Optimal. Leaf size=134 \[ \frac{2 i a^2 d \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac{2 i a^2 d \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac{4 i a^2 (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{a^2 (c+d x) \tan (e+f x)}{f}+\frac{a^2 (c+d x)^2}{2 d}+\frac{a^2 d \log (\cos (e+f x))}{f^2} \]
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Rubi [A] time = 0.114632, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4190, 4181, 2279, 2391, 4184, 3475} \[ \frac{2 i a^2 d \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac{2 i a^2 d \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac{4 i a^2 (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{a^2 (c+d x) \tan (e+f x)}{f}+\frac{a^2 (c+d x)^2}{2 d}+\frac{a^2 d \log (\cos (e+f x))}{f^2} \]
Antiderivative was successfully verified.
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Rule 4190
Rule 4181
Rule 2279
Rule 2391
Rule 4184
Rule 3475
Rubi steps
\begin{align*} \int (c+d x) (a+a \sec (e+f x))^2 \, dx &=\int \left (a^2 (c+d x)+2 a^2 (c+d x) \sec (e+f x)+a^2 (c+d x) \sec ^2(e+f x)\right ) \, dx\\ &=\frac{a^2 (c+d x)^2}{2 d}+a^2 \int (c+d x) \sec ^2(e+f x) \, dx+\left (2 a^2\right ) \int (c+d x) \sec (e+f x) \, dx\\ &=\frac{a^2 (c+d x)^2}{2 d}-\frac{4 i a^2 (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{a^2 (c+d x) \tan (e+f x)}{f}-\frac{\left (a^2 d\right ) \int \tan (e+f x) \, dx}{f}-\frac{\left (2 a^2 d\right ) \int \log \left (1-i e^{i (e+f x)}\right ) \, dx}{f}+\frac{\left (2 a^2 d\right ) \int \log \left (1+i e^{i (e+f x)}\right ) \, dx}{f}\\ &=\frac{a^2 (c+d x)^2}{2 d}-\frac{4 i a^2 (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{a^2 d \log (\cos (e+f x))}{f^2}+\frac{a^2 (c+d x) \tan (e+f x)}{f}+\frac{\left (2 i a^2 d\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^2}-\frac{\left (2 i a^2 d\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^2}\\ &=\frac{a^2 (c+d x)^2}{2 d}-\frac{4 i a^2 (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac{a^2 d \log (\cos (e+f x))}{f^2}+\frac{2 i a^2 d \text{Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac{2 i a^2 d \text{Li}_2\left (i e^{i (e+f x)}\right )}{f^2}+\frac{a^2 (c+d x) \tan (e+f x)}{f}\\ \end{align*}
Mathematica [B] time = 5.42315, size = 330, normalized size = 2.46 \[ \frac{a^2 (\cos (e+f x)+1)^2 \sec ^4\left (\frac{1}{2} (e+f x)\right ) \left (-\frac{4 d \csc (e) \left (i \text{PolyLog}\left (2,-e^{i \left (f x-\tan ^{-1}(\cot (e))\right )}\right )-i \text{PolyLog}\left (2,e^{i \left (f x-\tan ^{-1}(\cot (e))\right )}\right )+\left (f x-\tan ^{-1}(\cot (e))\right ) \left (\log \left (1-e^{i \left (f x-\tan ^{-1}(\cot (e))\right )}\right )-\log \left (1+e^{i \left (f x-\tan ^{-1}(\cot (e))\right )}\right )\right )\right )}{\sqrt{\csc ^2(e)}}+f x (2 c f+2 d \tan (e)+d f x)+\frac{2 f (c+d x) \sin \left (\frac{f x}{2}\right )}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{2 f (c+d x) \sin \left (\frac{f x}{2}\right )}{\left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}+8 c f \tanh ^{-1}\left (\cos (e) \tan \left (\frac{f x}{2}\right )+\sin (e)\right )-2 d f x \tan (e)+2 d (f x \tan (e)+\log (\cos (e+f x)))+8 d \tan ^{-1}(\cot (e)) \tanh ^{-1}\left (\cos (e) \tan \left (\frac{f x}{2}\right )+\sin (e)\right )\right )}{8 f^2} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.137, size = 274, normalized size = 2. \begin{align*}{\frac{{a}^{2}d{x}^{2}}{2}}+{a}^{2}cx+{\frac{2\,i{a}^{2} \left ( dx+c \right ) }{f \left ( 1+{{\rm e}^{2\,i \left ( fx+e \right ) }} \right ) }}+{\frac{{a}^{2}d\ln \left ( 1+{{\rm e}^{2\,i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-2\,{\frac{{a}^{2}d\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-{\frac{4\,i{a}^{2}c\arctan \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{f}}+{\frac{4\,i{a}^{2}de\arctan \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-2\,{\frac{{a}^{2}\ln \left ( 1+i{{\rm e}^{i \left ( fx+e \right ) }} \right ) dx}{f}}-2\,{\frac{{a}^{2}\ln \left ( 1+i{{\rm e}^{i \left ( fx+e \right ) }} \right ) de}{{f}^{2}}}+2\,{\frac{{a}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) dx}{f}}+2\,{\frac{{a}^{2}\ln \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) de}{{f}^{2}}}+{\frac{2\,i{a}^{2}d{\it dilog} \left ( 1+i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-{\frac{2\,i{a}^{2}d{\it dilog} \left ( 1-i{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.19364, size = 1378, normalized size = 10.28 \begin{align*} \frac{-2 i \, a^{2} d \cos \left (f x + e\right ){\rm Li}_2\left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) - 2 i \, a^{2} d \cos \left (f x + e\right ){\rm Li}_2\left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) + 2 i \, a^{2} d \cos \left (f x + e\right ){\rm Li}_2\left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) + 2 i \, a^{2} d \cos \left (f x + e\right ){\rm Li}_2\left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) -{\left (2 \, a^{2} d e - 2 \, a^{2} c f - a^{2} d\right )} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) +{\left (2 \, a^{2} d e - 2 \, a^{2} c f + a^{2} d\right )} \cos \left (f x + e\right ) \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right ) + 2 \,{\left (a^{2} d f x + a^{2} d e\right )} \cos \left (f x + e\right ) \log \left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (a^{2} d f x + a^{2} d e\right )} \cos \left (f x + e\right ) \log \left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (a^{2} d f x + a^{2} d e\right )} \cos \left (f x + e\right ) \log \left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (a^{2} d f x + a^{2} d e\right )} \cos \left (f x + e\right ) \log \left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) -{\left (2 \, a^{2} d e - 2 \, a^{2} c f - a^{2} d\right )} \cos \left (f x + e\right ) \log \left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) +{\left (2 \, a^{2} d e - 2 \, a^{2} c f + a^{2} d\right )} \cos \left (f x + e\right ) \log \left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right ) +{\left (a^{2} d f^{2} x^{2} + 2 \, a^{2} c f^{2} x\right )} \cos \left (f x + e\right ) + 2 \,{\left (a^{2} d f x + a^{2} c f\right )} \sin \left (f x + e\right )}{2 \, f^{2} \cos \left (f x + e\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*} a^{2} \left (\int c\, dx + \int 2 c \sec{\left (e + f x \right )}\, dx + \int c \sec ^{2}{\left (e + f x \right )}\, dx + \int d x\, dx + \int 2 d x \sec{\left (e + f x \right )}\, dx + \int d x \sec ^{2}{\left (e + f x \right )}\, dx\right ) \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{\left (d x + c\right )}{\left (a \sec \left (f x + e\right ) + a\right )}^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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