Optimal. Leaf size=65 \[ \frac{a^2}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac{2 a}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}+\frac{1}{f \sqrt{a \cos ^2(e+f x)}} \]
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Rubi [A] time = 0.11512, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3176, 3205, 16, 43} \[ \frac{a^2}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac{2 a}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}+\frac{1}{f \sqrt{a \cos ^2(e+f x)}} \]
Antiderivative was successfully verified.
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Rule 3176
Rule 3205
Rule 16
Rule 43
Rubi steps
\begin{align*} \int \frac{\tan ^5(e+f x)}{\sqrt{a-a \sin ^2(e+f x)}} \, dx &=\int \frac{\tan ^5(e+f x)}{\sqrt{a \cos ^2(e+f x)}} \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(1-x)^2}{x^3 \sqrt{a x}} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{a^3 \operatorname{Subst}\left (\int \frac{(1-x)^2}{(a x)^{7/2}} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{a^3 \operatorname{Subst}\left (\int \left (\frac{1}{(a x)^{7/2}}-\frac{2}{a (a x)^{5/2}}+\frac{1}{a^2 (a x)^{3/2}}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac{a^2}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac{2 a}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}+\frac{1}{f \sqrt{a \cos ^2(e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.0811033, size = 43, normalized size = 0.66 \[ \frac{3 \sec ^4(e+f x)-10 \sec ^2(e+f x)+15}{15 f \sqrt{a \cos ^2(e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 1.846, size = 51, normalized size = 0.8 \begin{align*}{\frac{15\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-10\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+3}{15\,a \left ( \cos \left ( fx+e \right ) \right ) ^{6}f}\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.03155, size = 93, normalized size = 1.43 \begin{align*} \frac{15 \,{\left (a \sin \left (f x + e\right )^{2} - a\right )}^{2} a^{3} + 10 \,{\left (a \sin \left (f x + e\right )^{2} - a\right )} a^{4} + 3 \, a^{5}}{15 \,{\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}} a^{3} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.61127, size = 127, normalized size = 1.95 \begin{align*} \frac{{\left (15 \, \cos \left (f x + e\right )^{4} - 10 \, \cos \left (f x + e\right )^{2} + 3\right )} \sqrt{a \cos \left (f x + e\right )^{2}}}{15 \, a f \cos \left (f x + e\right )^{6}} \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{\tan ^{5}{\left (e + f x \right )}}{\sqrt{- a \left (\sin{\left (e + f x \right )} - 1\right ) \left (\sin{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 2.91545, size = 96, normalized size = 1.48 \begin{align*} \frac{16 \,{\left (10 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 5 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}}{15 \,{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{5} \sqrt{a} f \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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