\(\int \frac {(c+d x)^3}{(a+b \sin (e+f x))^2} \, dx\) [168]

Optimal result

Integrand size = 20, antiderivative size = 925 \[ \int \frac {(c+d x)^3}{(a+b \sin (e+f x))^2} \, dx=\frac {i (c+d x)^3}{\left (a^2-b^2\right ) f}-\frac {3 d (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}-\frac {i a (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}-\frac {3 d (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^2}+\frac {i a (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}-\frac {3 a d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}+\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^3}+\frac {3 a d (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^2}-\frac {6 d^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^4}-\frac {6 i a d^2 (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^3}-\frac {6 d^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right ) f^4}+\frac {6 i a d^2 (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^3}+\frac {6 a d^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^4}-\frac {6 a d^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f^4}+\frac {b (c+d x)^3 \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 3.84 (sec) , antiderivative size = 742, normalized size of antiderivative = 0.80 \[ \int \frac {(c+d x)^3}{(a+b \sin (e+f x))^2} \, dx=\frac {i f^3 (c+d x)^3-3 d f^2 (c+d x)^2 \log \left (1+\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )-3 d f^2 (c+d x)^2 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )+6 i d^2 \left (f (c+d x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )+i d \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )\right )+6 i d^2 \left (f (c+d x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )+i d \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )-\frac {i a \left (f^3 (c+d x)^3 \log \left (1+\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )-f^3 (c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )-3 i d \left (f^2 (c+d x)^2 \operatorname {PolyLog}\left (2,-\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )+2 i d f (c+d x) \operatorname {PolyLog}\left (3,-\frac {i b e^{i (e+f x)}}{-a+\sqrt {a^2-b^2}}\right )-2 d^2 \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )\right )+3 i d \left (f^2 (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )+2 i d f (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )-2 d^2 \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )\right )\right )}{\sqrt {a^2-b^2}}+\frac {b f^3 (c+d x)^3 \cos (e+f x)}{a+b \sin (e+f x)}}{\left (a^2-b^2\right ) f^4} \] Input:

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

Output:

(I*f^3*(c + d*x)^3 - 3*d*f^2*(c + d*x)^2*Log[1 + (I*b*E^(I*(e + f*x)))/(-a 
 + Sqrt[a^2 - b^2])] - 3*d*f^2*(c + d*x)^2*Log[1 - (I*b*E^(I*(e + f*x)))/( 
a + Sqrt[a^2 - b^2])] + (6*I)*d^2*(f*(c + d*x)*PolyLog[2, (I*b*E^(I*(e + f 
*x)))/(a - Sqrt[a^2 - b^2])] + I*d*PolyLog[3, (I*b*E^(I*(e + f*x)))/(a - S 
qrt[a^2 - b^2])]) + (6*I)*d^2*(f*(c + d*x)*PolyLog[2, (I*b*E^(I*(e + f*x)) 
)/(a + Sqrt[a^2 - b^2])] + I*d*PolyLog[3, (I*b*E^(I*(e + f*x)))/(a + Sqrt[ 
a^2 - b^2])]) - (I*a*(f^3*(c + d*x)^3*Log[1 + (I*b*E^(I*(e + f*x)))/(-a + 
Sqrt[a^2 - b^2])] - f^3*(c + d*x)^3*Log[1 - (I*b*E^(I*(e + f*x)))/(a + Sqr 
t[a^2 - b^2])] - (3*I)*d*(f^2*(c + d*x)^2*PolyLog[2, ((-I)*b*E^(I*(e + f*x 
)))/(-a + Sqrt[a^2 - b^2])] + (2*I)*d*f*(c + d*x)*PolyLog[3, ((-I)*b*E^(I* 
(e + f*x)))/(-a + Sqrt[a^2 - b^2])] - 2*d^2*PolyLog[4, (I*b*E^(I*(e + f*x) 
))/(a - Sqrt[a^2 - b^2])]) + (3*I)*d*(f^2*(c + d*x)^2*PolyLog[2, (I*b*E^(I 
*(e + f*x)))/(a + Sqrt[a^2 - b^2])] + (2*I)*d*f*(c + d*x)*PolyLog[3, (I*b* 
E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])] - 2*d^2*PolyLog[4, (I*b*E^(I*(e + 
f*x)))/(a + Sqrt[a^2 - b^2])])))/Sqrt[a^2 - b^2] + (b*f^3*(c + d*x)^3*Cos[ 
e + f*x])/(a + b*Sin[e + f*x]))/((a^2 - b^2)*f^4)
 

Rubi [A] (verified)

Time = 4.30 (sec) , antiderivative size = 857, normalized size of antiderivative = 0.93, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 3805, 3042, 3804, 2694, 27, 2620, 3011, 5030, 2620, 3011, 2720, 7143, 7163, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3805

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3804

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 5030

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

\(\displaystyle \frac {b \cos (e+f x) (c+d x)^3}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac {3 b d \left (-\frac {i (c+d x)^3}{3 b d}+\frac {\log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right ) (c+d x)^2}{b f}+\frac {\log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right ) (c+d x)^2}{b f}-\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f}-\frac {d \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2}\right )}{b f}-\frac {2 d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f}-\frac {d \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^2}\right )}{b f}\right )}{\left (a^2-b^2\right ) f}+\frac {2 a \left (\frac {i b \left (\frac {(c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{b f}-\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f}-\frac {2 i d \left (\frac {d \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a+\sqrt {a^2-b^2}}\right )}{f}\right )}{f}\right )}{b f}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(c+d x)^3 \log \left (1-\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{b f}-\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f}-\frac {2 i d \left (\frac {d \operatorname {PolyLog}\left (4,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (e+f x)}}{a-\sqrt {a^2-b^2}}\right )}{f}\right )}{f}\right )}{b f}\right )}{2 \sqrt {a^2-b^2}}\right )}{a^2-b^2}\)

Input:

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

Output:

(-3*b*d*(((-1/3*I)*(c + d*x)^3)/(b*d) + ((c + d*x)^2*Log[1 - (I*b*E^(I*(e 
+ f*x)))/(a - Sqrt[a^2 - b^2])])/(b*f) + ((c + d*x)^2*Log[1 - (I*b*E^(I*(e 
 + f*x)))/(a + Sqrt[a^2 - b^2])])/(b*f) - (2*d*((I*(c + d*x)*PolyLog[2, (I 
*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 - b^2])])/f - (d*PolyLog[3, (I*b*E^(I*(e 
 + f*x)))/(a - Sqrt[a^2 - b^2])])/f^2))/(b*f) - (2*d*((I*(c + d*x)*PolyLog 
[2, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/f - (d*PolyLog[3, (I*b*E 
^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/f^2))/(b*f)))/((a^2 - b^2)*f) + (2 
*a*(((-1/2*I)*b*(((c + d*x)^3*Log[1 - (I*b*E^(I*(e + f*x)))/(a - Sqrt[a^2 
- b^2])])/(b*f) - (3*d*((I*(c + d*x)^2*PolyLog[2, (I*b*E^(I*(e + f*x)))/(a 
 - Sqrt[a^2 - b^2])])/f - ((2*I)*d*(((-I)*(c + d*x)*PolyLog[3, (I*b*E^(I*( 
e + f*x)))/(a - Sqrt[a^2 - b^2])])/f + (d*PolyLog[4, (I*b*E^(I*(e + f*x))) 
/(a - Sqrt[a^2 - b^2])])/f^2))/f))/(b*f)))/Sqrt[a^2 - b^2] + ((I/2)*b*(((c 
 + d*x)^3*Log[1 - (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])])/(b*f) - (3 
*d*((I*(c + d*x)^2*PolyLog[2, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2])] 
)/f - ((2*I)*d*(((-I)*(c + d*x)*PolyLog[3, (I*b*E^(I*(e + f*x)))/(a + Sqrt 
[a^2 - b^2])])/f + (d*PolyLog[4, (I*b*E^(I*(e + f*x)))/(a + Sqrt[a^2 - b^2 
])])/f^2))/f))/(b*f)))/Sqrt[a^2 - b^2]))/(a^2 - b^2) + (b*(c + d*x)^3*Cos[ 
e + f*x])/((a^2 - b^2)*f*(a + b*Sin[e + f*x]))
 

Defintions of rubi rules used

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 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 3805

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

rule 5030

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

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]
 

rule 7163

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

Maple [F]

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

Input:

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

Output:

int((d*x+c)^3/(a+b*sin(f*x+e))^2,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. 5112 vs. \(2 (807) = 1614\).

Time = 0.40 (sec) , antiderivative size = 5112, normalized size of antiderivative = 5.53 \[ \int \frac {(c+d x)^3}{(a+b \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-2)]

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

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

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(c+d x)^3}{(a+b \sin (e+f x))^2} \, dx=\text {too large to display} \] Input:

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

Output:

(8*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin( 
e + f*x)*a**5*b*c**3*f**3 - 96*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a 
+ b)/sqrt(a**2 - b**2))*sin(e + f*x)*a**5*b*c*d**2*f - 48*sqrt(a**2 - b**2 
)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)*a**4*b**2* 
c**2*d*f**2 + 384*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a** 
2 - b**2))*sin(e + f*x)*a**4*b**2*d**3 + 288*sqrt(a**2 - b**2)*atan((tan(( 
e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)*a**3*b**3*c*d**2*f + 48 
*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e 
+ f*x)*a**2*b**4*c**2*d*f**2 - 960*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2 
)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)*a**2*b**4*d**3 - 192*sqrt(a**2 - 
b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)*a*b**5 
*c*d**2*f + 576*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 
- b**2))*sin(e + f*x)*b**6*d**3 + 8*sqrt(a**2 - b**2)*atan((tan((e + f*x)/ 
2)*a + b)/sqrt(a**2 - b**2))*a**6*c**3*f**3 - 96*sqrt(a**2 - b**2)*atan((t 
an((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*a**6*c*d**2*f - 48*sqrt(a**2 - b 
**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*a**5*b*c**2*d*f**2 + 
 384*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*a* 
*5*b*d**3 + 288*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 
- b**2))*a**4*b**2*c*d**2*f + 48*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)* 
a + b)/sqrt(a**2 - b**2))*a**3*b**3*c**2*d*f**2 - 960*sqrt(a**2 - b**2)...