Integrand size = 25, antiderivative size = 364 \[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=\frac {a b x}{c^3 d^2}-\frac {i b^2}{2 c^4 d^2 (i-c x)}+\frac {i b^2 \arctan (c x)}{2 c^4 d^2}+\frac {b^2 x \arctan (c x)}{c^3 d^2}+\frac {b (a+b \arctan (c x))}{c^4 d^2 (i-c x)}+\frac {(a+b \arctan (c x))^2}{c^4 d^2}-\frac {2 i x (a+b \arctan (c x))^2}{c^3 d^2}-\frac {x^2 (a+b \arctan (c x))^2}{2 c^2 d^2}+\frac {i (a+b \arctan (c x))^2}{c^4 d^2 (i-c x)}-\frac {4 i b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c^4 d^2}-\frac {3 (a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^4 d^2}-\frac {b^2 \log \left (1+c^2 x^2\right )}{2 c^4 d^2}+\frac {2 b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^4 d^2}-\frac {3 i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^4 d^2}-\frac {3 b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^4 d^2} \]
a*b*x/c^3/d^2-1/2*I*b^2/c^4/d^2/(I-c*x)+1/2*I*b^2*arctan(c*x)/c^4/d^2+b^2* x*arctan(c*x)/c^3/d^2+b*(a+b*arctan(c*x))/c^4/d^2/(I-c*x)+(a+b*arctan(c*x) )^2/c^4/d^2-2*I*x*(a+b*arctan(c*x))^2/c^3/d^2-1/2*x^2*(a+b*arctan(c*x))^2/ c^2/d^2+I*(a+b*arctan(c*x))^2/c^4/d^2/(I-c*x)-4*I*b*(a+b*arctan(c*x))*ln(2 /(1+I*c*x))/c^4/d^2-3*(a+b*arctan(c*x))^2*ln(2/(1+I*c*x))/c^4/d^2-1/2*b^2* ln(c^2*x^2+1)/c^4/d^2+2*b^2*polylog(2,1-2/(1+I*c*x))/c^4/d^2-3*I*b*(a+b*ar ctan(c*x))*polylog(2,1-2/(1+I*c*x))/c^4/d^2-3/2*b^2*polylog(3,1-2/(1+I*c*x ))/c^4/d^2
Time = 1.61 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.18 \[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=-\frac {8 i a^2 c x+2 a^2 c^2 x^2+\frac {4 i a^2}{-i+c x}-12 i a^2 \arctan (c x)-6 a^2 \log \left (1+c^2 x^2\right )+2 a b \left (-2 c x-12 i \arctan (c x)^2+i \cos (2 \arctan (c x))-4 i \log \left (1+c^2 x^2\right )-6 i \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+2 \arctan (c x) \left (1+4 i c x+c^2 x^2-\cos (2 \arctan (c x))+6 \log \left (1+e^{2 i \arctan (c x)}\right )+i \sin (2 \arctan (c x))\right )+\sin (2 \arctan (c x))\right )+b^2 \left (-4 c x \arctan (c x)+10 \arctan (c x)^2+8 i c x \arctan (c x)^2+2 c^2 x^2 \arctan (c x)^2-8 i \arctan (c x)^3+\cos (2 \arctan (c x))+2 i \arctan (c x) \cos (2 \arctan (c x))-2 \arctan (c x)^2 \cos (2 \arctan (c x))+16 i \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )+12 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+2 \log \left (1+c^2 x^2\right )+4 (2-3 i \arctan (c x)) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+6 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )-i \sin (2 \arctan (c x))+2 \arctan (c x) \sin (2 \arctan (c x))+2 i \arctan (c x)^2 \sin (2 \arctan (c x))\right )}{4 c^4 d^2} \]
-1/4*((8*I)*a^2*c*x + 2*a^2*c^2*x^2 + ((4*I)*a^2)/(-I + c*x) - (12*I)*a^2* ArcTan[c*x] - 6*a^2*Log[1 + c^2*x^2] + 2*a*b*(-2*c*x - (12*I)*ArcTan[c*x]^ 2 + I*Cos[2*ArcTan[c*x]] - (4*I)*Log[1 + c^2*x^2] - (6*I)*PolyLog[2, -E^(( 2*I)*ArcTan[c*x])] + 2*ArcTan[c*x]*(1 + (4*I)*c*x + c^2*x^2 - Cos[2*ArcTan [c*x]] + 6*Log[1 + E^((2*I)*ArcTan[c*x])] + I*Sin[2*ArcTan[c*x]]) + Sin[2* ArcTan[c*x]]) + b^2*(-4*c*x*ArcTan[c*x] + 10*ArcTan[c*x]^2 + (8*I)*c*x*Arc Tan[c*x]^2 + 2*c^2*x^2*ArcTan[c*x]^2 - (8*I)*ArcTan[c*x]^3 + Cos[2*ArcTan[ c*x]] + (2*I)*ArcTan[c*x]*Cos[2*ArcTan[c*x]] - 2*ArcTan[c*x]^2*Cos[2*ArcTa n[c*x]] + (16*I)*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + 12*ArcTan[c* x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + 2*Log[1 + c^2*x^2] + 4*(2 - (3*I)*Ar cTan[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 6*PolyLog[3, -E^((2*I)*Arc Tan[c*x])] - I*Sin[2*ArcTan[c*x]] + 2*ArcTan[c*x]*Sin[2*ArcTan[c*x]] + (2* I)*ArcTan[c*x]^2*Sin[2*ArcTan[c*x]]))/(c^4*d^2)
Time = 0.83 (sec) , antiderivative size = 364, 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.
| \(\displaystyle \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx\) |
| \(\Big \downarrow \) 5411 |
| \(\displaystyle \int \left (\frac {3 (a+b \arctan (c x))^2}{c^3 d^2 (c x-i)}-\frac {2 i (a+b \arctan (c x))^2}{c^3 d^2}+\frac {i (a+b \arctan (c x))^2}{c^3 d^2 (c x-i)^2}-\frac {x (a+b \arctan (c x))^2}{c^2 d^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 i b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{c^4 d^2}+\frac {b (a+b \arctan (c x))}{c^4 d^2 (-c x+i)}+\frac {i (a+b \arctan (c x))^2}{c^4 d^2 (-c x+i)}+\frac {(a+b \arctan (c x))^2}{c^4 d^2}-\frac {4 i b \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c^4 d^2}-\frac {3 \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c^4 d^2}-\frac {2 i x (a+b \arctan (c x))^2}{c^3 d^2}-\frac {x^2 (a+b \arctan (c x))^2}{2 c^2 d^2}+\frac {a b x}{c^3 d^2}+\frac {i b^2 \arctan (c x)}{2 c^4 d^2}+\frac {b^2 x \arctan (c x)}{c^3 d^2}+\frac {2 b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c^4 d^2}-\frac {3 b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{2 c^4 d^2}-\frac {i b^2}{2 c^4 d^2 (-c x+i)}-\frac {b^2 \log \left (c^2 x^2+1\right )}{2 c^4 d^2}\) |
(a*b*x)/(c^3*d^2) - ((I/2)*b^2)/(c^4*d^2*(I - c*x)) + ((I/2)*b^2*ArcTan[c* x])/(c^4*d^2) + (b^2*x*ArcTan[c*x])/(c^3*d^2) + (b*(a + b*ArcTan[c*x]))/(c ^4*d^2*(I - c*x)) + (a + b*ArcTan[c*x])^2/(c^4*d^2) - ((2*I)*x*(a + b*ArcT an[c*x])^2)/(c^3*d^2) - (x^2*(a + b*ArcTan[c*x])^2)/(2*c^2*d^2) + (I*(a + b*ArcTan[c*x])^2)/(c^4*d^2*(I - c*x)) - ((4*I)*b*(a + b*ArcTan[c*x])*Log[2 /(1 + I*c*x)])/(c^4*d^2) - (3*(a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/(c ^4*d^2) - (b^2*Log[1 + c^2*x^2])/(2*c^4*d^2) + (2*b^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^4*d^2) - ((3*I)*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^4*d^2) - (3*b^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/(2*c^4*d^2)
3.2.4.3.1 Defintions of rubi rules used
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 = 21.60 (sec) , antiderivative size = 1104, normalized size of antiderivative = 3.03
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1104\) |
| default | \(\text {Expression too large to display}\) | \(1104\) |
| parts | \(\text {Expression too large to display}\) | \(1158\) |
1/c^4*(3/2*I*a*b/d^2*ln(c^2*x^2+1)-1/2*a^2/d^2*c^2*x^2-3*I*a*b/d^2*dilog(- 1/2*I*(c*x+I))+3/2*a^2/d^2*ln(c^2*x^2+1)-2*I*a^2/d^2*c*x+b^2/d^2*(3*I*Pi*c sgn((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*c^2*x^2*arctan(c*x)^2+I*arctan(c*x)*(c*x+I)/(2*c*x-2*I)+3*arctan(c*x)^2 *ln(c*x-I)-3*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-4*I*arctan(c*x) *ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-3*I*Pi*arctan(c*x)^2-3/2*polylog(3,-( 1+I*c*x)^2/(c^2*x^2+1))-4*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-4*dilog(1 -I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-3*arctan(c*x)^2+3*I*arctan(c*x)*polylog(2, -(1+I*c*x)^2/(c^2*x^2+1))+arctan(c*x)*(c*x-I)+3/2*I*Pi*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+ln(1+(1+I*c*x)^2/(c ^2*x^2+1))-I*arctan(c*x)^2/(c*x-I)+2*I*arctan(c*x)^3-4*I*arctan(c*x)*ln(1- I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/2*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+3/2*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 +3/2*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)^2-2*I*arctan(c*x)^2*c*x+1/4*(c*x+I)/ (c*x-I))-I*a^2/d^2/(c*x-I)-a*b/d^2*arctan(c*x)*c^2*x^2-2*I*a*b/d^2*arctan( c*x)/(c*x-I)+6*a*b/d^2*arctan(c*x)*ln(c*x-I)+3/2*I*a*b/d^2*ln(c*x-I)^2+1/4 *I*a*b/d^2*ln(c^4*x^4+10*c^2*x^2+9)-3*I*a*b/d^2*ln(-1/2*I*(c*x+I))*ln(c...
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{{\left (i \, c d x + d\right )}^{2}} \,d x } \]
integral(1/4*(b^2*x^3*log(-(c*x + I)/(c*x - I))^2 - 4*I*a*b*x^3*log(-(c*x + I)/(c*x - I)) - 4*a^2*x^3)/(c^2*d^2*x^2 - 2*I*c*d^2*x - d^2), x)
Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=\text {Timed out} \]
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{{\left (i \, c d x + d\right )}^{2}} \,d x } \]
-1/2*a^2*(2*I/(c^5*d^2*x - I*c^4*d^2) + (c*x^2 + 4*I*x)/(c^3*d^2) - 6*log( c*x - I)/(c^4*d^2)) + 1/32*(24*(I*b^2*c*x + b^2)*arctan(c*x)^3 - 3*(b^2*c* x - I*b^2)*log(c^2*x^2 + 1)^3 - 4*(b^2*c^3*x^3 + 3*I*b^2*c^2*x^2 + 4*b^2*c *x + 2*I*b^2)*arctan(c*x)^2 + (b^2*c^3*x^3 + 3*I*b^2*c^2*x^2 + 4*b^2*c*x + 2*I*b^2 + 6*(I*b^2*c*x + b^2)*arctan(c*x))*log(c^2*x^2 + 1)^2 - 2*(c^5*d^ 2*x - I*c^4*d^2)*(192*b^2*c^5*integrate(1/16*x^5*arctan(c*x)^2/(c^7*d^2*x^ 4 + 2*c^5*d^2*x^2 + c^3*d^2), x) + 16*b^2*c^5*integrate(1/16*x^5*log(c^2*x ^2 + 1)^2/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x) + 512*a*b*c^5*integr ate(1/16*x^5*arctan(c*x)/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x) + 32* b^2*c^5*integrate(1/16*x^5*log(c^2*x^2 + 1)/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x) + 128*b^2*c^4*integrate(1/16*x^4*arctan(c*x)*log(c^2*x^2 + 1 )/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x) + 192*b^2*c^4*integrate(1/16 *x^4*arctan(c*x)/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x) + 6*b^2*c^3*( c^2/(c^11*d^2*x^2 + c^9*d^2) + log(c^2*x^2 + 1)/(c^9*d^2*x^2 + c^7*d^2)) - 576*b^2*c^3*integrate(1/16*x^3*arctan(c*x)^2/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x) - 112*b^2*c^3*integrate(1/16*x^3*log(c^2*x^2 + 1)^2/(c^7*d ^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x) - 512*a*b*c^3*integrate(1/16*x^3*arc tan(c*x)/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x) + 160*b^2*c^3*integra te(1/16*x^3*log(c^2*x^2 + 1)/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x) + 320*b^2*c^2*integrate(1/16*x^2*arctan(c*x)/(c^7*d^2*x^4 + 2*c^5*d^2*x^...
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{{\left (i \, c d x + d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=\int \frac {x^3\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^2} \,d x \]