Integrand size = 25, antiderivative size = 273 \[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=-\frac {b c (a+b \arctan (c x))}{d x}-\frac {3 c^2 (a+b \arctan (c x))^2}{2 d}-\frac {(a+b \arctan (c x))^2}{2 d x^2}+\frac {i c (a+b \arctan (c x))^2}{d x}+\frac {b^2 c^2 \log (x)}{d}-\frac {b^2 c^2 \log \left (1+c^2 x^2\right )}{2 d}-\frac {2 i b c^2 (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {c^2 (a+b \arctan (c x))^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {b^2 c^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )}{d}-\frac {i b c^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d}-\frac {b^2 c^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d} \]
-b*c*(a+b*arctan(c*x))/d/x-3/2*c^2*(a+b*arctan(c*x))^2/d-1/2*(a+b*arctan(c *x))^2/d/x^2+I*c*(a+b*arctan(c*x))^2/d/x+b^2*c^2*ln(x)/d-1/2*b^2*c^2*ln(c^ 2*x^2+1)/d-2*I*b*c^2*(a+b*arctan(c*x))*ln(2-2/(1-I*c*x))/d-c^2*(a+b*arctan (c*x))^2*ln(2-2/(1+I*c*x))/d-b^2*c^2*polylog(2,-1+2/(1-I*c*x))/d-I*b*c^2*( a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))/d-1/2*b^2*c^2*polylog(3,-1+2/(1 +I*c*x))/d
Time = 1.23 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.37 \[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=\frac {-\frac {a^2}{x^2}+\frac {2 i a^2 c}{x}+2 i a^2 c^2 \arctan (c x)-2 a^2 c^2 \log (x)+a^2 c^2 \log \left (1+c^2 x^2\right )+\frac {2 i a b \left (2 c^2 x^2 \arctan (c x)^2+\arctan (c x) \left (i+2 c x+i c^2 x^2+2 i c^2 x^2 \log \left (1-e^{2 i \arctan (c x)}\right )\right )+c x \left (i-2 c x \log (c x)+c x \log \left (1+c^2 x^2\right )\right )+c^2 x^2 \operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )\right )}{x^2}+2 b^2 c^2 \left (\frac {i \pi ^3}{24}-\frac {\arctan (c x)}{c x}-\frac {3}{2} \arctan (c x)^2-\frac {\arctan (c x)^2}{2 c^2 x^2}+\frac {i \arctan (c x)^2}{c x}-\arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )-2 i \arctan (c x) \log \left (1-e^{2 i \arctan (c x)}\right )+\log (c x)-\frac {1}{2} \log \left (1+c^2 x^2\right )-i \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )-\operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )\right )}{2 d} \]
(-(a^2/x^2) + ((2*I)*a^2*c)/x + (2*I)*a^2*c^2*ArcTan[c*x] - 2*a^2*c^2*Log[ x] + a^2*c^2*Log[1 + c^2*x^2] + ((2*I)*a*b*(2*c^2*x^2*ArcTan[c*x]^2 + ArcT an[c*x]*(I + 2*c*x + I*c^2*x^2 + (2*I)*c^2*x^2*Log[1 - E^((2*I)*ArcTan[c*x ])]) + c*x*(I - 2*c*x*Log[c*x] + c*x*Log[1 + c^2*x^2]) + c^2*x^2*PolyLog[2 , E^((2*I)*ArcTan[c*x])]))/x^2 + 2*b^2*c^2*((I/24)*Pi^3 - ArcTan[c*x]/(c*x ) - (3*ArcTan[c*x]^2)/2 - ArcTan[c*x]^2/(2*c^2*x^2) + (I*ArcTan[c*x]^2)/(c *x) - ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - (2*I)*ArcTan[c*x]*Lo g[1 - E^((2*I)*ArcTan[c*x])] + Log[c*x] - Log[1 + c^2*x^2]/2 - I*ArcTan[c* x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] - PolyLog[2, E^((2*I)*ArcTan[c*x])] - PolyLog[3, E^((-2*I)*ArcTan[c*x])]/2))/(2*d)
Time = 2.54 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.01, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {5405, 27, 5361, 5405, 5361, 5403, 5453, 5361, 243, 47, 14, 16, 5419, 5459, 5403, 2897, 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.
| \(\displaystyle \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx\) |
| \(\Big \downarrow \) 5405 |
| \(\displaystyle \frac {\int \frac {(a+b \arctan (c x))^2}{x^3}dx}{d}-i c \int \frac {(a+b \arctan (c x))^2}{d x^2 (i c x+1)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(a+b \arctan (c x))^2}{x^3}dx}{d}-\frac {i c \int \frac {(a+b \arctan (c x))^2}{x^2 (i c x+1)}dx}{d}\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {b c \int \frac {a+b \arctan (c x)}{x^2 \left (c^2 x^2+1\right )}dx-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \int \frac {(a+b \arctan (c x))^2}{x^2 (i c x+1)}dx}{d}\) |
| \(\Big \downarrow \) 5405 |
| \(\displaystyle \frac {b c \int \frac {a+b \arctan (c x)}{x^2 \left (c^2 x^2+1\right )}dx-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (\int \frac {(a+b \arctan (c x))^2}{x^2}dx-i c \int \frac {(a+b \arctan (c x))^2}{x (i c x+1)}dx\right )}{d}\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {b c \int \frac {a+b \arctan (c x)}{x^2 \left (c^2 x^2+1\right )}dx-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (2 b c \int \frac {a+b \arctan (c x)}{x \left (c^2 x^2+1\right )}dx-i c \int \frac {(a+b \arctan (c x))^2}{x (i c x+1)}dx-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
| \(\Big \downarrow \) 5403 |
| \(\displaystyle \frac {b c \int \frac {a+b \arctan (c x)}{x^2 \left (c^2 x^2+1\right )}dx-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (2 b c \int \frac {a+b \arctan (c x)}{x \left (c^2 x^2+1\right )}dx-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-2 b c \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
| \(\Big \downarrow \) 5453 |
| \(\displaystyle \frac {b c \left (\int \frac {a+b \arctan (c x)}{x^2}dx-c^2 \int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (2 b c \int \frac {a+b \arctan (c x)}{x \left (c^2 x^2+1\right )}dx-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-2 b c \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
| \(\Big \downarrow \) 5361 |
| \(\displaystyle \frac {b c \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )+b c \int \frac {1}{x \left (c^2 x^2+1\right )}dx-\frac {a+b \arctan (c x)}{x}\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (2 b c \int \frac {a+b \arctan (c x)}{x \left (c^2 x^2+1\right )}dx-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-2 b c \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {b c \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )+\frac {1}{2} b c \int \frac {1}{x^2 \left (c^2 x^2+1\right )}dx^2-\frac {a+b \arctan (c x)}{x}\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (2 b c \int \frac {a+b \arctan (c x)}{x \left (c^2 x^2+1\right )}dx-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-2 b c \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
| \(\Big \downarrow \) 47 |
| \(\displaystyle \frac {b c \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )+\frac {1}{2} b c \left (\int \frac {1}{x^2}dx^2-c^2 \int \frac {1}{c^2 x^2+1}dx^2\right )-\frac {a+b \arctan (c x)}{x}\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (2 b c \int \frac {a+b \arctan (c x)}{x \left (c^2 x^2+1\right )}dx-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-2 b c \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
| \(\Big \downarrow \) 14 |
| \(\displaystyle \frac {b c \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )+\frac {1}{2} b c \left (\log \left (x^2\right )-c^2 \int \frac {1}{c^2 x^2+1}dx^2\right )-\frac {a+b \arctan (c x)}{x}\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (2 b c \int \frac {a+b \arctan (c x)}{x \left (c^2 x^2+1\right )}dx-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-2 b c \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {b c \left (c^2 \left (-\int \frac {a+b \arctan (c x)}{c^2 x^2+1}dx\right )-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (2 b c \int \frac {a+b \arctan (c x)}{x \left (c^2 x^2+1\right )}dx-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-2 b c \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
| \(\Big \downarrow \) 5419 |
| \(\displaystyle \frac {b c \left (-\frac {c (a+b \arctan (c x))^2}{2 b}-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (2 b c \int \frac {a+b \arctan (c x)}{x \left (c^2 x^2+1\right )}dx-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-2 b c \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
| \(\Big \downarrow \) 5459 |
| \(\displaystyle \frac {b c \left (-\frac {c (a+b \arctan (c x))^2}{2 b}-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-2 b c \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )+2 b c \left (i \int \frac {a+b \arctan (c x)}{x (c x+i)}dx-\frac {i (a+b \arctan (c x))^2}{2 b}\right )-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
| \(\Big \downarrow \) 5403 |
| \(\displaystyle \frac {b c \left (-\frac {c (a+b \arctan (c x))^2}{2 b}-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (2 b c \left (i \left (i b c \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{c^2 x^2+1}dx-i \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))\right )-\frac {i (a+b \arctan (c x))^2}{2 b}\right )-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-2 b c \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
| \(\Big \downarrow \) 2897 |
| \(\displaystyle \frac {b c \left (-\frac {c (a+b \arctan (c x))^2}{2 b}-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-2 b c \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )+2 b c \left (i \left (-i \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right )\right )-\frac {i (a+b \arctan (c x))^2}{2 b}\right )-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
| \(\Big \downarrow \) 5529 |
| \(\displaystyle \frac {b c \left (-\frac {c (a+b \arctan (c x))^2}{2 b}-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-2 b c \left (\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right )}{c^2 x^2+1}dx-\frac {i \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))}{2 c}\right )\right )+2 b c \left (i \left (-i \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right )\right )-\frac {i (a+b \arctan (c x))^2}{2 b}\right )-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
| \(\Big \downarrow \) 7164 |
| \(\displaystyle \frac {b c \left (-\frac {c (a+b \arctan (c x))^2}{2 b}-\frac {a+b \arctan (c x)}{x}+\frac {1}{2} b c \left (\log \left (x^2\right )-\log \left (c^2 x^2+1\right )\right )\right )-\frac {(a+b \arctan (c x))^2}{2 x^2}}{d}-\frac {i c \left (2 b c \left (i \left (-i \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))-\frac {1}{2} b \operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right )\right )-\frac {i (a+b \arctan (c x))^2}{2 b}\right )-i c \left (\log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-2 b c \left (-\frac {i \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))}{2 c}-\frac {b \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right )}{4 c}\right )\right )-\frac {(a+b \arctan (c x))^2}{x}\right )}{d}\) |
(-1/2*(a + b*ArcTan[c*x])^2/x^2 + b*c*(-((a + b*ArcTan[c*x])/x) - (c*(a + b*ArcTan[c*x])^2)/(2*b) + (b*c*(Log[x^2] - Log[1 + c^2*x^2]))/2))/d - (I*c *(-((a + b*ArcTan[c*x])^2/x) + 2*b*c*(((-1/2*I)*(a + b*ArcTan[c*x])^2)/b + I*((-I)*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)] - (b*PolyLog[2, -1 + 2 /(1 - I*c*x)])/2)) - I*c*((a + b*ArcTan[c*x])^2*Log[2 - 2/(1 + I*c*x)] - 2 *b*c*(((-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)))))/d
3.2.1.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
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]
Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, x][[2]], Expon[Pq, x]]
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]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_ Symbol] :> Simp[(a + b*ArcTan[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Si mp[b*c*(p/d) Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 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]
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e _.)*(x_)), x_Symbol] :> Simp[1/d Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x ] - Simp[e/(d*f) 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] && LtQ[m, -1]
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]
Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e _.)*(x_)^2), x_Symbol] :> Simp[1/d Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Simp[e/(d*f^2) 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] && LtQ[m, -1]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(-I)*((a + b*ArcTan[c*x])^(p + 1)/(b*d*(p + 1))), x] + Si mp[I/d Int[(a + b*ArcTan[c*x])^p/(x*(I + c*x)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 0]
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 30.95 (sec) , antiderivative size = 1873, normalized size of antiderivative = 6.86
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1873\) |
| default | \(\text {Expression too large to display}\) | \(1873\) |
| parts | \(\text {Expression too large to display}\) | \(1877\) |
c^2*(-1/2*a^2/d/c^2/x^2+I*a^2/d/c/x-a^2/d*ln(c*x)+1/2*a^2/d*ln(c^2*x^2+1)+ I*a^2/d*arctan(c*x)+b^2/d*(-1/2/c^2/x^2*arctan(c*x)^2-2*polylog(3,(1+I*c*x )/(c^2*x^2+1)^(1/2))+ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*polylog(3,-(1+I*c *x)/(c^2*x^2+1)^(1/2))+ln((1+I*c*x)/(c^2*x^2+1)^(1/2)-1)-3/2*arctan(c*x)^2 +arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)-arctan(c*x)^2*ln(c*x)-arctan( c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))-arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2 *x^2+1)^(1/2))-2*dilog(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*dilog((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(I*((1+I *c*x)^2/(c^2*x^2+1)-1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/ (c^2*x^2+1)))*arctan(c*x)^2+2*I*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1 )^(1/2))+2*I*arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/2*I*Pi* csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(((1+I *c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2-1/2*I* Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn((( 1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*arctan(c*x)^2+1/2*I *Pi*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1) /(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2+1/2*I*Pi*csgn(I*((1+I*c*x)^2 /(c^2*x^2+1)-1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^ 2+1)))^2*arctan(c*x)^2-1/2*I*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c *x)^2/(c^2*x^2+1)))^3*arctan(c*x)^2+1/2*I*Pi*csgn(((1+I*c*x)^2/(c^2*x^2...
\[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )} x^{3}} \,d x } \]
1/8*(2*b^2*c^2*x^2*log(2*c*x/(c*x - I))*log(-(c*x + I)/(c*x - I))^2 + 4*b^ 2*c^2*x^2*dilog(-2*c*x/(c*x - I) + 1)*log(-(c*x + I)/(c*x - I)) - 4*b^2*c^ 2*x^2*polylog(3, -(c*x + I)/(c*x - I)) + 8*d*x^2*integral(1/2*(-2*I*a^2*c* x + 2*a^2 + (2*b^2*c^2*x^2 + (2*a*b + I*b^2)*c*x + 2*I*a*b)*log(-(c*x + I) /(c*x - I)))/(c^2*d*x^5 + d*x^3), x) + (-2*I*b^2*c*x + b^2)*log(-(c*x + I) /(c*x - I))^2)/(d*x^2)
\[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=- \frac {i \left (\int \frac {a^{2}}{c x^{4} - i x^{3}}\, dx + \int \frac {b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{c x^{4} - i x^{3}}\, dx + \int \frac {2 a b \operatorname {atan}{\left (c x \right )}}{c x^{4} - i x^{3}}\, dx\right )}{d} \]
-I*(Integral(a**2/(c*x**4 - I*x**3), x) + Integral(b**2*atan(c*x)**2/(c*x* *4 - I*x**3), x) + Integral(2*a*b*atan(c*x)/(c*x**4 - I*x**3), x))/d
\[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )} x^{3}} \,d x } \]
1/2*(2*c^2*log(I*c*x + 1)/d - 2*c^2*log(x)/d + (2*I*c*x - 1)/(d*x^2))*a^2 - 1/96*(-24*I*b^2*c^2*x^2*arctan(c*x)^3 + 3*b^2*c^2*x^2*log(c^2*x^2 + 1)^3 - 2*(384*b^2*c^4*integrate(1/16*x^4*arctan(c*x)^2/(c^2*d*x^5 + d*x^3), x) + b^2*c^2*log(c^2*x^2 + 1)^3/d + 12*b^2*c^2*arctan(c*x)^2/d + 96*b^2*c^2* integrate(1/16*x^2*log(c^2*x^2 + 1)/(c^2*d*x^5 + d*x^3), x) - 192*b^2*c*in tegrate(1/16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*d*x^5 + d*x^3), x) + 192* b^2*c*integrate(1/16*x*arctan(c*x)/(c^2*d*x^5 + d*x^3), x) + 576*b^2*integ rate(1/16*arctan(c*x)^2/(c^2*d*x^5 + d*x^3), x) + 48*b^2*integrate(1/16*lo g(c^2*x^2 + 1)^2/(c^2*d*x^5 + d*x^3), x) + 1536*a*b*integrate(1/16*arctan( c*x)/(c^2*d*x^5 + d*x^3), x))*d*x^2 + 16*I*(b^2*c^2*arctan(c*x)^3/d + 12*b ^2*c^3*integrate(1/16*x^3*log(c^2*x^2 + 1)^2/(c^2*d*x^5 + d*x^3), x) - 24* b^2*c^3*integrate(1/16*x^3*log(c^2*x^2 + 1)/(c^2*d*x^5 + d*x^3), x) + 24*b ^2*c^2*integrate(1/16*x^2*arctan(c*x)/(c^2*d*x^5 + d*x^3), x) + 72*b^2*c*i ntegrate(1/16*x*arctan(c*x)^2/(c^2*d*x^5 + d*x^3), x) + 6*b^2*c*integrate( 1/16*x*log(c^2*x^2 + 1)^2/(c^2*d*x^5 + d*x^3), x) + 192*a*b*c*integrate(1/ 16*x*arctan(c*x)/(c^2*d*x^5 + d*x^3), x) - 12*b^2*c*integrate(1/16*x*log(c ^2*x^2 + 1)/(c^2*d*x^5 + d*x^3), x) + 24*b^2*integrate(1/16*arctan(c*x)*lo g(c^2*x^2 + 1)/(c^2*d*x^5 + d*x^3), x))*d*x^2 + 12*(-2*I*b^2*c*x + b^2)*ar ctan(c*x)^2 - 3*(2*I*b^2*c^2*x^2*arctan(c*x) - 2*I*b^2*c*x + b^2)*log(c^2* x^2 + 1)^2 + 12*(b^2*c^2*x^2*arctan(c*x)^2 + (2*b^2*c*x + I*b^2)*arctan...
\[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )} x^{3}} \,d x } \]
Timed out. \[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{x^3\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )} \,d x \]