Optimal. Leaf size=132 \[ -\frac {2 i \text {ArcTan}(a+b x) \text {ArcTan}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}+\frac {i \text {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}-\frac {i \text {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b} \]
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Rubi [A]
time = 0.07, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps
used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {5163, 5006}
\begin {gather*} -\frac {2 i \text {ArcTan}(a+b x) \text {ArcTan}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}+\frac {i \text {Li}_2\left (-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{b}-\frac {i \text {Li}_2\left (\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{b} \end {gather*}
Antiderivative was successfully verified.
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Rule 5006
Rule 5163
Rubi steps
\begin {align*} \int \frac {\tan ^{-1}(a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\text {Subst}\left (\int \frac {\tan ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b}\\ &=-\frac {2 i \tan ^{-1}(a+b x) \tan ^{-1}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}+\frac {i \text {Li}_2\left (-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}-\frac {i \text {Li}_2\left (\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}\\ \end {align*}
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Mathematica [A]
time = 0.08, size = 97, normalized size = 0.73 \begin {gather*} \frac {\text {ArcTan}(a+b x) \left (\log \left (1-i e^{i \text {ArcTan}(a+b x)}\right )-\log \left (1+i e^{i \text {ArcTan}(a+b x)}\right )\right )+i \text {PolyLog}\left (2,-i e^{i \text {ArcTan}(a+b x)}\right )-i \text {PolyLog}\left (2,i e^{i \text {ArcTan}(a+b x)}\right )}{b} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.16, size = 135, normalized size = 1.02
method | result | size |
default | \(\frac {-\arctan \left (b x +a \right ) \ln \left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )+\arctan \left (b x +a \right ) \ln \left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )+i \dilog \left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-i \dilog \left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{b}\) | \(135\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {atan}{\left (a + b x \right )}}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx \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 )}{\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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