Optimal. Leaf size=183 \[ -\frac {\tan (e+f x)}{b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {x \tan ^2(e+f x)}{b^2 \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot (e+f x)}{3 b^2 f \sqrt {b \tan ^4(e+f x)}} \]
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Rubi [A] time = 0.06, antiderivative size = 183, normalized size of antiderivative = 1.00, 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 8
Rule 3473
Rule 3658
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*}
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Mathematica [C] time = 0.03, size = 45, normalized size = 0.25 \[ -\frac {\tan (e+f x) \, _2F_1\left (-\frac {9}{2},1;-\frac {7}{2};-\tan ^2(e+f x)\right )}{9 f \left (b \tan ^4(e+f x)\right )^{5/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.97, size = 82, normalized size = 0.45 \[ -\frac {{\left (315 \, f x \tan \left (f x + e\right )^{9} + 315 \, \tan \left (f x + e\right )^{8} - 105 \, \tan \left (f x + e\right )^{6} + 63 \, \tan \left (f x + e\right )^{4} - 45 \, \tan \left (f x + e\right )^{2} + 35\right )} \sqrt {b \tan \left (f x + e\right )^{4}}}{315 \, b^{3} f \tan \left (f x + e\right )^{11}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 4.88, size = 196, normalized size = 1.07 \[ -\frac {\frac {161280 \, {\left (f x + e\right )}}{b^{\frac {5}{2}}} + \frac {121590 \, \sqrt {b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 18480 \, \sqrt {b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 3528 \, \sqrt {b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 495 \, \sqrt {b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 35 \, \sqrt {b}}{b^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9}} - \frac {35 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} - 495 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 3528 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 18480 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 121590 \, b^{\frac {49}{2}} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{b^{27}}}{161280 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 83, normalized size = 0.45 \[ -\frac {\tan \left (f x +e \right ) \left (315 \arctan \left (\tan \left (f x +e \right )\right ) \left (\tan ^{9}\left (f x +e \right )\right )+315 \left (\tan ^{8}\left (f x +e \right )\right )-105 \left (\tan ^{6}\left (f x +e \right )\right )+63 \left (\tan ^{4}\left (f x +e \right )\right )-45 \left (\tan ^{2}\left (f x +e \right )\right )+35\right )}{315 f \left (b \left (\tan ^{4}\left (f x +e \right )\right )\right )^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.68, size = 70, normalized size = 0.38 \[ -\frac {\frac {315 \, {\left (f x + e\right )}}{b^{\frac {5}{2}}} + \frac {315 \, \tan \left (f x + e\right )^{8} - 105 \, \tan \left (f x + e\right )^{6} + 63 \, \tan \left (f x + e\right )^{4} - 45 \, \tan \left (f x + e\right )^{2} + 35}{b^{\frac {5}{2}} \tan \left (f x + e\right )^{9}}}{315 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tan ^{4}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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