Optimal. Leaf size=99 \[ \frac{1}{f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}+\frac{1}{3 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{5/2}} \]
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Rubi [A] time = 0.107685, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3670, 444, 51, 63, 208} \[ \frac{1}{f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}+\frac{1}{3 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3670
Rule 444
Rule 51
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x}{\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{1}{(1+x) (a+b x)^{5/2}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{1}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) (a+b x)^{3/2}} \, dx,x,\tan ^2(e+f x)\right )}{2 (a-b) f}\\ &=\frac{1}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{1}{(a-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{1}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{1}{(a-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{1}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{1}{(a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}
Mathematica [C] time = 0.16327, size = 58, normalized size = 0.59 \[ \frac{\text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{a+b \tan ^2(e+f x)}{a-b}\right )}{3 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.016, size = 94, normalized size = 1. \begin{align*}{\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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )}{{\left (b \tan \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 [B] time = 2.51089, size = 1230, normalized size = 12.42 \begin{align*} \left [\frac{3 \,{\left (b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{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 (3 \,{\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + 4 \, a^{2} - 5 \, a b + b^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{12 \,{\left ({\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} f\right )}}, \frac{3 \,{\left (b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{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 (3 \,{\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + 4 \, a^{2} - 5 \, a b + b^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{6 \,{\left ({\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} f\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 19.7342, size = 83, normalized size = 0.84 \begin{align*} \frac{1}{3 f \left (a - b\right ) \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}} + \frac{1}{f \left (a - b\right )^{2} \sqrt{a + b \tan ^{2}{\left (e + f x \right )}}} + \frac{\operatorname{atan}{\left (\frac{\sqrt{a + b \tan ^{2}{\left (e + f x \right )}}}{\sqrt{- a + b}} \right )}}{f \sqrt{- a + b} \left (a - b\right )^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.34103, size = 142, normalized size = 1.43 \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 \, b \tan \left (f x + e\right )^{2} + 4 \, a - b}{3 \,{\left (a^{2} f - 2 \, a b f + b^{2} 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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