Optimal. Leaf size=52 \[ \frac{\tan (e+f x)}{b f}-\frac{a \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{b^{3/2} f \sqrt{a+b}} \]
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Rubi [A] time = 0.0676325, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4146, 388, 205} \[ \frac{\tan (e+f x)}{b f}-\frac{a \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{b^{3/2} f \sqrt{a+b}} \]
Antiderivative was successfully verified.
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Rule 4146
Rule 388
Rule 205
Rubi steps
\begin{align*} \int \frac{\sec ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\tan (e+f x)}{b f}-\frac{a \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{b f}\\ &=-\frac{a \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{b^{3/2} \sqrt{a+b} f}+\frac{\tan (e+f x)}{b f}\\ \end{align*}
Mathematica [C] time = 0.804315, size = 192, normalized size = 3.69 \[ \frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\sqrt{a+b} \sec (e) \sin (f x) \sqrt{b (\sin (e)+i \cos (e))^4} \sec (e+f x)+a (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )\right )}{2 b f \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4} \left (a+b \sec ^2(e+f x)\right )} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.059, size = 47, normalized size = 0.9 \begin{align*}{\frac{\tan \left ( fx+e \right ) }{fb}}-{\frac{a}{fb}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \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 [B] time = 0.542119, size = 678, normalized size = 13.04 \begin{align*} \left [-\frac{\sqrt{-a b - b^{2}} a \cos \left (f x + e\right ) \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt{-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) - 4 \,{\left (a b + b^{2}\right )} \sin \left (f x + e\right )}{4 \,{\left (a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )}, \frac{\sqrt{a b + b^{2}} a \arctan \left (\frac{{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt{a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) + 2 \,{\left (a b + b^{2}\right )} \sin \left (f x + e\right )}{2 \,{\left (a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )}\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 ^{4}{\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 = 1.18399, size = 93, normalized size = 1.79 \begin{align*} -\frac{\frac{{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )} a}{\sqrt{a b + b^{2}} b} - \frac{\tan \left (f x + e\right )}{b}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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