Optimal. Leaf size=115 \[ \frac{a^2}{3 b^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{a (a-2 b)}{b^2 f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{5/2}} \]
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Rubi [A] time = 0.203918, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3670, 446, 87, 63, 208} \[ \frac{a^2}{3 b^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{a (a-2 b)}{b^2 f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3670
Rule 446
Rule 87
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{(1+x) (a+b x)^{5/2}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a^2}{(a-b) b (a+b x)^{5/2}}+\frac{a (a-2 b)}{(a-b)^2 b (a+b x)^{3/2}}+\frac{1}{(a-b)^2 (1+x) \sqrt{a+b x}}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{a^2}{3 (a-b) b^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{a (a-2 b)}{(a-b)^2 b^2 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 (a-b)^2 f}\\ &=\frac{a^2}{3 (a-b) b^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{a (a-2 b)}{(a-b)^2 b^2 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \tan ^2(e+f x)}\right )}{(a-b)^2 b f}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{(a-b)^{5/2} f}+\frac{a^2}{3 (a-b) b^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{a (a-2 b)}{(a-b)^2 b^2 f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}
Mathematica [C] time = 0.452866, size = 91, normalized size = 0.79 \[ \frac{b^2 \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{a+b \tan ^2(e+f x)}{a-b}\right )-(a-b) \left (2 a+3 b \tan ^2(e+f x)-b\right )}{3 b^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.027, size = 169, normalized size = 1.5 \begin{align*} -{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{fb} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,a}{3\,f{b}^{2}} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{1}{3\,fb} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{1}{ \left ( a-b \right ) ^{2}f}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}+{\frac{1}{ \left ( a-b \right ) ^{2}f}\arctan \left ({\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}+{\frac{1}{ \left ( 3\,a-3\,b \right ) f} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.50223, size = 1338, normalized size = 11.63 \begin{align*} \left [\frac{3 \,{\left (b^{4} \tan \left (f x + e\right )^{4} + 2 \, a b^{3} \tan \left (f x + e\right )^{2} + a^{2} b^{2}\right )} \sqrt{a - b} \log \left (-\frac{b^{2} \tan \left (f x + e\right )^{4} + 2 \,{\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \,{\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) - 4 \,{\left (2 \, a^{4} - 7 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \,{\left (a^{3} b - 3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{12 \,{\left ({\left (a^{3} b^{4} - 3 \, a^{2} b^{5} + 3 \, a b^{6} - b^{7}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b^{3} - 3 \, a^{3} b^{4} + 3 \, a^{2} b^{5} - a b^{6}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f\right )}}, \frac{3 \,{\left (b^{4} \tan \left (f x + e\right )^{4} + 2 \, a b^{3} \tan \left (f x + e\right )^{2} + a^{2} b^{2}\right )} \sqrt{-a + b} \arctan \left (\frac{2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) - 2 \,{\left (2 \, a^{4} - 7 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \,{\left (a^{3} b - 3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{6 \,{\left ({\left (a^{3} b^{4} - 3 \, a^{2} b^{5} + 3 \, a b^{6} - b^{7}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b^{3} - 3 \, a^{3} b^{4} + 3 \, a^{2} b^{5} - a b^{6}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} f\right )}}\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{\tan ^{5}{\left (e + f x \right )}}{\left (a + b \tan ^{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 [A] time = 1.49219, size = 185, normalized size = 1.61 \begin{align*} \frac{\arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{-a + b}}\right )}{{\left (a^{2} f - 2 \, a b f + b^{2} f\right )} \sqrt{-a + b}} - \frac{3 \,{\left (b \tan \left (f x + e\right )^{2} + a\right )} a^{2} - a^{3} - 6 \,{\left (b \tan \left (f x + e\right )^{2} + a\right )} a b + a^{2} b}{3 \,{\left (a^{2} b^{2} f - 2 \, a b^{3} f + b^{4} f\right )}{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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