Optimal. Leaf size=319 \[ \frac{\sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \text{EllipticF}\left (\sin ^{-1}(\sin (e+f x)),\frac{a}{a+b}\right )}{3 a f (a+b) \sqrt{\cos ^2(e+f x)} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}-\frac{(a-b) \sin (e+f x)}{3 b f (a+b)^2 \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}+\frac{\sin (e+f x)}{3 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right ) \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}+\frac{(a-b) \left (-a \sin ^2(e+f x)+a+b\right ) E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{3 a b f (a+b)^2 \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}} \]
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Rubi [A] time = 0.564733, antiderivative size = 381, 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, 412, 527, 524, 426, 424, 421, 419} \[ -\frac{(a-b) \sin (e+f x) \sqrt{a \cos ^2(e+f x)+b}}{3 b f (a+b)^2 \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}+\frac{\sin (e+f x) \sqrt{a \cos ^2(e+f x)+b}}{3 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^{3/2} \sqrt{a+b \sec ^2(e+f x)}}+\frac{\sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a \cos ^2(e+f x)+b} F\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{3 a f (a+b) \sqrt{\cos ^2(e+f x)} \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}+\frac{(a-b) \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a \cos ^2(e+f x)+b} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{3 a b f (a+b)^2 \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a+b \sec ^2(e+f x)}} \]
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 \frac{\sec ^3(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 )^2 \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{\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{\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{\sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 (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{-2+x^2}{\sqrt{1-x^2} \left (a+b-a x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b) f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=\frac{\sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 (a+b) f \sqrt{a+b \sec ^2(e+f x)} \left (a+b-a \sin ^2(e+f x)\right )^{3/2}}-\frac{(a-b) \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 b (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+(-a+b) x^2}{\sqrt{1-x^2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{3 b (a+b)^2 f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=\frac{\sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 (a+b) f \sqrt{a+b \sec ^2(e+f x)} \left (a+b-a \sin ^2(e+f x)\right )^{3/2}}-\frac{(a-b) \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 b (a+b)^2 f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}-\frac{\left ((-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 )}{3 a b (a+b)^2 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} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{3 a (a+b) f \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}}\\ &=\frac{\sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 (a+b) f \sqrt{a+b \sec ^2(e+f x)} \left (a+b-a \sin ^2(e+f x)\right )^{3/2}}-\frac{(a-b) \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 b (a+b)^2 f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}-\frac{\left ((-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 )}{3 a b (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 a (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{\sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 (a+b) f \sqrt{a+b \sec ^2(e+f x)} \left (a+b-a \sin ^2(e+f x)\right )^{3/2}}-\frac{(a-b) \sqrt{b+a \cos ^2(e+f x)} \sin (e+f x)}{3 b (a+b)^2 f \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}+\frac{(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)}}{3 a b (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 a (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 [C] time = 10.2832, size = 1156, normalized size = 3.62 \[ \frac{(\cos (2 e+2 f x) a+a+2 b)^{5/2} \sec ^5(e+f x) \left (\frac{\cos (2 (e+f x)) \left (2 i b \left (a^2+b a+b^2\right ) \sqrt{\frac{a-a \cos (2 e+2 f x)}{a+b}} \sqrt{4-\frac{2 (\cos (2 e+2 f x) a+a+2 b)}{b}} E\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{1}{a+b}} \sqrt{\cos (2 e+2 f x) a+a+2 b}}{\sqrt{2}}\right )|\frac{a+b}{b}\right ) (\cos (2 e+2 f x) a+a+2 b)^{3/2}+i a b (b-a) \sqrt{\frac{a-a \cos (2 e+2 f x)}{a+b}} \sqrt{4-\frac{2 (\cos (2 e+2 f x) a+a+2 b)}{b}} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{1}{a+b}} \sqrt{\cos (2 e+2 f x) a+a+2 b}}{\sqrt{2}}\right ),\frac{a+b}{b}\right ) (\cos (2 e+2 f x) a+a+2 b)^{3/2}-2 \sqrt{-\frac{1}{a+b}} (-\cos (2 e+2 f x) a-a) \left (4 b^4-(\cos (2 e+2 f x) a+a+2 b)^2 b^2+a \left (10 b^2+(\cos (2 e+2 f x) a+a+2 b) b-(\cos (2 e+2 f x) a+a+2 b)^2\right ) b+2 a^3 (\cos (2 e+2 f x) a+a+3 b)+a^2 \left (8 b^2+3 (\cos (2 e+2 f x) a+a+2 b) b-(\cos (2 e+2 f x) a+a+2 b)^2\right )\right )\right ) \sec \left (2 \left (e+\frac{1}{2} \left (\cos ^{-1}(\cos (2 e+2 f x))-2 e\right )\right )\right ) \sin (2 e+2 f x)}{24 a^2 b^2 \sqrt{-\frac{1}{a+b}} (a+b)^2 f \sqrt{\frac{(a-a \cos (2 e+2 f x)) (\cos (2 e+2 f x) a+a)}{a^2}} (\cos (2 e+2 f x) a+a+2 b)^{3/2} \sqrt{1-\cos ^2(2 e+2 f x)}}-\frac{\left (2 i b (a+2 b) \sqrt{\frac{a-a \cos (2 e+2 f x)}{a+b}} \sqrt{4-\frac{2 (\cos (2 e+2 f x) a+a+2 b)}{b}} E\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{1}{a+b}} \sqrt{\cos (2 e+2 f x) a+a+2 b}}{\sqrt{2}}\right )|\frac{a+b}{b}\right ) (\cos (2 e+2 f x) a+a+2 b)^{3/2}-i b (a+3 b) \sqrt{\frac{a-a \cos (2 e+2 f x)}{a+b}} \sqrt{4-\frac{2 (\cos (2 e+2 f x) a+a+2 b)}{b}} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{1}{a+b}} \sqrt{\cos (2 e+2 f x) a+a+2 b}}{\sqrt{2}}\right ),\frac{a+b}{b}\right ) (\cos (2 e+2 f x) a+a+2 b)^{3/2}-2 \sqrt{-\frac{1}{a+b}} (-\cos (2 e+2 f x) a-a) \left (2 (\cos (2 e+2 f x) a+a+3 b) a^2+\left (4 b^2+5 (\cos (2 e+2 f x) a+a+2 b) b-(\cos (2 e+2 f x) a+a+2 b)^2\right ) a+b \left (2 b^2+3 (\cos (2 e+2 f x) a+a+2 b) b-2 (\cos (2 e+2 f x) a+a+2 b)^2\right )\right )\right ) \sin (2 e+2 f x)}{24 a b^2 \sqrt{-\frac{1}{a+b}} (a+b)^2 f \sqrt{\frac{(a-a \cos (2 e+2 f x)) (\cos (2 e+2 f x) a+a)}{a^2}} (\cos (2 e+2 f x) a+a+2 b)^{3/2} \sqrt{1-\cos ^2(2 e+2 f x)}}\right )}{2 \left (b \sec ^2(e+f x)+a\right )^{5/2}} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.604, size = 10271, normalized size = 32.2 \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 )^{3}}{{\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 )^{3}}{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 ^{3}{\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 )^{3}}{{\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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