Optimal. Leaf size=165 \[ \frac{\left (48 a^2+80 a b+35 b^2\right ) \tanh ^{-1}(\sin (e+f x))}{128 f}+\frac{\left (48 a^2+80 a b+35 b^2\right ) \tan (e+f x) \sec ^3(e+f x)}{192 f}+\frac{\left (48 a^2+80 a b+35 b^2\right ) \tan (e+f x) \sec (e+f x)}{128 f}+\frac{b (10 a+7 b) \tan (e+f x) \sec ^5(e+f x)}{48 f}+\frac{b \tan (e+f x) \sec ^7(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{8 f} \]
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Rubi [A] time = 0.141082, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4147, 413, 385, 199, 206} \[ \frac{\left (48 a^2+80 a b+35 b^2\right ) \tanh ^{-1}(\sin (e+f x))}{128 f}+\frac{\left (48 a^2+80 a b+35 b^2\right ) \tan (e+f x) \sec ^3(e+f x)}{192 f}+\frac{\left (48 a^2+80 a b+35 b^2\right ) \tan (e+f x) \sec (e+f x)}{128 f}+\frac{b (10 a+7 b) \tan (e+f x) \sec ^5(e+f x)}{48 f}+\frac{b \tan (e+f x) \sec ^7(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{8 f} \]
Antiderivative was successfully verified.
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Rule 4147
Rule 413
Rule 385
Rule 199
Rule 206
Rubi steps
\begin{align*} \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b-a x^2\right )^2}{\left (1-x^2\right )^5} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{b \sec ^7(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{8 f}-\frac{\operatorname{Subst}\left (\int \frac{-(a+b) (8 a+7 b)+a (8 a+5 b) x^2}{\left (1-x^2\right )^4} \, dx,x,\sin (e+f x)\right )}{8 f}\\ &=\frac{b (10 a+7 b) \sec ^5(e+f x) \tan (e+f x)}{48 f}+\frac{b \sec ^7(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{8 f}+\frac{\left (48 a^2+80 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{48 f}\\ &=\frac{\left (48 a^2+80 a b+35 b^2\right ) \sec ^3(e+f x) \tan (e+f x)}{192 f}+\frac{b (10 a+7 b) \sec ^5(e+f x) \tan (e+f x)}{48 f}+\frac{b \sec ^7(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{8 f}+\frac{\left (48 a^2+80 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{64 f}\\ &=\frac{\left (48 a^2+80 a b+35 b^2\right ) \sec (e+f x) \tan (e+f x)}{128 f}+\frac{\left (48 a^2+80 a b+35 b^2\right ) \sec ^3(e+f x) \tan (e+f x)}{192 f}+\frac{b (10 a+7 b) \sec ^5(e+f x) \tan (e+f x)}{48 f}+\frac{b \sec ^7(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{8 f}+\frac{\left (48 a^2+80 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (e+f x)\right )}{128 f}\\ &=\frac{\left (48 a^2+80 a b+35 b^2\right ) \tanh ^{-1}(\sin (e+f x))}{128 f}+\frac{\left (48 a^2+80 a b+35 b^2\right ) \sec (e+f x) \tan (e+f x)}{128 f}+\frac{\left (48 a^2+80 a b+35 b^2\right ) \sec ^3(e+f x) \tan (e+f x)}{192 f}+\frac{b (10 a+7 b) \sec ^5(e+f x) \tan (e+f x)}{48 f}+\frac{b \sec ^7(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{8 f}\\ \end{align*}
Mathematica [A] time = 0.535401, size = 119, normalized size = 0.72 \[ \frac{3 \left (48 a^2+80 a b+35 b^2\right ) \tanh ^{-1}(\sin (e+f x))+\tan (e+f x) \sec (e+f x) \left (2 \left (48 a^2+80 a b+35 b^2\right ) \sec ^2(e+f x)+3 \left (48 a^2+80 a b+35 b^2\right )+8 b (16 a+7 b) \sec ^4(e+f x)+48 b^2 \sec ^6(e+f x)\right )}{384 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.041, size = 256, normalized size = 1.6 \begin{align*}{\frac{{a}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{4\,f}}+{\frac{3\,{a}^{2}\tan \left ( fx+e \right ) \sec \left ( fx+e \right ) }{8\,f}}+{\frac{3\,{a}^{2}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}}+{\frac{ab\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{5}}{3\,f}}+{\frac{5\,ab\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{12\,f}}+{\frac{5\,ab\tan \left ( fx+e \right ) \sec \left ( fx+e \right ) }{8\,f}}+{\frac{5\,ab\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}}+{\frac{{b}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{7}}{8\,f}}+{\frac{7\,{b}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{5}}{48\,f}}+{\frac{35\,{b}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{192\,f}}+{\frac{35\,{b}^{2}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{128\,f}}+{\frac{35\,{b}^{2}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{128\,f}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.02563, size = 270, normalized size = 1.64 \begin{align*} \frac{3 \,{\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (f x + e\right ) - 1\right ) - \frac{2 \,{\left (3 \,{\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \sin \left (f x + e\right )^{7} - 11 \,{\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \sin \left (f x + e\right )^{5} +{\left (624 \, a^{2} + 1168 \, a b + 511 \, b^{2}\right )} \sin \left (f x + e\right )^{3} - 3 \,{\left (80 \, a^{2} + 176 \, a b + 93 \, b^{2}\right )} \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{8} - 4 \, \sin \left (f x + e\right )^{6} + 6 \, \sin \left (f x + e\right )^{4} - 4 \, \sin \left (f x + e\right )^{2} + 1}}{768 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.548484, size = 428, normalized size = 2.59 \begin{align*} \frac{3 \,{\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \cos \left (f x + e\right )^{8} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \cos \left (f x + e\right )^{8} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (3 \,{\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + 2 \,{\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \,{\left (16 \, a b + 7 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 48 \, b^{2}\right )} \sin \left (f x + e\right )}{768 \, f \cos \left (f x + e\right )^{8}} \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 \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \sec ^{5}{\left (e + f x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.34345, size = 319, normalized size = 1.93 \begin{align*} \frac{3 \,{\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (48 \, a^{2} + 80 \, a b + 35 \, b^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - \frac{2 \,{\left (144 \, a^{2} \sin \left (f x + e\right )^{7} + 240 \, a b \sin \left (f x + e\right )^{7} + 105 \, b^{2} \sin \left (f x + e\right )^{7} - 528 \, a^{2} \sin \left (f x + e\right )^{5} - 880 \, a b \sin \left (f x + e\right )^{5} - 385 \, b^{2} \sin \left (f x + e\right )^{5} + 624 \, a^{2} \sin \left (f x + e\right )^{3} + 1168 \, a b \sin \left (f x + e\right )^{3} + 511 \, b^{2} \sin \left (f x + e\right )^{3} - 240 \, a^{2} \sin \left (f x + e\right ) - 528 \, a b \sin \left (f x + e\right ) - 279 \, b^{2} \sin \left (f x + e\right )\right )}}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{4}}}{768 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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