Optimal. Leaf size=103 \[ -\frac{a}{3 b f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{1}{f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{5/2}} \]
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Rubi [A] time = 0.155672, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3670, 446, 78, 51, 63, 208} \[ -\frac{a}{3 b f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{1}{f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3670
Rule 446
Rule 78
Rule 51
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^3}{\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}{(1+x) (a+b x)^{5/2}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac{a}{3 (a-b) 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{a}{3 (a-b) 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{a}{3 (a-b) 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{a}{3 (a-b) 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 = 0.297833, size = 84, normalized size = 0.82 \[ \frac{a (b-a)-3 b \left (a+b \tan ^2(e+f x)\right ) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan ^2(e+f x)}{a-b}\right )}{3 b f (a-b)^2 \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.018, size = 118, normalized size = 1.2 \begin{align*} -{\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 = 2.50577, size = 1269, normalized size = 12.32 \begin{align*} \left [\frac{3 \,{\left (b^{3} \tan \left (f x + e\right )^{4} + 2 \, a b^{2} \tan \left (f x + e\right )^{2} + a^{2} b\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 (a^{3} + a^{2} b - 2 \, a b^{2} + 3 \,{\left (a b^{2} - 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^{3} - 3 \, a^{2} b^{4} + 3 \, a b^{5} - b^{6}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b^{2} - 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} - a b^{5}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{5} b - 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - a^{2} b^{4}\right )} f\right )}}, -\frac{3 \,{\left (b^{3} \tan \left (f x + e\right )^{4} + 2 \, a b^{2} \tan \left (f x + e\right )^{2} + a^{2} b\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 (a^{3} + a^{2} b - 2 \, a b^{2} + 3 \,{\left (a b^{2} - 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^{3} - 3 \, a^{2} b^{4} + 3 \, a b^{5} - b^{6}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b^{2} - 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} - a b^{5}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{5} b - 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - a^{2} b^{4}\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 ^{3}{\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 = 1.44239, size = 157, normalized size = 1.52 \begin{align*} -\frac{\frac{3 \, b \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{a^{2} + 3 \,{\left (b \tan \left (f x + e\right )^{2} + a\right )} b - a b}{{\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}}}}{3 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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