Optimal. Leaf size=100 \[ \frac{2 (a c-b d) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{f (a-b)^{3/2} (a+b)^{3/2}}-\frac{(b c-a d) \tan (e+f x)}{f \left (a^2-b^2\right ) (a+b \sec (e+f x))} \]
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Rubi [A] time = 0.136388, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {4003, 12, 3831, 2659, 208} \[ \frac{2 (a c-b d) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{f (a-b)^{3/2} (a+b)^{3/2}}-\frac{(b c-a d) \tan (e+f x)}{f \left (a^2-b^2\right ) (a+b \sec (e+f x))} \]
Antiderivative was successfully verified.
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Rule 4003
Rule 12
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\sec (e+f x) (c+d \sec (e+f x))}{(a+b \sec (e+f x))^2} \, dx &=-\frac{(b c-a d) \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))}+\frac{\int \frac{(-a c+b d) \sec (e+f x)}{a+b \sec (e+f x)} \, dx}{-a^2+b^2}\\ &=-\frac{(b c-a d) \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))}+\frac{(a c-b d) \int \frac{\sec (e+f x)}{a+b \sec (e+f x)} \, dx}{a^2-b^2}\\ &=-\frac{(b c-a d) \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))}+\frac{(a c-b d) \int \frac{1}{1+\frac{a \cos (e+f x)}{b}} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{(b c-a d) \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))}+\frac{(2 (a c-b d)) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{b \left (a^2-b^2\right ) f}\\ &=\frac{2 (a c-b d) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} f}-\frac{(b c-a d) \tan (e+f x)}{\left (a^2-b^2\right ) f (a+b \sec (e+f x))}\\ \end{align*}
Mathematica [A] time = 0.36182, size = 97, normalized size = 0.97 \[ \frac{\frac{(a d-b c) \sin (e+f x)}{(a-b) (a+b) (a \cos (e+f x)+b)}-\frac{2 (a c-b d) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}}{f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.068, size = 132, normalized size = 1.3 \begin{align*}{\frac{1}{f} \left ( -2\,{\frac{ \left ( ad-bc \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{ \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}b-a-b \right ) }}+2\,{\frac{ac-db}{ \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \right ) } \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 = 0.527234, size = 861, normalized size = 8.61 \begin{align*} \left [\frac{{\left (a b c - b^{2} d +{\left (a^{2} c - a b d\right )} \cos \left (f x + e\right )\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (f x + e\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (f x + e\right )^{2} + 2 \, a b \cos \left (f x + e\right ) + b^{2}}\right ) - 2 \,{\left ({\left (a^{2} b - b^{3}\right )} c -{\left (a^{3} - a b^{2}\right )} d\right )} \sin \left (f x + e\right )}{2 \,{\left ({\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f \cos \left (f x + e\right ) +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} f\right )}}, \frac{{\left (a b c - b^{2} d +{\left (a^{2} c - a b d\right )} \cos \left (f x + e\right )\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (f x + e\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (f x + e\right )}\right ) -{\left ({\left (a^{2} b - b^{3}\right )} c -{\left (a^{3} - a b^{2}\right )} d\right )} \sin \left (f x + e\right )}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f \cos \left (f x + e\right ) +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} f}\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{\left (c + d \sec{\left (e + f x \right )}\right ) \sec{\left (e + f x \right )}}{\left (a + b \sec{\left (e + f x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.29692, size = 243, normalized size = 2.43 \begin{align*} -\frac{2 \,{\left (\frac{{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}{\left (a c - b d\right )}}{{\left (a^{2} - b^{2}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{b c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - a d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a - b\right )}{\left (a^{2} - b^{2}\right )}}\right )}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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