Integrand size = 13, antiderivative size = 122 \[ \int -\frac {\arctan (a-x)}{a+x} \, dx=\arctan (a-x) \log \left (\frac {2}{1-i (a-x)}\right )-\arctan (a-x) \log \left (-\frac {2 (a+x)}{(i-2 a) (1-i (a-x))}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (a-x)}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1+\frac {2 (a+x)}{(i-2 a) (1-i (a-x))}\right ) \]
[Out]
Time = 0.07 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {5155, 4966, 2449, 2352, 2497} \[ \int -\frac {\arctan (a-x)}{a+x} \, dx=\arctan (a-x) \log \left (\frac {2}{1-i (a-x)}\right )-\arctan (a-x) \log \left (-\frac {2 (a+x)}{(-2 a+i) (1-i (a-x))}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (a-x)}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,\frac {2 (a+x)}{(i-2 a) (1-i (a-x))}+1\right ) \]
[In]
[Out]
Rule 2352
Rule 2449
Rule 2497
Rule 4966
Rule 5155
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {\arctan (x)}{2 a-x} \, dx,x,a-x\right ) \\ & = \arctan (a-x) \log \left (\frac {2}{1-i (a-x)}\right )-\arctan (a-x) \log \left (-\frac {2 (a+x)}{(i-2 a) (1-i (a-x))}\right )-\text {Subst}\left (\int \frac {\log \left (\frac {2}{1-i x}\right )}{1+x^2} \, dx,x,a-x\right )+\text {Subst}\left (\int \frac {\log \left (\frac {2 (2 a-x)}{(-i+2 a) (1-i x)}\right )}{1+x^2} \, dx,x,a-x\right ) \\ & = \arctan (a-x) \log \left (\frac {2}{1-i (a-x)}\right )-\arctan (a-x) \log \left (-\frac {2 (a+x)}{(i-2 a) (1-i (a-x))}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1+\frac {2 (a+x)}{(i-2 a) (1-i (a-x))}\right )-i \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-i (a-x)}\right ) \\ & = \arctan (a-x) \log \left (\frac {2}{1-i (a-x)}\right )-\arctan (a-x) \log \left (-\frac {2 (a+x)}{(i-2 a) (1-i (a-x))}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i (a-x)}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1+\frac {2 (a+x)}{(i-2 a) (1-i (a-x))}\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.86 \[ \int -\frac {\arctan (a-x)}{a+x} \, dx=-\frac {1}{2} i \left (-\log (1+i (a-x)) \log \left (\frac {a+x}{-i+2 a}\right )+\log (1-i a+i x) \log \left (\frac {a+x}{i+2 a}\right )+\operatorname {PolyLog}\left (2,\frac {i+a-x}{i+2 a}\right )-\operatorname {PolyLog}\left (2,\frac {i-a+x}{i-2 a}\right )\right ) \]
[In]
[Out]
Time = 0.39 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.84
method | result | size |
derivativedivides | \(-\ln \left (a +x \right ) \arctan \left (a -x \right )+\frac {i \ln \left (a +x \right ) \ln \left (\frac {i+a -x}{2 a +i}\right )}{2}-\frac {i \ln \left (a +x \right ) \ln \left (\frac {i-a +x}{i-2 a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+a -x}{2 a +i}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-a +x}{i-2 a}\right )}{2}\) | \(102\) |
default | \(-\ln \left (a +x \right ) \arctan \left (a -x \right )+\frac {i \ln \left (a +x \right ) \ln \left (\frac {i+a -x}{2 a +i}\right )}{2}-\frac {i \ln \left (a +x \right ) \ln \left (\frac {i-a +x}{i-2 a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+a -x}{2 a +i}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-a +x}{i-2 a}\right )}{2}\) | \(102\) |
parts | \(-\ln \left (a +x \right ) \arctan \left (a -x \right )+\frac {i \ln \left (a +x \right ) \ln \left (\frac {i+a -x}{2 a +i}\right )}{2}-\frac {i \ln \left (a +x \right ) \ln \left (\frac {i-a +x}{i-2 a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i+a -x}{2 a +i}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i-a +x}{i-2 a}\right )}{2}\) | \(102\) |
risch | \(-\frac {i \operatorname {dilog}\left (\frac {i a +i x}{2 i a -1}\right )}{2}-\frac {i \ln \left (-i a +i x +1\right ) \ln \left (\frac {i a +i x}{2 i a -1}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {-i a -i x}{-2 i a -1}\right )}{2}+\frac {i \ln \left (i a -i x +1\right ) \ln \left (\frac {-i a -i x}{-2 i a -1}\right )}{2}\) | \(112\) |
[In]
[Out]
\[ \int -\frac {\arctan (a-x)}{a+x} \, dx=\int { -\frac {\arctan \left (a - x\right )}{a + x} \,d x } \]
[In]
[Out]
\[ \int -\frac {\arctan (a-x)}{a+x} \, dx=- \int \frac {\operatorname {atan}{\left (a - x \right )}}{a + x}\, dx \]
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.97 \[ \int -\frac {\arctan (a-x)}{a+x} \, dx=-\frac {1}{2} \, \arctan \left (\frac {a + x}{4 \, a^{2} + 1}, \frac {2 \, {\left (a^{2} + a x\right )}}{4 \, a^{2} + 1}\right ) \log \left (a^{2} - 2 \, a x + x^{2} + 1\right ) + \frac {1}{2} \, \arctan \left (-a + x\right ) \log \left (\frac {a^{2} + 2 \, a x + x^{2}}{4 \, a^{2} + 1}\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-\frac {-i \, a + i \, x + 1}{2 i \, a - 1}\right ) + \frac {1}{2} i \, {\rm Li}_2\left (-\frac {-i \, a + i \, x - 1}{2 i \, a + 1}\right ) \]
[In]
[Out]
\[ \int -\frac {\arctan (a-x)}{a+x} \, dx=\int { -\frac {\arctan \left (a - x\right )}{a + x} \,d x } \]
[In]
[Out]
Timed out. \[ \int -\frac {\arctan (a-x)}{a+x} \, dx=-\int \frac {\mathrm {atan}\left (a-x\right )}{a+x} \,d x \]
[In]
[Out]