Optimal. Leaf size=231 \[ -\frac{2 c^{3/2} g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \sqrt{g} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} d f (c-d) \sqrt{c+d}}+\frac{\sqrt{2} g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} f (c-d)}+\frac{2 g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} d f} \]
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Rubi [A] time = 0.815862, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3978, 3965, 208, 4023, 3808, 3802} \[ -\frac{2 c^{3/2} g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \sqrt{g} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} d f (c-d) \sqrt{c+d}}+\frac{\sqrt{2} g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} f (c-d)}+\frac{2 g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} d f} \]
Antiderivative was successfully verified.
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Rule 3978
Rule 3965
Rule 208
Rule 4023
Rule 3808
Rule 3802
Rubi steps
\begin{align*} \int \frac{(g \sec (e+f x))^{5/2}}{\sqrt{a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx &=\frac{g^2 \int \frac{\sqrt{g \sec (e+f x)} (a c+(a c-a d) \sec (e+f x))}{\sqrt{a+a \sec (e+f x)}} \, dx}{a (c-d) d}-\frac{\left (c^2 g^2\right ) \int \frac{\sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx}{a (c-d) d}\\ &=\frac{g^2 \int \frac{\sqrt{g \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}} \, dx}{c-d}+\frac{g^2 \int \sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)} \, dx}{a d}+\frac{\left (2 c^2 g^3\right ) \operatorname{Subst}\left (\int \frac{1}{a c+a d-c g x^2} \, dx,x,-\frac{a \tan (e+f x)}{\sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\right )}{(c-d) d f}\\ &=-\frac{2 c^{3/2} g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \sqrt{g} \tan (e+f x)}{\sqrt{c+d} \sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{a} (c-d) d \sqrt{c+d} f}-\frac{\left (2 g^3\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-g x^2} \, dx,x,-\frac{a \tan (e+f x)}{\sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\right )}{(c-d) f}-\frac{\left (2 g^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-g x^2} \, dx,x,-\frac{a \tan (e+f x)}{\sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\right )}{d f}\\ &=\frac{2 g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{a} d f}+\frac{\sqrt{2} g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{2} \sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{a} (c-d) f}-\frac{2 c^{3/2} g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \sqrt{g} \tan (e+f x)}{\sqrt{c+d} \sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{a} (c-d) d \sqrt{c+d} f}\\ \end{align*}
Mathematica [A] time = 0.391366, size = 155, normalized size = 0.67 \[ \frac{2 g^2 \cos \left (\frac{1}{2} (e+f x)\right ) \sqrt{g \sec (e+f x)} \left (\sqrt{2} \left ((c-d) \sqrt{c+d} \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (e+f x)\right )\right )-c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sin \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )\right )+d \sqrt{c+d} \tanh ^{-1}\left (\sin \left (\frac{1}{2} (e+f x)\right )\right )\right )}{d f (c-d) \sqrt{c+d} \sqrt{a (\sec (e+f x)+1)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.346, size = 726, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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