Integrand size = 20, antiderivative size = 1039 \[ \int \frac {(a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\frac {i c (a+b \arctan (c x))^2}{2 d \left (c^2 d-e\right )}-\frac {(a+b \arctan (c x))^2}{4 d \sqrt {e} \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {(a+b \arctan (c x))^2}{4 d \sqrt {e} \left (\sqrt {-d}+\sqrt {e} x\right )}-\frac {b c (a+b \arctan (c x)) \log \left (\frac {2}{1-i c x}\right )}{d \left (c^2 d-e\right )}+\frac {b c (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{d \left (c^2 d-e\right )}+\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 d \left (c^2 d-e\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 (-d)^{3/2} \sqrt {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 d \left (c^2 d-e\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 (-d)^{3/2} \sqrt {e}}+\frac {i b^2 c \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{2 d \left (c^2 d-e\right )}+\frac {i b^2 c \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 d \left (c^2 d-e\right )}-\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 d \left (c^2 d-e\right )}+\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 (-d)^{3/2} \sqrt {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 d \left (c^2 d-e\right )}-\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 (-d)^{3/2} \sqrt {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 )}{8 (-d)^{3/2} \sqrt {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 )}{8 (-d)^{3/2} \sqrt {e}} \]
1/2*I*b^2*c*polylog(2,1-2/(1+I*c*x))/d/(c^2*d-e)-b*c*(a+b*arctan(c*x))*ln( 2/(1-I*c*x))/d/(c^2*d-e)+b*c*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/d/(c^2*d-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)))/d/(c^2*d-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)))/d/(c^2*d-e)-1/4*I*b^2*c*p olylog(2,1-2*c*((-d)^(1/2)+x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)+I*e^(1/2)))/ d/(c^2*d-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)))/(-d)^(3/2)/e^(1/2)-1/4*I*b^2*c*poly log(2,1-2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2)))/d/( c^2*d-e)+1/2*I*c*(a+b*arctan(c*x))^2/d/(c^2*d-e)-1/4*(a+b*arctan(c*x))^2*l n(2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2)))/(-d)^(3/2 )/e^(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)))/(-d)^(3/2)/e^(1/2)+1/2*I*b^2*c*polylog(2,1-2/(1- I*c*x))/d/(c^2*d-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)))/(-d)^(3/2)/e^(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))) /(-d)^(3/2)/e^(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)))/(-d)^(3/2)/e^(1/2)-1/4*(a+b*arctan(c*x))^2/d/ e^(1/2)/((-d)^(1/2)-x*e^(1/2))+1/4*(a+b*arctan(c*x))^2/d/e^(1/2)/((-d)^(1/ 2)+x*e^(1/2))
\[ \int \frac {(a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\int \frac {(a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx \]
Time = 1.53 (sec) , antiderivative size = 1039, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5449, 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.
| \(\displaystyle \int \frac {(a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 5449 |
| \(\displaystyle \int \left (-\frac {e (a+b \arctan (c x))^2}{2 d \left (-d e-e^2 x^2\right )}-\frac {e (a+b \arctan (c x))^2}{4 d \left (\sqrt {-d} \sqrt {e}-e x\right )^2}-\frac {e (a+b \arctan (c x))^2}{4 d \left (\sqrt {-d} \sqrt {e}+e x\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {i c \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) b^2}{2 d \left (c^2 d-e\right )}+\frac {i c \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) b^2}{2 d \left (c^2 d-e\right )}-\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 d \left (c^2 d-e\right )}-\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 d \left (c^2 d-e\right )}-\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 (-d)^{3/2} \sqrt {e}}+\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 (-d)^{3/2} \sqrt {e}}-\frac {c (a+b \arctan (c x)) \log \left (\frac {2}{1-i c x}\right ) b}{d \left (c^2 d-e\right )}+\frac {c (a+b \arctan (c x)) \log \left (\frac {2}{i c x+1}\right ) b}{d \left (c^2 d-e\right )}+\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 d \left (c^2 d-e\right )}+\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 d \left (c^2 d-e\right )}+\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 (-d)^{3/2} \sqrt {e}}-\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 (-d)^{3/2} \sqrt {e}}+\frac {i c (a+b \arctan (c x))^2}{2 d \left (c^2 d-e\right )}-\frac {(a+b \arctan (c x))^2}{4 d \sqrt {e} \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {(a+b \arctan (c x))^2}{4 d \sqrt {e} \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 (-d)^{3/2} \sqrt {e}}+\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 (-d)^{3/2} \sqrt {e}}\) |
((I/2)*c*(a + b*ArcTan[c*x])^2)/(d*(c^2*d - e)) - (a + b*ArcTan[c*x])^2/(4 *d*Sqrt[e]*(Sqrt[-d] - Sqrt[e]*x)) + (a + b*ArcTan[c*x])^2/(4*d*Sqrt[e]*(S qrt[-d] + Sqrt[e]*x)) - (b*c*(a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/(d*(c ^2*d - e)) + (b*c*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(d*(c^2*d - 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*d*(c^2*d - 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*( -d)^(3/2)*Sqrt[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*d*(c^2*d - 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*(-d)^(3/2)*Sqrt[e]) + ((I/2)*b^2*c*PolyLog[2, 1 - 2/(1 - I*c*x)])/(d*(c^2*d - e)) + ((I/2)*b^2*c*PolyLog[2, 1 - 2/(1 + I*c*x)])/(d *(c^2*d - e)) - ((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))])/(d*(c^2*d - 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))])/((-d)^(3/2)*Sqrt[e]) - ((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))])/(d* (c^2*d - 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))])/((-d)^(3/2)*Sqrt[e] ) - (b^2*PolyLog[3, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*S...
3.13.71.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_.), x _Symbol] :> Int[ExpandIntegrand[(a + b*ArcTan[c*x])^p, (d + e*x^2)^q, x], x ] /; FreeQ[{a, b, c, d, e}, x] && IntegerQ[q] && IGtQ[p, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 44.79 (sec) , antiderivative size = 6565, normalized size of antiderivative = 6.32
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(6565\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(6570\) |
| default | \(\text {Expression too large to display}\) | \(6570\) |
\[ \int \frac {(a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(a+b \arctan (c x))^2}{\left (d+e x^2\right )^2} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (e\,x^2+d\right )}^2} \,d x \]