Optimal. Leaf size=241 \[ \frac{\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 a f (a+b)^3}-\frac{\left (55 a^2 b+15 a^3+73 a b^2-15 b^3\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 a f (a+b)^4}-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2} f}-\frac{(a-5 b) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{5 a f (a+b)^2}-\frac{b \cot ^5(e+f x)}{a f (a+b) \sqrt{a+b \tan ^2(e+f x)+b}} \]
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Rubi [A] time = 0.471311, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {4141, 1975, 472, 583, 12, 377, 203} \[ \frac{\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 a f (a+b)^3}-\frac{\left (55 a^2 b+15 a^3+73 a b^2-15 b^3\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 a f (a+b)^4}-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2} f}-\frac{(a-5 b) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{5 a f (a+b)^2}-\frac{b \cot ^5(e+f x)}{a f (a+b) \sqrt{a+b \tan ^2(e+f x)+b}} \]
Antiderivative was successfully verified.
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Rule 4141
Rule 1975
Rule 472
Rule 583
Rule 12
Rule 377
Rule 203
Rubi steps
\begin{align*} \int \frac{\cot ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{b \cot ^5(e+f x)}{a (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{a-5 b-6 b x^2}{x^6 \left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a (a+b) f}\\ &=-\frac{b \cot ^5(e+f x)}{a (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{(a-5 b) \cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 a (a+b)^2 f}-\frac{\operatorname{Subst}\left (\int \frac{5 a^2+14 a b-15 b^2+4 (a-5 b) b x^2}{x^4 \left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 a (a+b)^2 f}\\ &=-\frac{b \cot ^5(e+f x)}{a (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 a (a+b)^3 f}-\frac{(a-5 b) \cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 a (a+b)^2 f}+\frac{\operatorname{Subst}\left (\int \frac{15 a^3+55 a^2 b+73 a b^2-15 b^3+2 b \left (5 a^2+14 a b-15 b^2\right ) x^2}{x^2 \left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a (a+b)^3 f}\\ &=-\frac{b \cot ^5(e+f x)}{a (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{\left (15 a^3+55 a^2 b+73 a b^2-15 b^3\right ) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 a (a+b)^4 f}+\frac{\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 a (a+b)^3 f}-\frac{(a-5 b) \cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 a (a+b)^2 f}-\frac{\operatorname{Subst}\left (\int \frac{15 (a+b)^4}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a (a+b)^4 f}\\ &=-\frac{b \cot ^5(e+f x)}{a (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{\left (15 a^3+55 a^2 b+73 a b^2-15 b^3\right ) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 a (a+b)^4 f}+\frac{\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 a (a+b)^3 f}-\frac{(a-5 b) \cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 a (a+b)^2 f}-\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a f}\\ &=-\frac{b \cot ^5(e+f x)}{a (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{\left (15 a^3+55 a^2 b+73 a b^2-15 b^3\right ) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 a (a+b)^4 f}+\frac{\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 a (a+b)^3 f}-\frac{(a-5 b) \cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 a (a+b)^2 f}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{a f}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{a^{3/2} f}-\frac{b \cot ^5(e+f x)}{a (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{\left (15 a^3+55 a^2 b+73 a b^2-15 b^3\right ) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 a (a+b)^4 f}+\frac{\left (5 a^2+14 a b-15 b^2\right ) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 a (a+b)^3 f}-\frac{(a-5 b) \cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 a (a+b)^2 f}\\ \end{align*}
Mathematica [A] time = 11.1437, size = 237, normalized size = 0.98 \[ \frac{\tan (e+f x) \sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b)^2 \left (-\left (23 a^2+80 a b+90 b^2\right ) \csc ^2(e+f x)+\frac{30 b^4}{a (a \cos (2 (e+f x))+a+2 b)}-3 (a+b)^2 \csc ^6(e+f x)+(a+b) (11 a+20 b) \csc ^4(e+f x)\right )}{60 f (a+b)^4 \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\sec ^3(e+f x) (a \cos (2 e+2 f x)+a+2 b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{-a \sin ^2(e+f x)+a+b}}\right )}{2 \sqrt{2} a^{3/2} f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.697, size = 14137, normalized size = 58.7 \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(-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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Fricas [B] time = 24.112, size = 3425, normalized size = 14.21 \begin{align*} \text{result too large to display} \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 \frac{\cot \left (f x + e\right )^{6}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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