Optimal. Leaf size=68 \[ \frac{d (2 c-d) \tanh ^{-1}(\sin (e+f x))}{a f}+\frac{(c-d)^2 \tan (e+f x)}{f (a \sec (e+f x)+a)}+\frac{d^2 \tan (e+f x)}{a f} \]
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Rubi [A] time = 0.14553, antiderivative size = 125, normalized size of antiderivative = 1.84, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3987, 89, 80, 63, 217, 203} \[ \frac{2 d (2 c-d) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a (\sec (e+f x)+1)}}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{(c-d)^2 \tan (e+f x)}{f (a \sec (e+f x)+a)}+\frac{d^2 \tan (e+f x)}{a f} \]
Antiderivative was successfully verified.
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Rule 3987
Rule 89
Rule 80
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \frac{\sec (e+f x) (c+d \sec (e+f x))^2}{a+a \sec (e+f x)} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^2}{\sqrt{a-a x} (a+a x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d)^2 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{a^3 (2 c-d) d+a^3 d^2 x}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{d^2 \tan (e+f x)}{a f}+\frac{(c-d)^2 \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{(a (2 c-d) d \tan (e+f x)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a 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{d^2 \tan (e+f x)}{a f}+\frac{(c-d)^2 \tan (e+f x)}{f (a+a \sec (e+f x))}+\frac{(2 (2 c-d) d \tan (e+f x)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 a-x^2}} \, 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{d^2 \tan (e+f x)}{a f}+\frac{(c-d)^2 \tan (e+f x)}{f (a+a \sec (e+f x))}+\frac{(2 (2 c-d) d \tan (e+f x)) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt{a-a \sec (e+f 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{d^2 \tan (e+f x)}{a f}+\frac{(c-d)^2 \tan (e+f x)}{f (a+a \sec (e+f x))}+\frac{2 (2 c-d) d \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}}\right ) \tan (e+f x)}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ \end{align*}
Mathematica [B] time = 1.56551, size = 237, normalized size = 3.49 \[ \frac{2 \cos \left (\frac{1}{2} (e+f x)\right ) \cos (e+f x) (c+d \sec (e+f x))^2 \left ((c-d)^2 \sec \left (\frac{e}{2}\right ) \sin \left (\frac{f x}{2}\right )+d \cos \left (\frac{1}{2} (e+f x)\right ) \left (\frac{d \sin (f x)}{\left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}-(2 c-d) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )\right )\right )}{a f (\sec (e+f x)+1) (c \cos (e+f x)+d)^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.051, size = 196, normalized size = 2.9 \begin{align*}{\frac{{c}^{2}}{fa}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }-2\,{\frac{cd\tan \left ( 1/2\,fx+e/2 \right ) }{fa}}+{\frac{{d}^{2}}{fa}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) }-{\frac{{d}^{2}}{fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}+2\,{\frac{d\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) c}{fa}}-{\frac{{d}^{2}}{fa}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{{d}^{2}}{fa} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{d\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) c}{fa}}+{\frac{{d}^{2}}{fa}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 0.968144, size = 301, normalized size = 4.43 \begin{align*} -\frac{d^{2}{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac{2 \, \sin \left (f x + e\right )}{{\left (a - \frac{a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}} - \frac{\sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - 2 \, c d{\left (\frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a} - \frac{\sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}\right )} - \frac{c^{2} \sin \left (f x + e\right )}{a{\left (\cos \left (f x + e\right ) + 1\right )}}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 0.480677, size = 370, normalized size = 5.44 \begin{align*} \frac{{\left ({\left (2 \, c d - d^{2}\right )} \cos \left (f x + e\right )^{2} +{\left (2 \, c d - d^{2}\right )} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left ({\left (2 \, c d - d^{2}\right )} \cos \left (f x + e\right )^{2} +{\left (2 \, c d - d^{2}\right )} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (d^{2} +{\left (c^{2} - 2 \, c d + 2 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \,{\left (a f \cos \left (f x + e\right )^{2} + a 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*} \frac{\int \frac{c^{2} \sec{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx + \int \frac{2 c d \sec ^{2}{\left (e + f x \right )}}{\sec{\left (e + f x \right )} + 1}\, dx}{a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.29584, size = 193, normalized size = 2.84 \begin{align*} \frac{\frac{{\left (2 \, c d - d^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a} - \frac{{\left (2 \, c d - d^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a} - \frac{2 \, d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} a} + \frac{c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a}}{f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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