Integrand size = 23, antiderivative size = 1033 \[ \int \frac {x^2 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=-\frac {i c (a+b \arctan (c x))^2}{2 \left (c^2 d-e\right ) e}+\frac {(a+b \arctan (c x))^2}{4 e^{3/2} \left (\sqrt {-d}-\sqrt {e} x\right )}-\frac {(a+b \arctan (c x))^2}{4 e^{3/2} \left (\sqrt {-d}+\sqrt {e} x\right )}+\frac {b c (a+b \arctan (c x)) \log \left (\frac {2}{1-i c x}\right )}{\left (c^2 d-e\right ) e}-\frac {b c (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{\left (c^2 d-e\right ) e}-\frac {b c (a+b \arctan (c x)) \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 \left (c^2 d-e\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 )}{4 \sqrt {-d} e^{3/2}}-\frac {b c (a+b \arctan (c x)) \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 \left (c^2 d-e\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 )}{4 \sqrt {-d} e^{3/2}}-\frac {i b^2 c \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 \left (c^2 d-e\right ) e}-\frac {i b^2 c \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 \left (c^2 d-e\right ) e}+\frac {i b^2 c \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 )}{4 \left (c^2 d-e\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 )}{4 \sqrt {-d} e^{3/2}}+\frac {i b^2 c \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 )}{4 \left (c^2 d-e\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 )}{4 \sqrt {-d} e^{3/2}}+\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 )}{8 \sqrt {-d} e^{3/2}}-\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 )}{8 \sqrt {-d} e^{3/2}} \]
1/4*I*b^2*c*polylog(2,1-2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2) -I*e^(1/2)))/(c^2*d-e)/e+b*c*(a+b*arctan(c*x))*ln(2/(1-I*c*x))/(c^2*d-e)/e -b*c*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/(c^2*d-e)/e-1/2*b*c*(a+b*arctan(c*x ))*ln(2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2)))/(c^2* d-e)/e-1/2*b*c*(a+b*arctan(c*x))*ln(2*c*((-d)^(1/2)+x*e^(1/2))/(1-I*c*x)/( c*(-d)^(1/2)+I*e^(1/2)))/(c^2*d-e)/e+1/4*I*b^2*c*polylog(2,1-2*c*((-d)^(1/ 2)+x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)+I*e^(1/2)))/(c^2*d-e)/e-1/2*I*b^2*c* polylog(2,1-2/(1-I*c*x))/(c^2*d-e)/e-1/4*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^(3/2)/(- d)^(1/2)+1/4*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^(3/2)/(-d)^(1/2)+1/4*(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^( 3/2)/(-d)^(1/2)-1/4*(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^(3/2)/(-d)^(1/2)-1/2*I*b^2*c*polylog(2,1 -2/(1+I*c*x))/(c^2*d-e)/e-1/2*I*c*(a+b*arctan(c*x))^2/(c^2*d-e)/e+1/8*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^(3/2)/(-d)^(1/2)-1/8*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^(3/2)/(-d)^(1/2)+1/4*(a+b*arctan(c*x))^2/e^ (3/2)/((-d)^(1/2)-x*e^(1/2))-1/4*(a+b*arctan(c*x))^2/e^(3/2)/((-d)^(1/2)+x *e^(1/2))
\[ \int \frac {x^2 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\int \frac {x^2 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx \]
Time = 1.90 (sec) , antiderivative size = 1033, 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.
| \(\displaystyle \int \frac {x^2 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 5515 |
| \(\displaystyle \int \left (\frac {(a+b \arctan (c x))^2}{e \left (d+e x^2\right )}-\frac {d (a+b \arctan (c x))^2}{e \left (d+e x^2\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {i c \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) b^2}{2 \left (c^2 d-e\right ) e}-\frac {i c \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) b^2}{2 \left (c^2 d-e\right ) e}+\frac {i c \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 ) b^2}{4 \left (c^2 d-e\right ) e}+\frac {i c \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 ) b^2}{4 \left (c^2 d-e\right ) e}+\frac {\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 ) b^2}{8 \sqrt {-d} e^{3/2}}-\frac {\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 ) b^2}{8 \sqrt {-d} e^{3/2}}+\frac {c (a+b \arctan (c x)) \log \left (\frac {2}{1-i c x}\right ) b}{\left (c^2 d-e\right ) e}-\frac {c (a+b \arctan (c x)) \log \left (\frac {2}{i c x+1}\right ) b}{\left (c^2 d-e\right ) e}-\frac {c (a+b \arctan (c x)) \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 ) b}{2 \left (c^2 d-e\right ) e}-\frac {c (a+b \arctan (c x)) \log \left (\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right ) b}{2 \left (c^2 d-e\right ) e}-\frac {i (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 ) b}{4 \sqrt {-d} e^{3/2}}+\frac {i (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 ) b}{4 \sqrt {-d} e^{3/2}}-\frac {i c (a+b \arctan (c x))^2}{2 \left (c^2 d-e\right ) e}+\frac {(a+b \arctan (c x))^2}{4 e^{3/2} \left (\sqrt {-d}-\sqrt {e} x\right )}-\frac {(a+b \arctan (c x))^2}{4 e^{3/2} \left (\sqrt {e} x+\sqrt {-d}\right )}+\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 )}{4 \sqrt {-d} e^{3/2}}-\frac {(a+b \arctan (c x))^2 \log \left (\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{4 \sqrt {-d} e^{3/2}}\) |
((-1/2*I)*c*(a + b*ArcTan[c*x])^2)/((c^2*d - e)*e) + (a + b*ArcTan[c*x])^2 /(4*e^(3/2)*(Sqrt[-d] - Sqrt[e]*x)) - (a + b*ArcTan[c*x])^2/(4*e^(3/2)*(Sq rt[-d] + Sqrt[e]*x)) + (b*c*(a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/((c^2* d - e)*e) - (b*c*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/((c^2*d - e)*e) - (b*c*(a + b*ArcTan[c*x])*Log[(2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(2*(c^2*d - e)*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))])/(4*Sq rt[-d]*e^(3/2)) - (b*c*(a + b*ArcTan[c*x])*Log[(2*c*(Sqrt[-d] + Sqrt[e]*x) )/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/(2*(c^2*d - e)*e) - ((a + b*Arc Tan[c*x])^2*Log[(2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/(4*Sqrt[-d]*e^(3/2)) - ((I/2)*b^2*c*PolyLog[2, 1 - 2/(1 - I*c* x)])/((c^2*d - e)*e) - ((I/2)*b^2*c*PolyLog[2, 1 - 2/(1 + I*c*x)])/((c^2*d - e)*e) + ((I/4)*b^2*c*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sq rt[-d] - I*Sqrt[e])*(1 - I*c*x))])/((c^2*d - e)*e) - ((I/4)*b*(a + b*ArcTa n[c*x])*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[ e])*(1 - I*c*x))])/(Sqrt[-d]*e^(3/2)) + ((I/4)*b^2*c*PolyLog[2, 1 - (2*c*( Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/((c^2*d - e)*e) + ((I/4)*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))])/(Sqrt[-d]*e^(3/2)) + (b^2* PolyLog[3, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(...
3.13.69.3.1 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 = 157.17 (sec) , antiderivative size = 6565, normalized size of antiderivative = 6.36
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(6565\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(6636\) |
| default | \(\text {Expression too large to display}\) | \(6636\) |
\[ \int \frac {x^2 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{2}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
integral((b^2*x^2*arctan(c*x)^2 + 2*a*b*x^2*arctan(c*x) + a^2*x^2)/(e^2*x^ 4 + 2*d*e*x^2 + d^2), x)
Timed out. \[ \int \frac {x^2 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {x^2 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {x^2 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{2}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^2 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\int \frac {x^2\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (e\,x^2+d\right )}^2} \,d x \]