Integrand size = 21, antiderivative size = 217 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=-\frac {a b e x}{c}-\frac {b^2 e x \arctan (c x)}{c}+\frac {e (a+b \arctan (c x))^2}{2 c^2}+\frac {1}{2} e x^2 (a+b \arctan (c x))^2+2 d (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+\frac {b^2 e \log \left (1+c^2 x^2\right )}{2 c^2}-i b d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b d (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]
-a*b*e*x/c-b^2*e*x*arctan(c*x)/c+1/2*e*(a+b*arctan(c*x))^2/c^2+1/2*e*x^2*( a+b*arctan(c*x))^2-2*d*(a+b*arctan(c*x))^2*arctanh(-1+2/(1+I*c*x))+1/2*b^2 *e*ln(c^2*x^2+1)/c^2-I*b*d*(a+b*arctan(c*x))*polylog(2,1-2/(1+I*c*x))+I*b* d*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))-1/2*b^2*d*polylog(3,1-2/(1+I *c*x))+1/2*b^2*d*polylog(3,-1+2/(1+I*c*x))
Time = 0.25 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.21 \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=\frac {1}{2} a^2 e x^2+\frac {a b e \left (-c x+\left (1+c^2 x^2\right ) \arctan (c x)\right )}{c^2}+a^2 d \log (x)+\frac {b^2 e \left (-2 c x \arctan (c x)+\left (1+c^2 x^2\right ) \arctan (c x)^2+\log \left (1+c^2 x^2\right )\right )}{2 c^2}+i a b d (\operatorname {PolyLog}(2,-i c x)-\operatorname {PolyLog}(2,i c x))+b^2 d \left (-\frac {i \pi ^3}{24}+\frac {2}{3} i \arctan (c x)^3+\arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )-\arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+i \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )+i \arctan (c x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+\frac {1}{2} \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right ) \]
(a^2*e*x^2)/2 + (a*b*e*(-(c*x) + (1 + c^2*x^2)*ArcTan[c*x]))/c^2 + a^2*d*L og[x] + (b^2*e*(-2*c*x*ArcTan[c*x] + (1 + c^2*x^2)*ArcTan[c*x]^2 + Log[1 + c^2*x^2]))/(2*c^2) + I*a*b*d*(PolyLog[2, (-I)*c*x] - PolyLog[2, I*c*x]) + b^2*d*((-1/24*I)*Pi^3 + ((2*I)/3)*ArcTan[c*x]^3 + ArcTan[c*x]^2*Log[1 - E ^((-2*I)*ArcTan[c*x])] - ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + I* ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + I*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + PolyLog[3, E^((-2*I)*ArcTan[c*x])]/2 - PolyLog[3 , -E^((2*I)*ArcTan[c*x])]/2)
Time = 0.64 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {5515, 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.
| \(\displaystyle \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx\) |
| \(\Big \downarrow \) 5515 |
| \(\displaystyle \int \left (\frac {d (a+b \arctan (c x))^2}{x}+e x (a+b \arctan (c x))^2\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 d \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2+\frac {e (a+b \arctan (c x))^2}{2 c^2}-i b d \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))+i b d \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))+\frac {1}{2} e x^2 (a+b \arctan (c x))^2-\frac {a b e x}{c}-\frac {b^2 e x \arctan (c x)}{c}+\frac {b^2 e \log \left (c^2 x^2+1\right )}{2 c^2}-\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )+\frac {1}{2} b^2 d \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right )\) |
-((a*b*e*x)/c) - (b^2*e*x*ArcTan[c*x])/c + (e*(a + b*ArcTan[c*x])^2)/(2*c^ 2) + (e*x^2*(a + b*ArcTan[c*x])^2)/2 + 2*d*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] + (b^2*e*Log[1 + c^2*x^2])/(2*c^2) - I*b*d*(a + b*ArcTan [c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)] + I*b*d*(a + b*ArcTan[c*x])*PolyLog[2 , -1 + 2/(1 + I*c*x)] - (b^2*d*PolyLog[3, 1 - 2/(1 + I*c*x)])/2 + (b^2*d*P olyLog[3, -1 + 2/(1 + I*c*x)])/2
3.13.51.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_)^2)^(q_.), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*ArcTan[c*x] )^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d , e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || IntegerQ[m])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 6.62 (sec) , antiderivative size = 1262, normalized size of antiderivative = 5.82
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1262\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1263\) |
| default | \(\text {Expression too large to display}\) | \(1263\) |
1/2*a^2*e*x^2+a^2*d*ln(x)+b^2*(1/2*arctan(c*x)^2*x^2*e+arctan(c*x)^2*d*ln( c*x)-1/c^2*(-1/2*I*c^2*d*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/ (c^2*x^2+1)+1))^3*arctan(c*x)^2+2*I*c^2*d*arctan(c*x)*polylog(2,(1+I*c*x)/ (c^2*x^2+1)^(1/2))+1/2*I*c^2*d*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+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^ 2-I*c^2*d*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))-1/2*I*c^2*d*Pi*c sgn(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))*csg n(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2 -1/2*arctan(c*x)^2*e-1/2*I*c^2*d*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((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+I*c*x)^2/(c ^2*x^2+1)+1))*arctan(c*x)^2+e*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+1/2*I*c^2*d*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+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+1/2*I*c^2*d*Pi*csgn(((1+I*c*x)^ 2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+e*arctan(c*x )*(c*x-I)-1/2*I*c^2*d*Pi*arctan(c*x)^2+ln((1+I*c*x)^2/(c^2*x^2+1)-1)*c^2*d *arctan(c*x)^2-ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))*c^2*d*arctan(c*x)^2-1/2*I *c^2*d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^ 3*arctan(c*x)^2-2*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))*c^2*d-ln(1-(1+I* c*x)/(c^2*x^2+1)^(1/2))*c^2*d*arctan(c*x)^2+2*I*c^2*d*arctan(c*x)*polylog( 2,-(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)...
\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x} \,d x } \]
integral((a^2*e*x^2 + a^2*d + (b^2*e*x^2 + b^2*d)*arctan(c*x)^2 + 2*(a*b*e *x^2 + a*b*d)*arctan(c*x))/x, x)
\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2} \left (d + e x^{2}\right )}{x}\, dx \]
\[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x} \,d x } \]
1/8*b^2*e*x^2*arctan(c*x)^2 - 1/32*b^2*e*x^2*log(c^2*x^2 + 1)^2 + 12*b^2*c ^2*e*integrate(1/16*x^4*arctan(c*x)^2/(c^2*x^3 + x), x) + b^2*c^2*e*integr ate(1/16*x^4*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 32*a*b*c^2*e*integrate (1/16*x^4*arctan(c*x)/(c^2*x^3 + x), x) + 2*b^2*c^2*e*integrate(1/16*x^4*l og(c^2*x^2 + 1)/(c^2*x^3 + x), x) + 12*b^2*c^2*d*integrate(1/16*x^2*arctan (c*x)^2/(c^2*x^3 + x), x) + 32*a*b*c^2*d*integrate(1/16*x^2*arctan(c*x)/(c ^2*x^3 + x), x) + 1/96*b^2*d*log(c^2*x^2 + 1)^3 + 1/2*a^2*e*x^2 - 4*b^2*c* e*integrate(1/16*x^3*arctan(c*x)/(c^2*x^3 + x), x) + 12*b^2*e*integrate(1/ 16*x^2*arctan(c*x)^2/(c^2*x^3 + x), x) + 32*a*b*e*integrate(1/16*x^2*arcta n(c*x)/(c^2*x^3 + x), x) + 12*b^2*d*integrate(1/16*arctan(c*x)^2/(c^2*x^3 + x), x) + b^2*d*integrate(1/16*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 32* a*b*d*integrate(1/16*arctan(c*x)/(c^2*x^3 + x), x) + 1/96*b^2*e*log(c^2*x^ 2 + 1)^3/c^2 + a^2*d*log(x)
Timed out. \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\left (d+e x^2\right ) (a+b \arctan (c x))^2}{x} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,\left (e\,x^2+d\right )}{x} \,d x \]