Optimal. Leaf size=145 \[ \frac{\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}-\frac{(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}+\frac{(a-b) \sqrt{a+b \tan ^2(e+f x)}}{f}-\frac{(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f} \]
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Rubi [A] time = 0.174963, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3670, 446, 88, 50, 63, 208} \[ \frac{\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}-\frac{(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}+\frac{(a-b) \sqrt{a+b \tan ^2(e+f x)}}{f}-\frac{(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f} \]
Antiderivative was successfully verified.
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Rule 3670
Rule 446
Rule 88
Rule 50
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \tan ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5 \left (a+b x^2\right )^{3/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2 (a+b x)^{3/2}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{(-a-b) (a+b x)^{3/2}}{b}+\frac{(a+b x)^{3/2}}{1+x}+\frac{(a+b x)^{5/2}}{b}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac{(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}+\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}+\frac{(a-b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{(a-b) \sqrt{a+b \tan ^2(e+f x)}}{f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}+\frac{(a-b)^2 \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{(a-b) \sqrt{a+b \tan ^2(e+f x)}}{f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}+\frac{(a-b)^2 \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 )}{b f}\\ &=-\frac{(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f}+\frac{(a-b) \sqrt{a+b \tan ^2(e+f x)}}{f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2 f}\\ \end{align*}
Mathematica [A] time = 1.33256, size = 139, normalized size = 0.96 \[ \frac{\frac{2 \left (a+b \tan ^2(e+f x)\right )^{7/2}}{7 b^2}-\frac{2 (a+b) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b^2}+\frac{2}{3} \left (a+b \tan ^2(e+f x)\right )^{3/2}+2 (a-b) \left (\sqrt{a+b \tan ^2(e+f x)}-\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )\right )}{2 f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.026, size = 256, normalized size = 1.8 \begin{align*}{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{7\,fb} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{2\,a}{35\,f{b}^{2}} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{1}{5\,fb} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{3\,f}\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}+{\frac{4\,a}{3\,f}\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}-{\frac{b}{f}\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}+{\frac{{b}^{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}}}}-2\,{\frac{ab}{f\sqrt{-a+b}}\arctan \left ({\frac{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}{\sqrt{-a+b}}} \right ) }+{\frac{{a}^{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}}}} \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{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \tan \left (f x + e\right )^{5}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.29907, size = 992, normalized size = 6.84 \begin{align*} \left [-\frac{105 \,{\left (a b^{2} - b^{3}\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 (15 \, b^{3} \tan \left (f x + e\right )^{6} + 3 \,{\left (8 \, a b^{2} - 7 \, b^{3}\right )} \tan \left (f x + e\right )^{4} - 6 \, a^{3} - 21 \, a^{2} b + 140 \, a b^{2} - 105 \, b^{3} +{\left (3 \, a^{2} b - 42 \, a b^{2} + 35 \, b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{420 \, b^{2} f}, \frac{105 \,{\left (a b^{2} - b^{3}\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 (15 \, b^{3} \tan \left (f x + e\right )^{6} + 3 \,{\left (8 \, a b^{2} - 7 \, b^{3}\right )} \tan \left (f x + e\right )^{4} - 6 \, a^{3} - 21 \, a^{2} b + 140 \, a b^{2} - 105 \, b^{3} +{\left (3 \, a^{2} b - 42 \, a b^{2} + 35 \, b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{210 \, b^{2} f}\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 \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}} \tan ^{5}{\left (e + f x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.28983, size = 265, normalized size = 1.83 \begin{align*} \frac{{\left (a^{2} - 2 \, a b + b^{2}\right )} \arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{-a + b}}\right )}{\sqrt{-a + b} f} + \frac{15 \,{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{7}{2}} b^{12} f^{6} - 21 \,{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}} a b^{12} f^{6} - 21 \,{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}} b^{13} f^{6} + 35 \,{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} b^{14} f^{6} + 105 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} a b^{14} f^{6} - 105 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} b^{15} f^{6}}{105 \, b^{14} f^{7}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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