3.3.33 \(\int \frac {(e+f x)^2 \csc (c+d x)}{a+b \sin (c+d x)} \, dx\) [233]

3.3.33.1 Optimal result

Integrand size = 26, antiderivative size = 528 \[ \int \frac {(e+f x)^2 \csc (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 (e+f x)^2 \text {arctanh}\left (e^{i (c+d x)}\right )}{a d}+\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}-\frac {i b (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d}+\frac {2 i f (e+f x) \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {2 i f (e+f x) \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}+\frac {2 b f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^2}-\frac {2 b f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^2}-\frac {2 f^2 \operatorname {PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac {2 f^2 \operatorname {PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac {2 i b f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^3}-\frac {2 i b f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d^3} \]

output

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

3.3.33.2 Mathematica [A] (verified)

Time = 1.42 (sec) , antiderivative size = 573, normalized size of antiderivative = 1.09 \[ \int \frac {(e+f x)^2 \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {d^2 (e+f x)^2 \log \left (1-e^{i (c+d x)}\right )-d^2 (e+f x)^2 \log \left (1+e^{i (c+d x)}\right )+2 i d f (e+f x) \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )-2 i d f (e+f x) \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right )-2 f^2 \operatorname {PolyLog}\left (3,-e^{i (c+d x)}\right )+2 f^2 \operatorname {PolyLog}\left (3,e^{i (c+d x)}\right )+\frac {b \left (2 d f (e+f x) \operatorname {PolyLog}\left (2,-\frac {i b e^{i (c+d x)}}{-a+\sqrt {a^2-b^2}}\right )+i \left (2 i d^2 e^2 \arctan \left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )+2 d^2 e f x \log \left (1+\frac {i b e^{i (c+d x)}}{-a+\sqrt {a^2-b^2}}\right )+d^2 f^2 x^2 \log \left (1+\frac {i b e^{i (c+d x)}}{-a+\sqrt {a^2-b^2}}\right )-2 d^2 e f x \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )-d^2 f^2 x^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )+2 i d f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )+2 f^2 \operatorname {PolyLog}\left (3,-\frac {i b e^{i (c+d x)}}{-a+\sqrt {a^2-b^2}}\right )-2 f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )\right )\right )}{\sqrt {a^2-b^2}}}{a d^3} \]

input

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

output

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

3.3.33.3 Rubi [A] (verified)

Time = 2.07 (sec) , antiderivative size = 487, normalized size of antiderivative = 0.92, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {5046, 3042, 3804, 2694, 27, 2620, 3011, 2720, 4671, 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.

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

\(\Big \downarrow \) 5046

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3804

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 4671

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

input

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

output

((-2*(e + f*x)^2*ArcTanh[E^(I*(c + d*x))])/d + (2*f*((I*(e + f*x)*PolyLog[ 
2, -E^(I*(c + d*x))])/d - (f*PolyLog[3, -E^(I*(c + d*x))])/d^2))/d - (2*f* 
((I*(e + f*x)*PolyLog[2, E^(I*(c + d*x))])/d - (f*PolyLog[3, E^(I*(c + d*x 
))])/d^2))/d)/a - (2*b*(((-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]))/a
 

3.3.33.3.1 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 4671

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[- 
2*(c + d*x)^m*(ArcTanh[E^(I*(e + f*x))]/f), x] + (-Simp[d*(m/f)   Int[(c + 
d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Simp[d*(m/f)   Int[(c + d*x 
)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IG 
tQ[m, 0]
 

rule 5046

Int[(Csc[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_. 
)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/a   Int[(e + f*x)^m*Csc[c + 
 d*x]^n, x], x] - Simp[b/a   Int[(e + f*x)^m*(Csc[c + d*x]^(n - 1)/(a + b*S 
in[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]
 

3.3.33.4 Maple [F]

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

input

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

output

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

3.3.33.5 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. 2412 vs. \(2 (454) = 908\).

Time = 0.53 (sec) , antiderivative size = 2412, normalized size of antiderivative = 4.57 \[ \int \frac {(e+f x)^2 \csc (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]

input

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

output

-1/2*(2*b^2*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(d*x + c) + a*s 
in(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/ 
b) - 2*b^2*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(d*x + c) + a*si 
n(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b 
) + 2*b^2*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(-I*a*cos(d*x + c) + a*si 
n(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b 
) - 2*b^2*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(-I*a*cos(d*x + c) + a*si 
n(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b 
) - 2*(a^2 - b^2)*f^2*polylog(3, cos(d*x + c) + I*sin(d*x + c)) - 2*(a^2 - 
 b^2)*f^2*polylog(3, cos(d*x + c) - I*sin(d*x + c)) + 2*(a^2 - b^2)*f^2*po 
lylog(3, -cos(d*x + c) + I*sin(d*x + c)) + 2*(a^2 - b^2)*f^2*polylog(3, -c 
os(d*x + c) - I*sin(d*x + c)) - 2*(-I*b^2*d*f^2*x - I*b^2*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) - 2*(I*b^2*d*f^2*x + 
 I*b^2*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) - 2*(I*b^2*d*f^2*x + I*b^2*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) - 2*(-I*b^2*d*f^2*x - I*b^2*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) ...
 

3.3.33.6 Sympy [F]

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

input

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

output

Integral((e + f*x)**2*csc(c + d*x)/(a + b*sin(c + d*x)), x)
 

3.3.33.7 Maxima [F(-2)]

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

input

integrate((f*x+e)^2*csc(d*x+c)/(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
 

3.3.33.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.3.33.9 Mupad [F(-1)]

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

input

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

output

\text{Hanged}