Integrand size = 18, antiderivative size = 131 \[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\frac {a^2 (c+d x)^2}{2 d}-\frac {4 i a b (c+d x) \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {b^2 d \log (\cos (e+f x))}{f^2}+\frac {2 i a b d \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a b d \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}+\frac {b^2 (c+d x) \tan (e+f x)}{f} \]
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Time = 0.15 (sec) , antiderivative size = 131, 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} \[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\frac {a^2 (c+d x)^2}{2 d}-\frac {4 i a b (c+d x) \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {2 i a b d \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a b d \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}+\frac {b^2 (c+d x) \tan (e+f x)}{f}+\frac {b^2 d \log (\cos (e+f x))}{f^2} \]
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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 b (c+d x) \sec (e+f x)+b^2 (c+d x) \sec ^2(e+f x)\right ) \, dx \\ & = \frac {a^2 (c+d x)^2}{2 d}+(2 a b) \int (c+d x) \sec (e+f x) \, dx+b^2 \int (c+d x) \sec ^2(e+f x) \, dx \\ & = \frac {a^2 (c+d x)^2}{2 d}-\frac {4 i a b (c+d x) \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {b^2 (c+d x) \tan (e+f x)}{f}-\frac {(2 a b d) \int \log \left (1-i e^{i (e+f x)}\right ) \, dx}{f}+\frac {(2 a b d) \int \log \left (1+i e^{i (e+f x)}\right ) \, dx}{f}-\frac {\left (b^2 d\right ) \int \tan (e+f x) \, dx}{f} \\ & = \frac {a^2 (c+d x)^2}{2 d}-\frac {4 i a b (c+d x) \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {b^2 d \log (\cos (e+f x))}{f^2}+\frac {b^2 (c+d x) \tan (e+f x)}{f}+\frac {(2 i a b d) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^2}-\frac {(2 i a b d) \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 b (c+d x) \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {b^2 d \log (\cos (e+f x))}{f^2}+\frac {2 i a b d \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i a b d \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}+\frac {b^2 (c+d x) \tan (e+f x)}{f} \\ \end{align*}
Time = 0.55 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.15 \[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\frac {2 a^2 c f^2 x+a^2 d f^2 x^2-8 i a b d f x \arctan \left (e^{i (e+f x)}\right )+4 a b c f \text {arctanh}(\sin (e+f x))+2 b^2 d \log (\cos (e+f x))+4 i a b d \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )-4 i a b d \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )+2 b^2 c f \tan (e+f x)+2 b^2 d f x \tan (e+f x)}{2 f^2} \]
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Time = 0.49 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.45
method | result | size |
parts | \(a^{2} \left (\frac {1}{2} d \,x^{2}+x c \right )+\frac {b^{2} d \tan \left (f x +e \right ) x}{f}+\frac {b^{2} d \ln \left (\cos \left (f x +e \right )\right )}{f^{2}}+\frac {b^{2} c \tan \left (f x +e \right )}{f}+\frac {2 a b \left (\frac {d \left (-\left (f x +e \right ) \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )+\left (f x +e \right ) \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )+i \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )-i \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )\right )}{f}+c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )-\frac {e d \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\right )}{f}\) | \(190\) |
derivativedivides | \(\frac {a^{2} c \left (f x +e \right )-\frac {a^{2} d e \left (f x +e \right )}{f}+\frac {a^{2} d \left (f x +e \right )^{2}}{2 f}+2 a b c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )-\frac {2 a b d e \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}+\frac {2 a b d \left (-\left (f x +e \right ) \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )+\left (f x +e \right ) \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )+i \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )-i \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )\right )}{f}+b^{2} c \tan \left (f x +e \right )-\frac {b^{2} d e \tan \left (f x +e \right )}{f}+\frac {b^{2} d \left (\left (f x +e \right ) \tan \left (f x +e \right )+\ln \left (\cos \left (f x +e \right )\right )\right )}{f}}{f}\) | \(232\) |
default | \(\frac {a^{2} c \left (f x +e \right )-\frac {a^{2} d e \left (f x +e \right )}{f}+\frac {a^{2} d \left (f x +e \right )^{2}}{2 f}+2 a b c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )-\frac {2 a b d e \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}+\frac {2 a b d \left (-\left (f x +e \right ) \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )+\left (f x +e \right ) \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )+i \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )-i \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )\right )}{f}+b^{2} c \tan \left (f x +e \right )-\frac {b^{2} d e \tan \left (f x +e \right )}{f}+\frac {b^{2} d \left (\left (f x +e \right ) \tan \left (f x +e \right )+\ln \left (\cos \left (f x +e \right )\right )\right )}{f}}{f}\) | \(232\) |
risch | \(\frac {a^{2} d \,x^{2}}{2}+a^{2} x c +\frac {2 i b^{2} \left (d x +c \right )}{f \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}+\frac {b^{2} d \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 b^{2} d \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {4 i b a c \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}+\frac {4 i b a d e \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 b a d \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}-\frac {2 b a d \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}+\frac {2 b a d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}+\frac {2 b a d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}+\frac {2 i b a d \operatorname {dilog}\left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 i b a d \operatorname {dilog}\left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}\) | \(266\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 505 vs. \(2 (111) = 222\).
Time = 0.33 (sec) , antiderivative size = 505, normalized size of antiderivative = 3.85 \[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\frac {-2 i \, a b 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 b 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 b 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 b 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 b d e - 2 \, a b c f - b^{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 b d e - 2 \, a b c f + b^{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 b d f x + a b 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 b d f x + a b 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 b d f x + a b 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 b d f x + a b 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 b d e - 2 \, a b c f - b^{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 b d e - 2 \, a b c f + b^{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 (b^{2} d f x + b^{2} c f\right )} \sin \left (f x + e\right )}{2 \, f^{2} \cos \left (f x + e\right )} \]
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\[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\int \left (a + b \sec {\left (e + f x \right )}\right )^{2} \left (c + d x\right )\, dx \]
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\[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\int { {\left (d x + c\right )} {\left (b \sec \left (f x + e\right ) + a\right )}^{2} \,d x } \]
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\[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\int { {\left (d x + c\right )} {\left (b \sec \left (f x + e\right ) + a\right )}^{2} \,d x } \]
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Timed out. \[ \int (c+d x) (a+b \sec (e+f x))^2 \, dx=\int {\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^2\,\left (c+d\,x\right ) \,d x \]
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