\(\int x^3 \cot ^3(a+b x) \, dx\) [11]

Optimal result

Integrand size = 12, antiderivative size = 202 \[ \int x^3 \cot ^3(a+b x) \, dx=-\frac {3 i x^2}{2 b^2}-\frac {x^3}{2 b}+\frac {i x^4}{4}-\frac {3 x^2 \cot (a+b x)}{2 b^2}-\frac {x^3 \cot ^2(a+b x)}{2 b}+\frac {3 x \log \left (1-e^{2 i (a+b x)}\right )}{b^3}-\frac {x^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {3 i \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^4}+\frac {3 i x^2 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 x \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 i \operatorname {PolyLog}\left (4,e^{2 i (a+b x)}\right )}{4 b^4} \] Output:

-3/2*I*x^2/b^2-1/2*x^3/b+1/4*I*x^4-3/2*x^2*cot(b*x+a)/b^2-1/2*x^3*cot(b*x+ 
a)^2/b+3*x*ln(1-exp(2*I*(b*x+a)))/b^3-x^3*ln(1-exp(2*I*(b*x+a)))/b-3/2*I*p 
olylog(2,exp(2*I*(b*x+a)))/b^4+3/2*I*x^2*polylog(2,exp(2*I*(b*x+a)))/b^2-3 
/2*x*polylog(3,exp(2*I*(b*x+a)))/b^3-3/4*I*polylog(4,exp(2*I*(b*x+a)))/b^4
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(491\) vs. \(2(202)=404\).

Time = 6.47 (sec) , antiderivative size = 491, normalized size of antiderivative = 2.43 \[ \int x^3 \cot ^3(a+b x) \, dx=-\frac {1}{4} x^4 \cot (a)-\frac {x^3 \csc ^2(a+b x)}{2 b}+\frac {e^{i a} \csc (a) \left (b^4 e^{-2 i a} x^4+2 i b^3 \left (1-e^{-2 i a}\right ) x^3 \log \left (1-e^{-i (a+b x)}\right )+2 i b^3 \left (1-e^{-2 i a}\right ) x^3 \log \left (1+e^{-i (a+b x)}\right )-6 b^2 \left (1-e^{-2 i a}\right ) x^2 \operatorname {PolyLog}\left (2,-e^{-i (a+b x)}\right )-6 b^2 \left (1-e^{-2 i a}\right ) x^2 \operatorname {PolyLog}\left (2,e^{-i (a+b x)}\right )+12 i b \left (1-e^{-2 i a}\right ) x \operatorname {PolyLog}\left (3,-e^{-i (a+b x)}\right )+12 i b \left (1-e^{-2 i a}\right ) x \operatorname {PolyLog}\left (3,e^{-i (a+b x)}\right )+12 \left (1-e^{-2 i a}\right ) \operatorname {PolyLog}\left (4,-e^{-i (a+b x)}\right )+12 \left (1-e^{-2 i a}\right ) \operatorname {PolyLog}\left (4,e^{-i (a+b x)}\right )\right )}{4 b^4}+\frac {3 x^2 \csc (a) \csc (a+b x) \sin (b x)}{2 b^2}-\frac {3 \csc (a) \sec (a) \left (b^2 e^{i \arctan (\tan (a))} x^2+\frac {\left (i b x (-\pi +2 \arctan (\tan (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x+\arctan (\tan (a))) \log \left (1-e^{2 i (b x+\arctan (\tan (a)))}\right )+\pi \log (\cos (b x))+2 \arctan (\tan (a)) \log (\sin (b x+\arctan (\tan (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x+\arctan (\tan (a)))}\right )\right ) \tan (a)}{\sqrt {1+\tan ^2(a)}}\right )}{2 b^4 \sqrt {\sec ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}} \] Input:

Integrate[x^3*Cot[a + b*x]^3,x]
 

Output:

-1/4*(x^4*Cot[a]) - (x^3*Csc[a + b*x]^2)/(2*b) + (E^(I*a)*Csc[a]*((b^4*x^4 
)/E^((2*I)*a) + (2*I)*b^3*(1 - E^((-2*I)*a))*x^3*Log[1 - E^((-I)*(a + b*x) 
)] + (2*I)*b^3*(1 - E^((-2*I)*a))*x^3*Log[1 + E^((-I)*(a + b*x))] - 6*b^2* 
(1 - E^((-2*I)*a))*x^2*PolyLog[2, -E^((-I)*(a + b*x))] - 6*b^2*(1 - E^((-2 
*I)*a))*x^2*PolyLog[2, E^((-I)*(a + b*x))] + (12*I)*b*(1 - E^((-2*I)*a))*x 
*PolyLog[3, -E^((-I)*(a + b*x))] + (12*I)*b*(1 - E^((-2*I)*a))*x*PolyLog[3 
, E^((-I)*(a + b*x))] + 12*(1 - E^((-2*I)*a))*PolyLog[4, -E^((-I)*(a + b*x 
))] + 12*(1 - E^((-2*I)*a))*PolyLog[4, E^((-I)*(a + b*x))]))/(4*b^4) + (3* 
x^2*Csc[a]*Csc[a + b*x]*Sin[b*x])/(2*b^2) - (3*Csc[a]*Sec[a]*(b^2*E^(I*Arc 
Tan[Tan[a]])*x^2 + ((I*b*x*(-Pi + 2*ArcTan[Tan[a]]) - Pi*Log[1 + E^((-2*I) 
*b*x)] - 2*(b*x + ArcTan[Tan[a]])*Log[1 - E^((2*I)*(b*x + ArcTan[Tan[a]])) 
] + Pi*Log[Cos[b*x]] + 2*ArcTan[Tan[a]]*Log[Sin[b*x + ArcTan[Tan[a]]]] + I 
*PolyLog[2, E^((2*I)*(b*x + ArcTan[Tan[a]]))])*Tan[a])/Sqrt[1 + Tan[a]^2]) 
)/(2*b^4*Sqrt[Sec[a]^2*(Cos[a]^2 + Sin[a]^2)])
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.34, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.750, Rules used = {3042, 25, 4203, 25, 3042, 25, 4202, 2620, 3011, 4203, 15, 25, 3042, 25, 4202, 2620, 2715, 2838, 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.

\(\displaystyle \int x^3 \cot ^3(a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -x^3 \tan \left (a+b x+\frac {\pi }{2}\right )^3dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int x^3 \tan \left (\frac {1}{2} (2 a+\pi )+b x\right )^3dx\)

\(\Big \downarrow \) 4203

\(\displaystyle \int -x^3 \cot (a+b x)dx+\frac {3 \int x^2 \cot ^2(a+b x)dx}{2 b}-\frac {x^3 \cot ^2(a+b x)}{2 b}\)

\(\Big \downarrow \) 25

\(\displaystyle -\int x^3 \cot (a+b x)dx+\frac {3 \int x^2 \cot ^2(a+b x)dx}{2 b}-\frac {x^3 \cot ^2(a+b x)}{2 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\int -x^3 \tan \left (a+b x+\frac {\pi }{2}\right )dx+\frac {3 \int x^2 \tan \left (a+b x+\frac {\pi }{2}\right )^2dx}{2 b}-\frac {x^3 \cot ^2(a+b x)}{2 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \int x^3 \tan \left (\frac {1}{2} (2 a+\pi )+b x\right )dx+\frac {3 \int x^2 \tan \left (a+b x+\frac {\pi }{2}\right )^2dx}{2 b}-\frac {x^3 \cot ^2(a+b x)}{2 b}\)

\(\Big \downarrow \) 4202

\(\displaystyle -2 i \int \frac {e^{i (2 a+2 b x+\pi )} x^3}{1+e^{i (2 a+2 b x+\pi )}}dx+\frac {3 \int x^2 \tan \left (a+b x+\frac {\pi }{2}\right )^2dx}{2 b}-\frac {x^3 \cot ^2(a+b x)}{2 b}+\frac {i x^4}{4}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {3 \int x^2 \tan \left (a+b x+\frac {\pi }{2}\right )^2dx}{2 b}-2 i \left (\frac {3 i \int x^2 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )dx}{2 b}-\frac {i x^3 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )-\frac {x^3 \cot ^2(a+b x)}{2 b}+\frac {i x^4}{4}\)

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 4203

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

\(\displaystyle -2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {i \left (\frac {\int e^{-i (2 a+2 b x+\pi )} \operatorname {PolyLog}\left (3,-e^{i (2 a+2 b x+\pi )}\right )de^{i (2 a+2 b x+\pi )}}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )+\frac {3 \left (-\frac {2 \left (\frac {i x^2}{2}-2 i \left (-\frac {\operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{4 b^2}-\frac {i x \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )\right )}{b}-\frac {x^2 \cot (a+b x)}{b}-\frac {x^3}{3}\right )}{2 b}-\frac {x^3 \cot ^2(a+b x)}{2 b}+\frac {i x^4}{4}\)

\(\Big \downarrow \) 7143

\(\displaystyle -2 i \left (\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {i \left (\frac {\operatorname {PolyLog}\left (4,-e^{i (2 a+2 b x+\pi )}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {i x^3 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )+\frac {3 \left (-\frac {2 \left (\frac {i x^2}{2}-2 i \left (-\frac {\operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{4 b^2}-\frac {i x \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )\right )}{b}-\frac {x^2 \cot (a+b x)}{b}-\frac {x^3}{3}\right )}{2 b}-\frac {x^3 \cot ^2(a+b x)}{2 b}+\frac {i x^4}{4}\)

Input:

Int[x^3*Cot[a + b*x]^3,x]
 

Output:

(I/4)*x^4 - (x^3*Cot[a + b*x]^2)/(2*b) + (3*(-1/3*x^3 - (x^2*Cot[a + b*x]) 
/b - (2*((I/2)*x^2 - (2*I)*(((-1/2*I)*x*Log[1 + E^(I*(2*a + Pi + 2*b*x))]) 
/b - PolyLog[2, -E^(I*(2*a + Pi + 2*b*x))]/(4*b^2))))/b))/(2*b) - (2*I)*(( 
(-1/2*I)*x^3*Log[1 + E^(I*(2*a + Pi + 2*b*x))])/b + (((3*I)/2)*(((I/2)*x^2 
*PolyLog[2, -E^(I*(2*a + Pi + 2*b*x))])/b - (I*(((-1/2*I)*x*PolyLog[3, -E^ 
(I*(2*a + Pi + 2*b*x))])/b + PolyLog[4, -E^(I*(2*a + Pi + 2*b*x))]/(4*b^2) 
))/b))/b)
 

Defintions of rubi rules used

rule 15

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

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 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 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 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

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 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 4203

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symb 
ol] :> Simp[b*(c + d*x)^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] + (-Si 
mp[b*d*(m/(f*(n - 1)))   Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1), x] 
, x] - Simp[b^2   Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; Free 
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 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]
 

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 [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 443 vs. \(2 (167 ) = 334\).

Time = 0.34 (sec) , antiderivative size = 444, normalized size of antiderivative = 2.20

method	result	size
risch	\(\frac {3 i \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {2 a^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 i a^{4}}{2 b^{4}}+\frac {x^{2} \left (2 b x \,{\mathrm e}^{2 i \left (b x +a \right )}-3 i {\mathrm e}^{2 i \left (b x +a \right )}+3 i\right )}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {3 i \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {6 i a x}{b^{3}}+\frac {2 i a^{3} x}{b^{3}}+\frac {i x^{4}}{4}-\frac {\ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{3}}{b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{3}}{b}-\frac {a^{3} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{3}}+\frac {3 \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{4}}-\frac {3 i \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {3 a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{4}}+\frac {6 a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {3 i x^{2}}{b^{2}}-\frac {6 i \operatorname {polylog}\left (4, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {6 i \operatorname {polylog}\left (4, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {3 i \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{4}}-\frac {6 \operatorname {polylog}\left (3, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}-\frac {6 \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {3 \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}-\frac {3 i a^{2}}{b^{4}}\)	\(444\)

Input:

int(x^3*cot(b*x+a)^3,x,method=_RETURNVERBOSE)
 

Output:

x^2*(2*b*x*exp(2*I*(b*x+a))-3*I*exp(2*I*(b*x+a))+3*I)/b^2/(exp(2*I*(b*x+a) 
)-1)^2-1/b*ln(1-exp(I*(b*x+a)))*x^3-1/b*ln(exp(I*(b*x+a))+1)*x^3-1/b^4*a^3 
*ln(1-exp(I*(b*x+a)))+3*I/b^2*polylog(2,exp(I*(b*x+a)))*x^2+3*I/b^2*polylo 
g(2,-exp(I*(b*x+a)))*x^2-6*I/b^3*a*x+2*I/b^3*a^3*x+3/b^3*ln(exp(I*(b*x+a)) 
+1)*x+3/b^4*ln(1-exp(I*(b*x+a)))*a-2/b^4*a^3*ln(exp(I*(b*x+a)))-3*I/b^4*po 
lylog(2,-exp(I*(b*x+a)))-3/b^4*a*ln(exp(I*(b*x+a))-1)+6/b^4*a*ln(exp(I*(b* 
x+a)))+1/b^4*a^3*ln(exp(I*(b*x+a))-1)-6/b^3*polylog(3,exp(I*(b*x+a)))*x-6/ 
b^3*polylog(3,-exp(I*(b*x+a)))*x+3/b^3*ln(1-exp(I*(b*x+a)))*x+3/2*I/b^4*a^ 
4-3*I/b^4*a^2-3*I/b^2*x^2-6*I/b^4*polylog(4,exp(I*(b*x+a)))-6*I/b^4*polylo 
g(4,-exp(I*(b*x+a)))-3*I/b^4*polylog(2,exp(I*(b*x+a)))+1/4*I*x^4
 

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. 569 vs. \(2 (160) = 320\).

Time = 0.11 (sec) , antiderivative size = 569, normalized size of antiderivative = 2.82 \[ \int x^3 \cot ^3(a+b x) \, dx =\text {Too large to display} \] Input:

integrate(x^3*cot(b*x+a)^3,x, algorithm="fricas")
 

Output:

1/8*(8*b^3*x^3 + 12*b^2*x^2*sin(2*b*x + 2*a) - 6*(I*b^2*x^2 + (-I*b^2*x^2 
+ I)*cos(2*b*x + 2*a) - I)*dilog(cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a)) - 
6*(-I*b^2*x^2 + (I*b^2*x^2 - I)*cos(2*b*x + 2*a) + I)*dilog(cos(2*b*x + 2* 
a) - I*sin(2*b*x + 2*a)) - 4*(a^3 - (a^3 - 3*a)*cos(2*b*x + 2*a) - 3*a)*lo 
g(-1/2*cos(2*b*x + 2*a) + 1/2*I*sin(2*b*x + 2*a) + 1/2) - 4*(a^3 - (a^3 - 
3*a)*cos(2*b*x + 2*a) - 3*a)*log(-1/2*cos(2*b*x + 2*a) - 1/2*I*sin(2*b*x + 
 2*a) + 1/2) + 4*(b^3*x^3 + a^3 - 3*b*x - (b^3*x^3 + a^3 - 3*b*x - 3*a)*co 
s(2*b*x + 2*a) - 3*a)*log(-cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + 1) + 4* 
(b^3*x^3 + a^3 - 3*b*x - (b^3*x^3 + a^3 - 3*b*x - 3*a)*cos(2*b*x + 2*a) - 
3*a)*log(-cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + 1) - 3*(I*cos(2*b*x + 2* 
a) - I)*polylog(4, cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a)) - 3*(-I*cos(2*b* 
x + 2*a) + I)*polylog(4, cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a)) - 6*(b*x*c 
os(2*b*x + 2*a) - b*x)*polylog(3, cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a)) - 
 6*(b*x*cos(2*b*x + 2*a) - b*x)*polylog(3, cos(2*b*x + 2*a) - I*sin(2*b*x 
+ 2*a)))/(b^4*cos(2*b*x + 2*a) - b^4)
 

Sympy [F]

\[ \int x^3 \cot ^3(a+b x) \, dx=\int x^{3} \cot ^{3}{\left (a + b x \right )}\, dx \] Input:

integrate(x**3*cot(b*x+a)**3,x)
 

Output:

Integral(x**3*cot(a + b*x)**3, x)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1960 vs. \(2 (160) = 320\).

Time = 0.31 (sec) , antiderivative size = 1960, normalized size of antiderivative = 9.70 \[ \int x^3 \cot ^3(a+b x) \, dx=\text {Too large to display} \] Input:

integrate(x^3*cot(b*x+a)^3,x, algorithm="maxima")
 

Output:

1/2*(a^3*(1/sin(b*x + a)^2 + log(sin(b*x + a)^2)) + 2*((b*x + a)^4 - 4*(b* 
x + a)^3*a + 6*(b*x + a)^2*a^2 + 12*a^2 - 4*((b*x + a)^3 - 3*(b*x + a)^2*a 
 + 3*(a^2 - 1)*(b*x + a) + ((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(a^2 - 1)*(b 
*x + a) + 3*a)*cos(4*b*x + 4*a) - 2*((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(a^ 
2 - 1)*(b*x + a) + 3*a)*cos(2*b*x + 2*a) - (-I*(b*x + a)^3 + 3*I*(b*x + a) 
^2*a + 3*(-I*a^2 + I)*(b*x + a) - 3*I*a)*sin(4*b*x + 4*a) - 2*(I*(b*x + a) 
^3 - 3*I*(b*x + a)^2*a + 3*(I*a^2 - I)*(b*x + a) + 3*I*a)*sin(2*b*x + 2*a) 
 + 3*a)*arctan2(sin(b*x + a), cos(b*x + a) + 1) - 12*(a*cos(4*b*x + 4*a) - 
 2*a*cos(2*b*x + 2*a) + I*a*sin(4*b*x + 4*a) - 2*I*a*sin(2*b*x + 2*a) + a) 
*arctan2(sin(b*x + a), cos(b*x + a) - 1) + 4*((b*x + a)^3 - 3*(b*x + a)^2* 
a + 3*(a^2 - 1)*(b*x + a) + ((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(a^2 - 1)*( 
b*x + a))*cos(4*b*x + 4*a) - 2*((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(a^2 - 1 
)*(b*x + a))*cos(2*b*x + 2*a) + (I*(b*x + a)^3 - 3*I*(b*x + a)^2*a + 3*(I* 
a^2 - I)*(b*x + a))*sin(4*b*x + 4*a) + 2*(-I*(b*x + a)^3 + 3*I*(b*x + a)^2 
*a + 3*(-I*a^2 + I)*(b*x + a))*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), -co 
s(b*x + a) + 1) + ((b*x + a)^4 - 4*(b*x + a)^3*a + 6*(a^2 - 2)*(b*x + a)^2 
 + 24*(b*x + a)*a)*cos(4*b*x + 4*a) - 2*((b*x + a)^4 - 4*(b*x + a)^3*(a - 
I) + 6*(a^2 - 2*I*a - 1)*(b*x + a)^2 - 12*(-I*a^2 - a)*(b*x + a) + 6*a^2)* 
cos(2*b*x + 2*a) + 12*((b*x + a)^2 - 2*(b*x + a)*a + a^2 + ((b*x + a)^2 - 
2*(b*x + a)*a + a^2 - 1)*cos(4*b*x + 4*a) - 2*((b*x + a)^2 - 2*(b*x + a...
 

Giac [F]

\[ \int x^3 \cot ^3(a+b x) \, dx=\int { x^{3} \cot \left (b x + a\right )^{3} \,d x } \] Input:

integrate(x^3*cot(b*x+a)^3,x, algorithm="giac")
 

Output:

integrate(x^3*cot(b*x + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \cot ^3(a+b x) \, dx=\int x^3\,{\mathrm {cot}\left (a+b\,x\right )}^3 \,d x \] Input:

int(x^3*cot(a + b*x)^3,x)
 

Output:

int(x^3*cot(a + b*x)^3, x)
 

Reduce [F]

\[ \int x^3 \cot ^3(a+b x) \, dx=\int \cot \left (b x +a \right )^{3} x^{3}d x \] Input:

int(x^3*cot(b*x+a)^3,x)
 

Output:

int(cot(a + b*x)**3*x**3,x)