Integrand size = 18, antiderivative size = 134 \[ \int (c+d x) (a+a \sec (e+f x))^2 \, dx=\frac {a^2 (c+d x)^2}{2 d}-\frac {4 i a^2 (c+d x) \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {a^2 d \log (\cos (e+f x))}{f^2}+\frac {2 i a^2 d \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a^2 d \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}+\frac {a^2 (c+d x) \tan (e+f x)}{f} \]
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Time = 0.16 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4275, 4266, 2317, 2438, 4269, 3556} \[ \int (c+d x) (a+a \sec (e+f x))^2 \, dx=-\frac {4 i a^2 (c+d x) \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {a^2 (c+d x) \tan (e+f x)}{f}+\frac {a^2 (c+d x)^2}{2 d}+\frac {2 i a^2 d \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a^2 d \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}+\frac {a^2 d \log (\cos (e+f x))}{f^2} \]
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Rule 2317
Rule 2438
Rule 3556
Rule 4266
Rule 4269
Rule 4275
Rubi steps \begin{align*} \text {integral}& = \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) \arctan \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) \arctan \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 ) \text {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 ) \text {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) \arctan \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 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a^2 d \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}+\frac {a^2 (c+d x) \tan (e+f x)}{f} \\ \end{align*}
Time = 1.11 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.92 \[ \int (c+d x) (a+a \sec (e+f x))^2 \, dx=\frac {a^2 \left (f^2 (c+d x)^2-8 i d f (c+d x) \arctan \left (e^{i (e+f x)}\right )+2 d^2 \log (\cos (e+f x))+4 i d^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )-4 i d^2 \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )+2 d f (c+d x) \tan (e+f x)\right )}{2 d f^2} \]
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Time = 0.42 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.43
method | result | size |
parts | \(a^{2} \left (\frac {1}{2} d \,x^{2}+x c \right )+\frac {a^{2} d \tan \left (f x +e \right ) x}{f}+\frac {a^{2} d \ln \left (\cos \left (f x +e \right )\right )}{f^{2}}+\frac {a^{2} c \tan \left (f x +e \right )}{f}+\frac {2 a^{2} \left (\frac {d \left (-\left (f x +e \right ) \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )+\left (f x +e \right ) \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )+i \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )-i \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )\right )}{f}+c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )-\frac {e d \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\right )}{f}\) | \(191\) |
derivativedivides | \(\frac {a^{2} c \tan \left (f x +e \right )-\frac {a^{2} d e \tan \left (f x +e \right )}{f}+\frac {a^{2} d \left (\left (f x +e \right ) \tan \left (f x +e \right )+\ln \left (\cos \left (f x +e \right )\right )\right )}{f}+2 a^{2} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )-\frac {2 a^{2} d e \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}+\frac {2 a^{2} d \left (-\left (f x +e \right ) \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )+\left (f x +e \right ) \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )+i \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )-i \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )\right )}{f}+a^{2} c \left (f x +e \right )-\frac {a^{2} d e \left (f x +e \right )}{f}+\frac {a^{2} d \left (f x +e \right )^{2}}{2 f}}{f}\) | \(235\) |
default | \(\frac {a^{2} c \tan \left (f x +e \right )-\frac {a^{2} d e \tan \left (f x +e \right )}{f}+\frac {a^{2} d \left (\left (f x +e \right ) \tan \left (f x +e \right )+\ln \left (\cos \left (f x +e \right )\right )\right )}{f}+2 a^{2} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )-\frac {2 a^{2} d e \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}+\frac {2 a^{2} d \left (-\left (f x +e \right ) \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )+\left (f x +e \right ) \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )+i \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )-i \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )\right )}{f}+a^{2} c \left (f x +e \right )-\frac {a^{2} d e \left (f x +e \right )}{f}+\frac {a^{2} d \left (f x +e \right )^{2}}{2 f}}{f}\) | \(235\) |
risch | \(\frac {a^{2} d \,x^{2}}{2}+a^{2} x c +\frac {2 i a^{2} \left (d x +c \right )}{f \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}+\frac {a^{2} d \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 a^{2} d \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {4 i a^{2} c \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}+\frac {4 i a^{2} d e \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 a^{2} d \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}-\frac {2 a^{2} d \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}+\frac {2 a^{2} d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}+\frac {2 a^{2} d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}+\frac {2 i a^{2} d \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 i a^{2} d \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}\) | \(274\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 525 vs. \(2 (114) = 228\).
Time = 0.31 (sec) , antiderivative size = 525, normalized size of antiderivative = 3.92 \[ \int (c+d x) (a+a \sec (e+f x))^2 \, dx=\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 )} \]
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\[ \int (c+d x) (a+a \sec (e+f x))^2 \, dx=a^{2} \left (\int c\, dx + \int 2 c \sec {\left (e + f x \right )}\, dx + \int c \sec ^{2}{\left (e + f x \right )}\, dx + \int d x\, dx + \int 2 d x \sec {\left (e + f x \right )}\, dx + \int d x \sec ^{2}{\left (e + f x \right )}\, dx\right ) \]
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\[ \int (c+d x) (a+a \sec (e+f x))^2 \, dx=\int { {\left (d x + c\right )} {\left (a \sec \left (f x + e\right ) + a\right )}^{2} \,d x } \]
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\[ \int (c+d x) (a+a \sec (e+f x))^2 \, dx=\int { {\left (d x + c\right )} {\left (a \sec \left (f x + e\right ) + a\right )}^{2} \,d x } \]
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Timed out. \[ \int (c+d x) (a+a \sec (e+f x))^2 \, dx=\int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^2\,\left (c+d\,x\right ) \,d x \]
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