Optimal. Leaf size=192 \[ -\frac{\left (15 a^2+20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right )}{15 f (a+b)^2}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{5 f (a+b)}-\frac{(10 a+b (2 p+7)) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{15 f (a+b)^2} \]
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Rubi [A] time = 0.173566, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4132, 462, 453, 365, 364} \[ -\frac{\left (15 a^2+20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \tan ^2(e+f x)}{a+b}\right )}{15 f (a+b)^2}-\frac{\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{5 f (a+b)}-\frac{(10 a+b (2 p+7)) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{15 f (a+b)^2} \]
Antiderivative was successfully verified.
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Rule 4132
Rule 462
Rule 453
Rule 365
Rule 364
Rubi steps
\begin{align*} \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^p}{x^6} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{5 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{\left (a+b+b x^2\right )^p \left (10 a+b (7+2 p)+5 (a+b) x^2\right )}{x^4} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f}\\ &=-\frac{(10 a+b (7+2 p)) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{5 (a+b) f}+\frac{\left (15 a^2+20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \operatorname{Subst}\left (\int \frac{\left (a+b+b x^2\right )^p}{x^2} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f}\\ &=-\frac{(10 a+b (7+2 p)) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{5 (a+b) f}+\frac{\left (\left (15 a^2+20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a+b}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{b x^2}{a+b}\right )^p}{x^2} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f}\\ &=-\frac{(10 a+b (7+2 p)) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{15 (a+b)^2 f}-\frac{\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{5 (a+b) f}-\frac{\left (15 a^2+20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \cot (e+f x) \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \tan ^2(e+f x)}{a+b}\right ) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a+b}\right )^{-p}}{15 (a+b)^2 f}\\ \end{align*}
Mathematica [A] time = 1.98683, size = 149, normalized size = 0.78 \[ -\frac{\cot (e+f x) \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \left (a+b \sec ^2(e+f x)\right )^p \left (15 \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right )+3 \cot ^4(e+f x) \text{Hypergeometric2F1}\left (-\frac{5}{2},-p,-\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right )+10 \cot ^2(e+f x) \text{Hypergeometric2F1}\left (-\frac{3}{2},-p,-\frac{1}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right )\right )}{15 f} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.454, size = 0, normalized size = 0. \begin{align*} \int \left ( \csc \left ( fx+e \right ) \right ) ^{6} \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \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 \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \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 [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{6}, x\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 \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{6}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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