Optimal. Leaf size=321 \[ -\frac{\sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \text{EllipticF}\left (\sin ^{-1}(\sin (e+f x)),\frac{a}{a+b}\right )}{3 b f (a+b) \sqrt{\cos ^2(e+f x)} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}-\frac{2 a (a+2 b) \sin (e+f x)}{3 b^2 f (a+b)^2 \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}+\frac{2 (a+2 b) \left (-a \sin ^2(e+f x)+a+b\right ) E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{3 b^2 f (a+b)^2 \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}-\frac{a \sin (e+f x)}{3 b f (a+b) \left (-a \sin ^2(e+f x)+a+b\right ) \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}} \]
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Rubi [A] time = 0.636959, antiderivative size = 383, normalized size of antiderivative = 1.19, 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} \[ -\frac{2 a (a+2 b) \sin (e+f x) \sqrt{a \cos ^2(e+f x)+b}}{3 b^2 f (a+b)^2 \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}+\frac{2 (a+2 b) \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a \cos ^2(e+f x)+b} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{3 b^2 f (a+b)^2 \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a+b \sec ^2(e+f x)}}-\frac{a \sin (e+f x) \sqrt{a \cos ^2(e+f x)+b}}{3 b f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^{3/2} \sqrt{a+b \sec ^2(e+f x)}}-\frac{\sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a \cos ^2(e+f x)+b} F\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{3 b f (a+b) \sqrt{\cos ^2(e+f x)} \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}} \]
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 )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^3 \left (a+\frac{b}{1-x^2}\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\sqrt{b+a \cos ^2(e+f x)} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \left (b+a \left (1-x^2\right )\right )^{5/2}} \, 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}{\sqrt{1-x^2} \left (a+b-a x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=-\frac{a \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 b (a+b) f \sqrt{a+b \sec ^2(e+f x)} \left (a+b-a \sin ^2(e+f x)\right )^{3/2}}-\frac{\sqrt{b+a \cos ^2(e+f x)} \operatorname{Subst}\left (\int \frac{-a-3 b-a x^2}{\sqrt{1-x^2} \left (a+b-a x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 b (a+b) f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=-\frac{a \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 b (a+b) f \sqrt{a+b \sec ^2(e+f x)} \left (a+b-a \sin ^2(e+f x)\right )^{3/2}}-\frac{2 a (a+2 b) \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 b^2 (a+b)^2 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+b) (2 a+3 b)-2 a (a+2 b) x^2}{\sqrt{1-x^2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{3 b^2 (a+b)^2 f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=-\frac{a \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 b (a+b) f \sqrt{a+b \sec ^2(e+f x)} \left (a+b-a \sin ^2(e+f x)\right )^{3/2}}-\frac{2 a (a+2 b) \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 b^2 (a+b)^2 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 )}{3 b (a+b) f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}+\frac{\left (2 (a+2 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 )}{3 b^2 (a+b)^2 f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=-\frac{a \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 b (a+b) f \sqrt{a+b \sec ^2(e+f x)} \left (a+b-a \sin ^2(e+f x)\right )^{3/2}}-\frac{2 a (a+2 b) \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 b^2 (a+b)^2 f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}+\frac{\left (2 (a+2 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 )}{3 b^2 (a+b)^2 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 )}{3 b (a+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 \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 b (a+b) f \sqrt{a+b \sec ^2(e+f x)} \left (a+b-a \sin ^2(e+f x)\right )^{3/2}}-\frac{2 a (a+2 b) \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 b^2 (a+b)^2 f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}+\frac{2 (a+2 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)}}{3 b^2 (a+b)^2 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}}}{3 b (a+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)}}\\ \end{align*}
Mathematica [A] time = 2.81984, size = 167, normalized size = 0.52 \[ \frac{\sec ^5(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\sqrt{2} (a+b)^2 \left (\frac{a \cos (2 (e+f x))+a+2 b}{a+b}\right )^{3/2} \left (2 (a+2 b) E\left (e+f x\left |\frac{a}{a+b}\right .\right )-b \text{EllipticF}\left (e+f x,\frac{a}{a+b}\right )\right )-2 a \sin (2 (e+f x)) \left (a^2+a (a+2 b) \cos (2 (e+f x))+5 a b+5 b^2\right )\right )}{24 b^2 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.614, size = 14357, normalized size = 44.7 \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{5}{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^{3} \sec \left (f x + e\right )^{6} + 3 \, a b^{2} \sec \left (f x + e\right )^{4} + 3 \, a^{2} b \sec \left (f x + e\right )^{2} + a^{3}}, 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{5}{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{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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