Optimal. Leaf size=86 \[ \frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{b^2 f \sqrt{a+b}}-\frac{(2 a-b) \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac{\tan (e+f x) \sec (e+f x)}{2 b f} \]
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Rubi [A] time = 0.120596, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4147, 414, 522, 206, 208} \[ \frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{b^2 f \sqrt{a+b}}-\frac{(2 a-b) \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac{\tan (e+f x) \sec (e+f x)}{2 b f} \]
Antiderivative was successfully verified.
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Rule 4147
Rule 414
Rule 522
Rule 206
Rule 208
Rubi steps
\begin{align*} \int \frac{\sec ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^2 \left (a+b-a x^2\right )} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\sec (e+f x) \tan (e+f x)}{2 b f}+\frac{\operatorname{Subst}\left (\int \frac{-a+b-a x^2}{\left (1-x^2\right ) \left (a+b-a x^2\right )} \, dx,x,\sin (e+f x)\right )}{2 b f}\\ &=\frac{\sec (e+f x) \tan (e+f x)}{2 b f}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{b^2 f}-\frac{(2 a-b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (e+f x)\right )}{2 b^2 f}\\ &=-\frac{(2 a-b) \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{b^2 \sqrt{a+b} f}+\frac{\sec (e+f x) \tan (e+f x)}{2 b f}\\ \end{align*}
Mathematica [C] time = 6.45738, size = 1195, normalized size = 13.9 \[ \frac{(\cos (2 (e+f x)) a+a+2 b) \sec ^2(e+f x) \left (\frac{2 i \tan ^{-1}\left (\frac{2 \sin (e) \left (\sin (2 e) a+i a-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt{a}-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt{a}+\sqrt{a+b} \cos (f x) \sqrt{(\cos (e)-i \sin (e))^2} \sqrt{a}-\sqrt{a+b} \cos (2 e+f x) \sqrt{(\cos (e)-i \sin (e))^2} \sqrt{a}+i b+i (a+b) \cos (2 e)+b \sin (2 e)\right )}{i (a+3 b) \cos (e)+i (a+b) \cos (3 e)+i a \cos (e+2 f x)+i a \cos (3 e+2 f x)+3 a \sin (e)+b \sin (e)+a \sin (3 e)+b \sin (3 e)+a \sin (e+2 f x)-a \sin (3 e+2 f x)}\right ) \sqrt{(\cos (e)-i \sin (e))^2} (\cos (e)+i \sin (e)) a^{3/2}}{\sqrt{a+b}}-\frac{i \log \left (-\cos (2 (e+f x)) a-2 i \sin (2 e) a+a+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt{a}+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt{a}+2 (a+b) \cos (2 e)-2 i b \sin (2 e)\right ) \sin (e) a^{3/2}}{\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}}+\frac{i \log \left (\cos (2 (e+f x)) a+2 i \sin (2 e) a-a+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt{a}+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt{a}-2 (a+b) \cos (2 e)+2 i b \sin (2 e)\right ) \sin (e) a^{3/2}}{\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}}+\frac{2 \tan ^{-1}\left (\frac{(a+b) \sin (e)}{(a+b) \cos (e)-\sqrt{a} \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} (\cos (2 e)+i \sin (2 e)) \sin (e+f x)}\right ) \sqrt{(\cos (e)-i \sin (e))^2} (\sin (e)-i \cos (e)) a^{3/2}}{\sqrt{a+b}}+\frac{\cos (e) \log \left (-\cos (2 (e+f x)) a-2 i \sin (2 e) a+a+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt{a}+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt{a}+2 (a+b) \cos (2 e)-2 i b \sin (2 e)\right ) a^{3/2}}{\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}}-\frac{\cos (e) \log \left (\cos (2 (e+f x)) a+2 i \sin (2 e) a-a+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt{a}+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt{a}-2 (a+b) \cos (2 e)+2 i b \sin (2 e)\right ) a^{3/2}}{\sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2}}+4 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) a-4 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )+\sin \left (\frac{1}{2} (e+f x)\right )\right ) a-2 b \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+2 b \log \left (\cos \left (\frac{1}{2} (e+f x)\right )+\sin \left (\frac{1}{2} (e+f x)\right )\right )+\frac{b}{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2}-\frac{b}{\left (\cos \left (\frac{1}{2} (e+f x)\right )+\sin \left (\frac{1}{2} (e+f x)\right )\right )^2}\right )}{8 b^2 f \left (b \sec ^2(e+f x)+a\right )} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.075, size = 141, normalized size = 1.6 \begin{align*} -{\frac{1}{4\,fb \left ( \sin \left ( fx+e \right ) +1 \right ) }}-{\frac{\ln \left ( \sin \left ( fx+e \right ) +1 \right ) a}{2\,f{b}^{2}}}+{\frac{\ln \left ( \sin \left ( fx+e \right ) +1 \right ) }{4\,fb}}+{\frac{{a}^{2}}{f{b}^{2}}{\it Artanh} \left ({a\sin \left ( fx+e \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}}-{\frac{1}{4\,fb \left ( \sin \left ( fx+e \right ) -1 \right ) }}+{\frac{\ln \left ( \sin \left ( fx+e \right ) -1 \right ) a}{2\,f{b}^{2}}}-{\frac{\ln \left ( \sin \left ( fx+e \right ) -1 \right ) }{4\,fb}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.602482, size = 687, normalized size = 7.99 \begin{align*} \left [\frac{2 \, a \sqrt{\frac{a}{a + b}} \cos \left (f x + e\right )^{2} \log \left (-\frac{a \cos \left (f x + e\right )^{2} - 2 \,{\left (a + b\right )} \sqrt{\frac{a}{a + b}} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) -{\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) +{\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, b \sin \left (f x + e\right )}{4 \, b^{2} f \cos \left (f x + e\right )^{2}}, -\frac{4 \, a \sqrt{-\frac{a}{a + b}} \arctan \left (\sqrt{-\frac{a}{a + b}} \sin \left (f x + e\right )\right ) \cos \left (f x + e\right )^{2} +{\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, b \sin \left (f x + e\right )}{4 \, b^{2} f \cos \left (f x + e\right )^{2}}\right ] \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 \frac{\sec ^{5}{\left (e + f x \right )}}{a + b \sec ^{2}{\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 = 1.25492, size = 159, normalized size = 1.85 \begin{align*} -\frac{\frac{4 \, a^{2} \arctan \left (\frac{a \sin \left (f x + e\right )}{\sqrt{-a^{2} - a b}}\right )}{\sqrt{-a^{2} - a b} b^{2}} + \frac{{\left (2 \, a - b\right )} \log \left (\sin \left (f x + e\right ) + 1\right )}{b^{2}} - \frac{{\left (2 \, a - b\right )} \log \left (-\sin \left (f x + e\right ) + 1\right )}{b^{2}} + \frac{2 \, \sin \left (f x + e\right )}{{\left (\sin \left (f x + e\right )^{2} - 1\right )} b}}{4 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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