Optimal. Leaf size=134 \[ -\frac {4 i a^2 (c+d x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {a^2 (c+d x) \tan (e+f x)}{f}+\frac {a^2 (c+d x)^2}{2 d}+\frac {2 i a^2 d \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a^2 d \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}+\frac {a^2 d \log (\cos (e+f x))}{f^2} \]
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Rubi [A] time = 0.11, antiderivative size = 134, normalized size of antiderivative = 1.00, 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 2279
Rule 2391
Rule 3475
Rule 4181
Rule 4184
Rule 4190
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*}
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Mathematica [B] time = 5.61, size = 330, normalized size = 2.46 \[ \frac {a^2 (\cos (e+f x)+1)^2 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \left (f x (2 c f+2 d \tan (e)+d f x)+\frac {2 f (c+d x) \sin \left (\frac {f x}{2}\right )}{\left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}+\frac {2 f (c+d x) \sin \left (\frac {f x}{2}\right )}{\left (\sin \left (\frac {e}{2}\right )+\cos \left (\frac {e}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}+8 c f \tanh ^{-1}\left (\cos (e) \tan \left (\frac {f x}{2}\right )+\sin (e)\right )-\frac {4 d \csc (e) \left (i \text {Li}_2\left (-e^{i \left (f x-\tan ^{-1}(\cot (e))\right )}\right )-i \text {Li}_2\left (e^{i \left (f x-\tan ^{-1}(\cot (e))\right )}\right )+\left (f x-\tan ^{-1}(\cot (e))\right ) \left (\log \left (1-e^{i \left (f x-\tan ^{-1}(\cot (e))\right )}\right )-\log \left (1+e^{i \left (f x-\tan ^{-1}(\cot (e))\right )}\right )\right )\right )}{\sqrt {\csc ^2(e)}}-2 d f x \tan (e)+2 d (f x \tan (e)+\log (\cos (e+f x)))+8 d \tan ^{-1}(\cot (e)) \tanh ^{-1}\left (\cos (e) \tan \left (\frac {f x}{2}\right )+\sin (e)\right )\right )}{8 f^2} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.85, size = 525, normalized size = 3.92 \[ \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 )} \]
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 \[ \int {\left (d x + c\right )} {\left (a \sec \left (f x + e\right ) + a\right )}^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.53, size = 279, normalized size = 2.08 \[ \frac {a^{2} d \,x^{2}}{2}-\frac {a^{2} d \,e^{2}}{2 f^{2}}+a^{2} c x +\frac {a^{2} c e}{f}-\frac {2 a^{2} d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) x}{f}+\frac {2 a^{2} d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}-\frac {2 i a^{2} d \dilog \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+\frac {2 i a^{2} d \dilog \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right )}{f^{2}}-\frac {2 a^{2} d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) e}{f^{2}}+\frac {2 a^{2} d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}+\frac {2 a^{2} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}-\frac {2 a^{2} d e \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f^{2}}+\frac {a^{2} d \tan \left (f x +e \right ) x}{f}+\frac {a^{2} d \ln \left (\cos \left (f x +e \right )\right )}{f^{2}}+\frac {a^{2} c \tan \left (f x +e \right )}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
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 \[ \int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^2\,\left (c+d\,x\right ) \,d x \]
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 \[ a^{2} \left (\int c\, dx + \int 2 c \sec {\left (e + f x \right )}\, dx + \int c \sec ^{2}{\left (e + f x \right )}\, dx + \int d x\, dx + \int 2 d x \sec {\left (e + f x \right )}\, dx + \int d x \sec ^{2}{\left (e + f x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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