Optimal. Leaf size=90 \[ \frac{a^2}{2 b^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac{a (a-2 b) \log \left (a+b \tan ^2(e+f x)\right )}{2 b^2 f (a-b)^2}-\frac{\log (\cos (e+f x))}{f (a-b)^2} \]
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Rubi [A] time = 0.121305, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3670, 446, 88} \[ \frac{a^2}{2 b^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac{a (a-2 b) \log \left (a+b \tan ^2(e+f x)\right )}{2 b^2 f (a-b)^2}-\frac{\log (\cos (e+f x))}{f (a-b)^2} \]
Antiderivative was successfully verified.
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Rule 3670
Rule 446
Rule 88
Rubi steps
\begin{align*} \int \frac{\tan ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^5}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{(1+x) (a+b x)^2} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{(a-b)^2 (1+x)}-\frac{a^2}{(a-b) b (a+b x)^2}+\frac{a (a-2 b)}{(a-b)^2 b (a+b x)}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac{\log (\cos (e+f x))}{(a-b)^2 f}+\frac{a (a-2 b) \log \left (a+b \tan ^2(e+f x)\right )}{2 (a-b)^2 b^2 f}+\frac{a^2}{2 (a-b) b^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 0.665275, size = 73, normalized size = 0.81 \[ \frac{\frac{a^2 (a-b)}{b^2 \left (a+b \tan ^2(e+f x)\right )}+\frac{a (a-2 b) \log \left (a+b \tan ^2(e+f x)\right )}{b^2}-2 \log (\cos (e+f x))}{2 f (a-b)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.026, size = 149, normalized size = 1.7 \begin{align*}{\frac{{a}^{2}\ln \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f \left ( a-b \right ) ^{2}{b}^{2}}}-{\frac{a\ln \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{f \left ( a-b \right ) ^{2}b}}+{\frac{{a}^{3}}{2\,f \left ( a-b \right ) ^{2}{b}^{2} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{{a}^{2}}{2\,f \left ( a-b \right ) ^{2}b \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f \left ( a-b \right ) ^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.14079, size = 173, normalized size = 1.92 \begin{align*} -\frac{\frac{a^{2}}{a^{3} b - 2 \, a^{2} b^{2} + a b^{3} -{\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} \sin \left (f x + e\right )^{2}} - \frac{{\left (a^{2} - 2 \, a b\right )} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{2} b^{2} - 2 \, a b^{3} + b^{4}} + \frac{\log \left (\sin \left (f x + e\right )^{2} - 1\right )}{b^{2}}}{2 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.26814, size = 423, normalized size = 4.7 \begin{align*} -\frac{a^{2} b \tan \left (f x + e\right )^{2} + a^{2} b -{\left (a^{3} - 2 \, a^{2} b +{\left (a^{2} b - 2 \, a b^{2}\right )} \tan \left (f x + e\right )^{2}\right )} \log \left (\frac{b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right ) +{\left (a^{3} - 2 \, a^{2} b + a b^{2} +{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \,{\left ({\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 118.299, size = 1583, normalized size = 17.59 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 2.97375, size = 536, normalized size = 5.96 \begin{align*} \frac{\frac{{\left (a^{3} - 2 \, a^{2} b\right )} \log \left ({\left | -a{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - 2 \, a + 4 \, b \right |}\right )}{a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}} + \frac{\log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right )}{a^{2} - 2 \, a b + b^{2}} - \frac{a^{3}{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - 2 \, a^{2} b{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 2 \, a^{3} - 12 \, a^{2} b + 12 \, a b^{2}}{{\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )}{\left (a{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 2 \, a - 4 \, b\right )}} - \frac{\log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2\right )}{b^{2}}}{2 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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