Optimal. Leaf size=125 \[ \frac{(a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}+\frac{b \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 f}+\frac{\sqrt{b} (3 a-2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f} \]
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Rubi [A] time = 0.099861, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {3661, 416, 523, 217, 206, 377, 203} \[ \frac{(a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}+\frac{b \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 f}+\frac{\sqrt{b} (3 a-2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f} \]
Antiderivative was successfully verified.
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Rule 3661
Rule 416
Rule 523
Rule 217
Rule 206
Rule 377
Rule 203
Rubi steps
\begin{align*} \int \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^{3/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{b \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 f}+\frac{\operatorname{Subst}\left (\int \frac{a (2 a-b)+(3 a-2 b) b x^2}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{b \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 f}+\frac{(a-b)^2 \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}+\frac{((3 a-2 b) b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{b \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 f}+\frac{(a-b)^2 \operatorname{Subst}\left (\int \frac{1}{1-(-a+b) x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}+\frac{((3 a-2 b) b) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f}\\ &=\frac{(a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}+\frac{(3 a-2 b) \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f}+\frac{b \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 f}\\ \end{align*}
Mathematica [C] time = 1.33439, size = 233, normalized size = 1.86 \[ \frac{b \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}-i (a-b)^{3/2} \log \left (-\frac{4 i \left (\sqrt{a-b} \sqrt{a+b \tan ^2(e+f x)}+a-i b \tan (e+f x)\right )}{(a-b)^{5/2} (\tan (e+f x)+i)}\right )+i (a-b)^{3/2} \log \left (\frac{4 i \left (\sqrt{a-b} \sqrt{a+b \tan ^2(e+f x)}+a+i b \tan (e+f x)\right )}{(a-b)^{5/2} (\tan (e+f x)-i)}\right )+\sqrt{b} (3 a-2 b) \log \left (\sqrt{b} \sqrt{a+b \tan ^2(e+f x)}+b \tan (e+f x)\right )}{2 f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.026, size = 297, normalized size = 2.4 \begin{align*}{\frac{b\tan \left ( fx+e \right ) }{2\,f}\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}+{\frac{3\,a}{2\,f}\sqrt{b}\ln \left ( \sqrt{b}\tan \left ( fx+e \right ) +\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}} \right ) }-{\frac{1}{f}{b}^{{\frac{3}{2}}}\ln \left ( \sqrt{b}\tan \left ( fx+e \right ) +\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}} \right ) }+{\frac{1}{f \left ( a-b \right ) }\sqrt{{b}^{4} \left ( a-b \right ) }\arctan \left ({ \left ( a-b \right ){b}^{2}\tan \left ( fx+e \right ){\frac{1}{\sqrt{{b}^{4} \left ( a-b \right ) }}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}} \right ) }-2\,{\frac{a\sqrt{{b}^{4} \left ( a-b \right ) }}{fb \left ( a-b \right ) }\arctan \left ({\frac{ \left ( a-b \right ){b}^{2}\tan \left ( fx+e \right ) }{\sqrt{{b}^{4} \left ( a-b \right ) }\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}} \right ) }+{\frac{{a}^{2}}{f{b}^{2} \left ( a-b \right ) }\sqrt{{b}^{4} \left ( a-b \right ) }\arctan \left ({ \left ( a-b \right ){b}^{2}\tan \left ( fx+e \right ){\frac{1}{\sqrt{{b}^{4} \left ( a-b \right ) }}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}} \right ) } \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}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 5.67117, size = 1382, normalized size = 11.06 \begin{align*} \left [-\frac{{\left (3 \, a - 2 \, b\right )} \sqrt{b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{b} \tan \left (f x + e\right ) + a\right ) + 2 \,{\left (a - b\right )} \sqrt{-a + b} \log \left (-\frac{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{4 \, f}, -\frac{{\left (3 \, a - 2 \, b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-b}}{b \tan \left (f x + e\right )}\right ) -{\left (-a + b\right )}^{\frac{3}{2}} \log \left (-\frac{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - \sqrt{b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{2 \, f}, \frac{4 \,{\left (a - b\right )}^{\frac{3}{2}} \arctan \left (-\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{a - b} \tan \left (f x + e\right )}\right ) -{\left (3 \, a - 2 \, b\right )} \sqrt{b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{b} \tan \left (f x + e\right ) + a\right ) + 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{4 \, f}, \frac{2 \,{\left (a - b\right )}^{\frac{3}{2}} \arctan \left (-\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{a - b} \tan \left (f x + e\right )}\right ) -{\left (3 \, a - 2 \, b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-b}}{b \tan \left (f x + e\right )}\right ) + \sqrt{b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{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}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [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}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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