Optimal. Leaf size=165 \[ -\frac{\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a f}-\frac{(a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}-\frac{a \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 f}+\frac{(5 a-6 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 f} \]
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Rubi [A] time = 0.244786, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3670, 474, 583, 12, 377, 203} \[ -\frac{\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a f}-\frac{(a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}-\frac{a \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 f}+\frac{(5 a-6 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 f} \]
Antiderivative was successfully verified.
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Rule 3670
Rule 474
Rule 583
Rule 12
Rule 377
Rule 203
Rubi steps
\begin{align*} \int \cot ^6(e+f x) \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}}{x^6 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 f}+\frac{\operatorname{Subst}\left (\int \frac{-a (5 a-6 b)-(4 a-5 b) b x^2}{x^4 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=\frac{(5 a-6 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 f}-\frac{a \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 f}-\frac{\operatorname{Subst}\left (\int \frac{-a \left (15 a^2-20 a b+3 b^2\right )-2 a (5 a-6 b) b x^2}{x^2 \left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a f}\\ &=-\frac{\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a f}+\frac{(5 a-6 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 f}-\frac{a \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 f}+\frac{\operatorname{Subst}\left (\int -\frac{15 a^2 (a-b)^2}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a^2 f}\\ &=-\frac{\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a f}+\frac{(5 a-6 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 f}-\frac{a \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 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{\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a f}+\frac{(5 a-6 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 f}-\frac{a \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 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{(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{\left (15 a^2-20 a b+3 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a f}+\frac{(5 a-6 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 f}-\frac{a \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 f}\\ \end{align*}
Mathematica [C] time = 8.50924, size = 140, normalized size = 0.85 \[ -\frac{\sin (e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)} \left (a \cot ^2(e+f x)+b\right )^2 \left (2 (a-b) ((a-b) \cos (2 (e+f x))+a+b) \text{Hypergeometric2F1}\left (2,2,\frac{1}{2},\frac{(a-b) \sin ^2(e+f x)}{a}\right )+a \left (3 a \cot ^2(e+f x)-2 b\right ) \text{Hypergeometric2F1}\left (1,1,-\frac{1}{2},\frac{(a-b) \sin ^2(e+f x)}{a}\right )\right )}{15 a^3 f} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.426, size = 10026, normalized size = 60.8 \begin{align*} \text{output too large to display} \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}} \cot \left (f x + e\right )^{6}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.47453, size = 927, normalized size = 5.62 \begin{align*} \left [-\frac{15 \,{\left (a^{2} - a b\right )} \sqrt{-a + b} \log \left (-\frac{{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \,{\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} + 4 \,{\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{5} + 4 \,{\left ({\left (15 \, a^{2} - 20 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{4} -{\left (5 \, a^{2} - 6 \, a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{60 \, a f \tan \left (f x + e\right )^{5}}, -\frac{15 \,{\left (a^{2} - a b\right )} \sqrt{a - b} \arctan \left (-\frac{2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) \tan \left (f x + e\right )^{5} + 2 \,{\left ({\left (15 \, a^{2} - 20 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{4} -{\left (5 \, a^{2} - 6 \, a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{30 \, a f \tan \left (f x + e\right )^{5}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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}} \cot \left (f x + e\right )^{6}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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