Optimal. Leaf size=187 \[ -\frac {\sqrt {1+(a+b x)^2}}{2 b}+\frac {(a+b x) \sqrt {1+(a+b x)^2} \text {ArcTan}(a+b x)}{2 b}+\frac {i \text {ArcTan}(a+b x) \text {ArcTan}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{b}-\frac {i \text {PolyLog}\left (2,-\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b}+\frac {i \text {PolyLog}\left (2,\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b} \]
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Rubi [A]
time = 0.16, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {5165, 5072,
267, 5006} \begin {gather*} \frac {i \text {ArcTan}\left (\frac {\sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right ) \text {ArcTan}(a+b x)}{b}+\frac {(a+b x) \sqrt {(a+b x)^2+1} \text {ArcTan}(a+b x)}{2 b}-\frac {i \text {Li}_2\left (-\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b}+\frac {i \text {Li}_2\left (\frac {i \sqrt {i (a+b x)+1}}{\sqrt {1-i (a+b x)}}\right )}{2 b}-\frac {\sqrt {(a+b x)^2+1}}{2 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 267
Rule 5006
Rule 5072
Rule 5165
Rubi steps
\begin {align*} \int \frac {(a+b x)^2 \tan ^{-1}(a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\text {Subst}\left (\int \frac {x^2 \tan ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) \sqrt {1+(a+b x)^2} \tan ^{-1}(a+b x)}{2 b}-\frac {\text {Subst}\left (\int \frac {x}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b}-\frac {\text {Subst}\left (\int \frac {\tan ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,a+b x\right )}{2 b}\\ &=-\frac {\sqrt {1+(a+b x)^2}}{2 b}+\frac {(a+b x) \sqrt {1+(a+b x)^2} \tan ^{-1}(a+b x)}{2 b}+\frac {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 )}{2 b}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1+i (a+b x)}}{\sqrt {1-i (a+b x)}}\right )}{2 b}\\ \end {align*}
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Mathematica [A]
time = 0.47, size = 145, normalized size = 0.78 \begin {gather*} \frac {-\sqrt {1+(a+b x)^2}+(a+b x) \sqrt {1+(a+b x)^2} \text {ArcTan}(a+b x)-\text {ArcTan}(a+b x) \log \left (1-i e^{i \text {ArcTan}(a+b x)}\right )+\text {ArcTan}(a+b x) \log \left (1+i e^{i \text {ArcTan}(a+b x)}\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 )}{2 b} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.46, size = 180, normalized size = 0.96
method | result | size |
default | \(\frac {\left (\arctan \left (b x +a \right ) b x +\arctan \left (b x +a \right ) a -1\right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b}+\frac {\arctan \left (b x +a \right ) \ln \left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-\arctan \left (b x +a \right ) \ln \left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )-i \dilog \left (1+\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )+i \dilog \left (1-\frac {i \left (1+i \left (b x +a \right )\right )}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{2 b}\) | \(180\) |
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 {\left (a + b x\right )^{2} \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 )\,{\left (a+b\,x\right )}^2}{\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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