3.4.29 \(\int \frac {(e+f x)^3 \cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\) [329]

3.4.29.1 Optimal result

Integrand size = 34, antiderivative size = 763 \[ \int \frac {(e+f x)^3 \cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {i (e+f x)^4}{4 a f}-\frac {i \left (a^2-b^2\right ) (e+f x)^4}{4 a b^2 f}+\frac {6 f^3 \cos (c+d x)}{b d^4}-\frac {3 f (e+f x)^2 \cos (c+d x)}{b d^2}+\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d}+\frac {(e+f x)^3 \log \left (1-e^{2 i (c+d x)}\right )}{a d}-\frac {3 i \left (a^2-b^2\right ) f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {3 i \left (a^2-b^2\right ) f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d^2}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,e^{2 i (c+d x)}\right )}{2 a d^2}+\frac {6 \left (a^2-b^2\right ) f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^3}+\frac {6 \left (a^2-b^2\right ) f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d^3}+\frac {3 f^2 (e+f x) \operatorname {PolyLog}\left (3,e^{2 i (c+d x)}\right )}{2 a d^3}+\frac {6 i \left (a^2-b^2\right ) f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^2 d^4}+\frac {6 i \left (a^2-b^2\right ) f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^2 d^4}+\frac {3 i f^3 \operatorname {PolyLog}\left (4,e^{2 i (c+d x)}\right )}{4 a d^4}+\frac {6 f^2 (e+f x) \sin (c+d x)}{b d^3}-\frac {(e+f x)^3 \sin (c+d x)}{b d} \]

output

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

3.4.29.2 Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(4052\) vs. \(2(763)=1526\).

Time = 9.84 (sec) , antiderivative size = 4052, normalized size of antiderivative = 5.31 \[ \int \frac {(e+f x)^3 \cos ^2(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Result too large to show} \]

input

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

output

-1/2*(e*E^(I*c)*f^2*Csc[c]*((2*d^3*x^3)/E^((2*I)*c) + (3*I)*d^2*(1 - E^((- 
2*I)*c))*x^2*Log[1 - E^((-I)*(c + d*x))] + (3*I)*d^2*(1 - E^((-2*I)*c))*x^ 
2*Log[1 + E^((-I)*(c + d*x))] - 6*d*(1 - E^((-2*I)*c))*x*PolyLog[2, -E^((- 
I)*(c + d*x))] - 6*d*(1 - E^((-2*I)*c))*x*PolyLog[2, E^((-I)*(c + d*x))] + 
 (6*I)*(1 - E^((-2*I)*c))*PolyLog[3, -E^((-I)*(c + d*x))] + (6*I)*(1 - E^( 
(-2*I)*c))*PolyLog[3, E^((-I)*(c + d*x))]))/(a*d^3) - (E^(I*c)*f^3*Csc[c]* 
((d^4*x^4)/E^((2*I)*c) + (2*I)*d^3*(1 - E^((-2*I)*c))*x^3*Log[1 - E^((-I)* 
(c + d*x))] + (2*I)*d^3*(1 - E^((-2*I)*c))*x^3*Log[1 + E^((-I)*(c + d*x))] 
 - 6*d^2*(1 - E^((-2*I)*c))*x^2*PolyLog[2, -E^((-I)*(c + d*x))] - 6*d^2*(1 
 - E^((-2*I)*c))*x^2*PolyLog[2, E^((-I)*(c + d*x))] + (12*I)*d*(1 - E^((-2 
*I)*c))*x*PolyLog[3, -E^((-I)*(c + d*x))] + (12*I)*d*(1 - E^((-2*I)*c))*x* 
PolyLog[3, E^((-I)*(c + d*x))] + 12*(1 - E^((-2*I)*c))*PolyLog[4, -E^((-I) 
*(c + d*x))] + 12*(1 - E^((-2*I)*c))*PolyLog[4, E^((-I)*(c + d*x))]))/(4*a 
*d^4) + ((a^2 - b^2)*((-4*I)*d^4*e^3*E^((2*I)*c)*x - (6*I)*d^4*e^2*E^((2*I 
)*c)*f*x^2 - (4*I)*d^4*e*E^((2*I)*c)*f^2*x^3 - I*d^4*E^((2*I)*c)*f^3*x^4 - 
 (2*I)*d^3*e^3*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x)))) 
] + (2*I)*d^3*e^3*E^((2*I)*c)*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2* 
I)*(c + d*x))))] - d^3*e^3*Log[4*a^2*E^((2*I)*(c + d*x)) + b^2*(-1 + E^((2 
*I)*(c + d*x)))^2] + d^3*e^3*E^((2*I)*c)*Log[4*a^2*E^((2*I)*(c + d*x)) + b 
^2*(-1 + E^((2*I)*(c + d*x)))^2] - 6*d^3*e^2*f*x*Log[1 + (b*E^(I*(2*c +...
 

3.4.29.3 Rubi [F]

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.

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

\(\Big \downarrow \) 5054

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

\(\Big \downarrow \) 4908

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4202

\(\displaystyle -\frac {b \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)}dx}{a}+\frac {2 i \int \frac {e^{i (2 c+2 d x+\pi )} (e+f x)^3}{1+e^{i (2 c+2 d x+\pi )}}dx-\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 2620

\(\displaystyle -\frac {b \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)}dx}{a}+\frac {2 i \left (\frac {3 i f \int (e+f x)^2 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )dx}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )-\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 4904

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3792

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

\(\Big \downarrow \) 17

\(\displaystyle -\frac {b \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)}dx}{a}+\frac {\frac {3 f \left (-\frac {f^2 \int \sin ^2(c+d x)dx}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}+2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {b \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)}dx}{a}+\frac {\frac {3 f \left (-\frac {f^2 \int \sin (c+d x)^2dx}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}+2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3115

\(\displaystyle -\frac {b \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)}dx}{a}+\frac {\frac {3 f \left (-\frac {f^2 \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}+2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 24

\(\displaystyle -\frac {b \int \frac {(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)}dx}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 5036

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \int (e+f x)^3 \cos (c+d x)dx}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \int (e+f x)^3 \sin \left (c+d x+\frac {\pi }{2}\right )dx}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3777

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {3 f \int -(e+f x)^2 \sin (c+d x)dx}{d}+\frac {(e+f x)^3 \sin (c+d x)}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \int (e+f x)^2 \sin (c+d x)dx}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \int (e+f x)^2 \sin (c+d x)dx}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3777

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \int (e+f x) \cos (c+d x)dx}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \int (e+f x) \sin \left (c+d x+\frac {\pi }{2}\right )dx}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3777

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \int -\sin (c+d x)dx}{d}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {(e+f x) \sin (c+d x)}{d}-\frac {f \int \sin (c+d x)dx}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {(e+f x) \sin (c+d x)}{d}-\frac {f \int \sin (c+d x)dx}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3118

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x)dx}{b}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 4904

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \int (e+f x)^2 \sin ^2(c+d x)dx}{2 d}}{b}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \int (e+f x)^2 \sin (c+d x)^2dx}{2 d}}{b}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3792

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (-\frac {f^2 \int \sin ^2(c+d x)dx}{2 d^2}+\frac {1}{2} \int (e+f x)^2dx+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d}}{b}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 17

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (-\frac {f^2 \int \sin ^2(c+d x)dx}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}}{b}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {b \left (-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)}dx}{b^2}-\frac {\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {3 f \left (-\frac {f^2 \int \sin (c+d x)^2dx}{2 d^2}+\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}}{b}+\frac {a \left (\frac {(e+f x)^3 \sin (c+d x)}{d}-\frac {3 f \left (\frac {2 f \left (\frac {f \cos (c+d x)}{d^2}+\frac {(e+f x) \sin (c+d x)}{d}\right )}{d}-\frac {(e+f x)^2 \cos (c+d x)}{d}\right )}{d}\right )}{b^2}\right )}{a}+\frac {2 i \left (\frac {3 i f \left (\frac {i (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )}{2 d}-\frac {i f \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (2 c+2 d x+\pi )}\right )dx}{d}\right )}{2 d}-\frac {i (e+f x)^3 \log \left (1+e^{i (2 c+2 d x+\pi )}\right )}{2 d}\right )+\frac {3 f \left (\frac {f (e+f x) \sin ^2(c+d x)}{2 d^2}-\frac {f^2 \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )}{2 d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {(e+f x)^3}{6 f}\right )}{2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 d}-\frac {i (e+f x)^4}{4 f}}{a}\)

input

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

output

$Aborted
 

3.4.29.3.1 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 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 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 4202

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

rule 4904

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

rule 4908

Int[Cos[(a_.) + (b_.)*(x_)]^(n_.)*Cot[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d 
_.)*(x_))^(m_.), x_Symbol] :> -Int[(c + d*x)^m*Cos[a + b*x]^n*Cot[a + b*x]^ 
(p - 2), x] + Int[(c + d*x)^m*Cos[a + b*x]^(n - 2)*Cot[a + b*x]^p, x] /; Fr 
eeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]
 

rule 5036

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

rule 5054

Int[(Cos[(c_.) + (d_.)*(x_)]^(p_.)*Cot[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + ( 
f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp 
[1/a   Int[(e + f*x)^m*Cos[c + d*x]^p*Cot[c + d*x]^n, x], x] - Simp[b/a   I 
nt[(e + f*x)^m*Cos[c + d*x]^(p + 1)*(Cot[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] && 
 IGtQ[p, 0]
 

3.4.29.4 Maple [F]

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

input

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

output

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

3.4.29.5 Fricas [F(-2)]

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

input

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

output

Exception raised: TypeError >>  Error detected within library code:   Too 
many variables
 

3.4.29.6 Sympy [F]

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

input

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

output

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

3.4.29.7 Maxima [F(-2)]

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

input

integrate((f*x+e)^3*cos(d*x+c)^2*cot(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
 

3.4.29.8 Giac [F]

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

input

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

output

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

3.4.29.9 Mupad [F(-1)]

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

input

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

output

\text{Hanged}