Optimal. Leaf size=131 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}+\frac{(a-4 b) \tan (e+f x)}{3 b f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}-\frac{a \tan (e+f x)}{3 b f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
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Rubi [A] time = 0.158553, antiderivative size = 131, 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, 470, 527, 12, 377, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}+\frac{(a-4 b) \tan (e+f x)}{3 b f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}-\frac{a \tan (e+f x)}{3 b f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3670
Rule 470
Rule 527
Rule 12
Rule 377
Rule 203
Rubi steps
\begin{align*} \int \frac{\tan ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a \tan (e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{a+(a-3 b) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 (a-b) b f}\\ &=-\frac{a \tan (e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{(a-4 b) \tan (e+f x)}{3 (a-b)^2 b f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{3 a b}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a (a-b)^2 b f}\\ &=-\frac{a \tan (e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{(a-4 b) \tan (e+f x)}{3 (a-b)^2 b f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{(a-b)^2 f}\\ &=-\frac{a \tan (e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{(a-4 b) \tan (e+f x)}{3 (a-b)^2 b f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\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 )}{(a-b)^2 f}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{(a-b)^{5/2} f}-\frac{a \tan (e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{(a-4 b) \tan (e+f x)}{3 (a-b)^2 b f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}
Mathematica [A] time = 6.03819, size = 260, normalized size = 1.98 \[ \frac{\tan ^5(e+f x) \left (\frac{b \tan ^2(e+f x)}{a}+1\right ) \left (-\frac{\left (\frac{b \tan ^2(e+f x)}{a}-\tan ^2(e+f x)\right )^2}{3 \left (\frac{b \tan ^2(e+f x)}{a}+1\right )^2}-\frac{\frac{b \tan ^2(e+f x)}{a}-\tan ^2(e+f x)}{\frac{b \tan ^2(e+f x)}{a}+1}+\frac{\sqrt{\frac{b \tan ^2(e+f x)}{a}-\tan ^2(e+f x)} \tanh ^{-1}\left (\frac{\sqrt{\frac{b \tan ^2(e+f x)}{a}-\tan ^2(e+f x)}}{\sqrt{\frac{b \tan ^2(e+f x)}{a}+1}}\right )}{\sqrt{\frac{b \tan ^2(e+f x)}{a}+1}}\right )}{a^2 f \sqrt{a+b \tan ^2(e+f x)} \left (\frac{b \tan ^2(e+f x)}{a}-\tan ^2(e+f x)\right )^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.046, size = 291, normalized size = 2.2 \begin{align*} -{\frac{\tan \left ( fx+e \right ) }{3\,fb} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{\tan \left ( fx+e \right ) }{3\,fab}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}-{\frac{\tan \left ( fx+e \right ) }{3\,fa} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,\tan \left ( fx+e \right ) }{3\,f{a}^{2}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}-{\frac{b\tan \left ( fx+e \right ) }{f \left ( a-b \right ) ^{2}a}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}+{\frac{1}{f \left ( a-b \right ) ^{3}{b}^{2}}\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 ) }-{\frac{b\tan \left ( fx+e \right ) }{3\,a \left ( a-b \right ) f} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,b\tan \left ( fx+e \right ) }{3\,f \left ( a-b \right ){a}^{2}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{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 [A] time = 2.19079, size = 1137, normalized size = 8.68 \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{{\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 \,{\left ({\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )^{3} - 3 \,{\left (a^{2} - a b\right )} \tan \left (f x + e\right )\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 )}}, \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{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{a - b} \tan \left (f x + e\right )}\right ) +{\left ({\left (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )^{3} - 3 \,{\left (a^{2} - a b\right )} \tan \left (f x + e\right )\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{3 \,{\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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )^{4}}{{\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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