Integrand size = 20, antiderivative size = 64 \[ \int \frac {\arctan (a x)}{x \left (c+a^2 c x^2\right )} \, dx=-\frac {i \arctan (a x)^2}{2 c}+\frac {\arctan (a x) \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {i \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{2 c} \]
Time = 0.02 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.61 \[ \int \frac {\arctan (a x)}{x \left (c+a^2 c x^2\right )} \, dx=\frac {i \arctan (a x)^2}{2 c}+\frac {\arctan (a x) \log \left (\frac {2 i}{i-a x}\right )}{c}+\frac {i \operatorname {PolyLog}(2,-i a x)}{2 c}-\frac {i \operatorname {PolyLog}(2,i a x)}{2 c}+\frac {i \operatorname {PolyLog}\left (2,-\frac {i+a x}{i-a x}\right )}{2 c} \]
((I/2)*ArcTan[a*x]^2)/c + (ArcTan[a*x]*Log[(2*I)/(I - a*x)])/c + ((I/2)*Po lyLog[2, (-I)*a*x])/c - ((I/2)*PolyLog[2, I*a*x])/c + ((I/2)*PolyLog[2, -( (I + a*x)/(I - a*x))])/c
Time = 0.32 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {5459, 5403, 2897}
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.
\(\displaystyle \int \frac {\arctan (a x)}{x \left (a^2 c x^2+c\right )} \, dx\) |
\(\Big \downarrow \) 5459 |
\(\displaystyle \frac {i \int \frac {\arctan (a x)}{x (a x+i)}dx}{c}-\frac {i \arctan (a x)^2}{2 c}\) |
\(\Big \downarrow \) 5403 |
\(\displaystyle \frac {i \left (i a \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{a^2 x^2+1}dx-i \arctan (a x) \log \left (2-\frac {2}{1-i a x}\right )\right )}{c}-\frac {i \arctan (a x)^2}{2 c}\) |
\(\Big \downarrow \) 2897 |
\(\displaystyle \frac {i \left (-i \arctan (a x) \log \left (2-\frac {2}{1-i a x}\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,\frac {2}{1-i a x}-1\right )\right )}{c}-\frac {i \arctan (a x)^2}{2 c}\) |
((-1/2*I)*ArcTan[a*x]^2)/c + (I*((-I)*ArcTan[a*x]*Log[2 - 2/(1 - I*a*x)] - PolyLog[2, -1 + 2/(1 - I*a*x)]/2))/c
3.2.78.3.1 Defintions of rubi rules used
Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, x][[2]], Expon[Pq, x]]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_ Symbol] :> Simp[(a + b*ArcTan[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Si mp[b*c*(p/d) Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))]/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2* d^2 + e^2, 0]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(-I)*((a + b*ArcTan[c*x])^(p + 1)/(b*d*(p + 1))), x] + Si mp[I/d Int[(a + b*ArcTan[c*x])^p/(x*(I + c*x)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (56 ) = 112\).
Time = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.09
method | result | size |
risch | \(-\frac {i \operatorname {dilog}\left (-i a x +1\right )}{2 c}-\frac {i \ln \left (-i a x +1\right )^{2}}{8 c}-\frac {i \ln \left (\frac {1}{2}+\frac {i a x}{2}\right ) \ln \left (-i a x +1\right )}{4 c}+\frac {i \operatorname {dilog}\left (\frac {1}{2}-\frac {i a x}{2}\right )}{4 c}+\frac {i \operatorname {dilog}\left (i a x +1\right )}{2 c}+\frac {i \ln \left (i a x +1\right )^{2}}{8 c}+\frac {i \ln \left (\frac {1}{2}-\frac {i a x}{2}\right ) \ln \left (i a x +1\right )}{4 c}-\frac {i \operatorname {dilog}\left (\frac {1}{2}+\frac {i a x}{2}\right )}{4 c}\) | \(134\) |
parts | \(-\frac {\arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{2 c}+\frac {\arctan \left (a x \right ) \ln \left (x \right )}{c}-\frac {a \left (-\frac {i \ln \left (x \right ) \left (\ln \left (i a x +1\right )-\ln \left (-i a x +1\right )\right )}{a}-\frac {i \left (\operatorname {dilog}\left (i a x +1\right )-\operatorname {dilog}\left (-i a x +1\right )\right )}{a}-\frac {\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (a^{2} \textit {\_Z}^{2}+1\right )}{\sum }\frac {2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (a^{2} x^{2}+1\right )-a^{2} \left (\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{a^{2} \underline {\hspace {1.25 ex}}\alpha }+2 \underline {\hspace {1.25 ex}}\alpha \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right )+2 \underline {\hspace {1.25 ex}}\alpha \operatorname {dilog}\left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right )\right )}{\underline {\hspace {1.25 ex}}\alpha }}{4 a^{2}}\right )}{2 c}\) | \(182\) |
derivativedivides | \(\frac {\arctan \left (a x \right ) \ln \left (a x \right )}{c}-\frac {\arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{2 c}-\frac {-i \ln \left (a x \right ) \ln \left (i a x +1\right )+i \ln \left (a x \right ) \ln \left (-i a x +1\right )-i \operatorname {dilog}\left (i a x +1\right )+i \operatorname {dilog}\left (-i a x +1\right )+\frac {i \left (\ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )-\operatorname {dilog}\left (-\frac {i \left (a x +i\right )}{2}\right )-\ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )-\frac {\ln \left (a x -i\right )^{2}}{2}\right )}{2}-\frac {i \left (\ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )-\operatorname {dilog}\left (\frac {i \left (a x -i\right )}{2}\right )-\ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )-\frac {\ln \left (a x +i\right )^{2}}{2}\right )}{2}}{2 c}\) | \(219\) |
default | \(\frac {\arctan \left (a x \right ) \ln \left (a x \right )}{c}-\frac {\arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{2 c}-\frac {-i \ln \left (a x \right ) \ln \left (i a x +1\right )+i \ln \left (a x \right ) \ln \left (-i a x +1\right )-i \operatorname {dilog}\left (i a x +1\right )+i \operatorname {dilog}\left (-i a x +1\right )+\frac {i \left (\ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )-\operatorname {dilog}\left (-\frac {i \left (a x +i\right )}{2}\right )-\ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )-\frac {\ln \left (a x -i\right )^{2}}{2}\right )}{2}-\frac {i \left (\ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )-\operatorname {dilog}\left (\frac {i \left (a x -i\right )}{2}\right )-\ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )-\frac {\ln \left (a x +i\right )^{2}}{2}\right )}{2}}{2 c}\) | \(219\) |
-1/2*I/c*dilog(1-I*a*x)-1/8*I/c*ln(1-I*a*x)^2-1/4*I/c*ln(1/2+1/2*I*a*x)*ln (1-I*a*x)+1/4*I/c*dilog(1/2-1/2*I*a*x)+1/2*I/c*dilog(1+I*a*x)+1/8*I/c*ln(1 +I*a*x)^2+1/4*I/c*ln(1/2-1/2*I*a*x)*ln(1+I*a*x)-1/4*I/c*dilog(1/2+1/2*I*a* x)
\[ \int \frac {\arctan (a x)}{x \left (c+a^2 c x^2\right )} \, dx=\int { \frac {\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )} x} \,d x } \]
\[ \int \frac {\arctan (a x)}{x \left (c+a^2 c x^2\right )} \, dx=\frac {\int \frac {\operatorname {atan}{\left (a x \right )}}{a^{2} x^{3} + x}\, dx}{c} \]
\[ \int \frac {\arctan (a x)}{x \left (c+a^2 c x^2\right )} \, dx=\int { \frac {\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )} x} \,d x } \]
\[ \int \frac {\arctan (a x)}{x \left (c+a^2 c x^2\right )} \, dx=\int { \frac {\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )} x} \,d x } \]
Timed out. \[ \int \frac {\arctan (a x)}{x \left (c+a^2 c x^2\right )} \, dx=\int \frac {\mathrm {atan}\left (a\,x\right )}{x\,\left (c\,a^2\,x^2+c\right )} \,d x \]