Optimal. Leaf size=176 \[ \frac{2 a^2 \tan (e+f x) \left (2 \left (6 c^2+25 c d+9 d^2\right )+d (4 c+9 d) \sec (e+f x)\right )}{15 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^{5/2} c^2 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 \tan (e+f x) (c+d \sec (e+f x))^2}{5 f \sqrt{a \sec (e+f x)+a}} \]
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Rubi [A] time = 0.143539, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {3940, 153, 147, 63, 206} \[ \frac{2 a^2 \tan (e+f x) \left (2 \left (6 c^2+25 c d+9 d^2\right )+d (4 c+9 d) \sec (e+f x)\right )}{15 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^{5/2} c^2 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 \tan (e+f x) (c+d \sec (e+f x))^2}{5 f \sqrt{a \sec (e+f x)+a}} \]
Antiderivative was successfully verified.
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Rule 3940
Rule 153
Rule 147
Rule 63
Rule 206
Rubi steps
\begin{align*} \int (a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^2 \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x) (c+d x)^2}{x \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt{a+a \sec (e+f x)}}+\frac{(2 a \tan (e+f x)) \operatorname{Subst}\left (\int \frac{(c+d x) \left (-\frac{5 a^2 c}{2}-\frac{1}{2} a^2 (4 c+9 d) x\right )}{x \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{5 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 \left (2 \left (6 c^2+25 c d+9 d^2\right )+d (4 c+9 d) \sec (e+f x)\right ) \tan (e+f x)}{15 f \sqrt{a+a \sec (e+f x)}}-\frac{\left (a^3 c^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 \left (2 \left (6 c^2+25 c d+9 d^2\right )+d (4 c+9 d) \sec (e+f x)\right ) \tan (e+f x)}{15 f \sqrt{a+a \sec (e+f x)}}+\frac{\left (2 a^2 c^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^2}{a}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a^{5/2} c^2 \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right ) \tan (e+f x)}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 \left (2 \left (6 c^2+25 c d+9 d^2\right )+d (4 c+9 d) \sec (e+f x)\right ) \tan (e+f x)}{15 f \sqrt{a+a \sec (e+f x)}}\\ \end{align*}
Mathematica [A] time = 1.33611, size = 145, normalized size = 0.82 \[ \frac{a \sec \left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x) \sqrt{a (\sec (e+f x)+1)} \left (2 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\left (15 c^2+50 c d+18 d^2\right ) \cos (2 (e+f x))+15 c^2+2 d (10 c+9 d) \cos (e+f x)+50 c d+24 d^2\right )+30 \sqrt{2} c^2 \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (e+f x)\right )\right ) \cos ^{\frac{5}{2}}(e+f x)\right )}{30 f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.271, size = 382, normalized size = 2.2 \begin{align*} -{\frac{a}{60\,f \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) }\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ( 15\,\sqrt{2}\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ){c}^{2}+30\,\sqrt{2}\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ){c}^{2}+15\,\sqrt{2} \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ){c}^{2}\sin \left ( fx+e \right ) +120\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}{c}^{2}+400\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}cd+144\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}{d}^{2}-120\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{c}^{2}-320\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}cd-72\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{d}^{2}-80\,\cos \left ( fx+e \right ) cd-48\,\cos \left ( fx+e \right ){d}^{2}-24\,{d}^{2} \right ) } \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 [A] time = 1.21746, size = 1006, normalized size = 5.72 \begin{align*} \left [\frac{15 \,{\left (a c^{2} \cos \left (f x + e\right )^{3} + a c^{2} \cos \left (f x + e\right )^{2}\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) + 2 \,{\left (3 \, a d^{2} +{\left (15 \, a c^{2} + 50 \, a c d + 18 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} +{\left (10 \, a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{15 \,{\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}, -\frac{2 \,{\left (15 \,{\left (a c^{2} \cos \left (f x + e\right )^{3} + a c^{2} \cos \left (f x + e\right )^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt{a} \sin \left (f x + e\right )}\right ) -{\left (3 \, a d^{2} +{\left (15 \, a c^{2} + 50 \, a c d + 18 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} +{\left (10 \, a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{15 \,{\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [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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Giac [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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