Optimal. Leaf size=183 \[ -\frac{x \tan ^2(e+f x)}{b^2 \sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot (e+f x)}{3 b^2 f \sqrt{b \tan ^4(e+f x)}} \]
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Rubi [A] time = 0.0646674, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3658, 3473, 8} \[ -\frac{x \tan ^2(e+f x)}{b^2 \sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot (e+f x)}{3 b^2 f \sqrt{b \tan ^4(e+f x)}} \]
Antiderivative was successfully verified.
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Rule 3658
Rule 3473
Rule 8
Rubi steps
\begin{align*} \int \frac{1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx &=\frac{\tan ^2(e+f x) \int \cot ^{10}(e+f x) \, dx}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ &=-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan ^2(e+f x) \int \cot ^8(e+f x) \, dx}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ &=\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\tan ^2(e+f x) \int \cot ^6(e+f x) \, dx}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ &=-\frac{\cot ^3(e+f x)}{5 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan ^2(e+f x) \int \cot ^4(e+f x) \, dx}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ &=\frac{\cot (e+f x)}{3 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\tan ^2(e+f x) \int \cot ^2(e+f x) \, dx}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ &=\frac{\cot (e+f x)}{3 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan ^2(e+f x) \int 1 \, dx}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ &=\frac{\cot (e+f x)}{3 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^3(e+f x)}{5 b^2 f \sqrt{b \tan ^4(e+f x)}}+\frac{\cot ^5(e+f x)}{7 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\cot ^7(e+f x)}{9 b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{\tan (e+f x)}{b^2 f \sqrt{b \tan ^4(e+f x)}}-\frac{x \tan ^2(e+f x)}{b^2 \sqrt{b \tan ^4(e+f x)}}\\ \end{align*}
Mathematica [C] time = 0.0326329, size = 45, normalized size = 0.25 \[ -\frac{\tan (e+f x) \text{Hypergeometric2F1}\left (-\frac{9}{2},1,-\frac{7}{2},-\tan ^2(e+f x)\right )}{9 f \left (b \tan ^4(e+f x)\right )^{5/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.021, size = 83, normalized size = 0.5 \begin{align*} -{\frac{\tan \left ( fx+e \right ) \left ( 315\,\arctan \left ( \tan \left ( fx+e \right ) \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{9}+315\, \left ( \tan \left ( fx+e \right ) \right ) ^{8}-105\, \left ( \tan \left ( fx+e \right ) \right ) ^{6}+63\, \left ( \tan \left ( fx+e \right ) \right ) ^{4}-45\, \left ( \tan \left ( fx+e \right ) \right ) ^{2}+35 \right ) }{315\,f} \left ( b \left ( \tan \left ( fx+e \right ) \right ) ^{4} \right ) ^{-{\frac{5}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.6532, size = 95, normalized size = 0.52 \begin{align*} -\frac{\frac{315 \,{\left (f x + e\right )}}{b^{\frac{5}{2}}} + \frac{315 \, \tan \left (f x + e\right )^{8} - 105 \, \tan \left (f x + e\right )^{6} + 63 \, \tan \left (f x + e\right )^{4} - 45 \, \tan \left (f x + e\right )^{2} + 35}{b^{\frac{5}{2}} \tan \left (f x + e\right )^{9}}}{315 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.92304, size = 225, normalized size = 1.23 \begin{align*} -\frac{{\left (315 \, f x \tan \left (f x + e\right )^{9} + 315 \, \tan \left (f x + e\right )^{8} - 105 \, \tan \left (f x + e\right )^{6} + 63 \, \tan \left (f x + e\right )^{4} - 45 \, \tan \left (f x + e\right )^{2} + 35\right )} \sqrt{b \tan \left (f x + e\right )^{4}}}{315 \, b^{3} f \tan \left (f x + e\right )^{11}} \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{1}{\left (b \tan ^{4}{\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 = 2.79148, size = 265, normalized size = 1.45 \begin{align*} -\frac{\frac{161280 \,{\left (f x + e\right )}}{b^{\frac{5}{2}}} + \frac{121590 \, \sqrt{b} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 18480 \, \sqrt{b} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 3528 \, \sqrt{b} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 495 \, \sqrt{b} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 35 \, \sqrt{b}}{b^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9}} - \frac{35 \, b^{\frac{49}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9} - 495 \, b^{\frac{49}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 3528 \, b^{\frac{49}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 18480 \, b^{\frac{49}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 121590 \, b^{\frac{49}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{b^{27}}}{161280 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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