Optimal. Leaf size=289 \[ \frac{\sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \text{EllipticF}\left (\sin ^{-1}(\sin (e+f x)),\frac{a}{a+b}\right )}{b f \sqrt{\cos ^2(e+f x)} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}+\frac{a (2 a+b) \sin (e+f x)}{b^2 f (a+b) \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}-\frac{(2 a+b) \left (-a \sin ^2(e+f x)+a+b\right ) E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{b^2 f (a+b) \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}+\frac{\tan (e+f x) \sec (e+f x)}{b f \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}} \]
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Rubi [A] time = 0.590233, antiderivative size = 367, normalized size of antiderivative = 1.27, number of steps used = 10, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4148, 6722, 1974, 414, 527, 524, 426, 424, 421, 419} \[ \frac{a (2 a+b) \sin (e+f x) \sqrt{a \cos ^2(e+f x)+b}}{b^2 f (a+b) \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}-\frac{(2 a+b) \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a \cos ^2(e+f x)+b} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{b^2 f (a+b) \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a+b \sec ^2(e+f x)}}+\frac{\tan (e+f x) \sec (e+f x) \sqrt{a \cos ^2(e+f x)+b}}{b f \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}+\frac{\sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a \cos ^2(e+f x)+b} F\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{b f \sqrt{\cos ^2(e+f x)} \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}} \]
Antiderivative was successfully verified.
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Rule 4148
Rule 6722
Rule 1974
Rule 414
Rule 527
Rule 524
Rule 426
Rule 424
Rule 421
Rule 419
Rubi steps
\begin{align*} \int \frac{\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^3 \left (a+\frac{b}{1-x^2}\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\sqrt{b+a \cos ^2(e+f x)} \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^{3/2} \left (b+a \left (1-x^2\right )\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=\frac{\sqrt{b+a \cos ^2(e+f x)} \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^{3/2} \left (a+b-a x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=\frac{\sqrt{b+a \cos ^2(e+f x)} \sec (e+f x) \tan (e+f x)}{b f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}+\frac{\sqrt{b+a \cos ^2(e+f x)} \operatorname{Subst}\left (\int \frac{-a-a x^2}{\sqrt{1-x^2} \left (a+b-a x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{b f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=\frac{a (2 a+b) \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{b^2 (a+b) f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}+\frac{\sqrt{b+a \cos ^2(e+f x)} \sec (e+f x) \tan (e+f x)}{b f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}-\frac{\sqrt{b+a \cos ^2(e+f x)} \operatorname{Subst}\left (\int \frac{2 a (a+b)-a (2 a+b) x^2}{\sqrt{1-x^2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{b^2 (a+b) f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=\frac{a (2 a+b) \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{b^2 (a+b) f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}+\frac{\sqrt{b+a \cos ^2(e+f x)} \sec (e+f x) \tan (e+f x)}{b f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}+\frac{\sqrt{b+a \cos ^2(e+f x)} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{b f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}-\frac{\left ((2 a+b) \sqrt{b+a \cos ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b-a x^2}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{b^2 (a+b) f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=\frac{a (2 a+b) \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{b^2 (a+b) f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}+\frac{\sqrt{b+a \cos ^2(e+f x)} \sec (e+f x) \tan (e+f x)}{b f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}-\frac{\left ((2 a+b) \sqrt{b+a \cos ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-\frac{a x^2}{a+b}}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{b^2 (a+b) f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}+\frac{\left (\sqrt{b+a \cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1-\frac{a x^2}{a+b}}} \, dx,x,\sin (e+f x)\right )}{b f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}\\ &=\frac{a (2 a+b) \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{b^2 (a+b) f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}-\frac{(2 a+b) \sqrt{b+a \cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right ) \sqrt{a+b-a \sin ^2(e+f x)}}{b^2 (a+b) f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}+\frac{\sqrt{b+a \cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right ) \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}{b f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}+\frac{\sqrt{b+a \cos ^2(e+f x)} \sec (e+f x) \tan (e+f x)}{b f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}\\ \end{align*}
Mathematica [F] time = 20.1776, size = 0, normalized size = 0. \[ \int \frac{\sec ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx \]
Verification is Not applicable to the result.
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Maple [C] time = 0.65, size = 12514, normalized size = 43.3 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )^{5}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sec \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{5}}{b^{2} \sec \left (f x + e\right )^{4} + 2 \, a b \sec \left (f x + e\right )^{2} + a^{2}}, x\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 ^{5}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (f x + e\right )^{5}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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