\(\int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [229]

Optimal result

Integrand size = 28, antiderivative size = 592 \[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {f^2 x}{4 b d^2}+\frac {a^2 (e+f x)^3}{3 b^3 f}+\frac {(e+f x)^3}{6 b f}-\frac {2 a f^2 \cos (c+d x)}{b^2 d^3}+\frac {a (e+f x)^2 \cos (c+d x)}{b^2 d}+\frac {i a^3 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}-\frac {i a^3 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}+\frac {2 a^3 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^2}-\frac {2 a^3 f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^2}+\frac {2 i a^3 f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^3}-\frac {2 i a^3 f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^3}-\frac {2 a f (e+f x) \sin (c+d x)}{b^2 d^2}+\frac {f^2 \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac {(e+f x)^2 \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {f (e+f x) \sin ^2(c+d x)}{2 b d^2} \] Output:

-1/4*f^2*x/b/d^2+1/3*a^2*(f*x+e)^3/b^3/f+1/6*(f*x+e)^3/b/f-2*a*f^2*cos(d*x 
+c)/b^2/d^3+a*(f*x+e)^2*cos(d*x+c)/b^2/d+I*a^3*(f*x+e)^2*ln(1-I*b*exp(I*(d 
*x+c))/(a-(a^2-b^2)^(1/2)))/b^3/(a^2-b^2)^(1/2)/d-I*a^3*(f*x+e)^2*ln(1-I*b 
*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/(a^2-b^2)^(1/2)/d+2*a^3*f*(f*x+e) 
*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b^3/(a^2-b^2)^(1/2)/d^2 
-2*a^3*f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/(a^ 
2-b^2)^(1/2)/d^2+2*I*a^3*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/ 
2)))/b^3/(a^2-b^2)^(1/2)/d^3-2*I*a^3*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a+( 
a^2-b^2)^(1/2)))/b^3/(a^2-b^2)^(1/2)/d^3-2*a*f*(f*x+e)*sin(d*x+c)/b^2/d^2+ 
1/4*f^2*cos(d*x+c)*sin(d*x+c)/b/d^3-1/2*(f*x+e)^2*cos(d*x+c)*sin(d*x+c)/b/ 
d+1/2*f*(f*x+e)*sin(d*x+c)^2/b/d^2
 

Mathematica [A] (warning: unable to verify)

Time = 5.68 (sec) , antiderivative size = 1166, normalized size of antiderivative = 1.97 \[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(24*a^2*Sqrt[-(a^2 - b^2)^2]*d^3*e^2*x + 12*b^2*Sqrt[-(-a^2 + b^2)^2]*d^3* 
e^2*x + 24*a^2*Sqrt[-(a^2 - b^2)^2]*d^3*e*f*x^2 + 12*b^2*Sqrt[-(-a^2 + b^2 
)^2]*d^3*e*f*x^2 + 8*a^2*Sqrt[-(a^2 - b^2)^2]*d^3*f^2*x^3 + 4*b^2*Sqrt[-(- 
a^2 + b^2)^2]*d^3*f^2*x^3 - 48*a^3*Sqrt[-a^2 + b^2]*d^2*e^2*ArcTan[(I*a + 
b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]] + 24*a*b*Sqrt[-(a^2 - b^2)^2]*d^2*e^2* 
Cos[c + d*x] - 48*a*b*Sqrt[-(a^2 - b^2)^2]*f^2*Cos[c + d*x] + 48*a*b*Sqrt[ 
-(a^2 - b^2)^2]*d^2*e*f*x*Cos[c + d*x] + 24*a*b*Sqrt[-(a^2 - b^2)^2]*d^2*f 
^2*x^2*Cos[c + d*x] - 6*b^2*Sqrt[-(a^2 - b^2)^2]*d*e*f*Cos[2*(c + d*x)] - 
6*b^2*Sqrt[-(a^2 - b^2)^2]*d*f^2*x*Cos[2*(c + d*x)] - 48*a^3*Sqrt[a^2 - b^ 
2]*d^2*e*f*x*Log[1 - (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - 24 
*a^3*Sqrt[a^2 - b^2]*d^2*f^2*x^2*Log[1 - (b*E^(I*(c + d*x)))/((-I)*a + Sqr 
t[-a^2 + b^2])] + 48*a^3*Sqrt[a^2 - b^2]*d^2*e*f*x*Log[1 + (b*E^(I*(c + d* 
x)))/(I*a + Sqrt[-a^2 + b^2])] + 24*a^3*Sqrt[a^2 - b^2]*d^2*f^2*x^2*Log[1 
+ (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])] + (48*I)*a^3*Sqrt[a^2 - b^ 
2]*d*f*(e + f*x)*PolyLog[2, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2] 
)] - (48*I)*a^3*Sqrt[a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, -((b*E^(I*(c + d* 
x)))/(I*a + Sqrt[-a^2 + b^2]))] - 48*a^3*Sqrt[a^2 - b^2]*f^2*PolyLog[3, (b 
*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + 48*a^3*Sqrt[a^2 - b^2]*f^ 
2*PolyLog[3, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))] - 48*a*b*Sqr 
t[-(a^2 - b^2)^2]*d*e*f*Sin[c + d*x] - 48*a*b*Sqrt[-(a^2 - b^2)^2]*d*f^...
 

Rubi [A] (verified)

Time = 3.18 (sec) , antiderivative size = 540, normalized size of antiderivative = 0.91, number of steps used = 26, number of rules used = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.893, Rules used = {5026, 3042, 3792, 17, 3042, 3115, 24, 5026, 3042, 3777, 3042, 3777, 25, 3042, 3118, 5026, 17, 3042, 3804, 2694, 27, 2620, 3011, 2720, 7143}

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.

\(\displaystyle \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

\(\Big \downarrow \) 5026

\(\displaystyle \frac {\int (e+f x)^2 \sin ^2(c+d x)dx}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (e+f x)^2 \sin (c+d x)^2dx}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\)

\(\Big \downarrow \) 3792

\(\displaystyle \frac {-\frac {f^2 \int \sin ^2(c+d x)dx}{2 d^2}+\frac {1}{2} \int (e+f x)^2dx+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\)

\(\Big \downarrow \) 17

\(\displaystyle \frac {-\frac {f^2 \int \sin ^2(c+d x)dx}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {f^2 \int \sin (c+d x)^2dx}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {-\frac {f^2 \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+b \sin (c+d x)}dx}{b}\)

\(\Big \downarrow \) 5026

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\int (e+f x)^2 \sin (c+d x)dx}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\int (e+f x)^2 \sin (c+d x)dx}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 3777

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \int (e+f x) \cos (c+d x)dx}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \int (e+f x) \sin \left (c+d x+\frac {\pi }{2}\right )dx}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 3777

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \int -\sin (c+d x)dx}{d}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {(e+f x) \sin (c+d x)}{d}-\frac {f \int \sin (c+d x)dx}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {(e+f x) \sin (c+d x)}{d}-\frac {f \int \sin (c+d x)dx}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 3118

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \int \frac {(e+f x)^2 \sin (c+d x)}{a+b \sin (c+d x)}dx}{b}\right )}{b}\)

\(\Big \downarrow \) 5026

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {\int (e+f x)^2dx}{b}-\frac {a \int \frac {(e+f x)^2}{a+b \sin (c+d x)}dx}{b}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 17

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {a \int \frac {(e+f x)^2}{a+b \sin (c+d x)}dx}{b}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {a \int \frac {(e+f x)^2}{a+b \sin (c+d x)}dx}{b}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 3804

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \int \frac {e^{i (c+d x)} (e+f x)^2}{2 e^{i (c+d x)} a-i b e^{2 i (c+d x)}+i b}dx}{b}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 2694

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{2 \left (a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{2 \left (a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}\right )}{b}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\right )}{b}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}}{b}-\frac {a \left (\frac {\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}}{b}-\frac {a \left (\frac {(e+f x)^3}{3 b f}-\frac {2 a \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b}\right )}{b}\right )}{b}\)

Input:

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

Output:

((e + f*x)^3/(6*f) - ((e + f*x)^2*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (f*(e 
 + f*x)*Sin[c + d*x]^2)/(2*d^2) - (f^2*(x/2 - (Cos[c + d*x]*Sin[c + d*x])/ 
(2*d)))/(2*d^2))/b - (a*(-((a*((e + f*x)^3/(3*b*f) - (2*a*(((-1/2*I)*b*((( 
e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d) - ( 
2*f*((I*(e + f*x)*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])]) 
/d - (f*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/d^2))/(b* 
d)))/Sqrt[a^2 - b^2] + ((I/2)*b*(((e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)) 
)/(a + Sqrt[a^2 - b^2])])/(b*d) - (2*f*((I*(e + f*x)*PolyLog[2, (I*b*E^(I* 
(c + d*x)))/(a + Sqrt[a^2 - b^2])])/d - (f*PolyLog[3, (I*b*E^(I*(c + d*x)) 
)/(a + Sqrt[a^2 - b^2])])/d^2))/(b*d)))/Sqrt[a^2 - b^2]))/b))/b) + (-(((e 
+ f*x)^2*Cos[c + d*x])/d) + (2*f*((f*Cos[c + d*x])/d^2 + ((e + f*x)*Sin[c 
+ d*x])/d))/d)/b))/b
 

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2694

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.) 
*(F_)^(v_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[2*(c/q)   Int[(f + g*x) 
^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[ 
v, 2*u] && LinearQ[u, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3118

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ 
[{c, d}, x]
 

rule 3777

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( 
-(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*C 
os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 3792

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo 
l] :> Simp[d*m*(c + d*x)^(m - 1)*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Sim 
p[b*(c + d*x)^m*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^ 
2*((n - 1)/n)   Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[d^2 
*m*((m - 1)/(f^2*n^2))   Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x]) 
/; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]
 

rule 3804

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[2   Int[(c + d*x)^m*(E^(I*(e + f*x))/(I*b + 2*a*E^(I*(e + f*x 
)) - I*b*E^(2*I*(e + f*x)))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ 
[a^2 - b^2, 0] && IGtQ[m, 0]
 

rule 5026

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

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [F]

\[\int \frac {\left (f x +e \right )^{2} \sin \left (d x +c \right )^{3}}{a +b \sin \left (d x +c \right )}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2050 vs. \(2 (522) = 1044\).

Time = 0.29 (sec) , antiderivative size = 2050, normalized size of antiderivative = 3.46 \[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/12*(2*(2*a^4 - a^2*b^2 - b^4)*d^3*f^2*x^3 + 6*(2*a^4 - a^2*b^2 - b^4)*d^ 
3*e*f*x^2 - 12*a^3*b*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(d*x + 
 c) + a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^ 
2)/b^2))/b) + 12*a^3*b*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(d*x 
 + c) + a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - 
b^2)/b^2))/b) - 12*a^3*b*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(-I*a*cos( 
d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 
 - b^2)/b^2))/b) + 12*a^3*b*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(-I*a*c 
os(d*x + c) + a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-( 
a^2 - b^2)/b^2))/b) - 6*((a^2*b^2 - b^4)*d*f^2*x + (a^2*b^2 - b^4)*d*e*f)* 
cos(d*x + c)^2 + 12*(-I*a^3*b*d*f^2*x - I*a^3*b*d*e*f)*sqrt(-(a^2 - b^2)/b 
^2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d 
*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 12*(I*a^3*b*d*f^2*x + I*a^3* 
b*d*e*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) - 
 (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 
12*(I*a^3*b*d*f^2*x + I*a^3*b*d*e*f)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*co 
s(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a 
^2 - b^2)/b^2) - b)/b + 1) + 12*(-I*a^3*b*d*f^2*x - I*a^3*b*d*e*f)*sqrt(-( 
a^2 - b^2)/b^2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c 
) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 6*(a^3*b*d^2...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{2} \sin \left (d x + c\right )^{3}}{b \sin \left (d x + c\right ) + a} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Hanged} \] Input:

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\cos \left (d x +c \right )^{2} a^{2} b^{2} d^{2} e f \,x^{2}+\sin \left (d x +c \right )^{2} a^{2} b^{2} d^{2} e f \,x^{2}-2 \left (\int \frac {\sin \left (d x +c \right )^{3} x^{2}}{\sin \left (d x +c \right ) b +a}d x \right ) b^{5} d^{2} f^{2}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2} d e f x +\cos \left (d x +c \right )^{2} b^{4} e f -b^{4} c d \,e^{2}-b^{4} d^{2} e^{2} x +2 a^{4} c d \,e^{2}+2 a^{4} d^{2} e^{2} x +2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4} d e f x +4 \cos \left (d x +c \right ) a^{3} b d e f x -4 \cos \left (d x +c \right ) a \,b^{3} d e f x -\sin \left (d x +c \right )^{2} b^{4} d^{2} e f \,x^{2}-a^{2} b^{2} c d \,e^{2}-a^{2} b^{2} d^{2} e^{2} x +4 \left (\int \frac {\sin \left (d x +c \right ) x}{\sin \left (d x +c \right ) b +a}d x \right ) a^{4} b \,d^{2} e f -4 \left (\int \frac {\sin \left (d x +c \right ) x}{\sin \left (d x +c \right ) b +a}d x \right ) a^{2} b^{3} d^{2} e f -\cos \left (d x +c \right )^{2} a^{2} b^{2} e f -\cos \left (d x +c \right )^{2} b^{4} d^{2} e f \,x^{2}+\cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4} d \,e^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2} d \,e^{2}-4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{3} d \,e^{2}+2 \cos \left (d x +c \right ) a^{3} b d \,e^{2}-2 \cos \left (d x +c \right ) a \,b^{3} d \,e^{2}-4 \sin \left (d x +c \right ) a^{3} b e f +4 \sin \left (d x +c \right ) a \,b^{3} e f +2 \left (\int \frac {\sin \left (d x +c \right )^{3} x^{2}}{\sin \left (d x +c \right ) b +a}d x \right ) a^{2} b^{3} d^{2} f^{2}}{2 b^{3} d^{2} \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

( - 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a 
**3*d*e**2 + cos(c + d*x)**2*a**2*b**2*d**2*e*f*x**2 - cos(c + d*x)**2*a** 
2*b**2*e*f - cos(c + d*x)**2*b**4*d**2*e*f*x**2 + cos(c + d*x)**2*b**4*e*f 
 - cos(c + d*x)*sin(c + d*x)*a**2*b**2*d*e**2 - 2*cos(c + d*x)*sin(c + d*x 
)*a**2*b**2*d*e*f*x + cos(c + d*x)*sin(c + d*x)*b**4*d*e**2 + 2*cos(c + d* 
x)*sin(c + d*x)*b**4*d*e*f*x + 2*cos(c + d*x)*a**3*b*d*e**2 + 4*cos(c + d* 
x)*a**3*b*d*e*f*x - 2*cos(c + d*x)*a*b**3*d*e**2 - 4*cos(c + d*x)*a*b**3*d 
*e*f*x + 2*int((sin(c + d*x)**3*x**2)/(sin(c + d*x)*b + a),x)*a**2*b**3*d* 
*2*f**2 - 2*int((sin(c + d*x)**3*x**2)/(sin(c + d*x)*b + a),x)*b**5*d**2*f 
**2 + 4*int((sin(c + d*x)*x)/(sin(c + d*x)*b + a),x)*a**4*b*d**2*e*f - 4*i 
nt((sin(c + d*x)*x)/(sin(c + d*x)*b + a),x)*a**2*b**3*d**2*e*f + sin(c + d 
*x)**2*a**2*b**2*d**2*e*f*x**2 - sin(c + d*x)**2*b**4*d**2*e*f*x**2 - 4*si 
n(c + d*x)*a**3*b*e*f + 4*sin(c + d*x)*a*b**3*e*f + 2*a**4*c*d*e**2 + 2*a* 
*4*d**2*e**2*x - a**2*b**2*c*d*e**2 - a**2*b**2*d**2*e**2*x - b**4*c*d*e** 
2 - b**4*d**2*e**2*x)/(2*b**3*d**2*(a**2 - b**2))