Integrand size = 23, antiderivative size = 355 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))^2}{x} \, dx=-\frac {2 a b d e x}{c}+\frac {a b e^2 x}{2 c^3}+\frac {b^2 e^2 x^2}{12 c^2}-\frac {2 b^2 d e x \arctan (c x)}{c}+\frac {b^2 e^2 x \arctan (c x)}{2 c^3}-\frac {b e^2 x^3 (a+b \arctan (c x))}{6 c}+\frac {d e (a+b \arctan (c x))^2}{c^2}-\frac {e^2 (a+b \arctan (c x))^2}{4 c^4}+d e x^2 (a+b \arctan (c x))^2+\frac {1}{4} e^2 x^4 (a+b \arctan (c x))^2+2 d^2 (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+\frac {b^2 d e \log \left (1+c^2 x^2\right )}{c^2}-\frac {b^2 e^2 \log \left (1+c^2 x^2\right )}{3 c^4}-i b d^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b d^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]
-2*a*b*d*e*x/c+1/2*a*b*e^2*x/c^3+1/12*b^2*e^2*x^2/c^2-2*b^2*d*e*x*arctan(c *x)/c+1/2*b^2*e^2*x*arctan(c*x)/c^3-1/6*b*e^2*x^3*(a+b*arctan(c*x))/c+d*e* (a+b*arctan(c*x))^2/c^2-1/4*e^2*(a+b*arctan(c*x))^2/c^4+d*e*x^2*(a+b*arcta n(c*x))^2+1/4*e^2*x^4*(a+b*arctan(c*x))^2-2*d^2*(a+b*arctan(c*x))^2*arctan h(-1+2/(1+I*c*x))+b^2*d*e*ln(c^2*x^2+1)/c^2-1/3*b^2*e^2*ln(c^2*x^2+1)/c^4+ I*b*d^2*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))-I*b*d^2*(a+b*arctan(c* x))*polylog(2,1-2/(1+I*c*x))-1/2*b^2*d^2*polylog(3,1-2/(1+I*c*x))+1/2*b^2* d^2*polylog(3,-1+2/(1+I*c*x))
Time = 0.47 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.10 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))^2}{x} \, dx=a^2 d e x^2+\frac {1}{4} a^2 e^2 x^4+\frac {2 a b d e \left (-c x+\left (1+c^2 x^2\right ) \arctan (c x)\right )}{c^2}+\frac {a b e^2 \left (3 c x-c^3 x^3+3 \left (-1+c^4 x^4\right ) \arctan (c x)\right )}{6 c^4}+a^2 d^2 \log (x)+\frac {b^2 e^2 \left (1+c^2 x^2+\left (6 c x-2 c^3 x^3\right ) \arctan (c x)+3 \left (-1+c^4 x^4\right ) \arctan (c x)^2-4 \log \left (1+c^2 x^2\right )\right )}{12 c^4}+\frac {b^2 d e \left (-2 c x \arctan (c x)+\left (1+c^2 x^2\right ) \arctan (c x)^2+\log \left (1+c^2 x^2\right )\right )}{c^2}+i a b d^2 (\operatorname {PolyLog}(2,-i c x)-\operatorname {PolyLog}(2,i c x))+b^2 d^2 \left (-\frac {i \pi ^3}{24}+\frac {2}{3} i \arctan (c x)^3+\arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )-\arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+i \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )+i \arctan (c x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+\frac {1}{2} \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right ) \]
a^2*d*e*x^2 + (a^2*e^2*x^4)/4 + (2*a*b*d*e*(-(c*x) + (1 + c^2*x^2)*ArcTan[ c*x]))/c^2 + (a*b*e^2*(3*c*x - c^3*x^3 + 3*(-1 + c^4*x^4)*ArcTan[c*x]))/(6 *c^4) + a^2*d^2*Log[x] + (b^2*e^2*(1 + c^2*x^2 + (6*c*x - 2*c^3*x^3)*ArcTa n[c*x] + 3*(-1 + c^4*x^4)*ArcTan[c*x]^2 - 4*Log[1 + c^2*x^2]))/(12*c^4) + (b^2*d*e*(-2*c*x*ArcTan[c*x] + (1 + c^2*x^2)*ArcTan[c*x]^2 + Log[1 + c^2*x ^2]))/c^2 + I*a*b*d^2*(PolyLog[2, (-I)*c*x] - PolyLog[2, I*c*x]) + b^2*d^2 *((-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 = 0.92 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, 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.
\(\displaystyle \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))^2}{x} \, dx\) |
\(\Big \downarrow \) 5515 |
\(\displaystyle \int \left (\frac {d^2 (a+b \arctan (c x))^2}{x}+2 d e x (a+b \arctan (c x))^2+e^2 x^3 (a+b \arctan (c x))^2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 d^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-\frac {e^2 (a+b \arctan (c x))^2}{4 c^4}+\frac {d e (a+b \arctan (c x))^2}{c^2}-i b d^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))+i b d^2 \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))+d e x^2 (a+b \arctan (c x))^2+\frac {1}{4} e^2 x^4 (a+b \arctan (c x))^2-\frac {b e^2 x^3 (a+b \arctan (c x))}{6 c}+\frac {a b e^2 x}{2 c^3}-\frac {2 a b d e x}{c}+\frac {b^2 e^2 x \arctan (c x)}{2 c^3}-\frac {2 b^2 d e x \arctan (c x)}{c}+\frac {b^2 d e \log \left (c^2 x^2+1\right )}{c^2}+\frac {b^2 e^2 x^2}{12 c^2}-\frac {b^2 e^2 \log \left (c^2 x^2+1\right )}{3 c^4}-\frac {1}{2} b^2 d^2 \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )+\frac {1}{2} b^2 d^2 \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right )\) |
(-2*a*b*d*e*x)/c + (a*b*e^2*x)/(2*c^3) + (b^2*e^2*x^2)/(12*c^2) - (2*b^2*d *e*x*ArcTan[c*x])/c + (b^2*e^2*x*ArcTan[c*x])/(2*c^3) - (b*e^2*x^3*(a + b* ArcTan[c*x]))/(6*c) + (d*e*(a + b*ArcTan[c*x])^2)/c^2 - (e^2*(a + b*ArcTan [c*x])^2)/(4*c^4) + d*e*x^2*(a + b*ArcTan[c*x])^2 + (e^2*x^4*(a + b*ArcTan [c*x])^2)/4 + 2*d^2*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] + (b^ 2*d*e*Log[1 + c^2*x^2])/c^2 - (b^2*e^2*Log[1 + c^2*x^2])/(3*c^4) - I*b*d^2 *(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)] + I*b*d^2*(a + b*ArcTan [c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)] - (b^2*d^2*PolyLog[3, 1 - 2/(1 + I*c *x)])/2 + (b^2*d^2*PolyLog[3, -1 + 2/(1 + I*c*x)])/2
3.13.58.3.1 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 = 13.60 (sec) , antiderivative size = 1524, normalized size of antiderivative = 4.29
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1524\) |
default | \(\text {Expression too large to display}\) | \(1524\) |
parts | \(\text {Expression too large to display}\) | \(1526\) |
a^2*d^2*ln(c*x)+b^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))*d^2*arctan(c*x)^2+2* I*b^2/c^2*arctan(c*x)*d*e-1/2*I*b^2*Pi*d^2*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+1/2*I*b^2*Pi*d^2*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+1/2*I* b^2*Pi*d^2*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+a^2*d*e*x^2+1/4*b^2*arctan(c*x)^2*e^2*x^4+1/12*b^2*e^2*x^ 2/c^2+1/2*b^2*e^2*x*arctan(c*x)/c^3+b^2/c^2*e*arctan(c*x)^2*d-2*b^2*d*e*x* arctan(c*x)/c+2*a*b/c^4*(arctan(c*x)*d*c^4*e*x^2+1/4*arctan(c*x)*e^2*c^4*x ^4+arctan(c*x)*c^4*d^2*ln(c*x)-1/4*e*(4*c^3*x*d+1/3*e*c^3*x^3-e*c*x+(-4*c^ 2*d+e)*arctan(c*x))-c^4*d^2*(-1/2*I*ln(c*x)*ln(1+I*c*x)+1/2*I*ln(c*x)*ln(1 -I*c*x)-1/2*I*dilog(1+I*c*x)+1/2*I*dilog(1-I*c*x)))-1/4*b^2/c^4*e^2*arctan (c*x)^2-1/6*b^2/c*e^2*arctan(c*x)*x^3+I*b^2*d^2*arctan(c*x)*polylog(2,-(1+ I*c*x)^2/(c^2*x^2+1))-2*b^2/c^2*d*e*ln((1+I*c*x)^2/(c^2*x^2+1)+1)-2*I*b^2* polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))*d^2*arctan(c*x)+1/2*I*b^2*Pi*d^2*ar ctan(c*x)^2-2*I*b^2*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))*d^2*arctan(c*x )+1/12*b^2/c^4*e^2+2*b^2*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))*d^2-1/2*b ^2*d^2*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+2*b^2*polylog(3,(1+I*c*x)/(c^2* x^2+1)^(1/2))*d^2+2/3*b^2/c^4*e^2*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+b^2*arctan (c*x)^2*d^2*ln(c*x)-b^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)*d^2*arctan(c*x)^2+b^ 2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))*d^2*arctan(c*x)^2+b^2*arctan(c*x)^2...
\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))^2}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x} \,d x } \]
integral((a^2*e^2*x^4 + 2*a^2*d*e*x^2 + a^2*d^2 + (b^2*e^2*x^4 + 2*b^2*d*e *x^2 + b^2*d^2)*arctan(c*x)^2 + 2*(a*b*e^2*x^4 + 2*a*b*d*e*x^2 + a*b*d^2)* arctan(c*x))/x, x)
\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))^2}{x} \, dx=\int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2} \left (d + e x^{2}\right )^{2}}{x}\, dx \]
\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))^2}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x} \,d x } \]
1/4*a^2*e^2*x^4 + 12*b^2*c^2*e^2*integrate(1/16*x^6*arctan(c*x)^2/(c^2*x^3 + x), x) + b^2*c^2*e^2*integrate(1/16*x^6*log(c^2*x^2 + 1)^2/(c^2*x^3 + x ), x) + 32*a*b*c^2*e^2*integrate(1/16*x^6*arctan(c*x)/(c^2*x^3 + x), x) + b^2*c^2*e^2*integrate(1/16*x^6*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + 24*b^2 *c^2*d*e*integrate(1/16*x^4*arctan(c*x)^2/(c^2*x^3 + x), x) + 2*b^2*c^2*d* e*integrate(1/16*x^4*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 64*a*b*c^2*d*e *integrate(1/16*x^4*arctan(c*x)/(c^2*x^3 + x), x) + 4*b^2*c^2*d*e*integrat e(1/16*x^4*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + 12*b^2*c^2*d^2*integrate(1 /16*x^2*arctan(c*x)^2/(c^2*x^3 + x), x) + 32*a*b*c^2*d^2*integrate(1/16*x^ 2*arctan(c*x)/(c^2*x^3 + x), x) + 1/96*b^2*d^2*log(c^2*x^2 + 1)^3 + a^2*d* e*x^2 - 2*b^2*c*e^2*integrate(1/16*x^5*arctan(c*x)/(c^2*x^3 + x), x) - 8*b ^2*c*d*e*integrate(1/16*x^3*arctan(c*x)/(c^2*x^3 + x), x) + 12*b^2*e^2*int egrate(1/16*x^4*arctan(c*x)^2/(c^2*x^3 + x), x) + b^2*e^2*integrate(1/16*x ^4*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 32*a*b*e^2*integrate(1/16*x^4*ar ctan(c*x)/(c^2*x^3 + x), x) + 24*b^2*d*e*integrate(1/16*x^2*arctan(c*x)^2/ (c^2*x^3 + x), x) + 64*a*b*d*e*integrate(1/16*x^2*arctan(c*x)/(c^2*x^3 + x ), x) + 12*b^2*d^2*integrate(1/16*arctan(c*x)^2/(c^2*x^3 + x), x) + b^2*d^ 2*integrate(1/16*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 32*a*b*d^2*integra te(1/16*arctan(c*x)/(c^2*x^3 + x), x) + 1/48*b^2*d*e*log(c^2*x^2 + 1)^3/c^ 2 + a^2*d^2*log(x) + 1/16*(b^2*e^2*x^4 + 4*b^2*d*e*x^2)*arctan(c*x)^2 -...
Timed out. \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))^2}{x} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))^2}{x} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,{\left (e\,x^2+d\right )}^2}{x} \,d x \]