d**2*x*sin(e + f*x)**2/2 + a**3*d**2*x*cos(e + f*x)*
*2/2 - a**3*d**2*sin(e + f*x)*cos(e + f*x)/(2*f) - 3*a**2*b*c**2*cos(e + f*x)/f + 3*a**2*b*c*d*x*sin(e + f*x)*
*2 + 3*a**2*b*c*d*x*cos(e + f*x)**2 - 3*a**2*b*c*d*sin(e + f*x)*cos(e + f*x)/f - 3*a**2*b*d**2*sin(e + f*x)**2
*cos(e + f*x)/f - 2*a**2*b*d**2*cos(e + f*x)**3/f + 3*a*b**2*c**2*x*sin(e + f*x)**2/2 + 3*a*b**2*c**2*x*cos(e
+ f*x)**2/2 - 3*a*b**2*c**2*sin(e + f*x)*cos(e + f*x)/(2*f) - 6*a*b**2*c*d*sin(e + f*x)**2*cos(e + f*x)/f - 4*
a*b**2*c*d*cos(e + f*x)**3/f + 9*a*b**2*d**2*x*sin(e + f*x)**4/8 + 9*a*b**2*d**2*x*sin(e + f*x)**2*cos(e + f*x
)**2/4 + 9*a*b**2*d**2*x*cos(e + f*x)**4/8 - 15*a*b**2*d**2*sin(e + f*x)**3*cos(e + f*x)/(8*f) - 9*a*b**2*d**2
*sin(e + f*x)*cos(e + f*x)**3/(8*f) - b**3*c**2*sin(e + f*x)**2*cos(e + f*x)/f - 2*b**3*c**2*cos(e + f*x)**3/(
3*f) + 3*b**3*c*d*x*sin(e + f*x)**4/4 + 3*b**3*c*d*x*sin(e + f*x)**2*cos(e + f*x)**2/2 + 3*b**3*c*d*x*cos(e +
f*x)**4/4 - 5*b**3*c*d*sin(e + f*x)**3*cos(e + f*x)/(4*f) - 3*b**3*c*d*sin(e + f*x)*cos(e + f*x)**3/(4*f) - b*
*3*d**2*sin(e + f*x)**4*cos(e + f*x)/f - 4*b**3*d**2*sin(e + f*x)**2*cos(e + f*x)**3/(3*f) - 8*b**3*d**2*cos(e
+ f*x)**5/(15*f), Ne(f, 0)), (x*(a + b*sin(e))**3*(c + d*sin(e))**2, True))

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Giac [A]  time = 1.38243, size = 370, normalized size = 1.17 \begin{align*} -\frac{b^{3} d^{2} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac{1}{8} \,{\left (8 \, a^{3} c^{2} + 12 \, a b^{2} c^{2} + 24 \, a^{2} b c d + 6 \, b^{3} c d + 4 \, a^{3} d^{2} + 9 \, a b^{2} d^{2}\right )} x + \frac{{\left (4 \, b^{3} c^{2} + 24 \, a b^{2} c d + 12 \, a^{2} b d^{2} + 5 \, b^{3} d^{2}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac{{\left (24 \, a^{2} b c^{2} + 6 \, b^{3} c^{2} + 16 \, a^{3} c d + 36 \, a b^{2} c d + 18 \, a^{2} b d^{2} + 5 \, b^{3} d^{2}\right )} \cos \left (f x + e\right )}{8 \, f} + \frac{{\left (2 \, b^{3} c d + 3 \, a b^{2} d^{2}\right )} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} - \frac{{\left (3 \, a b^{2} c^{2} + 6 \, a^{2} b c d + 2 \, b^{3} c d + a^{3} d^{2} + 3 \, a b^{2} 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))^3*(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-1/80*b^3*d^2*cos(5*f*x + 5*e)/f + 1/8*(8*a^3*c^2 + 12*a*b^2*c^2 + 24*a^2*b*c*d + 6*b^3*c*d + 4*a^3*d^2 + 9*a*
b^2*d^2)*x + 1/48*(4*b^3*c^2 + 24*a*b^2*c*d + 12*a^2*b*d^2 + 5*b^3*d^2)*cos(3*f*x + 3*e)/f - 1/8*(24*a^2*b*c^2
+ 6*b^3*c^2 + 16*a^3*c*d + 36*a*b^2*c*d + 18*a^2*b*d^2 + 5*b^3*d^2)*cos(f*x + e)/f + 1/32*(2*b^3*c*d + 3*a*b^
2*d^2)*sin(4*f*x + 4*e)/f - 1/4*(3*a*b^2*c^2 + 6*a^2*b*c*d + 2*b^3*c*d + a^3*d^2 + 3*a*b^2*d^2)*sin(2*f*x + 2*
e)/f