Optimal. Leaf size=121 \[ i \tan ^{-1}(x) \operatorname {PolyLog}\left (2,-i e^{i \tan ^{-1}(x)}\right )-i \tan ^{-1}(x) \operatorname {PolyLog}\left (2,i e^{i \tan ^{-1}(x)}\right )-\operatorname {PolyLog}\left (3,-i e^{i \tan ^{-1}(x)}\right )+\operatorname {PolyLog}\left (3,i e^{i \tan ^{-1}(x)}\right )+\frac {1}{2} x \sqrt {x^2+1} \tan ^{-1}(x)^2-\sqrt {x^2+1} \tan ^{-1}(x)-i \tan ^{-1}\left (e^{i \tan ^{-1}(x)}\right ) \tan ^{-1}(x)^2+\sinh ^{-1}(x) \]
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Rubi [A] time = 0.11, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4880, 4888, 4181, 2531, 2282, 6589, 215} \[ i \tan ^{-1}(x) \text {PolyLog}\left (2,-i e^{i \tan ^{-1}(x)}\right )-i \tan ^{-1}(x) \text {PolyLog}\left (2,i e^{i \tan ^{-1}(x)}\right )-\text {PolyLog}\left (3,-i e^{i \tan ^{-1}(x)}\right )+\text {PolyLog}\left (3,i e^{i \tan ^{-1}(x)}\right )+\frac {1}{2} x \sqrt {x^2+1} \tan ^{-1}(x)^2-\sqrt {x^2+1} \tan ^{-1}(x)-i \tan ^{-1}\left (e^{i \tan ^{-1}(x)}\right ) \tan ^{-1}(x)^2+\sinh ^{-1}(x) \]
Antiderivative was successfully verified.
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Rule 215
Rule 2282
Rule 2531
Rule 4181
Rule 4880
Rule 4888
Rule 6589
Rubi steps
\begin {align*} \int \sqrt {1+x^2} \tan ^{-1}(x)^2 \, dx &=-\sqrt {1+x^2} \tan ^{-1}(x)+\frac {1}{2} x \sqrt {1+x^2} \tan ^{-1}(x)^2+\frac {1}{2} \int \frac {\tan ^{-1}(x)^2}{\sqrt {1+x^2}} \, dx+\int \frac {1}{\sqrt {1+x^2}} \, dx\\ &=\sinh ^{-1}(x)-\sqrt {1+x^2} \tan ^{-1}(x)+\frac {1}{2} x \sqrt {1+x^2} \tan ^{-1}(x)^2+\frac {1}{2} \operatorname {Subst}\left (\int x^2 \sec (x) \, dx,x,\tan ^{-1}(x)\right )\\ &=\sinh ^{-1}(x)-\sqrt {1+x^2} \tan ^{-1}(x)+\frac {1}{2} x \sqrt {1+x^2} \tan ^{-1}(x)^2-i \tan ^{-1}\left (e^{i \tan ^{-1}(x)}\right ) \tan ^{-1}(x)^2-\operatorname {Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\tan ^{-1}(x)\right )+\operatorname {Subst}\left (\int x \log \left (1+i e^{i x}\right ) \, dx,x,\tan ^{-1}(x)\right )\\ &=\sinh ^{-1}(x)-\sqrt {1+x^2} \tan ^{-1}(x)+\frac {1}{2} x \sqrt {1+x^2} \tan ^{-1}(x)^2-i \tan ^{-1}\left (e^{i \tan ^{-1}(x)}\right ) \tan ^{-1}(x)^2+i \tan ^{-1}(x) \text {Li}_2\left (-i e^{i \tan ^{-1}(x)}\right )-i \tan ^{-1}(x) \text {Li}_2\left (i e^{i \tan ^{-1}(x)}\right )-i \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\tan ^{-1}(x)\right )+i \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\tan ^{-1}(x)\right )\\ &=\sinh ^{-1}(x)-\sqrt {1+x^2} \tan ^{-1}(x)+\frac {1}{2} x \sqrt {1+x^2} \tan ^{-1}(x)^2-i \tan ^{-1}\left (e^{i \tan ^{-1}(x)}\right ) \tan ^{-1}(x)^2+i \tan ^{-1}(x) \text {Li}_2\left (-i e^{i \tan ^{-1}(x)}\right )-i \tan ^{-1}(x) \text {Li}_2\left (i e^{i \tan ^{-1}(x)}\right )-\operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \tan ^{-1}(x)}\right )+\operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \tan ^{-1}(x)}\right )\\ &=\sinh ^{-1}(x)-\sqrt {1+x^2} \tan ^{-1}(x)+\frac {1}{2} x \sqrt {1+x^2} \tan ^{-1}(x)^2-i \tan ^{-1}\left (e^{i \tan ^{-1}(x)}\right ) \tan ^{-1}(x)^2+i \tan ^{-1}(x) \text {Li}_2\left (-i e^{i \tan ^{-1}(x)}\right )-i \tan ^{-1}(x) \text {Li}_2\left (i e^{i \tan ^{-1}(x)}\right )-\text {Li}_3\left (-i e^{i \tan ^{-1}(x)}\right )+\text {Li}_3\left (i e^{i \tan ^{-1}(x)}\right )\\ \end {align*}
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Mathematica [A] time = 0.22, size = 131, normalized size = 1.08 \[ i \tan ^{-1}(x) \operatorname {PolyLog}\left (2,-i e^{i \tan ^{-1}(x)}\right )-i \tan ^{-1}(x) \operatorname {PolyLog}\left (2,i e^{i \tan ^{-1}(x)}\right )-\operatorname {PolyLog}\left (3,-i e^{i \tan ^{-1}(x)}\right )+\operatorname {PolyLog}\left (3,i e^{i \tan ^{-1}(x)}\right )+\frac {1}{2} x \sqrt {x^2+1} \tan ^{-1}(x)^2-\sqrt {x^2+1} \tan ^{-1}(x)+\tanh ^{-1}\left (\frac {x}{\sqrt {x^2+1}}\right )-i \tan ^{-1}\left (e^{i \tan ^{-1}(x)}\right ) \tan ^{-1}(x)^2 \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {x^{2} + 1} \arctan \relax (x)^{2}, x\right ) \]
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 \[ \int \sqrt {x^{2} + 1} \arctan \relax (x)^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.38, size = 171, normalized size = 1.41 \[ \frac {\arctan \relax (x )^{2} \ln \left (1-\frac {i \left (i x +1\right )}{\sqrt {x^{2}+1}}\right )}{2}-\frac {\arctan \relax (x )^{2} \ln \left (1+\frac {i \left (i x +1\right )}{\sqrt {x^{2}+1}}\right )}{2}+i \polylog \left (2, -\frac {i \left (i x +1\right )}{\sqrt {x^{2}+1}}\right ) \arctan \relax (x )-i \polylog \left (2, \frac {i \left (i x +1\right )}{\sqrt {x^{2}+1}}\right ) \arctan \relax (x )-\polylog \left (3, -\frac {i \left (i x +1\right )}{\sqrt {x^{2}+1}}\right )+\polylog \left (3, \frac {i \left (i x +1\right )}{\sqrt {x^{2}+1}}\right )+\frac {\sqrt {x^{2}+1}\, \left (x \arctan \relax (x )-2\right ) \arctan \relax (x )}{2}-2 i \arctan \left (\frac {i x +1}{\sqrt {x^{2}+1}}\right ) \]
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 \[ \int \sqrt {x^{2} + 1} \arctan \relax (x)^{2}\,{d x} \]
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 \[ \int {\mathrm {atan}\relax (x)}^2\,\sqrt {x^2+1} \,d x \]
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 \[ \int \sqrt {x^{2} + 1} \operatorname {atan}^{2}{\relax (x )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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