Integrand size = 25, antiderivative size = 462 \[ \int \frac {x^4 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=-\frac {i a b x}{c^4 d^3}+\frac {i b^2}{16 c^5 d^3 (i-c x)^2}-\frac {29 b^2}{16 c^5 d^3 (i-c x)}+\frac {29 b^2 \arctan (c x)}{16 c^5 d^3}-\frac {i b^2 x \arctan (c x)}{c^4 d^3}-\frac {b (a+b \arctan (c x))}{4 c^5 d^3 (i-c x)^2}-\frac {15 i b (a+b \arctan (c x))}{4 c^5 d^3 (i-c x)}-\frac {5 i (a+b \arctan (c x))^2}{8 c^5 d^3}-\frac {3 x (a+b \arctan (c x))^2}{c^4 d^3}+\frac {i x^2 (a+b \arctan (c x))^2}{2 c^3 d^3}-\frac {i (a+b \arctan (c x))^2}{2 c^5 d^3 (i-c x)^2}+\frac {4 (a+b \arctan (c x))^2}{c^5 d^3 (i-c x)}-\frac {6 b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c^5 d^3}+\frac {6 i (a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^5 d^3}+\frac {i b^2 \log \left (1+c^2 x^2\right )}{2 c^5 d^3}-\frac {3 i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^5 d^3}-\frac {6 b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^5 d^3}+\frac {3 i b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{c^5 d^3} \] Output:
-I*a*b*x/c^4/d^3+1/2*I*b^2*ln(c^2*x^2+1)/c^5/d^3-29/16*b^2/c^5/d^3/(I-c*x) +29/16*b^2*arctan(c*x)/c^5/d^3+3*I*b^2*polylog(3,1-2/(1+I*c*x))/c^5/d^3-1/ 4*b*(a+b*arctan(c*x))/c^5/d^3/(I-c*x)^2-5/8*I*(a+b*arctan(c*x))^2/c^5/d^3- 3*I*b^2*polylog(2,1-2/(1+I*c*x))/c^5/d^3-3*x*(a+b*arctan(c*x))^2/c^4/d^3+1 /2*I*x^2*(a+b*arctan(c*x))^2/c^3/d^3-I*b^2*x*arctan(c*x)/c^4/d^3+4*(a+b*ar ctan(c*x))^2/c^5/d^3/(I-c*x)-6*b*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c^5/d^3 +1/16*I*b^2/c^5/d^3/(I-c*x)^2-15/4*I*b*(a+b*arctan(c*x))/c^5/d^3/(I-c*x)-1 /2*I*(a+b*arctan(c*x))^2/c^5/d^3/(I-c*x)^2-6*b*(a+b*arctan(c*x))*polylog(2 ,1-2/(1+I*c*x))/c^5/d^3+6*I*(a+b*arctan(c*x))^2*ln(2/(1+I*c*x))/c^5/d^3
Time = 1.76 (sec) , antiderivative size = 578, normalized size of antiderivative = 1.25 \[ \int \frac {x^4 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\frac {-48 a^2 c x+8 i a^2 c^2 x^2-\frac {8 i a^2}{(-i+c x)^2}-\frac {64 a^2}{-i+c x}+96 a^2 \arctan (c x)-48 i a^2 \log \left (1+c^2 x^2\right )+a b \left (-16 i c x+192 \arctan (c x)^2-28 \cos (2 \arctan (c x))+\cos (4 \arctan (c x))+48 \log \left (1+c^2 x^2\right )+96 \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+28 i \sin (2 \arctan (c x))+4 i \arctan (c x) \left (4+24 i c x+4 c^2 x^2-14 \cos (2 \arctan (c x))+\cos (4 \arctan (c x))+48 \log \left (1+e^{2 i \arctan (c x)}\right )+14 i \sin (2 \arctan (c x))-i \sin (4 \arctan (c x))\right )-i \sin (4 \arctan (c x))\right )+16 i b^2 \left (-c x \arctan (c x)+3 \arctan (c x)^2+3 i c x \arctan (c x)^2+\frac {1}{2} \left (1+c^2 x^2\right ) \arctan (c x)^2-4 i \arctan (c x)^3-\frac {7}{8} \left (-1-2 i \arctan (c x)+2 \arctan (c x)^2\right ) \cos (2 \arctan (c x))-\frac {1}{64} \cos (4 \arctan (c x))-\frac {1}{16} i \arctan (c x) \cos (4 \arctan (c x))+\frac {1}{8} \arctan (c x)^2 \cos (4 \arctan (c x))+6 i \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )+6 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+\frac {1}{2} \log \left (1+c^2 x^2\right )+(3-6 i \arctan (c x)) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+3 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )-\frac {7}{8} i \sin (2 \arctan (c x))+\frac {7}{4} \arctan (c x) \sin (2 \arctan (c x))+\frac {7}{4} i \arctan (c x)^2 \sin (2 \arctan (c x))+\frac {1}{64} i \sin (4 \arctan (c x))-\frac {1}{16} \arctan (c x) \sin (4 \arctan (c x))-\frac {1}{8} i \arctan (c x)^2 \sin (4 \arctan (c x))\right )}{16 c^5 d^3} \] Input:
Integrate[(x^4*(a + b*ArcTan[c*x])^2)/(d + I*c*d*x)^3,x]
Output:
(-48*a^2*c*x + (8*I)*a^2*c^2*x^2 - ((8*I)*a^2)/(-I + c*x)^2 - (64*a^2)/(-I + c*x) + 96*a^2*ArcTan[c*x] - (48*I)*a^2*Log[1 + c^2*x^2] + a*b*((-16*I)* c*x + 192*ArcTan[c*x]^2 - 28*Cos[2*ArcTan[c*x]] + Cos[4*ArcTan[c*x]] + 48* Log[1 + c^2*x^2] + 96*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + (28*I)*Sin[2*Ar cTan[c*x]] + (4*I)*ArcTan[c*x]*(4 + (24*I)*c*x + 4*c^2*x^2 - 14*Cos[2*ArcT an[c*x]] + Cos[4*ArcTan[c*x]] + 48*Log[1 + E^((2*I)*ArcTan[c*x])] + (14*I) *Sin[2*ArcTan[c*x]] - I*Sin[4*ArcTan[c*x]]) - I*Sin[4*ArcTan[c*x]]) + (16* I)*b^2*(-(c*x*ArcTan[c*x]) + 3*ArcTan[c*x]^2 + (3*I)*c*x*ArcTan[c*x]^2 + ( (1 + c^2*x^2)*ArcTan[c*x]^2)/2 - (4*I)*ArcTan[c*x]^3 - (7*(-1 - (2*I)*ArcT an[c*x] + 2*ArcTan[c*x]^2)*Cos[2*ArcTan[c*x]])/8 - Cos[4*ArcTan[c*x]]/64 - (I/16)*ArcTan[c*x]*Cos[4*ArcTan[c*x]] + (ArcTan[c*x]^2*Cos[4*ArcTan[c*x]] )/8 + (6*I)*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + 6*ArcTan[c*x]^2*L og[1 + E^((2*I)*ArcTan[c*x])] + Log[1 + c^2*x^2]/2 + (3 - (6*I)*ArcTan[c*x ])*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 3*PolyLog[3, -E^((2*I)*ArcTan[c*x] )] - ((7*I)/8)*Sin[2*ArcTan[c*x]] + (7*ArcTan[c*x]*Sin[2*ArcTan[c*x]])/4 + ((7*I)/4)*ArcTan[c*x]^2*Sin[2*ArcTan[c*x]] + (I/64)*Sin[4*ArcTan[c*x]] - (ArcTan[c*x]*Sin[4*ArcTan[c*x]])/16 - (I/8)*ArcTan[c*x]^2*Sin[4*ArcTan[c*x ]]))/(16*c^5*d^3)
Time = 1.09 (sec) , antiderivative size = 462, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {5411, 2009}
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.
\(\displaystyle \int \frac {x^4 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx\) |
\(\Big \downarrow \) 5411 |
\(\displaystyle \int \left (-\frac {6 i (a+b \arctan (c x))^2}{c^4 d^3 (c x-i)}+\frac {4 (a+b \arctan (c x))^2}{c^4 d^3 (c x-i)^2}-\frac {3 (a+b \arctan (c x))^2}{c^4 d^3}+\frac {i (a+b \arctan (c x))^2}{c^4 d^3 (c x-i)^3}+\frac {i x (a+b \arctan (c x))^2}{c^3 d^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {6 b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{c^5 d^3}-\frac {15 i b (a+b \arctan (c x))}{4 c^5 d^3 (-c x+i)}-\frac {b (a+b \arctan (c x))}{4 c^5 d^3 (-c x+i)^2}+\frac {4 (a+b \arctan (c x))^2}{c^5 d^3 (-c x+i)}-\frac {i (a+b \arctan (c x))^2}{2 c^5 d^3 (-c x+i)^2}-\frac {5 i (a+b \arctan (c x))^2}{8 c^5 d^3}-\frac {6 b \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c^5 d^3}+\frac {6 i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c^5 d^3}-\frac {3 x (a+b \arctan (c x))^2}{c^4 d^3}+\frac {i x^2 (a+b \arctan (c x))^2}{2 c^3 d^3}-\frac {i a b x}{c^4 d^3}+\frac {29 b^2 \arctan (c x)}{16 c^5 d^3}-\frac {i b^2 x \arctan (c x)}{c^4 d^3}-\frac {3 i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c^5 d^3}+\frac {3 i b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{c^5 d^3}-\frac {29 b^2}{16 c^5 d^3 (-c x+i)}+\frac {i b^2}{16 c^5 d^3 (-c x+i)^2}+\frac {i b^2 \log \left (c^2 x^2+1\right )}{2 c^5 d^3}\) |
Input:
Int[(x^4*(a + b*ArcTan[c*x])^2)/(d + I*c*d*x)^3,x]
Output:
((-I)*a*b*x)/(c^4*d^3) + ((I/16)*b^2)/(c^5*d^3*(I - c*x)^2) - (29*b^2)/(16 *c^5*d^3*(I - c*x)) + (29*b^2*ArcTan[c*x])/(16*c^5*d^3) - (I*b^2*x*ArcTan[ c*x])/(c^4*d^3) - (b*(a + b*ArcTan[c*x]))/(4*c^5*d^3*(I - c*x)^2) - (((15* I)/4)*b*(a + b*ArcTan[c*x]))/(c^5*d^3*(I - c*x)) - (((5*I)/8)*(a + b*ArcTa n[c*x])^2)/(c^5*d^3) - (3*x*(a + b*ArcTan[c*x])^2)/(c^4*d^3) + ((I/2)*x^2* (a + b*ArcTan[c*x])^2)/(c^3*d^3) - ((I/2)*(a + b*ArcTan[c*x])^2)/(c^5*d^3* (I - c*x)^2) + (4*(a + b*ArcTan[c*x])^2)/(c^5*d^3*(I - c*x)) - (6*b*(a + b *ArcTan[c*x])*Log[2/(1 + I*c*x)])/(c^5*d^3) + ((6*I)*(a + b*ArcTan[c*x])^2 *Log[2/(1 + I*c*x)])/(c^5*d^3) + ((I/2)*b^2*Log[1 + c^2*x^2])/(c^5*d^3) - ((3*I)*b^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^5*d^3) - (6*b*(a + b*ArcTan[c *x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^5*d^3) + ((3*I)*b^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/(c^5*d^3)
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f* x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p, 0] & & IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 14.61 (sec) , antiderivative size = 1215, normalized size of antiderivative = 2.63
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1215\) |
default | \(\text {Expression too large to display}\) | \(1215\) |
parts | \(\text {Expression too large to display}\) | \(1279\) |
Input:
int(x^4*(a+b*arctan(c*x))^2/(d+I*c*d*x)^3,x,method=_RETURNVERBOSE)
Output:
1/c^5*(I*a*b/d^3*arctan(c*x)*c^2*x^2-12*I*a*b/d^3*arctan(c*x)*ln(c*x-I)-I* a*b/d^3*arctan(c*x)/(c*x-I)^2-6*a*b/d^3*arctan(c*x)*c*x-I*a*b/d^3*c*x-a*b/ d^3-6*a*b/d^3*ln(c*x-I)*ln(-1/2*I*(c*x+I))-5/8*I*a*b/d^3*arctan(1/2*c*x-1/ 2*I)+5/16*I*a*b/d^3*arctan(1/2*c*x)+15/4*I*a*b/d^3/(c*x-I)+43/8*I*a*b/d^3* arctan(c*x)-5/16*I*a*b/d^3*arctan(1/6*c^3*x^3+7/6*c*x)-3*I*a^2/d^3*ln(c^2* x^2+1)-1/2*I*a^2/d^3/(c*x-I)^2-3*a^2/d^3*c*x+5/32*a*b/d^3*ln(c^4*x^4+10*c^ 2*x^2+9)-1/4*a*b/d^3/(c*x-I)^2+43/16*a*b/d^3*ln(c^2*x^2+1)+3*a*b/d^3*ln(c* x-I)^2-6*a*b/d^3*dilog(-1/2*I*(c*x+I))+b^2/d^3*(7/4*(c*x+I)*arctan(c*x)/(c *x-I)-3*c*x*arctan(c*x)^2+4*arctan(c*x)^3-4*arctan(c*x)^2/(c*x-I)+6*Pi*csg n((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*P i*csgn((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+1/2*I*arctan(c*x)^2*c^2*x^2+3*Pi*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csg n((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)^2-6*Pi*arctan(c*x)^2-3*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*arc tan(c*x)^2+3*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+6*I*arctan(c*x)^2*ln(2*I*(1+ I*c*x)^2/(c^2*x^2+1))-I*arctan(c*x)*(c*x-I)-6*I*arctan(c*x)^2*ln(c*x-I)-1/ 2*I*arctan(c*x)^2/(c*x-I)^2+43/8*I*arctan(c*x)^2+6*I*dilog(1+I*(1+I*c*x)/( c^2*x^2+1)^(1/2))-I*ln(1+(1+I*c*x)^2/(c^2*x^2+1))+6*arctan(c*x)*polylog...
\[ \int \frac {x^4 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{4}}{{\left (i \, c d x + d\right )}^{3}} \,d x } \] Input:
integrate(x^4*(a+b*arctan(c*x))^2/(d+I*c*d*x)^3,x, algorithm="fricas")
Output:
integral(1/4*(-I*b^2*x^4*log(-(c*x + I)/(c*x - I))^2 - 4*a*b*x^4*log(-(c*x + I)/(c*x - I)) + 4*I*a^2*x^4)/(c^3*d^3*x^3 - 3*I*c^2*d^3*x^2 - 3*c*d^3*x + I*d^3), x)
Timed out. \[ \int \frac {x^4 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\text {Timed out} \] Input:
integrate(x**4*(a+b*atan(c*x))**2/(d+I*c*d*x)**3,x)
Output:
Timed out
\[ \int \frac {x^4 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{4}}{{\left (i \, c d x + d\right )}^{3}} \,d x } \] Input:
integrate(x^4*(a+b*arctan(c*x))^2/(d+I*c*d*x)^3,x, algorithm="maxima")
Output:
1/128*(64*I*a^2*c^4*x^4 - 256*a^2*c^3*x^3 - 32*a^2*c^2*x^2*(15*arctan2(1, c*x) - 22*I) + 64*a^2*c*x*(15*I*arctan2(1, c*x) - 2) + 192*(b^2*c^2*x^2 - 2*I*b^2*c*x - b^2)*arctan(c*x)^3 + 24*(I*b^2*c^2*x^2 + 2*b^2*c*x - I*b^2)* log(c^2*x^2 + 1)^3 + 32*a^2*(15*arctan2(1, c*x) + 14*I) + 16*(I*b^2*c^4*x^ 4 - 4*b^2*c^3*x^3 + 11*I*b^2*c^2*x^2 - 2*b^2*c*x + 7*I*b^2)*arctan(c*x)^2 - 4*(I*b^2*c^4*x^4 - 4*b^2*c^3*x^3 + 11*I*b^2*c^2*x^2 - 2*b^2*c*x + 7*I*b^ 2 - 12*(b^2*c^2*x^2 - 2*I*b^2*c*x - b^2)*arctan(c*x))*log(c^2*x^2 + 1)^2 - 36*(I*b^2*c^8*d^3*x^2 + 2*b^2*c^7*d^3*x - I*b^2*c^6*d^3)*(((8*c^2*x^2 + 7 )*c^2/(c^16*d^3*x^4 + 2*c^14*d^3*x^2 + c^12*d^3) + 2*(4*c^2*x^2 + 3)*log(c ^2*x^2 + 1)/(c^14*d^3*x^4 + 2*c^12*d^3*x^2 + c^10*d^3))*c^4 + 2*(2*c^2*x^2 + 1)*c^2*log(c^2*x^2 + 1)^2/(c^12*d^3*x^4 + 2*c^10*d^3*x^2 + c^8*d^3) - c ^2*(c^2/(c^14*d^3*x^4 + 2*c^12*d^3*x^2 + c^10*d^3) + 2*log(c^2*x^2 + 1)/(c ^12*d^3*x^4 + 2*c^10*d^3*x^2 + c^8*d^3)) - 2048*c^2*integrate(1/64*x^3*arc tan(c*x)^2/(c^10*d^3*x^6 + 3*c^8*d^3*x^4 + 3*c^6*d^3*x^2 + c^4*d^3), x) - 2*log(c^2*x^2 + 1)^2/(c^10*d^3*x^4 + 2*c^8*d^3*x^2 + c^6*d^3) + 2048*integ rate(1/64*x*arctan(c*x)^2/(c^10*d^3*x^6 + 3*c^8*d^3*x^4 + 3*c^6*d^3*x^2 + c^4*d^3), x)) + 12*(-I*b^2*c^10*d^3*x^2 - 2*b^2*c^9*d^3*x + I*b^2*c^8*d^3) *(((8*c^2*x^2 + 7)*c^2/(c^16*d^3*x^4 + 2*c^14*d^3*x^2 + c^12*d^3) + 2*(4*c ^2*x^2 + 3)*log(c^2*x^2 + 1)/(c^14*d^3*x^4 + 2*c^12*d^3*x^2 + c^10*d^3))*c ^2 + 2048*c^2*integrate(1/64*x^5*arctan(c*x)^2/(c^10*d^3*x^6 + 3*c^8*d^...
\[ \int \frac {x^4 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{4}}{{\left (i \, c d x + d\right )}^{3}} \,d x } \] Input:
integrate(x^4*(a+b*arctan(c*x))^2/(d+I*c*d*x)^3,x, algorithm="giac")
Output:
integrate((b*arctan(c*x) + a)^2*x^4/(I*c*d*x + d)^3, x)
Timed out. \[ \int \frac {x^4 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\int \frac {x^4\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3} \,d x \] Input:
int((x^4*(a + b*atan(c*x))^2)/(d + c*d*x*1i)^3,x)
Output:
int((x^4*(a + b*atan(c*x))^2)/(d + c*d*x*1i)^3, x)
\[ \int \frac {x^4 (a+b \arctan (c x))^2}{(d+i c d x)^3} \, dx=\frac {-4 \left (\int \frac {\mathit {atan} \left (c x \right ) x^{4}}{c^{3} i \,x^{3}+3 c^{2} x^{2}-3 c i x -1}d x \right ) a b \,c^{5}-2 \left (\int \frac {\mathit {atan} \left (c x \right )^{2} x^{4}}{c^{3} i \,x^{3}+3 c^{2} x^{2}-3 c i x -1}d x \right ) b^{2} c^{5}+8 \left (\int \frac {x}{c^{3} i \,x^{3}+3 c^{2} x^{2}-3 c i x -1}d x \right ) a^{2} c^{2} i +6 \left (\int \frac {1}{c^{3} i \,x^{3}+3 c^{2} x^{2}-3 c i x -1}d x \right ) a^{2} c -4 \,\mathrm {log}\left (c^{3} i \,x^{3}+3 c^{2} x^{2}-3 c i x -1\right ) a^{2} i +a^{2} c^{2} i \,x^{2}-6 a^{2} c x}{2 c^{5} d^{3}} \] Input:
int(x^4*(a+b*atan(c*x))^2/(d+I*c*d*x)^3,x)
Output:
( - 4*int((atan(c*x)*x**4)/(c**3*i*x**3 + 3*c**2*x**2 - 3*c*i*x - 1),x)*a* b*c**5 - 2*int((atan(c*x)**2*x**4)/(c**3*i*x**3 + 3*c**2*x**2 - 3*c*i*x - 1),x)*b**2*c**5 + 8*int(x/(c**3*i*x**3 + 3*c**2*x**2 - 3*c*i*x - 1),x)*a** 2*c**2*i + 6*int(1/(c**3*i*x**3 + 3*c**2*x**2 - 3*c*i*x - 1),x)*a**2*c - 4 *log(c**3*i*x**3 + 3*c**2*x**2 - 3*c*i*x - 1)*a**2*i + a**2*c**2*i*x**2 - 6*a**2*c*x)/(2*c**5*d**3)