Optimal. Leaf size=372 \[ -\frac{(a-8 b) (a+b) \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} \text{EllipticF}\left (\sin ^{-1}(\sin (e+f x)),\frac{a}{a+b}\right )}{15 b f \left (-a \sin ^2(e+f x)+a+b\right )}-\frac{\left (2 a^2-3 a b-8 b^2\right ) \sin (e+f x) \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{15 b^2 f}+\frac{\left (2 a^2-3 a b-8 b^2\right ) \sqrt{\cos ^2(e+f x)} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{15 b^2 f \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}+\frac{\tan (e+f x) \sec ^3(e+f x) \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{5 f}+\frac{(a+4 b) \tan (e+f x) \sec (e+f x) \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{15 b f} \]
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Rubi [A] time = 0.688479, antiderivative size = 471, normalized size of antiderivative = 1.27, number of steps used = 11, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4148, 6722, 1974, 412, 527, 524, 426, 424, 421, 419} \[ -\frac{\left (2 a^2-3 a b-8 b^2\right ) \sin (e+f x) \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}{15 b^2 f \sqrt{a \cos ^2(e+f x)+b}}+\frac{\left (2 a^2-3 a b-8 b^2\right ) \sqrt{\cos ^2(e+f x)} \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{15 b^2 f \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a \cos ^2(e+f x)+b}}+\frac{\tan (e+f x) \sec ^3(e+f x) \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}{5 f \sqrt{a \cos ^2(e+f x)+b}}+\frac{(a+4 b) \tan (e+f x) \sec (e+f x) \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}{15 b f \sqrt{a \cos ^2(e+f x)+b}}-\frac{(a-8 b) (a+b) \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a+b \sec ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{15 b f \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a \cos ^2(e+f x)+b}} \]
Antiderivative was successfully verified.
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Rule 4148
Rule 6722
Rule 1974
Rule 412
Rule 527
Rule 524
Rule 426
Rule 424
Rule 421
Rule 419
Rubi steps
\begin{align*} \int \sec ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+\frac{b}{1-x^2}}}{\left (1-x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\left (\sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b+a \left (1-x^2\right )}}{\left (1-x^2\right )^{7/2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{b+a \cos ^2(e+f x)}}\\ &=\frac{\left (\sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b-a x^2}}{\left (1-x^2\right )^{7/2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{b+a \cos ^2(e+f x)}}\\ &=\frac{\sec ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt{b+a \cos ^2(e+f x)}}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{-4 (a+b)+3 a x^2}{\left (1-x^2\right )^{5/2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{5 f \sqrt{b+a \cos ^2(e+f x)}}\\ &=\frac{(a+4 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 b f \sqrt{b+a \cos ^2(e+f x)}}+\frac{\sec ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt{b+a \cos ^2(e+f x)}}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{(a-8 b) (a+b)+a (a+4 b) x^2}{\left (1-x^2\right )^{3/2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{15 b f \sqrt{b+a \cos ^2(e+f x)}}\\ &=-\frac{\left (2 a^2-3 a b-8 b^2\right ) \sqrt{a+b \sec ^2(e+f x)} \sin (e+f x) \sqrt{a+b-a \sin ^2(e+f x)}}{15 b^2 f \sqrt{b+a \cos ^2(e+f x)}}+\frac{(a+4 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 b f \sqrt{b+a \cos ^2(e+f x)}}+\frac{\sec ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt{b+a \cos ^2(e+f x)}}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{-2 a (a-2 b) (a+b)+a \left (2 a^2-3 a b-8 b^2\right ) x^2}{\sqrt{1-x^2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f \sqrt{b+a \cos ^2(e+f x)}}\\ &=-\frac{\left (2 a^2-3 a b-8 b^2\right ) \sqrt{a+b \sec ^2(e+f x)} \sin (e+f x) \sqrt{a+b-a \sin ^2(e+f x)}}{15 b^2 f \sqrt{b+a \cos ^2(e+f x)}}+\frac{(a+4 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 b f \sqrt{b+a \cos ^2(e+f x)}}+\frac{\sec ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt{b+a \cos ^2(e+f x)}}-\frac{\left ((a-8 b) (a+b) \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{15 b f \sqrt{b+a \cos ^2(e+f x)}}-\frac{\left (\left (-2 a^2+3 a b+8 b^2\right ) \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^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 )}{15 b^2 f \sqrt{b+a \cos ^2(e+f x)}}\\ &=-\frac{\left (2 a^2-3 a b-8 b^2\right ) \sqrt{a+b \sec ^2(e+f x)} \sin (e+f x) \sqrt{a+b-a \sin ^2(e+f x)}}{15 b^2 f \sqrt{b+a \cos ^2(e+f x)}}+\frac{(a+4 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 b f \sqrt{b+a \cos ^2(e+f x)}}+\frac{\sec ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt{b+a \cos ^2(e+f x)}}-\frac{\left (\left (-2 a^2+3 a b+8 b^2\right ) \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^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 )}{15 b^2 f \sqrt{b+a \cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}-\frac{\left ((a-8 b) (a+b) \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}}\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 )}{15 b f \sqrt{b+a \cos ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}\\ &=-\frac{\left (2 a^2-3 a b-8 b^2\right ) \sqrt{a+b \sec ^2(e+f x)} \sin (e+f x) \sqrt{a+b-a \sin ^2(e+f x)}}{15 b^2 f \sqrt{b+a \cos ^2(e+f x)}}+\frac{\left (2 a^2-3 a b-8 b^2\right ) \sqrt{\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right ) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}{15 b^2 f \sqrt{b+a \cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}-\frac{(a-8 b) (a+b) \sqrt{\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right ) \sqrt{a+b \sec ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}{15 b f \sqrt{b+a \cos ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}+\frac{(a+4 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 b f \sqrt{b+a \cos ^2(e+f x)}}+\frac{\sec ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt{b+a \cos ^2(e+f x)}}\\ \end{align*}
Mathematica [F] time = 26.6862, size = 0, normalized size = 0. \[ \int \sec ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)} \, dx \]
Verification is Not applicable to the result.
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Maple [C] time = 0.766, size = 6562, normalized size = 17.6 \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 \sqrt{b \sec \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{5}\,{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 (\sqrt{b \sec \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{5}, 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 \sqrt{a + b \sec ^{2}{\left (e + f x \right )}} \sec ^{5}{\left (e + f x \right )}\, 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 \sqrt{b \sec \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{5}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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