3.1.25 \(\int x^4 (a+b \arctan (c x))^3 \, dx\) [25]

3.1.25.1 Optimal result

Integrand size = 14, antiderivative size = 271 \[ \int x^4 (a+b \arctan (c x))^3 \, dx=-\frac {9 a b^2 x}{10 c^4}-\frac {b^3 x^2}{20 c^3}-\frac {9 b^3 x \arctan (c x)}{10 c^4}+\frac {b^2 x^3 (a+b \arctan (c x))}{10 c^2}+\frac {9 b (a+b \arctan (c x))^2}{20 c^5}+\frac {3 b x^2 (a+b \arctan (c x))^2}{10 c^3}-\frac {3 b x^4 (a+b \arctan (c x))^2}{20 c}+\frac {i (a+b \arctan (c x))^3}{5 c^5}+\frac {1}{5} x^5 (a+b \arctan (c x))^3+\frac {3 b (a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{5 c^5}+\frac {b^3 \log \left (1+c^2 x^2\right )}{2 c^5}+\frac {3 i b^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{5 c^5}+\frac {3 b^3 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{10 c^5} \]

output

-9/10*a*b^2*x/c^4-1/20*b^3*x^2/c^3-9/10*b^3*x*arctan(c*x)/c^4+1/10*b^2*x^3 
*(a+b*arctan(c*x))/c^2+9/20*b*(a+b*arctan(c*x))^2/c^5+3/10*b*x^2*(a+b*arct 
an(c*x))^2/c^3-3/20*b*x^4*(a+b*arctan(c*x))^2/c+1/5*I*(a+b*arctan(c*x))^3/ 
c^5+1/5*x^5*(a+b*arctan(c*x))^3+3/5*b*(a+b*arctan(c*x))^2*ln(2/(1+I*c*x))/ 
c^5+1/2*b^3*ln(c^2*x^2+1)/c^5+3/5*I*b^2*(a+b*arctan(c*x))*polylog(2,1-2/(1 
+I*c*x))/c^5+3/10*b^3*polylog(3,1-2/(1+I*c*x))/c^5
 

3.1.25.2 Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.46 \[ \int x^4 (a+b \arctan (c x))^3 \, dx=\frac {-b^3-18 a b^2 c x+6 a^2 b c^2 x^2-b^3 c^2 x^2+2 a b^2 c^3 x^3-3 a^2 b c^4 x^4+4 a^3 c^5 x^5+18 a b^2 \arctan (c x)-18 b^3 c x \arctan (c x)+12 a b^2 c^2 x^2 \arctan (c x)+2 b^3 c^3 x^3 \arctan (c x)-6 a b^2 c^4 x^4 \arctan (c x)+12 a^2 b c^5 x^5 \arctan (c x)-12 i a b^2 \arctan (c x)^2+9 b^3 \arctan (c x)^2+6 b^3 c^2 x^2 \arctan (c x)^2-3 b^3 c^4 x^4 \arctan (c x)^2+12 a b^2 c^5 x^5 \arctan (c x)^2-4 i b^3 \arctan (c x)^3+4 b^3 c^5 x^5 \arctan (c x)^3+24 a b^2 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )+12 b^3 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )-6 a^2 b \log \left (1+c^2 x^2\right )+10 b^3 \log \left (1+c^2 x^2\right )-12 i b^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+6 b^3 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )}{20 c^5} \]

input

Integrate[x^4*(a + b*ArcTan[c*x])^3,x]
 

output

(-b^3 - 18*a*b^2*c*x + 6*a^2*b*c^2*x^2 - b^3*c^2*x^2 + 2*a*b^2*c^3*x^3 - 3 
*a^2*b*c^4*x^4 + 4*a^3*c^5*x^5 + 18*a*b^2*ArcTan[c*x] - 18*b^3*c*x*ArcTan[ 
c*x] + 12*a*b^2*c^2*x^2*ArcTan[c*x] + 2*b^3*c^3*x^3*ArcTan[c*x] - 6*a*b^2* 
c^4*x^4*ArcTan[c*x] + 12*a^2*b*c^5*x^5*ArcTan[c*x] - (12*I)*a*b^2*ArcTan[c 
*x]^2 + 9*b^3*ArcTan[c*x]^2 + 6*b^3*c^2*x^2*ArcTan[c*x]^2 - 3*b^3*c^4*x^4* 
ArcTan[c*x]^2 + 12*a*b^2*c^5*x^5*ArcTan[c*x]^2 - (4*I)*b^3*ArcTan[c*x]^3 + 
 4*b^3*c^5*x^5*ArcTan[c*x]^3 + 24*a*b^2*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTa 
n[c*x])] + 12*b^3*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] - 6*a^2*b*L 
og[1 + c^2*x^2] + 10*b^3*Log[1 + c^2*x^2] - (12*I)*b^2*(a + b*ArcTan[c*x]) 
*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 6*b^3*PolyLog[3, -E^((2*I)*ArcTan[c* 
x])])/(20*c^5)
 

3.1.25.3 Rubi [A] (verified)

Time = 2.38 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.37, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.071, Rules used = {5361, 5451, 5361, 5451, 5361, 243, 49, 2009, 5451, 2009, 5419, 5455, 5379, 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.

\(\displaystyle \int x^4 (a+b \arctan (c x))^3 \, dx\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \int \frac {x^5 (a+b \arctan (c x))^2}{c^2 x^2+1}dx\)

\(\Big \downarrow \) 5451

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\int x^3 (a+b \arctan (c x))^2dx}{c^2}-\frac {\int \frac {x^3 (a+b \arctan (c x))^2}{c^2 x^2+1}dx}{c^2}\right )\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} b c \int \frac {x^4 (a+b \arctan (c x))}{c^2 x^2+1}dx}{c^2}-\frac {\int \frac {x^3 (a+b \arctan (c x))^2}{c^2 x^2+1}dx}{c^2}\right )\)

\(\Big \downarrow \) 5451

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} b c \left (\frac {\int x^2 (a+b \arctan (c x))dx}{c^2}-\frac {\int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx}{c^2}\right )}{c^2}-\frac {\frac {\int x (a+b \arctan (c x))^2dx}{c^2}-\frac {\int \frac {x (a+b \arctan (c x))^2}{c^2 x^2+1}dx}{c^2}}{c^2}\right )\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} b c \left (\frac {\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{3} b c \int \frac {x^3}{c^2 x^2+1}dx}{c^2}-\frac {\int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx}{c^2}\right )}{c^2}-\frac {\frac {\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx}{c^2}-\frac {\int \frac {x (a+b \arctan (c x))^2}{c^2 x^2+1}dx}{c^2}}{c^2}\right )\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} b c \left (\frac {\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \int \frac {x^2}{c^2 x^2+1}dx^2}{c^2}-\frac {\int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx}{c^2}\right )}{c^2}-\frac {\frac {\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx}{c^2}-\frac {\int \frac {x (a+b \arctan (c x))^2}{c^2 x^2+1}dx}{c^2}}{c^2}\right )\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} b c \left (\frac {\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (c^2 x^2+1\right )}\right )dx^2}{c^2}-\frac {\int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx}{c^2}\right )}{c^2}-\frac {\frac {\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx}{c^2}-\frac {\int \frac {x (a+b \arctan (c x))^2}{c^2 x^2+1}dx}{c^2}}{c^2}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} b c \left (\frac {\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )}{c^2}-\frac {\int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx}{c^2}\right )}{c^2}-\frac {\frac {\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \int \frac {x^2 (a+b \arctan (c x))}{c^2 x^2+1}dx}{c^2}-\frac {\int \frac {x (a+b \arctan (c x))^2}{c^2 x^2+1}dx}{c^2}}{c^2}\right )\)

\(\Big \downarrow \) 5451

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} b c \left (\frac {\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )}{c^2}-\frac {\frac {\int (a+b \arctan (c x))dx}{c^2}-\frac {\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx}{c^2}}{c^2}\right )}{c^2}-\frac {\frac {\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \left (\frac {\int (a+b \arctan (c x))dx}{c^2}-\frac {\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx}{c^2}\right )}{c^2}-\frac {\int \frac {x (a+b \arctan (c x))^2}{c^2 x^2+1}dx}{c^2}}{c^2}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} b c \left (\frac {\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )}{c^2}-\frac {\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx}{c^2}}{c^2}\right )}{c^2}-\frac {\frac {\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \left (\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx}{c^2}\right )}{c^2}-\frac {\int \frac {x (a+b \arctan (c x))^2}{c^2 x^2+1}dx}{c^2}}{c^2}\right )\)

\(\Big \downarrow \) 5419

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} b c \left (\frac {\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )}{c^2}-\frac {\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {(a+b \arctan (c x))^2}{2 b c^3}}{c^2}\right )}{c^2}-\frac {\frac {\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \left (\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {(a+b \arctan (c x))^2}{2 b c^3}\right )}{c^2}-\frac {\int \frac {x (a+b \arctan (c x))^2}{c^2 x^2+1}dx}{c^2}}{c^2}\right )\)

\(\Big \downarrow \) 5455

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} b c \left (\frac {\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )}{c^2}-\frac {\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {(a+b \arctan (c x))^2}{2 b c^3}}{c^2}\right )}{c^2}-\frac {\frac {\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \left (\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {(a+b \arctan (c x))^2}{2 b c^3}\right )}{c^2}-\frac {-\frac {\int \frac {(a+b \arctan (c x))^2}{i-c x}dx}{c}-\frac {i (a+b \arctan (c x))^3}{3 b c^2}}{c^2}}{c^2}\right )\)

\(\Big \downarrow \) 5379

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} b c \left (\frac {\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )}{c^2}-\frac {\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {(a+b \arctan (c x))^2}{2 b c^3}}{c^2}\right )}{c^2}-\frac {\frac {\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \left (\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {(a+b \arctan (c x))^2}{2 b c^3}\right )}{c^2}-\frac {-\frac {\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 b \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx}{c}-\frac {i (a+b \arctan (c x))^3}{3 b c^2}}{c^2}}{c^2}\right )\)

\(\Big \downarrow \) 5529

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} b c \left (\frac {\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )}{c^2}-\frac {\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {(a+b \arctan (c x))^2}{2 b c^3}}{c^2}\right )}{c^2}-\frac {\frac {\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \left (\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {(a+b \arctan (c x))^2}{2 b c^3}\right )}{c^2}-\frac {-\frac {\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 b \left (\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{2 c}\right )}{c}-\frac {i (a+b \arctan (c x))^3}{3 b c^2}}{c^2}}{c^2}\right )\)

\(\Big \downarrow \) 7164

\(\displaystyle \frac {1}{5} x^5 (a+b \arctan (c x))^3-\frac {3}{5} b c \left (\frac {\frac {1}{4} x^4 (a+b \arctan (c x))^2-\frac {1}{2} b c \left (\frac {\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )}{c^2}-\frac {\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {(a+b \arctan (c x))^2}{2 b c^3}}{c^2}\right )}{c^2}-\frac {\frac {\frac {1}{2} x^2 (a+b \arctan (c x))^2-b c \left (\frac {a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}}{c^2}-\frac {(a+b \arctan (c x))^2}{2 b c^3}\right )}{c^2}-\frac {-\frac {i (a+b \arctan (c x))^3}{3 b c^2}-\frac {\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 b \left (-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{2 c}-\frac {b \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{4 c}\right )}{c}}{c^2}}{c^2}\right )\)

input

Int[x^4*(a + b*ArcTan[c*x])^3,x]
 

output

(x^5*(a + b*ArcTan[c*x])^3)/5 - (3*b*c*(((x^4*(a + b*ArcTan[c*x])^2)/4 - ( 
b*c*(((x^3*(a + b*ArcTan[c*x]))/3 - (b*c*(x^2/c^2 - Log[1 + c^2*x^2]/c^4)) 
/6)/c^2 - (-1/2*(a + b*ArcTan[c*x])^2/(b*c^3) + (a*x + b*x*ArcTan[c*x] - ( 
b*Log[1 + c^2*x^2])/(2*c))/c^2)/c^2))/2)/c^2 - (((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*ArcTan[c*x] - 
(b*Log[1 + c^2*x^2])/(2*c))/c^2))/c^2 - (((-1/3*I)*(a + b*ArcTan[c*x])^3)/ 
(b*c^2) - (((a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/c - 2*b*(((-1/2*I)*( 
a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/c - (b*PolyLog[3, 1 - 2/ 
(1 + I*c*x)])/(4*c)))/c)/c^2)/c^2))/5
 

3.1.25.3.1 Defintions of rubi rules used

rule 49

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

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

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 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]
 

3.1.25.4 Maple [C] (warning: unable to verify)

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

Time = 6.28 (sec) , antiderivative size = 1185, normalized size of antiderivative = 4.37

\[\text {Expression too large to display}\]

input

int(x^4*(a+b*arctan(c*x))^3,x)
 

output

1/c^5*(1/5*a^3*c^5*x^5+b^3*(1/5*c^5*x^5*arctan(c*x)^3-3/20*c^4*x^4*arctan( 
c*x)^2+3/10*c^2*x^2*arctan(c*x)^2-3/10*arctan(c*x)^2*ln(c^2*x^2+1)+3/5*arc 
tan(c*x)^2*ln((1+I*c*x)/(c^2*x^2+1)^(1/2))-1/20*I*(3*Pi*csgn(I*(1+I*c*x)^2 
/(c^2*x^2+1))^3*arctan(c*x)^2-6*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^2*csgn( 
I*(1+I*c*x)/(c^2*x^2+1)^(1/2))*arctan(c*x)^2+3*Pi*csgn(I*(1+I*c*x)^2/(c^2* 
x^2+1))*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1))^2)*csgn(I*(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)^2-3*Pi*csgn(I*(1+I*c*x)^2/(c 
^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1))^2)^2 
*arctan(c*x)^2+3*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)/(c^2* 
x^2+1)^(1/2))^2*arctan(c*x)^2-3*Pi*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1))^2)*c 
sgn(I*(1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1))^2)^2*arctan(c*x) 
^2+3*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1))^2)^3*ar 
ctan(c*x)^2-3*Pi*csgn(I*(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*csgn(I*(1+(1+I*c*x) 
^2/(c^2*x^2+1))^2)*arctan(c*x)^2+6*Pi*csgn(I*(1+(1+I*c*x)^2/(c^2*x^2+1)))* 
csgn(I*(1+(1+I*c*x)^2/(c^2*x^2+1))^2)^2*arctan(c*x)^2-3*Pi*csgn(I*(1+(1+I* 
c*x)^2/(c^2*x^2+1))^2)^3*arctan(c*x)^2+4*arctan(c*x)^3+2*I*arctan(c*x)*c^3 
*x^3-I*c^2*x^2-20*arctan(c*x)-18*I*arctan(c*x)*c*x+12*I*ln(2)*arctan(c*x)^ 
2+9*I*arctan(c*x)^2-I)-ln(1+(1+I*c*x)^2/(c^2*x^2+1))-3/5*I*arctan(c*x)*pol 
ylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+3/10*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))) 
+3*a*b^2*(1/5*c^5*x^5*arctan(c*x)^2-1/10*c^4*x^4*arctan(c*x)+1/5*c^2*x^...
 

3.1.25.5 Fricas [F]

\[ \int x^4 (a+b \arctan (c x))^3 \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )}^{3} x^{4} \,d x } \]

input

integrate(x^4*(a+b*arctan(c*x))^3,x, algorithm="fricas")
 

output

integral(b^3*x^4*arctan(c*x)^3 + 3*a*b^2*x^4*arctan(c*x)^2 + 3*a^2*b*x^4*a 
rctan(c*x) + a^3*x^4, x)
 

3.1.25.6 Sympy [F]

\[ \int x^4 (a+b \arctan (c x))^3 \, dx=\int x^{4} \left (a + b \operatorname {atan}{\left (c x \right )}\right )^{3}\, dx \]

input

integrate(x**4*(a+b*atan(c*x))**3,x)
 

output

Integral(x**4*(a + b*atan(c*x))**3, x)
 

3.1.25.7 Maxima [F]

\[ \int x^4 (a+b \arctan (c x))^3 \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )}^{3} x^{4} \,d x } \]

input

integrate(x^4*(a+b*arctan(c*x))^3,x, algorithm="maxima")
 

output

1/40*b^3*x^5*arctan(c*x)^3 - 3/160*b^3*x^5*arctan(c*x)*log(c^2*x^2 + 1)^2 
+ 1/5*a^3*x^5 + 3/20*(4*x^5*arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log 
(c^2*x^2 + 1)/c^6))*a^2*b + integrate(1/160*(12*b^3*c^2*x^6*arctan(c*x)*lo 
g(c^2*x^2 + 1) + 140*(b^3*c^2*x^6 + b^3*x^4)*arctan(c*x)^3 + 12*(40*a*b^2* 
c^2*x^6 - b^3*c*x^5 + 40*a*b^2*x^4)*arctan(c*x)^2 + 3*(b^3*c*x^5 + 5*(b^3* 
c^2*x^6 + b^3*x^4)*arctan(c*x))*log(c^2*x^2 + 1)^2)/(c^2*x^2 + 1), x)
 

3.1.25.8 Giac [F]

\[ \int x^4 (a+b \arctan (c x))^3 \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )}^{3} x^{4} \,d x } \]

input

integrate(x^4*(a+b*arctan(c*x))^3,x, algorithm="giac")
 

output

sage0*x
 

3.1.25.9 Mupad [F(-1)]

Timed out. \[ \int x^4 (a+b \arctan (c x))^3 \, dx=\int x^4\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3 \,d x \]

input

int(x^4*(a + b*atan(c*x))^3,x)
 

output

int(x^4*(a + b*atan(c*x))^3, x)