Integrand size = 21, antiderivative size = 492 \[ \int \frac {x (a+b \arctan (c x))^2}{d+e x^2} \, dx=-\frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{1-i c x}\right )}{e}+\frac {(a+b \arctan (c x))^2 \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 e}+\frac {(a+b \arctan (c x))^2 \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 e}+\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{e}-\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 e}-\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{2 e}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 e}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 e}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{\left (c \sqrt {-d}+i \sqrt {e}\right ) (1-i c x)}\right )}{4 e} \]
-(a+b*arctan(c*x))^2*ln(2/(1-I*c*x))/e+1/2*(a+b*arctan(c*x))^2*ln(2*c*((-d )^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2)))/e+1/2*(a+b*arctan(c *x))^2*ln(2*c*((-d)^(1/2)+x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)+I*e^(1/2)))/e +I*b*(a+b*arctan(c*x))*polylog(2,1-2/(1-I*c*x))/e-1/2*I*b*(a+b*arctan(c*x) )*polylog(2,1-2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2) ))/e-1/2*I*b*(a+b*arctan(c*x))*polylog(2,1-2*c*((-d)^(1/2)+x*e^(1/2))/(1-I *c*x)/(c*(-d)^(1/2)+I*e^(1/2)))/e-1/2*b^2*polylog(3,1-2/(1-I*c*x))/e+1/4*b ^2*polylog(3,1-2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2 )))/e+1/4*b^2*polylog(3,1-2*c*((-d)^(1/2)+x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/ 2)+I*e^(1/2)))/e
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1322\) vs. \(2(492)=984\).
Time = 13.38 (sec) , antiderivative size = 1322, normalized size of antiderivative = 2.69 \[ \int \frac {x (a+b \arctan (c x))^2}{d+e x^2} \, dx =\text {Too large to display} \]
((8*I)*a*b*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[(c*e*x)/Sqrt[c^2*d*e]] - 8*a*b*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] - 4*b^2*ArcTan[c*x]^2* Log[1 + E^((2*I)*ArcTan[c*x])] - 4*a*b*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*L og[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e)] + 4*a*b*ArcTan[c*x]*Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan [c*x]))/(c^2*d - e)] - 4*b^2*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[c*x] *Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e) ] + 4*b^2*ArcTan[c*x]^2*Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*Ar cTan[c*x]))/(c^2*d - e)] + 4*a*b*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*Log[(-2 *Sqrt[c^2*d*e]*E^((2*I)*ArcTan[c*x]) + e*(-1 + E^((2*I)*ArcTan[c*x])) + c^ 2*d*(1 + E^((2*I)*ArcTan[c*x])))/(c^2*d - e)] + 4*a*b*ArcTan[c*x]*Log[(-2* Sqrt[c^2*d*e]*E^((2*I)*ArcTan[c*x]) + e*(-1 + E^((2*I)*ArcTan[c*x])) + c^2 *d*(1 + E^((2*I)*ArcTan[c*x])))/(c^2*d - e)] + 4*b^2*ArcSin[Sqrt[(c^2*d)/( c^2*d - e)]]*ArcTan[c*x]*Log[(-2*Sqrt[c^2*d*e]*E^((2*I)*ArcTan[c*x]) + e*( -1 + E^((2*I)*ArcTan[c*x])) + c^2*d*(1 + E^((2*I)*ArcTan[c*x])))/(c^2*d - e)] + 4*b^2*ArcTan[c*x]^2*Log[(-2*Sqrt[c^2*d*e]*E^((2*I)*ArcTan[c*x]) + e* (-1 + E^((2*I)*ArcTan[c*x])) + c^2*d*(1 + E^((2*I)*ArcTan[c*x])))/(c^2*d - e)] - 4*b^2*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[c*x]*Log[((2*I)*c^2* d - (2*I)*Sqrt[c^2*d*e] + 2*c*(-e + Sqrt[c^2*d*e])*x)/((c^2*d - e)*(I + c* x))] - 2*b^2*ArcTan[c*x]^2*Log[((2*I)*c^2*d - (2*I)*Sqrt[c^2*d*e] + 2*c...
Time = 0.54 (sec) , antiderivative size = 492, 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.
| \(\displaystyle \int \frac {x (a+b \arctan (c x))^2}{d+e x^2} \, dx\) |
| \(\Big \downarrow \) 5515 |
| \(\displaystyle \int \left (\frac {(a+b \arctan (c x))^2}{2 \sqrt {e} \left (\sqrt {-d}+\sqrt {e} x\right )}-\frac {(a+b \arctan (c x))^2}{2 \sqrt {e} \left (\sqrt {-d}-\sqrt {e} x\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 e}-\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{2 e}+\frac {(a+b \arctan (c x))^2 \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 e}+\frac {(a+b \arctan (c x))^2 \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 e}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{e}-\frac {\log \left (\frac {2}{1-i c x}\right ) (a+b \arctan (c x))^2}{e}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 e}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{4 e}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 e}\) |
-(((a + b*ArcTan[c*x])^2*Log[2/(1 - I*c*x)])/e) + ((a + b*ArcTan[c*x])^2*L og[(2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/( 2*e) + ((a + b*ArcTan[c*x])^2*Log[(2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d ] + I*Sqrt[e])*(1 - I*c*x))])/(2*e) + (I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])/e - ((I/2)*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*c*( Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/e - ((I/2) *b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqr t[-d] + I*Sqrt[e])*(1 - I*c*x))])/e - (b^2*PolyLog[3, 1 - 2/(1 - I*c*x)])/ (2*e) + (b^2*PolyLog[3, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I* Sqrt[e])*(1 - I*c*x))])/(4*e) + (b^2*PolyLog[3, 1 - (2*c*(Sqrt[-d] + Sqrt[ e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/(4*e)
3.13.63.3.1 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])
\[\int \frac {x \left (a +b \arctan \left (c x \right )\right )^{2}}{e \,x^{2}+d}d x\]
\[ \int \frac {x (a+b \arctan (c x))^2}{d+e x^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x}{e x^{2} + d} \,d x } \]
\[ \int \frac {x (a+b \arctan (c x))^2}{d+e x^2} \, dx=\int \frac {x \left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2}}{d + e x^{2}}\, dx \]
\[ \int \frac {x (a+b \arctan (c x))^2}{d+e x^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x}{e x^{2} + d} \,d x } \]
1/2*a^2*log(e*x^2 + d)/e + integrate((b^2*x*arctan(c*x)^2 + 2*a*b*x*arctan (c*x))/(e*x^2 + d), x)
\[ \int \frac {x (a+b \arctan (c x))^2}{d+e x^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x}{e x^{2} + d} \,d x } \]
Timed out. \[ \int \frac {x (a+b \arctan (c x))^2}{d+e x^2} \, dx=\int \frac {x\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{e\,x^2+d} \,d x \]