Integrand size = 25, antiderivative size = 292 \[ \int \frac {x^2 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=-\frac {b^2}{2 c^3 d^2 (i-c x)}+\frac {b^2 \arctan (c x)}{2 c^3 d^2}-\frac {i b (a+b \arctan (c x))}{c^3 d^2 (i-c x)}-\frac {i (a+b \arctan (c x))^2}{2 c^3 d^2}-\frac {x (a+b \arctan (c x))^2}{c^2 d^2}+\frac {(a+b \arctan (c x))^2}{c^3 d^2 (i-c x)}-\frac {2 b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c^3 d^2}+\frac {2 i (a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d^2}-\frac {i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^3 d^2}-\frac {2 b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^3 d^2}+\frac {i b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{c^3 d^2} \]
-1/2*b^2/c^3/d^2/(I-c*x)+1/2*b^2*arctan(c*x)/c^3/d^2-I*b*(a+b*arctan(c*x)) /c^3/d^2/(I-c*x)-1/2*I*(a+b*arctan(c*x))^2/c^3/d^2-x*(a+b*arctan(c*x))^2/c ^2/d^2+(a+b*arctan(c*x))^2/c^3/d^2/(I-c*x)-2*b*(a+b*arctan(c*x))*ln(2/(1+I *c*x))/c^3/d^2+2*I*(a+b*arctan(c*x))^2*ln(2/(1+I*c*x))/c^3/d^2-I*b^2*polyl og(2,1-2/(1+I*c*x))/c^3/d^2-2*b*(a+b*arctan(c*x))*polylog(2,1-2/(1+I*c*x)) /c^3/d^2+I*b^2*polylog(3,1-2/(1+I*c*x))/c^3/d^2
Time = 1.18 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.24 \[ \int \frac {x^2 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=-\frac {12 a^2 c x+\frac {12 a^2}{-i+c x}-24 a^2 \arctan (c x)+12 i a^2 \log \left (1+c^2 x^2\right )+b^2 \left (-12 i \arctan (c x)^2+12 c x \arctan (c x)^2-16 \arctan (c x)^3-3 i \cos (2 \arctan (c x))+6 \arctan (c x) \cos (2 \arctan (c x))+6 i \arctan (c x)^2 \cos (2 \arctan (c x))+24 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )-24 i \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )-12 (i+2 \arctan (c x)) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )-12 i \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )-3 \sin (2 \arctan (c x))-6 i \arctan (c x) \sin (2 \arctan (c x))+6 \arctan (c x)^2 \sin (2 \arctan (c x))\right )+6 a b \left (-8 \arctan (c x)^2+\cos (2 \arctan (c x))-2 \log \left (1+c^2 x^2\right )-4 \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )-i \sin (2 \arctan (c x))+2 \arctan (c x) \left (2 c x+i \cos (2 \arctan (c x))-4 i \log \left (1+e^{2 i \arctan (c x)}\right )+\sin (2 \arctan (c x))\right )\right )}{12 c^3 d^2} \]
-1/12*(12*a^2*c*x + (12*a^2)/(-I + c*x) - 24*a^2*ArcTan[c*x] + (12*I)*a^2* Log[1 + c^2*x^2] + b^2*((-12*I)*ArcTan[c*x]^2 + 12*c*x*ArcTan[c*x]^2 - 16* ArcTan[c*x]^3 - (3*I)*Cos[2*ArcTan[c*x]] + 6*ArcTan[c*x]*Cos[2*ArcTan[c*x] ] + (6*I)*ArcTan[c*x]^2*Cos[2*ArcTan[c*x]] + 24*ArcTan[c*x]*Log[1 + E^((2* I)*ArcTan[c*x])] - (24*I)*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] - 1 2*(I + 2*ArcTan[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c*x])] - (12*I)*PolyLog[ 3, -E^((2*I)*ArcTan[c*x])] - 3*Sin[2*ArcTan[c*x]] - (6*I)*ArcTan[c*x]*Sin[ 2*ArcTan[c*x]] + 6*ArcTan[c*x]^2*Sin[2*ArcTan[c*x]]) + 6*a*b*(-8*ArcTan[c* x]^2 + Cos[2*ArcTan[c*x]] - 2*Log[1 + c^2*x^2] - 4*PolyLog[2, -E^((2*I)*Ar cTan[c*x])] - I*Sin[2*ArcTan[c*x]] + 2*ArcTan[c*x]*(2*c*x + I*Cos[2*ArcTan [c*x]] - (4*I)*Log[1 + E^((2*I)*ArcTan[c*x])] + Sin[2*ArcTan[c*x]])))/(c^3 *d^2)
Time = 0.70 (sec) , antiderivative size = 292, 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.
| \(\displaystyle \int \frac {x^2 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx\) |
| \(\Big \downarrow \) 5411 |
| \(\displaystyle \int \left (-\frac {2 i (a+b \arctan (c x))^2}{c^2 d^2 (c x-i)}-\frac {(a+b \arctan (c x))^2}{c^2 d^2}+\frac {(a+b \arctan (c x))^2}{c^2 d^2 (c x-i)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{c^3 d^2}-\frac {i b (a+b \arctan (c x))}{c^3 d^2 (-c x+i)}+\frac {(a+b \arctan (c x))^2}{c^3 d^2 (-c x+i)}-\frac {i (a+b \arctan (c x))^2}{2 c^3 d^2}-\frac {2 b \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c^3 d^2}+\frac {2 i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c^3 d^2}-\frac {x (a+b \arctan (c x))^2}{c^2 d^2}+\frac {b^2 \arctan (c x)}{2 c^3 d^2}-\frac {i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c^3 d^2}+\frac {i b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{c^3 d^2}-\frac {b^2}{2 c^3 d^2 (-c x+i)}\) |
-1/2*b^2/(c^3*d^2*(I - c*x)) + (b^2*ArcTan[c*x])/(2*c^3*d^2) - (I*b*(a + b *ArcTan[c*x]))/(c^3*d^2*(I - c*x)) - ((I/2)*(a + b*ArcTan[c*x])^2)/(c^3*d^ 2) - (x*(a + b*ArcTan[c*x])^2)/(c^2*d^2) + (a + b*ArcTan[c*x])^2/(c^3*d^2* (I - c*x)) - (2*b*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(c^3*d^2) + ((2* I)*(a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/(c^3*d^2) - (I*b^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^3*d^2) - (2*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2 /(1 + I*c*x)])/(c^3*d^2) + (I*b^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/(c^3*d^2)
3.2.5.3.1 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 = 10.00 (sec) , antiderivative size = 4182, normalized size of antiderivative = 14.32
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(4182\) |
| default | \(\text {Expression too large to display}\) | \(4182\) |
| parts | \(\text {Expression too large to display}\) | \(4230\) |
1/c^3*(-a^2/d^2*c*x-a^2/d^2/(c*x-I)+2*a^2/d^2*arctan(c*x)+1/4*I*a*b/d^2*ar ctan(1/2*c*x)+b^2/d^2*(4/3*arctan(c*x)^3-arctan(c*x)^2*c*x-3/2*arctan(c*x) *ln(1+(1+I*c*x)^2/(c^2*x^2+1))+2*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x ^2+1))-1/2*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2*arctan(c*x) *ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*Pi*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^ (1/2))+2*Pi*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*Pi*arctan(c*x)^2-Pi*p olylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+I*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+3 /2*I*arctan(c*x)^2+2*Pi*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+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+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-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+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+1/2*arctan(c*x)*(c*x+I)/(c*x-I)-arctan(c*x)^2/(c*x-I)-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)*ln(1+(1+I*c*x)^2/(c^2*x^2+1))-2*I*Pi* arctan(c*x)*ln(1+(1+I*c*x)^2/(c^2*x^2+1))+2*I*Pi*arctan(c*x)*ln(1+I*(1+I*c *x)/(c^2*x^2+1)^(1/2))+2*I*Pi*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/ 2))-1/2*Pi*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn((1+I*c*x)^2/(c^2*x^...
\[ \int \frac {x^2 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{2}}{{\left (i \, c d x + d\right )}^{2}} \,d x } \]
integral(1/4*(b^2*x^2*log(-(c*x + I)/(c*x - I))^2 - 4*I*a*b*x^2*log(-(c*x + I)/(c*x - I)) - 4*a^2*x^2)/(c^2*d^2*x^2 - 2*I*c*d^2*x - d^2), x)
Timed out. \[ \int \frac {x^2 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=\text {Timed out} \]
\[ \int \frac {x^2 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{2}}{{\left (i \, c d x + d\right )}^{2}} \,d x } \]
-a^2*(1/(c^4*d^2*x - I*c^3*d^2) + x/(c^2*d^2) + 2*I*log(c*x - I)/(c^3*d^2) ) + 1/16*(8*(b^2*c*x - I*b^2)*arctan(c*x)^3 - (-I*b^2*c*x - b^2)*log(c^2*x ^2 + 1)^3 - 4*(b^2*c^2*x^2 - I*b^2*c*x + b^2)*arctan(c*x)^2 + (b^2*c^2*x^2 - I*b^2*c*x + b^2 + 2*(b^2*c*x - I*b^2)*arctan(c*x))*log(c^2*x^2 + 1)^2 - 2*(c^4*d^2*x - I*c^3*d^2)*(96*b^2*c^4*integrate(1/16*x^4*arctan(c*x)^2/(c ^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) + 8*b^2*c^4*integrate(1/16*x^4*l og(c^2*x^2 + 1)^2/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) + 256*a*b*c^ 4*integrate(1/16*x^4*arctan(c*x)/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) + 32*b^2*c^4*integrate(1/16*x^4*log(c^2*x^2 + 1)/(c^6*d^2*x^4 + 2*c^4*d ^2*x^2 + c^2*d^2), x) + 64*b^2*c^3*integrate(1/16*x^3*arctan(c*x)*log(c^2* x^2 + 1)/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) - 64*b^2*c^3*integrat e(1/16*x^3*arctan(c*x)/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) + 32*b^ 2*c^2*integrate(1/16*x^2*arctan(c*x)^2/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2* d^2), x) + 24*b^2*c^2*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) - 256*a*b*c^2*integrate(1/16*x^2*arctan(c*x) /(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) + 64*b^2*c^2*integrate(1/16*x ^2*log(c^2*x^2 + 1)/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) - (c*(x/(c ^6*d^2*x^2 + c^4*d^2) + arctan(c*x)/(c^5*d^2)) - 2*arctan(c*x)/(c^6*d^2*x^ 2 + c^4*d^2))*b^2*c + 128*b^2*integrate(1/16*arctan(c*x)^2/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) + 32*b^2*integrate(1/16*log(c^2*x^2 + 1)^2...
\[ \int \frac {x^2 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{2}}{{\left (i \, c d x + d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^2 (a+b \arctan (c x))^2}{(d+i c d x)^2} \, dx=\int \frac {x^2\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^2} \,d x \]