Optimal. Leaf size=45 \[ -i \text{PolyLog}\left (2,-e^{2 i x}\right )+x \log \left (a \sec ^2(x)\right )-i x^2+2 x \log \left (1+e^{2 i x}\right ) \]
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Rubi [A] time = 0.0543116, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.857, Rules used = {2548, 12, 3719, 2190, 2279, 2391} \[ -i \text{PolyLog}\left (2,-e^{2 i x}\right )+x \log \left (a \sec ^2(x)\right )-i x^2+2 x \log \left (1+e^{2 i x}\right ) \]
Antiderivative was successfully verified.
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Rule 2548
Rule 12
Rule 3719
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \log \left (a \sec ^2(x)\right ) \, dx &=x \log \left (a \sec ^2(x)\right )-\int 2 x \tan (x) \, dx\\ &=x \log \left (a \sec ^2(x)\right )-2 \int x \tan (x) \, dx\\ &=-i x^2+x \log \left (a \sec ^2(x)\right )+4 i \int \frac{e^{2 i x} x}{1+e^{2 i x}} \, dx\\ &=-i x^2+2 x \log \left (1+e^{2 i x}\right )+x \log \left (a \sec ^2(x)\right )-2 \int \log \left (1+e^{2 i x}\right ) \, dx\\ &=-i x^2+2 x \log \left (1+e^{2 i x}\right )+x \log \left (a \sec ^2(x)\right )+i \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i x}\right )\\ &=-i x^2+2 x \log \left (1+e^{2 i x}\right )+x \log \left (a \sec ^2(x)\right )-i \text{Li}_2\left (-e^{2 i x}\right )\\ \end{align*}
Mathematica [A] time = 0.02009, size = 43, normalized size = 0.96 \[ x \left (\log \left (a \sec ^2(x)\right )-i x+2 \log \left (1+e^{2 i x}\right )\right )-i \text{PolyLog}\left (2,-e^{2 i x}\right ) \]
Antiderivative was successfully verified.
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Maple [B] time = 0.105, size = 118, normalized size = 2.6 \begin{align*} i \left ( \ln \left ({{\rm e}^{ix}} \right ) \right ) ^{2}-i\ln \left ({{\rm e}^{ix}} \right ) \ln \left ({\frac{a{{\rm e}^{2\,ix}}}{ \left ( 1+{{\rm e}^{2\,ix}} \right ) ^{2}}} \right ) -2\,i\ln \left ({{\rm e}^{ix}} \right ) \ln \left ( 1+i{{\rm e}^{ix}} \right ) -2\,i\ln \left ({{\rm e}^{ix}} \right ) \ln \left ( 1-i{{\rm e}^{ix}} \right ) -2\,i\ln \left ({{\rm e}^{ix}} \right ) \ln \left ( 2 \right ) -2\,i{\it dilog} \left ( 1+i{{\rm e}^{ix}} \right ) -2\,i{\it dilog} \left ( 1-i{{\rm e}^{ix}} \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.93225, size = 82, normalized size = 1.82 \begin{align*} -i \, x^{2} + 2 i \, x \arctan \left (\sin \left (2 \, x\right ), \cos \left (2 \, x\right ) + 1\right ) + x \log \left (a \sec \left (x\right )^{2}\right ) + x \log \left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right ) - i \,{\rm Li}_2\left (-e^{\left (2 i \, x\right )}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.30184, size = 355, normalized size = 7.89 \begin{align*} x \log \left (\frac{a}{\cos \left (x\right )^{2}}\right ) + x \log \left (i \, \cos \left (x\right ) + \sin \left (x\right ) + 1\right ) + x \log \left (i \, \cos \left (x\right ) - \sin \left (x\right ) + 1\right ) + x \log \left (-i \, \cos \left (x\right ) + \sin \left (x\right ) + 1\right ) + x \log \left (-i \, \cos \left (x\right ) - \sin \left (x\right ) + 1\right ) + i \,{\rm Li}_2\left (i \, \cos \left (x\right ) + \sin \left (x\right )\right ) - i \,{\rm Li}_2\left (i \, \cos \left (x\right ) - \sin \left (x\right )\right ) - i \,{\rm Li}_2\left (-i \, \cos \left (x\right ) + \sin \left (x\right )\right ) + i \,{\rm Li}_2\left (-i \, \cos \left (x\right ) - \sin \left (x\right )\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \log{\left (a \sec ^{2}{\left (x \right )} \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \log \left (a \sec \left (x\right )^{2}\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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