Integrand size = 25, antiderivative size = 337 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))^2}{x^3} \, dx=-\frac {b c d^2 (a+b \arctan (c x))}{x}+\frac {3}{2} c^2 d^2 (a+b \arctan (c x))^2-\frac {d^2 (a+b \arctan (c x))^2}{2 x^2}-\frac {2 i c d^2 (a+b \arctan (c x))^2}{x}-2 c^2 d^2 (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+b^2 c^2 d^2 \log (x)-\frac {1}{2} b^2 c^2 d^2 \log \left (1+c^2 x^2\right )+4 i b c^2 d^2 (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )+2 b^2 c^2 d^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )+i b c^2 d^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-i b c^2 d^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 c^2 d^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 c^2 d^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]
-b*c*d^2*(a+b*arctan(c*x))/x+3/2*c^2*d^2*(a+b*arctan(c*x))^2-1/2*d^2*(a+b* arctan(c*x))^2/x^2-2*I*c*d^2*(a+b*arctan(c*x))^2/x+2*c^2*d^2*(a+b*arctan(c *x))^2*arctanh(-1+2/(1+I*c*x))+b^2*c^2*d^2*ln(x)-1/2*b^2*c^2*d^2*ln(c^2*x^ 2+1)+4*I*b*c^2*d^2*(a+b*arctan(c*x))*ln(2-2/(1-I*c*x))+2*b^2*c^2*d^2*polyl og(2,-1+2/(1-I*c*x))+I*b*c^2*d^2*(a+b*arctan(c*x))*polylog(2,1-2/(1+I*c*x) )-I*b*c^2*d^2*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))+1/2*b^2*c^2*d^2* polylog(3,1-2/(1+I*c*x))-1/2*b^2*c^2*d^2*polylog(3,-1+2/(1+I*c*x))
Time = 0.67 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.15 \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))^2}{x^3} \, dx=-\frac {d^2 \left (a^2+4 i a^2 c x+2 a b (\arctan (c x)+c x (1+c x \arctan (c x)))+2 a^2 c^2 x^2 \log (x)+b^2 \left (2 c x \arctan (c x)+\left (1+c^2 x^2\right ) \arctan (c x)^2-2 c^2 x^2 \log \left (\frac {c x}{\sqrt {1+c^2 x^2}}\right )\right )+4 i a b c x \left (2 \arctan (c x)+c x \left (-2 \log (c x)+\log \left (1+c^2 x^2\right )\right )\right )+4 i b^2 c x \left (\arctan (c x)^2-2 c x \arctan (c x) \log \left (1-e^{2 i \arctan (c x)}\right )+i c x \left (\arctan (c x)^2+\operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )\right )\right )+2 i a b c^2 x^2 (\operatorname {PolyLog}(2,-i c x)-\operatorname {PolyLog}(2,i c x))+\frac {1}{12} b^2 c^2 x^2 \left (-i \pi ^3+16 i \arctan (c x)^3+24 \arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )-24 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+24 i \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )+24 i \arctan (c x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+12 \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )-12 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right )\right )}{2 x^2} \]
-1/2*(d^2*(a^2 + (4*I)*a^2*c*x + 2*a*b*(ArcTan[c*x] + c*x*(1 + c*x*ArcTan[ c*x])) + 2*a^2*c^2*x^2*Log[x] + b^2*(2*c*x*ArcTan[c*x] + (1 + c^2*x^2)*Arc Tan[c*x]^2 - 2*c^2*x^2*Log[(c*x)/Sqrt[1 + c^2*x^2]]) + (4*I)*a*b*c*x*(2*Ar cTan[c*x] + c*x*(-2*Log[c*x] + Log[1 + c^2*x^2])) + (4*I)*b^2*c*x*(ArcTan[ c*x]^2 - 2*c*x*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] + I*c*x*(ArcTan[ c*x]^2 + PolyLog[2, E^((2*I)*ArcTan[c*x])])) + (2*I)*a*b*c^2*x^2*(PolyLog[ 2, (-I)*c*x] - PolyLog[2, I*c*x]) + (b^2*c^2*x^2*((-I)*Pi^3 + (16*I)*ArcTa n[c*x]^3 + 24*ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - 24*ArcTan[c* x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + (24*I)*ArcTan[c*x]*PolyLog[2, E^((-2 *I)*ArcTan[c*x])] + (24*I)*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 12*PolyLog[3, E^((-2*I)*ArcTan[c*x])] - 12*PolyLog[3, -E^((2*I)*ArcTan[c *x])]))/12))/x^2
Time = 0.83 (sec) , antiderivative size = 337, 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.
| \(\displaystyle \int \frac {(d+i c d x)^2 (a+b \arctan (c x))^2}{x^3} \, dx\) |
| \(\Big \downarrow \) 5411 |
| \(\displaystyle \int \left (-\frac {c^2 d^2 (a+b \arctan (c x))^2}{x}+\frac {d^2 (a+b \arctan (c x))^2}{x^3}+\frac {2 i c d^2 (a+b \arctan (c x))^2}{x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 c^2 d^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2+i b c^2 d^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))-i b c^2 d^2 \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))+\frac {3}{2} c^2 d^2 (a+b \arctan (c x))^2+4 i b c^2 d^2 \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))-\frac {d^2 (a+b \arctan (c x))^2}{2 x^2}-\frac {2 i c d^2 (a+b \arctan (c x))^2}{x}-\frac {b c d^2 (a+b \arctan (c x))}{x}+2 b^2 c^2 d^2 \operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right )+\frac {1}{2} b^2 c^2 d^2 \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )-\frac {1}{2} b^2 c^2 d^2 \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right )-\frac {1}{2} b^2 c^2 d^2 \log \left (c^2 x^2+1\right )+b^2 c^2 d^2 \log (x)\) |
-((b*c*d^2*(a + b*ArcTan[c*x]))/x) + (3*c^2*d^2*(a + b*ArcTan[c*x])^2)/2 - (d^2*(a + b*ArcTan[c*x])^2)/(2*x^2) - ((2*I)*c*d^2*(a + b*ArcTan[c*x])^2) /x - 2*c^2*d^2*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] + b^2*c^2* d^2*Log[x] - (b^2*c^2*d^2*Log[1 + c^2*x^2])/2 + (4*I)*b*c^2*d^2*(a + b*Arc Tan[c*x])*Log[2 - 2/(1 - I*c*x)] + 2*b^2*c^2*d^2*PolyLog[2, -1 + 2/(1 - I* c*x)] + I*b*c^2*d^2*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)] - I* b*c^2*d^2*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)] + (b^2*c^2*d^ 2*PolyLog[3, 1 - 2/(1 + I*c*x)])/2 - (b^2*c^2*d^2*PolyLog[3, -1 + 2/(1 + I *c*x)])/2
3.1.82.3.1 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 = 26.52 (sec) , antiderivative size = 1381, normalized size of antiderivative = 4.10
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1381\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1383\) |
| default | \(\text {Expression too large to display}\) | \(1383\) |
a^2*d^2*(-1/2/x^2-c^2*ln(x)-2*I*c/x)+b^2*d^2*c^2*(-1/2/c^2/x^2*arctan(c*x) ^2+1/2*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))-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))+4*dilog(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-4*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-I*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+2*I*arc tan(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*arctan(c*x)^2+4*I*arctan(c*x)*ln(1+ (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*arct an(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*c...
\[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))^2}{x^3} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{3}} \,d x } \]
integral(-1/4*(4*a^2*c^2*d^2*x^2 - 8*I*a^2*c*d^2*x - 4*a^2*d^2 - (b^2*c^2* d^2*x^2 - 2*I*b^2*c*d^2*x - b^2*d^2)*log(-(c*x + I)/(c*x - I))^2 + 4*(I*a* b*c^2*d^2*x^2 + 2*a*b*c*d^2*x - I*a*b*d^2)*log(-(c*x + I)/(c*x - I)))/x^3, x)
Timed out. \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))^2}{x^3} \, dx=\text {Timed out} \]
\[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))^2}{x^3} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{3}} \,d x } \]
-a^2*c^2*d^2*log(x) - 2*I*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x) /x)*a*b*c*d^2 - ((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*a*b*d^2 - 2*I* a^2*c*d^2/x - 1/2*a^2*d^2/x^2 + 1/96*(48*I*(b^2*c^2*d^2*arctan(c*x)^3 + 4* b^2*c^4*d^2*integrate(1/8*x^4*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^5 + x^3) , x) + 2*b^2*c^3*d^2*integrate(1/8*x^3*log(c^2*x^2 + 1)^2/(c^2*x^5 + x^3), x) - 8*b^2*c^3*d^2*integrate(1/8*x^3*log(c^2*x^2 + 1)/(c^2*x^5 + x^3), x) + 20*b^2*c^2*d^2*integrate(1/8*x^2*arctan(c*x)/(c^2*x^5 + x^3), x) + 24*b ^2*c*d^2*integrate(1/8*x*arctan(c*x)^2/(c^2*x^5 + x^3), x) + 2*b^2*c*d^2*i ntegrate(1/8*x*log(c^2*x^2 + 1)^2/(c^2*x^5 + x^3), x) + 2*b^2*c*d^2*integr ate(1/8*x*log(c^2*x^2 + 1)/(c^2*x^5 + x^3), x) - 4*b^2*d^2*integrate(1/8*a rctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^5 + x^3), x))*x^2 - (1152*b^2*c^4*d^2*i ntegrate(1/16*x^4*arctan(c*x)^2/(c^2*x^5 + x^3), x) + 3072*a*b*c^4*d^2*int egrate(1/16*x^4*arctan(c*x)/(c^2*x^5 + x^3), x) + b^2*c^2*d^2*log(c^2*x^2 + 1)^3 + 48*b^2*c^2*d^2*arctan(c*x)^2 - 768*b^2*c^3*d^2*integrate(1/16*x^3 *arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^5 + x^3), x) + 3072*a*b*c^2*d^2*integ rate(1/16*x^2*arctan(c*x)/(c^2*x^5 + x^3), x) + 960*b^2*c^2*d^2*integrate( 1/16*x^2*log(c^2*x^2 + 1)/(c^2*x^5 + x^3), x) - 768*b^2*c*d^2*integrate(1/ 16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^5 + x^3), x) - 384*b^2*c*d^2*inte grate(1/16*x*arctan(c*x)/(c^2*x^5 + x^3), x) - 1152*b^2*d^2*integrate(1/16 *arctan(c*x)^2/(c^2*x^5 + x^3), x) - 96*b^2*d^2*integrate(1/16*log(c^2*...
Timed out. \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))^2}{x^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(d+i c d x)^2 (a+b \arctan (c x))^2}{x^3} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^2}{x^3} \,d x \]