Optimal. Leaf size=217 \[ -\frac{\left (2 a^2-2 a b p-b^2 (1-p) p\right ) \left (a+b \tan ^2(e+f x)\right )^{p+1} \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{b \tan ^2(e+f x)}{a}+1\right )}{4 a^3 f (p+1)}+\frac{\left (a+b \tan ^2(e+f x)\right )^{p+1} \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{a+b \tan ^2(e+f x)}{a-b}\right )}{2 f (p+1) (a-b)}+\frac{(2 a-b p+b) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{4 a^2 f}-\frac{\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{4 a f} \]
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Rubi [A] time = 0.253299, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3670, 446, 103, 151, 156, 65, 68} \[ -\frac{\left (2 a^2-2 a b p-b^2 (1-p) p\right ) \left (a+b \tan ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \tan ^2(e+f x)}{a}+1\right )}{4 a^3 f (p+1)}+\frac{(2 a-b p+b) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{4 a^2 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \tan ^2(e+f x)+a}{a-b}\right )}{2 f (p+1) (a-b)}-\frac{\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{4 a f} \]
Antiderivative was successfully verified.
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Rule 3670
Rule 446
Rule 103
Rule 151
Rule 156
Rule 65
Rule 68
Rubi steps
\begin{align*} \int \cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^p}{x^5 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p}{x^3 (1+x)} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac{\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a f}-\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p (2 a+b-b p+b (1-p) x)}{x^2 (1+x)} \, dx,x,\tan ^2(e+f x)\right )}{4 a f}\\ &=\frac{(2 a+b-b p) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^2 f}-\frac{\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a f}+\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p \left (2 a^2-2 a b p-b^2 (1-p) p-b p (2 a+b-b p) x\right )}{x (1+x)} \, dx,x,\tan ^2(e+f x)\right )}{4 a^2 f}\\ &=\frac{(2 a+b-b p) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^2 f}-\frac{\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a f}-\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}+\frac{\left (2 a^2-2 a b p-b^2 (1-p) p\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^p}{x} \, dx,x,\tan ^2(e+f x)\right )}{4 a^2 f}\\ &=\frac{(2 a+b-b p) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^2 f}-\frac{\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a f}+\frac{\, _2F_1\left (1,1+p;2+p;\frac{a+b \tan ^2(e+f x)}{a-b}\right ) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{2 (a-b) f (1+p)}-\frac{\left (2 a^2-2 a b p-b^2 (1-p) p\right ) \, _2F_1\left (1,1+p;2+p;1+\frac{b \tan ^2(e+f x)}{a}\right ) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^3 f (1+p)}\\ \end{align*}
Mathematica [A] time = 2.54215, size = 172, normalized size = 0.79 \[ -\frac{\tan ^2(e+f x) \left (a \cot ^2(e+f x)+b\right ) \left (a+b \tan ^2(e+f x)\right )^p \left ((a-b) \left (\left (2 a^2-2 a b p+b^2 (p-1) p\right ) \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{b \tan ^2(e+f x)}{a}+1\right )+a (p+1) \cot ^2(e+f x) \left (a \cot ^2(e+f x)-2 a+b (p-1)\right )\right )-2 a^3 \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{a+b \tan ^2(e+f x)}{a-b}\right )\right )}{4 a^3 f (p+1) (a-b)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.301, size = 0, normalized size = 0. \begin{align*} \int \left ( \cot \left ( fx+e \right ) \right ) ^{5} \left ( a+b \left ( \tan \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(-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 [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{5}, 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 \tan \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{5}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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