Integrand size = 23, antiderivative size = 943 \[ \int \frac {x^3 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=-\frac {c^2 d (a+b \arctan (c x))^2}{2 \left (c^2 d-e\right ) e^2}+\frac {(a+b \arctan (c x))^2}{4 e^2 \left (1-\frac {\sqrt {e} x}{\sqrt {-d}}\right )}+\frac {(a+b \arctan (c x))^2}{4 e^2 \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )}-\frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{1-i c x}\right )}{e^2}-\frac {b c \sqrt {-d} (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^{3/2}}+\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^2}+\frac {b c \sqrt {-d} (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^{3/2}}+\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^2}+\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{e^2}+\frac {i b^2 c \sqrt {-d} \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^{3/2}}-\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^2}-\frac {i b^2 c \sqrt {-d} \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^{3/2}}-\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^2}-\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 e^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 )}{4 e^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 )}{4 e^2} \]
-1/2*c^2*d*(a+b*arctan(c*x))^2/(c^2*d-e)/e^2-(a+b*arctan(c*x))^2*ln(2/(1-I *c*x))/e^2+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^2+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^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)))*(-d)^(1/2) /(c^2*d-e)/e^(3/2)-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^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)))*(-d)^(1/2)/ (c^2*d-e)/e^(3/2)-1/2*b^2*polylog(3,1-2/(1-I*c*x))/e^2+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^2+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^2-1/2*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)))*(-d)^(1/2)/(c^2*d-e)/e^(3/2)+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)))*(- d)^(1/2)/(c^2*d-e)/e^(3/2)+I*b*(a+b*arctan(c*x))*polylog(2,1-2/(1-I*c*x))/ e^2-1/2*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^2+1/4*(a+b*arctan(c*x))^2/e^2/(1-x*e^(1/2 )/(-d)^(1/2))+1/4*(a+b*arctan(c*x))^2/e^2/(1+x*e^(1/2)/(-d)^(1/2))
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\int \frac {x^3 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx \]
Time = 1.86 (sec) , antiderivative size = 943, 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.
| \(\displaystyle \int \frac {x^3 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 5515 |
| \(\displaystyle \int \left (\frac {x (a+b \arctan (c x))^2}{e \left (d+e x^2\right )}-\frac {d x (a+b \arctan (c x))^2}{e \left (d+e x^2\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {i c \sqrt {-d} \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^{3/2}}-\frac {i c \sqrt {-d} \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^{3/2}}-\frac {\operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right ) b^2}{2 e^2}+\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}{4 e^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}{4 e^2}-\frac {c \sqrt {-d} (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^{3/2}}+\frac {c \sqrt {-d} (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^{3/2}}+\frac {i (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) b}{e^2}-\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}{2 e^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}{2 e^2}+\frac {(a+b \arctan (c x))^2}{4 e^2 \left (1-\frac {\sqrt {e} x}{\sqrt {-d}}\right )}+\frac {(a+b \arctan (c x))^2}{4 e^2 \left (\frac {\sqrt {e} x}{\sqrt {-d}}+1\right )}-\frac {c^2 d (a+b \arctan (c x))^2}{2 \left (c^2 d-e\right ) e^2}-\frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{1-i c x}\right )}{e^2}+\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^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 )}{2 e^2}\) |
-1/2*(c^2*d*(a + b*ArcTan[c*x])^2)/((c^2*d - e)*e^2) + (a + b*ArcTan[c*x]) ^2/(4*e^2*(1 - (Sqrt[e]*x)/Sqrt[-d])) + (a + b*ArcTan[c*x])^2/(4*e^2*(1 + (Sqrt[e]*x)/Sqrt[-d])) - ((a + b*ArcTan[c*x])^2*Log[2/(1 - I*c*x)])/e^2 - (b*c*Sqrt[-d]*(a + b*ArcTan[c*x])*Log[(2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqr t[-d] - I*Sqrt[e])*(1 - I*c*x))])/(2*(c^2*d - e)*e^(3/2)) + ((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^2) + (b*c*Sqrt[-d]*(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^(3/2 )) + ((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^2) + (I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])/e^2 + ((I/4)*b^2*c*Sqrt[-d]*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^ (3/2)) - ((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^2 - ((I/4)*b^2*c*Sqrt[- d]*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^(3/2)) - ((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^2 - (b^2*PolyLog[3, 1 - 2/(1 - I*c*x)])/(2*e^2) + (b^2*PolyLog[3, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(4 *e^2) + (b^2*PolyLog[3, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] +...
3.13.68.3.1 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])
\[\int \frac {x^{3} \left (a +b \arctan \left (c x \right )\right )^{2}}{\left (e \,x^{2}+d \right )^{2}}d x\]
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
integral((b^2*x^3*arctan(c*x)^2 + 2*a*b*x^3*arctan(c*x) + a^2*x^3)/(e^2*x^ 4 + 2*d*e*x^2 + d^2), x)
Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
1/2*a^2*(d/(e^3*x^2 + d*e^2) + log(e*x^2 + d)/e^2) + integrate((b^2*x^3*ar ctan(c*x)^2 + 2*a*b*x^3*arctan(c*x))/(e^2*x^4 + 2*d*e*x^2 + d^2), x)
\[ \int \frac {x^3 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\int \frac {x^3\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (e\,x^2+d\right )}^2} \,d x \]