Optimal. Leaf size=131 \[ \frac{a (2 c-d) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f (c-d)^{3/2} (c+d)^{5/2}}+\frac{a (c-2 d) \tan (e+f x)}{2 f (c-d) (c+d)^2 (c+d \sec (e+f x))}+\frac{a \tan (e+f x)}{2 f (c+d) (c+d \sec (e+f x))^2} \]
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Rubi [A] time = 0.270406, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{a (2 c-d) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f (c-d)^{3/2} (c+d)^{5/2}}+\frac{a (c-2 d) \tan (e+f x)}{2 f (c-d) (c+d)^2 (c+d \sec (e+f x))}+\frac{a \tan (e+f x)}{2 f (c+d) (c+d \sec (e+f x))^2} \]
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) (a+a \sec (e+f x))}{(c+d \sec (e+f x))^3} \, dx &=\frac{a \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}-\frac{\int \frac{\sec (e+f x) (-2 a (c-d)-a (c-d) \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx}{2 \left (c^2-d^2\right )}\\ &=\frac{a \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}+\frac{a (c-2 d) \tan (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sec (e+f x))}+\frac{\int \frac{a (c-d) (2 c-d) \sec (e+f x)}{c+d \sec (e+f x)} \, dx}{2 \left (c^2-d^2\right )^2}\\ &=\frac{a \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}+\frac{a (c-2 d) \tan (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sec (e+f x))}+\frac{(a (2 c-d)) \int \frac{\sec (e+f x)}{c+d \sec (e+f x)} \, dx}{2 (c-d) (c+d)^2}\\ &=\frac{a \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}+\frac{a (c-2 d) \tan (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sec (e+f x))}+\frac{(a (2 c-d)) \int \frac{1}{1+\frac{c \cos (e+f x)}{d}} \, dx}{2 (c-d) d (c+d)^2}\\ &=\frac{a \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}+\frac{a (c-2 d) \tan (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sec (e+f x))}+\frac{(a (2 c-d)) \operatorname{Subst}\left (\int \frac{1}{1+\frac{c}{d}+\left (1-\frac{c}{d}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{(c-d) d (c+d)^2 f}\\ &=\frac{a (2 c-d) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{(c-d)^{3/2} (c+d)^{5/2} f}+\frac{a \tan (e+f x)}{2 (c+d) f (c+d \sec (e+f x))^2}+\frac{a (c-2 d) \tan (e+f x)}{2 (c-d) (c+d)^2 f (c+d \sec (e+f x))}\\ \end{align*}
Mathematica [A] time = 1.17543, size = 167, normalized size = 1.27 \[ \frac{a (\cos (e+f x)+1) \sec ^2\left (\frac{1}{2} (e+f x)\right ) \left (\sqrt{c^2-d^2} \sin (e+f x) \left (\left (2 c^2-2 c d-d^2\right ) \cos (e+f x)+d (c-2 d)\right )-2 (2 c-d) (c \cos (e+f x)+d)^2 \tanh ^{-1}\left (\frac{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )\right )}{4 f (c-d) (c+d)^2 \sqrt{c^2-d^2} (c \cos (e+f x)+d)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.102, size = 178, normalized size = 1.4 \begin{align*} 4\,{\frac{a}{f} \left ({\frac{1}{ \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}d-c-d \right ) ^{2}} \left ( -1/4\,{\frac{ \left ( 2\,c-d \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{3}}{{c}^{2}+2\,cd+{d}^{2}}}+1/4\,{\frac{ \left ( 2\,c-3\,d \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{ \left ( c+d \right ) \left ( c-d \right ) }} \right ) }+1/4\,{\frac{2\,c-d}{ \left ({c}^{3}+{c}^{2}d-{d}^{2}c-{d}^{3} \right ) \sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}{\it Artanh} \left ({\frac{\tan \left ( 1/2\,fx+e/2 \right ) \left ( c-d \right ) }{\sqrt{ \left ( c+d \right ) \left ( c-d \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 [B] time = 0.579293, size = 1604, normalized size = 12.24 \begin{align*} \text{result too large to display} \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*} a \left (\int \frac{\sec{\left (e + f x \right )}}{c^{3} + 3 c^{2} d \sec{\left (e + f x \right )} + 3 c d^{2} \sec ^{2}{\left (e + f x \right )} + d^{3} \sec ^{3}{\left (e + f x \right )}}\, dx + \int \frac{\sec ^{2}{\left (e + f x \right )}}{c^{3} + 3 c^{2} d \sec{\left (e + f x \right )} + 3 c d^{2} \sec ^{2}{\left (e + f x \right )} + d^{3} \sec ^{3}{\left (e + f x \right )}}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.39031, size = 370, normalized size = 2.82 \begin{align*} \frac{\frac{{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, c + 2 \, d\right ) + \arctan \left (-\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-c^{2} + d^{2}}}\right )\right )}{\left (2 \, a c - a d\right )}}{{\left (c^{3} + c^{2} d - c d^{2} - d^{3}\right )} \sqrt{-c^{2} + d^{2}}} - \frac{2 \, a c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 3 \, a c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + a d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, a c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 3 \, a d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (c^{3} + c^{2} d - c d^{2} - d^{3}\right )}{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c - d\right )}^{2}}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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