Integrand size = 23, antiderivative size = 277 \[ \int \frac {x (a+b \arctan (c x))^3}{d+i c d x} \, dx=\frac {(a+b \arctan (c x))^3}{c^2 d}-\frac {i x (a+b \arctan (c x))^3}{c d}-\frac {3 i b (a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^2 d}-\frac {(a+b \arctan (c x))^3 \log \left (\frac {2}{1+i c x}\right )}{c^2 d}+\frac {3 b^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^2 d}-\frac {3 i b (a+b \arctan (c x))^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c^2 d}-\frac {3 i b^3 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^2 d}-\frac {3 b^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^2 d}+\frac {3 i b^3 \operatorname {PolyLog}\left (4,1-\frac {2}{1+i c x}\right )}{4 c^2 d} \]
(a+b*arctan(c*x))^3/c^2/d-I*x*(a+b*arctan(c*x))^3/c/d-3*I*b*(a+b*arctan(c* x))^2*ln(2/(1+I*c*x))/c^2/d-(a+b*arctan(c*x))^3*ln(2/(1+I*c*x))/c^2/d+3*b^ 2*(a+b*arctan(c*x))*polylog(2,1-2/(1+I*c*x))/c^2/d-3/2*I*b*(a+b*arctan(c*x ))^2*polylog(2,1-2/(1+I*c*x))/c^2/d-3/2*I*b^3*polylog(3,1-2/(1+I*c*x))/c^2 /d-3/2*b^2*(a+b*arctan(c*x))*polylog(3,1-2/(1+I*c*x))/c^2/d+3/4*I*b^3*poly log(4,1-2/(1+I*c*x))/c^2/d
Time = 0.98 (sec) , antiderivative size = 393, normalized size of antiderivative = 1.42 \[ \int \frac {x (a+b \arctan (c x))^3}{d+i c d x} \, dx=-\frac {i \left (4 a^3 c x-4 a^3 \arctan (c x)+12 a^2 b c x \arctan (c x)-12 a^2 b \arctan (c x)^2-12 i a b^2 \arctan (c x)^2+12 a b^2 c x \arctan (c x)^2-8 a b^2 \arctan (c x)^3-4 i b^3 \arctan (c x)^3+4 b^3 c x \arctan (c x)^3-2 b^3 \arctan (c x)^4-12 i a^2 b \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )+24 a b^2 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )-12 i a b^2 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+12 b^3 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )-4 i b^3 \arctan (c x)^3 \log \left (1+e^{2 i \arctan (c x)}\right )+2 i a^3 \log \left (1+c^2 x^2\right )-6 a^2 b \log \left (1+c^2 x^2\right )-6 b \left (a (a+2 i b)+2 (a+i b) b \arctan (c x)+b^2 \arctan (c x)^2\right ) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+6 b^2 (-i a+b-i b \arctan (c x)) \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )+3 b^3 \operatorname {PolyLog}\left (4,-e^{2 i \arctan (c x)}\right )\right )}{4 c^2 d} \]
((-1/4*I)*(4*a^3*c*x - 4*a^3*ArcTan[c*x] + 12*a^2*b*c*x*ArcTan[c*x] - 12*a ^2*b*ArcTan[c*x]^2 - (12*I)*a*b^2*ArcTan[c*x]^2 + 12*a*b^2*c*x*ArcTan[c*x] ^2 - 8*a*b^2*ArcTan[c*x]^3 - (4*I)*b^3*ArcTan[c*x]^3 + 4*b^3*c*x*ArcTan[c* x]^3 - 2*b^3*ArcTan[c*x]^4 - (12*I)*a^2*b*ArcTan[c*x]*Log[1 + E^((2*I)*Arc Tan[c*x])] + 24*a*b^2*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] - (12*I)* a*b^2*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + 12*b^3*ArcTan[c*x]^2* Log[1 + E^((2*I)*ArcTan[c*x])] - (4*I)*b^3*ArcTan[c*x]^3*Log[1 + E^((2*I)* ArcTan[c*x])] + (2*I)*a^3*Log[1 + c^2*x^2] - 6*a^2*b*Log[1 + c^2*x^2] - 6* b*(a*(a + (2*I)*b) + 2*(a + I*b)*b*ArcTan[c*x] + b^2*ArcTan[c*x]^2)*PolyLo g[2, -E^((2*I)*ArcTan[c*x])] + 6*b^2*((-I)*a + b - I*b*ArcTan[c*x])*PolyLo g[3, -E^((2*I)*ArcTan[c*x])] + 3*b^3*PolyLog[4, -E^((2*I)*ArcTan[c*x])]))/ (c^2*d)
Time = 1.62 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {5401, 27, 5345, 5379, 5455, 5379, 5529, 5533, 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.
| \(\displaystyle \int \frac {x (a+b \arctan (c x))^3}{d+i c d x} \, dx\) |
| \(\Big \downarrow \) 5401 |
| \(\displaystyle \frac {i \int \frac {(a+b \arctan (c x))^3}{d (i c x+1)}dx}{c}-\frac {i \int (a+b \arctan (c x))^3dx}{c d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {i \int \frac {(a+b \arctan (c x))^3}{i c x+1}dx}{c d}-\frac {i \int (a+b \arctan (c x))^3dx}{c d}\) |
| \(\Big \downarrow \) 5345 |
| \(\displaystyle \frac {i \int \frac {(a+b \arctan (c x))^3}{i c x+1}dx}{c d}-\frac {i \left (x (a+b \arctan (c x))^3-3 b c \int \frac {x (a+b \arctan (c x))^2}{c^2 x^2+1}dx\right )}{c d}\) |
| \(\Big \downarrow \) 5379 |
| \(\displaystyle \frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^3}{c}-3 i b \int \frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )}{c d}-\frac {i \left (x (a+b \arctan (c x))^3-3 b c \int \frac {x (a+b \arctan (c x))^2}{c^2 x^2+1}dx\right )}{c d}\) |
| \(\Big \downarrow \) 5455 |
| \(\displaystyle \frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^3}{c}-3 i b \int \frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )}{c d}-\frac {i \left (x (a+b \arctan (c x))^3-3 b c \left (-\frac {\int \frac {(a+b \arctan (c x))^2}{i-c x}dx}{c}-\frac {i (a+b \arctan (c x))^3}{3 b c^2}\right )\right )}{c d}\) |
| \(\Big \downarrow \) 5379 |
| \(\displaystyle \frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^3}{c}-3 i b \int \frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )}{c d}-\frac {i \left (x (a+b \arctan (c x))^3-3 b c \left (-\frac {\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 b \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx}{c}-\frac {i (a+b \arctan (c x))^3}{3 b c^2}\right )\right )}{c d}\) |
| \(\Big \downarrow \) 5529 |
| \(\displaystyle \frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^3}{c}-3 i b \left (i b \int \frac {(a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))^2}{2 c}\right )\right )}{c d}-\frac {i \left (x (a+b \arctan (c x))^3-3 b c \left (-\frac {\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 b \left (\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{2 c}\right )}{c}-\frac {i (a+b \arctan (c x))^3}{3 b c^2}\right )\right )}{c d}\) |
| \(\Big \downarrow \) 5533 |
| \(\displaystyle \frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^3}{c}-3 i b \left (i b \left (\frac {i \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{2 c}-\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))^2}{2 c}\right )\right )}{c d}-\frac {i \left (x (a+b \arctan (c x))^3-3 b c \left (-\frac {\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 b \left (\frac {1}{2} i b \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{2 c}\right )}{c}-\frac {i (a+b \arctan (c x))^3}{3 b c^2}\right )\right )}{c d}\) |
| \(\Big \downarrow \) 7164 |
| \(\displaystyle \frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^3}{c}-3 i b \left (i b \left (\frac {i \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{2 c}+\frac {b \operatorname {PolyLog}\left (4,1-\frac {2}{i c x+1}\right )}{4 c}\right )-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))^2}{2 c}\right )\right )}{c d}-\frac {i \left (x (a+b \arctan (c x))^3-3 b c \left (-\frac {i (a+b \arctan (c x))^3}{3 b c^2}-\frac {\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c}-2 b \left (-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{2 c}-\frac {b \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{4 c}\right )}{c}\right )\right )}{c d}\) |
((-I)*(x*(a + b*ArcTan[c*x])^3 - 3*b*c*(((-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*d) + (I*((I*(a + b*ArcTan[c*x])^3*Log[2/(1 + I*c*x)])/c - (3*I)*b*(((-1/2*I)*(a + b*ArcTan[c*x])^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/c + I*b*(((I/2)*(a + b*ArcTan[c*x])*PolyLog[3, 1 - 2/(1 + I*c*x )])/c + (b*PolyLog[4, 1 - 2/(1 + I*c*x)])/(4*c)))))/(c*d)
3.2.28.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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 ]
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]
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]
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[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_. )*(x_)^2), x_Symbol] :> Simp[I*(a + b*ArcTan[c*x])^p*(PolyLog[k + 1, u]/(2* c*d)), x] - Simp[b*p*(I/2) Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[k + 1 , u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[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 = 1.89 (sec) , antiderivative size = 3018, normalized size of antiderivative = 10.90
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(3018\) |
| default | \(\text {Expression too large to display}\) | \(3018\) |
| parts | \(\text {Expression too large to display}\) | \(3054\) |
1/c^2*(-I*a^3/d*c*x+1/2*a^3/d*ln(c^2*x^2+1)+I*a^3/d*arctan(c*x)+b^3/d*(1/2 *I*arctan(c*x)^4+ln(c*x-I)*arctan(c*x)^3-arctan(c*x)^3*ln(2*I*(1+I*c*x)^2/ (c^2*x^2+1))-3/2*I*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))-3/2*arctan(c*x)*pol ylog(3,-(1+I*c*x)^2/(c^2*x^2+1))-3*I*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2 +1)+1)+1/2*I*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^ 3*arctan(c*x)^3+I*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1) +1))^2*arctan(c*x)^3-1/2*I*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+ I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^3-I*arctan (c*x)^3*c*x+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+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^3+1/2*I*Pi*csgn(I/((1+I*c *x)^2/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2* x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^3-3/4*I*polylog(4,-(1+I*c* x)^2/(c^2*x^2+1))+3/2*I*arctan(c*x)^2*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))- I*Pi*arctan(c*x)^3-arctan(c*x)^3-3*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2 *x^2+1)))+3*a*b^2/d*(-1/2*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))-dilog(1+I*(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))*cs gn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2 *x^2+1)+1))*(I*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*arctan(c* x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/ 2))+dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2)))+1/4*I*Pi*csgn(I/((1+I*c*x)^...
\[ \int \frac {x (a+b \arctan (c x))^3}{d+i c d x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{3} x}{i \, c d x + d} \,d x } \]
integral(-1/8*(b^3*x*log(-(c*x + I)/(c*x - I))^3 - 6*I*a*b^2*x*log(-(c*x + I)/(c*x - I))^2 - 12*a^2*b*x*log(-(c*x + I)/(c*x - I)) + 8*I*a^3*x)/(c*d* x - I*d), x)
Timed out. \[ \int \frac {x (a+b \arctan (c x))^3}{d+i c d x} \, dx=\text {Timed out} \]
\[ \int \frac {x (a+b \arctan (c x))^3}{d+i c d x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{3} x}{i \, c d x + d} \,d x } \]
a^3*(-I*x/(c*d) + log(I*c*x + 1)/(c^2*d)) + 1/128*(-16*I*b^3*c*x*arctan(c* x)^3 + 12*I*b^3*c*x*arctan(c*x)*log(c^2*x^2 + 1)^2 + 16*I*b^3*arctan(c*x)^ 4 - I*b^3*log(c^2*x^2 + 1)^4 - 4*I*(896*b^3*c^2*integrate(1/32*x^2*arctan( c*x)^3/(c^3*d*x^2 + c*d), x) + 96*b^3*c^2*integrate(1/32*x^2*arctan(c*x)*l og(c^2*x^2 + 1)^2/(c^3*d*x^2 + c*d), x) + 3072*a*b^2*c^2*integrate(1/32*x^ 2*arctan(c*x)^2/(c^3*d*x^2 + c*d), x) + 384*b^3*c^2*integrate(1/32*x^2*arc tan(c*x)*log(c^2*x^2 + 1)/(c^3*d*x^2 + c*d), x) + 3072*a^2*b*c^2*integrate (1/32*x^2*arctan(c*x)/(c^3*d*x^2 + c*d), x) - 64*b^3*c*integrate(1/32*x*lo g(c^2*x^2 + 1)^3/(c^3*d*x^2 + c*d), x) - 384*b^3*c*integrate(1/32*x*arctan (c*x)^2/(c^3*d*x^2 + c*d), x) + 96*b^3*c*integrate(1/32*x*log(c^2*x^2 + 1) ^2/(c^3*d*x^2 + c*d), x) + 3*b^3*arctan(c*x)^4/(c^2*d) + 96*b^3*integrate( 1/32*arctan(c*x)*log(c^2*x^2 + 1)^2/(c^3*d*x^2 + c*d), x))*c^2*d + 128*c^2 *d*integrate(1/64*(80*b^3*c*x*arctan(c*x)^3 + 192*a^2*b*c*x*arctan(c*x) + (b^3*c^2*x^2 + 3*b^3)*log(c^2*x^2 + 1)^3 - 24*(b^3*c^2*x^2 - 8*a*b^2*c*x)* arctan(c*x)^2 + 6*(b^3*c^2*x^2 + 2*b^3*c*x*arctan(c*x))*log(c^2*x^2 + 1)^2 - 12*(2*b^3*c*x*arctan(c*x) + (b^3*c^2*x^2 - b^3)*arctan(c*x)^2)*log(c^2* x^2 + 1))/(c^3*d*x^2 + c*d), x) - 2*(b^3*c*x + 2*b^3*arctan(c*x))*log(c^2* x^2 + 1)^3 + 8*(3*b^3*c*x*arctan(c*x)^2 - 2*b^3*arctan(c*x)^3)*log(c^2*x^2 + 1))/(c^2*d)
\[ \int \frac {x (a+b \arctan (c x))^3}{d+i c d x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{3} x}{i \, c d x + d} \,d x } \]
Timed out. \[ \int \frac {x (a+b \arctan (c x))^3}{d+i c d x} \, dx=\int \frac {x\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3}{d+c\,d\,x\,1{}\mathrm {i}} \,d x \]