Optimal. Leaf size=167 \[ -\frac{\left (15 a^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac{\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 f}+\frac{(5 a-b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a f} \]
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Rubi [A] time = 0.230896, antiderivative size = 167, 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, 475, 583, 12, 377, 203} \[ -\frac{\left (15 a^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac{\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 f}+\frac{(5 a-b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a f} \]
Antiderivative was successfully verified.
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Rule 3670
Rule 475
Rule 583
Rule 12
Rule 377
Rule 203
Rubi steps
\begin{align*} \int \cot ^6(e+f x) \sqrt{a+b \tan ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{x^6 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 f}+\frac{\operatorname{Subst}\left (\int \frac{-5 a+b-4 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-b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 f}-\frac{\operatorname{Subst}\left (\int \frac{-15 a^2+5 a b+2 b^2-2 (5 a-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^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}+\frac{(5 a-b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a f}-\frac{\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)}{\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^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}+\frac{(5 a-b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 f}-\frac{(a-b) \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^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}+\frac{(5 a-b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 f}-\frac{(a-b) \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{\sqrt{a-b} \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^2-5 a b-2 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}+\frac{(5 a-b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 f}\\ \end{align*}
Mathematica [C] time = 14.2186, size = 339, normalized size = 2.03 \[ -\frac{\cos ^4(e+f x) \cot ^5(e+f x) \left (\frac{b \tan ^2(e+f x)}{a}+1\right ) \left (8 (a-b) \tan ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^3 \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{3}{2}\right \},\frac{(a-b) \sin ^2(e+f x)}{a}\right )+8 \tan ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \left (-2 a^2+a b \left (3 \tan ^2(e+f x)+2\right )-3 b^2 \tan ^2(e+f x)\right ) \text{Hypergeometric2F1}\left (2,2,\frac{3}{2},\frac{(a-b) \sin ^2(e+f x)}{a}\right )+\frac{a^2 \sec ^4(e+f x) \left (3 a^2-4 a b \tan ^2(e+f x)+8 b^2 \tan ^4(e+f x)\right ) \left (\sqrt{\frac{(a-b) \sin ^2(e+f x)}{a}} \sin ^{-1}\left (\sqrt{\frac{(a-b) \sin ^2(e+f x)}{a}}\right )+\sqrt{\frac{\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}\right )}{\sqrt{\frac{\cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a}}}\right )}{15 a^3 f \sqrt{a+b \tan ^2(e+f x)}} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.401, size = 6894, normalized size = 41.3 \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 \sqrt{b \tan \left (f x + e\right )^{2} + a} \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 = 2.42358, size = 902, normalized size = 5.4 \begin{align*} \left [\frac{15 \, a^{2} \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} - 5 \, a b - 2 \, b^{2}\right )} \tan \left (f x + e\right )^{4} -{\left (5 \, a^{2} - 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^{2} f \tan \left (f x + e\right )^{5}}, -\frac{15 \, \sqrt{a - b} a^{2} \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} - 5 \, a b - 2 \, b^{2}\right )} \tan \left (f x + e\right )^{4} -{\left (5 \, a^{2} - 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^{2} 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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \tan ^{2}{\left (e + f x \right )}} \cot ^{6}{\left (e + f x \right )}\, 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 \sqrt{b \tan \left (f x + e\right )^{2} + a} \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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