Optimal. Leaf size=120 \[ -\text {ArcTan}(a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )+\text {ArcTan}(a+b x) \log \left (\frac {2 b x}{(i-a) (1-i (a+b x))}\right )+\frac {1}{2} i \text {PolyLog}\left (2,1-\frac {2}{1-i (a+b x)}\right )-\frac {1}{2} i \text {PolyLog}\left (2,1-\frac {2 b x}{(i-a) (1-i (a+b x))}\right ) \]
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Rubi [A]
time = 0.08, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5155, 4966,
2449, 2352, 2497} \begin {gather*} -\text {ArcTan}(a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )+\text {ArcTan}(a+b x) \log \left (\frac {2 b x}{(-a+i) (1-i (a+b x))}\right )+\frac {1}{2} i \text {Li}_2\left (1-\frac {2}{1-i (a+b x)}\right )-\frac {1}{2} i \text {Li}_2\left (1-\frac {2 b x}{(i-a) (1-i (a+b x))}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 2352
Rule 2449
Rule 2497
Rule 4966
Rule 5155
Rubi steps
\begin {align*} \int \frac {\tan ^{-1}(a+b x)}{x} \, dx &=\frac {\text {Subst}\left (\int \frac {\tan ^{-1}(x)}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b}\\ &=-\tan ^{-1}(a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )+\tan ^{-1}(a+b x) \log \left (\frac {2 b x}{(i-a) (1-i (a+b x))}\right )+\text {Subst}\left (\int \frac {\log \left (\frac {2}{1-i x}\right )}{1+x^2} \, dx,x,a+b x\right )-\text {Subst}\left (\int \frac {\log \left (\frac {2 \left (-\frac {a}{b}+\frac {x}{b}\right )}{\left (\frac {i}{b}-\frac {a}{b}\right ) (1-i x)}\right )}{1+x^2} \, dx,x,a+b x\right )\\ &=-\tan ^{-1}(a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )+\tan ^{-1}(a+b x) \log \left (\frac {2 b x}{(i-a) (1-i (a+b x))}\right )-\frac {1}{2} i \text {Li}_2\left (1-\frac {2 b x}{(i-a) (1-i (a+b x))}\right )+i \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-i (a+b x)}\right )\\ &=-\tan ^{-1}(a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )+\tan ^{-1}(a+b x) \log \left (\frac {2 b x}{(i-a) (1-i (a+b x))}\right )+\frac {1}{2} i \text {Li}_2\left (1-\frac {2}{1-i (a+b x)}\right )-\frac {1}{2} i \text {Li}_2\left (1-\frac {2 b x}{(i-a) (1-i (a+b x))}\right )\\ \end {align*}
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Mathematica [A]
time = 0.01, size = 171, normalized size = 1.42 \begin {gather*} -\frac {1}{2} i \log (1+i (a+b x)) \log \left (\frac {i \left (-\frac {a}{b}+\frac {a+b x}{b}\right )}{-\frac {1}{b}-\frac {i a}{b}}\right )+\frac {1}{2} i \log (1-i (a+b x)) \log \left (-\frac {i \left (-\frac {a}{b}+\frac {a+b x}{b}\right )}{-\frac {1}{b}+\frac {i a}{b}}\right )+\frac {1}{2} i \text {PolyLog}\left (2,\frac {i (1-i (a+b x))}{i+a}\right )-\frac {1}{2} i \text {PolyLog}\left (2,-\frac {i (1+i (a+b x))}{-i+a}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.05, size = 106, normalized size = 0.88
method | result | size |
risch | \(\frac {i \dilog \left (-\frac {i x b}{i a -1}\right )}{2}+\frac {i \ln \left (-i b x -i a +1\right ) \ln \left (-\frac {i x b}{i a -1}\right )}{2}-\frac {i \dilog \left (\frac {i x b}{-i a -1}\right )}{2}-\frac {i \ln \left (i b x +i a +1\right ) \ln \left (\frac {i x b}{-i a -1}\right )}{2}\) | \(94\) |
derivativedivides | \(\ln \left (-b x \right ) \arctan \left (b x +a \right )-\frac {i \ln \left (-b x \right ) \ln \left (\frac {b x +a +i}{i+a}\right )}{2}+\frac {i \ln \left (-b x \right ) \ln \left (\frac {-b x -a +i}{i-a}\right )}{2}-\frac {i \dilog \left (\frac {b x +a +i}{i+a}\right )}{2}+\frac {i \dilog \left (\frac {-b x -a +i}{i-a}\right )}{2}\) | \(106\) |
default | \(\ln \left (-b x \right ) \arctan \left (b x +a \right )-\frac {i \ln \left (-b x \right ) \ln \left (\frac {b x +a +i}{i+a}\right )}{2}+\frac {i \ln \left (-b x \right ) \ln \left (\frac {-b x -a +i}{i-a}\right )}{2}-\frac {i \dilog \left (\frac {b x +a +i}{i+a}\right )}{2}+\frac {i \dilog \left (\frac {-b x -a +i}{i-a}\right )}{2}\) | \(106\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.51, size = 134, normalized size = 1.12 \begin {gather*} -\frac {1}{2} \, \arctan \left (\frac {b x}{a^{2} + 1}, -\frac {a b x}{a^{2} + 1}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) + \frac {1}{2} \, \arctan \left (b x + a\right ) \log \left (\frac {b^{2} x^{2}}{a^{2} + 1}\right ) + \arctan \left (b x + a\right ) \log \left (x\right ) - \arctan \left (\frac {b^{2} x + a b}{b}\right ) \log \left (x\right ) - \frac {1}{2} i \, {\rm Li}_2\left (\frac {i \, b x + i \, a + 1}{i \, a + 1}\right ) + \frac {1}{2} i \, {\rm Li}_2\left (\frac {i \, b x + i \, a - 1}{i \, a - 1}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {atan}\left (a+b\,x\right )}{x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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