\(\int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx\) [95]

Optimal result

Integrand size = 25, antiderivative size = 356 \[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx=-\frac {a b x}{c^3 d}-\frac {i b^2 x}{3 c^3 d}+\frac {i b^2 \arctan (c x)}{3 c^4 d}-\frac {b^2 x \arctan (c x)}{c^3 d}+\frac {i b x^2 (a+b \arctan (c x))}{3 c^2 d}-\frac {5 (a+b \arctan (c x))^2}{6 c^4 d}+\frac {i x (a+b \arctan (c x))^2}{c^3 d}+\frac {x^2 (a+b \arctan (c x))^2}{2 c^2 d}-\frac {i x^3 (a+b \arctan (c x))^2}{3 c d}+\frac {8 i b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{3 c^4 d}+\frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^4 d}+\frac {b^2 \log \left (1+c^2 x^2\right )}{2 c^4 d}-\frac {4 b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{3 c^4 d}+\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^4 d}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^4 d} \] Output:

-a*b*x/c^3/d-1/3*I*b^2*x/c^3/d+1/3*I*b^2*arctan(c*x)/c^4/d-b^2*x*arctan(c* 
x)/c^3/d+1/3*I*b*x^2*(a+b*arctan(c*x))/c^2/d-5/6*(a+b*arctan(c*x))^2/c^4/d 
+I*x*(a+b*arctan(c*x))^2/c^3/d+1/2*x^2*(a+b*arctan(c*x))^2/c^2/d-1/3*I*x^3 
*(a+b*arctan(c*x))^2/c/d+8/3*I*b*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c^4/d+( 
a+b*arctan(c*x))^2*ln(2/(1+I*c*x))/c^4/d+1/2*b^2*ln(c^2*x^2+1)/c^4/d-4/3*b 
^2*polylog(2,1-2/(1+I*c*x))/c^4/d+I*b*(a+b*arctan(c*x))*polylog(2,1-2/(1+I 
*c*x))/c^4/d+1/2*b^2*polylog(3,1-2/(1+I*c*x))/c^4/d
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.18 \[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx=\frac {i a^2 x}{c^3 d}+\frac {a^2 x^2}{2 c^2 d}-\frac {i a^2 x^3}{3 c d}-\frac {i a^2 \arctan (c x)}{c^4 d}-\frac {a^2 \log \left (1+c^2 x^2\right )}{2 c^4 d}-\frac {i a b \left (-3 i c x-8 c x \arctan (c x)+6 \arctan (c x)^2+\left (1+c^2 x^2\right ) (-1+3 i \arctan (c x)+2 c x \arctan (c x))+6 i \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )-8 \log \left (\frac {1}{\sqrt {1+c^2 x^2}}\right )+3 \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )\right )}{3 c^4 d}-\frac {i b^2 \left (2 c x-6 i c x \arctan (c x)-2 \left (1+c^2 x^2\right ) \arctan (c x)+8 i \arctan (c x)^2-8 c x \arctan (c x)^2+3 i \left (1+c^2 x^2\right ) \arctan (c x)^2+2 c x \left (1+c^2 x^2\right ) \arctan (c x)^2+4 \arctan (c x)^3-16 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )+6 i \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )-6 i \log \left (\frac {1}{\sqrt {1+c^2 x^2}}\right )+(8 i+6 \arctan (c x)) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+3 i \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right )}{6 c^4 d} \] Input:

Integrate[(x^3*(a + b*ArcTan[c*x])^2)/(d + I*c*d*x),x]
 

Output:

(I*a^2*x)/(c^3*d) + (a^2*x^2)/(2*c^2*d) - ((I/3)*a^2*x^3)/(c*d) - (I*a^2*A 
rcTan[c*x])/(c^4*d) - (a^2*Log[1 + c^2*x^2])/(2*c^4*d) - ((I/3)*a*b*((-3*I 
)*c*x - 8*c*x*ArcTan[c*x] + 6*ArcTan[c*x]^2 + (1 + c^2*x^2)*(-1 + (3*I)*Ar 
cTan[c*x] + 2*c*x*ArcTan[c*x]) + (6*I)*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan 
[c*x])] - 8*Log[1/Sqrt[1 + c^2*x^2]] + 3*PolyLog[2, -E^((2*I)*ArcTan[c*x]) 
]))/(c^4*d) - ((I/6)*b^2*(2*c*x - (6*I)*c*x*ArcTan[c*x] - 2*(1 + c^2*x^2)* 
ArcTan[c*x] + (8*I)*ArcTan[c*x]^2 - 8*c*x*ArcTan[c*x]^2 + (3*I)*(1 + c^2*x 
^2)*ArcTan[c*x]^2 + 2*c*x*(1 + c^2*x^2)*ArcTan[c*x]^2 + 4*ArcTan[c*x]^3 - 
16*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + (6*I)*ArcTan[c*x]^2*Log[1 
+ E^((2*I)*ArcTan[c*x])] - (6*I)*Log[1/Sqrt[1 + c^2*x^2]] + (8*I + 6*ArcTa 
n[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + (3*I)*PolyLog[3, -E^((2*I)*Ar 
cTan[c*x])]))/(c^4*d)
 

Rubi [A] (verified)

Time = 4.43 (sec) , antiderivative size = 469, normalized size of antiderivative = 1.32, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5401, 27, 5361, 5401, 5361, 5401, 5345, 5379, 5451, 2009, 5361, 262, 216, 5419, 5455, 5379, 2849, 2752, 5529, 7164}

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.

\(\displaystyle \int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx\)

\(\Big \downarrow \) 5401

\(\displaystyle \frac {i \int \frac {x^2 (a+b \arctan (c x))^2}{d (i c x+1)}dx}{c}-\frac {i \int x^2 (a+b \arctan (c x))^2dx}{c d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {i \int \frac {x^2 (a+b \arctan (c x))^2}{i c x+1}dx}{c d}-\frac {i \int x^2 (a+b \arctan (c x))^2dx}{c d}\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {i \int \frac {x^2 (a+b \arctan (c x))^2}{i c x+1}dx}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))^2-\frac {2}{3} b c \int \frac {x^3 (a+b \arctan (c x))}{c^2 x^2+1}dx\right )}{c d}\)

\(\Big \downarrow \) 5401

\(\displaystyle \frac {i \left (\frac {i \int \frac {x (a+b \arctan (c x))^2}{i c x+1}dx}{c}-\frac {i \int x (a+b \arctan (c x))^2dx}{c}\right )}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))^2-\frac {2}{3} b c \int \frac {x^3 (a+b \arctan (c x))}{c^2 x^2+1}dx\right )}{c d}\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {i \left (\frac {i \int \frac {x (a+b \arctan (c x))^2}{i c x+1}dx}{c}-\frac {i \left (\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx\right )}{c}\right )}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))^2-\frac {2}{3} b c \int \frac {x^3 (a+b \arctan (c x))}{c^2 x^2+1}dx\right )}{c d}\)

\(\Big \downarrow \) 5401

\(\displaystyle \frac {i \left (\frac {i \left (\frac {i \int \frac {(a+b \arctan (c x))^2}{i c x+1}dx}{c}-\frac {i \int (a+b \arctan (c x))^2dx}{c}\right )}{c}-\frac {i \left (\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx\right )}{c}\right )}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))^2-\frac {2}{3} b c \int \frac {x^3 (a+b \arctan (c x))}{c^2 x^2+1}dx\right )}{c d}\)

\(\Big \downarrow \) 5345

\(\displaystyle \frac {i \left (\frac {i \left (\frac {i \int \frac {(a+b \arctan (c x))^2}{i c x+1}dx}{c}-\frac {i \left (x (a+b \arctan (c x))^2-2 b c \int \frac {x (a+b \arctan (c x))}{c^2 x^2+1}dx\right )}{c}\right )}{c}-\frac {i \left (\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx\right )}{c}\right )}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))^2-\frac {2}{3} b c \int \frac {x^3 (a+b \arctan (c x))}{c^2 x^2+1}dx\right )}{c d}\)

\(\Big \downarrow \) 5379

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

\(\Big \downarrow \) 5451

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 5419

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

\(\Big \downarrow \) 5455

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

\(\Big \downarrow \) 5379

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

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

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

\(\Big \downarrow \) 5529

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

\(\Big \downarrow \) 7164

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

Input:

Int[(x^3*(a + b*ArcTan[c*x])^2)/(d + I*c*d*x),x]
 

Output:

((-I)*((x^3*(a + b*ArcTan[c*x])^2)/3 - (2*b*c*(((x^2*(a + b*ArcTan[c*x]))/ 
2 - (b*c*(x/c^2 - ArcTan[c*x]/c^3))/2)/c^2 - (((-1/2*I)*(a + b*ArcTan[c*x] 
)^2)/(b*c^2) - (((a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c + ((I/2)*b*Poly 
Log[2, 1 - 2/(1 + I*c*x)])/c)/c)/c^2))/3))/(c*d) + (I*(((-I)*((x^2*(a + b* 
ArcTan[c*x])^2)/2 - b*c*(-1/2*(a + b*ArcTan[c*x])^2/(b*c^3) + (a*x + b*x*A 
rcTan[c*x] - (b*Log[1 + c^2*x^2])/(2*c))/c^2)))/c + (I*(((-I)*(x*(a + b*Ar 
cTan[c*x])^2 - 2*b*c*(((-1/2*I)*(a + b*ArcTan[c*x])^2)/(b*c^2) - (((a + b* 
ArcTan[c*x])*Log[2/(1 + I*c*x)])/c + ((I/2)*b*PolyLog[2, 1 - 2/(1 + I*c*x) 
])/c)/c)))/c + (I*((I*(a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/c - (2*I)* 
b*(((-1/2*I)*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/c - (b*Pol 
yLog[3, 1 - 2/(1 + I*c*x)])/(4*c))))/c))/c))/(c*d)
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2752

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo 
g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
 

rule 2849

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp 
[-e/g   Subst[Int[Log[2*d*x]/(1 - 2*d*x), x], x, 1/(d + e*x)], x] /; FreeQ[ 
{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]
 

rule 5345

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a 
+ b*ArcTan[c*x^n])^p, x] - Simp[b*c*n*p   Int[x^n*((a + b*ArcTan[c*x^n])^(p 
 - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0] && 
 (EqQ[n, 1] || EqQ[p, 1])
 

rule 5361

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> 
 Simp[x^(m + 1)*((a + b*ArcTan[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 
1))   Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], 
x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] & 
& IntegerQ[m])) && NeQ[m, -1]
 

rule 5379

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] 
 :> Simp[(-(a + b*ArcTan[c*x])^p)*(Log[2/(1 + e*(x/d))]/e), x] + Simp[b*c*( 
p/e)   Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 + c^2*x^2)) 
, x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0 
]
 

rule 5401

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

rule 5419

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbo 
l] :> Simp[(a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, 
c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]
 

rule 5451

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e 
_.)*(x_)^2), x_Symbol] :> Simp[f^2/e   Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x] 
)^p, x], x] - Simp[d*(f^2/e)   Int[(f*x)^(m - 2)*((a + b*ArcTan[c*x])^p/(d 
+ e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]
 

rule 5455

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), 
x_Symbol] :> Simp[(-I)*((a + b*ArcTan[c*x])^(p + 1)/(b*e*(p + 1))), x] - Si 
mp[1/(c*d)   Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b, c, 
 d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]
 

rule 5529

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2 
), x_Symbol] :> Simp[(-I)*(a + b*ArcTan[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)) 
, x] + Simp[b*p*(I/2)   Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[2, 1 - u]/ 
(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c 
^2*d] && EqQ[(1 - u)^2 - (1 - 2*(I/(I - c*x)))^2, 0]
 

rule 7164

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, 
x]}, Simp[w*PolyLog[n + 1, v], x] /;  !FalseQ[w]] /; FreeQ[n, x]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 11.44 (sec) , antiderivative size = 1105, normalized size of antiderivative = 3.10

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1105\)
default	\(\text {Expression too large to display}\)	\(1105\)
parts	\(\text {Expression too large to display}\)	\(1149\)

Input:

int(x^3*(a+b*arctan(c*x))^2/(d+I*c*d*x),x,method=_RETURNVERBOSE)
 

Output:

1/c^4*(I*a*b/d*ln(c*x-I)*ln(-1/2*I*(c*x+I))+2*I*a*b/d*arctan(c*x)*c*x+1/2* 
a^2/d*c^2*x^2-1/2*a^2/d*ln(c^2*x^2+1)+1/3*I*a*b/d*c^2*x^2+b^2/d*(-1/3*I*ar 
ctan(c*x)^2*c^3*x^3-1/3*I*(c*x+I)+1/2*c^2*x^2*arctan(c*x)^2-arctan(c*x)^2* 
ln(c*x-I)+arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))+8/3*I*arctan(c*x)* 
ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*arctan(c*x)^2*c*x+1/2*polylog(3,-(1+ 
I*c*x)^2/(c^2*x^2+1))+1/2*I*Pi*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn((1 
+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2-ln(1+(1 
+I*c*x)^2/(c^2*x^2+1))-1/2*I*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c* 
x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2-1/2*I*Pi*csg 
n((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3*arctan(c*x)^2-2/3 
*I*arctan(c*x)^3+11/6*arctan(c*x)^2+8/3*I*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^ 
2*x^2+1)^(1/2))-1/2*I*Pi*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn((1+I*c*x 
)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1))) 
*arctan(c*x)^2+I*Pi*arctan(c*x)^2-I*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^ 
2*x^2+1))-1/3*I*arctan(c*x)*(c*x-I)^2-1/3*arctan(c*x)*(c*x-I)+2/3*I*arctan 
(c*x)*(c*x-I)*(c*x+I)-I*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^ 
2*x^2+1)))^2*arctan(c*x)^2+8/3*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+8/3* 
dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2)))-1/3*I*a^2/d*c^3*x^3-I*a^2/d*arctan 
(c*x)+a*b/d*arctan(c*x)*c^2*x^2-2*a*b/d*arctan(c*x)*ln(c*x-I)-5/24*I*a*b/d 
*ln(c^4*x^4+10*c^2*x^2+9)-1/2*I*a*b/d*ln(c*x-I)^2-11/12*I*a*b/d*ln(c^2*...
 

Fricas [F]

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

integrate(x^3*(a+b*arctan(c*x))^2/(d+I*c*d*x),x, algorithm="fricas")
 

Output:

integral(1/4*(I*b^2*x^3*log(-(c*x + I)/(c*x - I))^2 + 4*a*b*x^3*log(-(c*x 
+ I)/(c*x - I)) - 4*I*a^2*x^3)/(c*d*x - I*d), x)
 

Sympy [F(-1)]

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

integrate(x**3*(a+b*atan(c*x))**2/(d+I*c*d*x),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(x^3*(a+b*arctan(c*x))^2/(d+I*c*d*x),x, algorithm="maxima")
 

Output:

-1/6*a^2*(I*(2*c^2*x^3 + 3*I*c*x^2 - 6*x)/(c^3*d) + 6*log(I*c*x + 1)/(c^4* 
d)) - 1/96*(16*I*(216*b^2*c^4*integrate(1/48*x^4*arctan(c*x)^2/(c^5*d*x^2 
+ c^3*d), x) + 18*b^2*c^4*integrate(1/48*x^4*log(c^2*x^2 + 1)^2/(c^5*d*x^2 
 + c^3*d), x) + 576*a*b*c^4*integrate(1/48*x^4*arctan(c*x)/(c^5*d*x^2 + c^ 
3*d), x) + 24*b^2*c^4*integrate(1/48*x^4*log(c^2*x^2 + 1)/(c^5*d*x^2 + c^3 
*d), x) + 72*b^2*c^3*integrate(1/48*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c^5* 
d*x^2 + c^3*d), x) + 24*b^2*c^3*integrate(1/48*x^3*arctan(c*x)/(c^5*d*x^2 
+ c^3*d), x) - 36*b^2*c^2*integrate(1/48*x^2*log(c^2*x^2 + 1)/(c^5*d*x^2 + 
 c^3*d), x) + 144*b^2*c*integrate(1/48*x*arctan(c*x)/(c^5*d*x^2 + c^3*d), 
x) - 36*b^2*integrate(1/48*log(c^2*x^2 + 1)^2/(c^5*d*x^2 + c^3*d), x) - b^ 
2*arctan(c*x)^3/(c^4*d))*c^4*d - 96*c^4*d*integrate(1/48*(12*(3*b^2*c^2*x^ 
3 - 2*b^2*x)*arctan(c*x)^2 + 3*(b^2*c^2*x^3 - 2*b^2*x)*log(c^2*x^2 + 1)^2 
- 4*(2*b^2*c^3*x^4 - 24*a*b*c^2*x^3 - 3*b^2*c*x^2)*arctan(c*x) - 2*(6*b^2* 
c^3*x^4*arctan(c*x) - b^2*c^2*x^3 - 6*b^2*x)*log(c^2*x^2 + 1))/(c^4*d*x^2 
+ c^2*d), x) + 24*I*b^2*arctan(c*x)^3 - 3*b^2*log(c^2*x^2 + 1)^3 - 4*(-2*I 
*b^2*c^3*x^3 + 3*b^2*c^2*x^2 + 6*I*b^2*c*x)*arctan(c*x)^2 + (-2*I*b^2*c^3* 
x^3 + 3*b^2*c^2*x^2 + 6*I*b^2*c*x + 6*I*b^2*arctan(c*x))*log(c^2*x^2 + 1)^ 
2 - 4*(3*b^2*arctan(c*x)^2 + (2*b^2*c^3*x^3 + 3*I*b^2*c^2*x^2 - 6*b^2*c*x) 
*arctan(c*x))*log(c^2*x^2 + 1))/(c^4*d)
 

Giac [F]

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

integrate(x^3*(a+b*arctan(c*x))^2/(d+I*c*d*x),x, algorithm="giac")
 

Output:

integrate((b*arctan(c*x) + a)^2*x^3/(I*c*d*x + d), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx=\int \frac {x^3\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{d+c\,d\,x\,1{}\mathrm {i}} \,d x \] Input:

int((x^3*(a + b*atan(c*x))^2)/(d + c*d*x*1i),x)
 

Output:

int((x^3*(a + b*atan(c*x))^2)/(d + c*d*x*1i), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx=\frac {12 \left (\int \frac {\mathit {atan} \left (c x \right ) x^{3}}{c i x +1}d x \right ) a b \,c^{4}+6 \left (\int \frac {\mathit {atan} \left (c x \right )^{2} x^{3}}{c i x +1}d x \right ) b^{2} c^{4}-6 \,\mathrm {log}\left (c i x +1\right ) a^{2}-2 a^{2} c^{3} i \,x^{3}+3 a^{2} c^{2} x^{2}+6 a^{2} c i x}{6 c^{4} d} \] Input:

int(x^3*(a+b*atan(c*x))^2/(d+I*c*d*x),x)
 

Output:

(12*int((atan(c*x)*x**3)/(c*i*x + 1),x)*a*b*c**4 + 6*int((atan(c*x)**2*x** 
3)/(c*i*x + 1),x)*b**2*c**4 - 6*log(c*i*x + 1)*a**2 - 2*a**2*c**3*i*x**3 + 
 3*a**2*c**2*x**2 + 6*a**2*c*i*x)/(6*c**4*d)