Optimal. Leaf size=226 \[ \frac{\sec ^4(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac{b (c \sec (e+f x))^n}{a}+1\right )^{-p} \text{Hypergeometric2F1}\left (\frac{4}{n},-p,\frac{n+4}{n},-\frac{b (c \sec (e+f x))^n}{a}\right )}{4 f}-\frac{\sec ^2(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac{b (c \sec (e+f x))^n}{a}+1\right )^{-p} \text{Hypergeometric2F1}\left (\frac{2}{n},-p,\frac{n+2}{n},-\frac{b (c \sec (e+f x))^n}{a}\right )}{f}-\frac{\left (a+b (c \sec (e+f x))^n\right )^{p+1} \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{b (c \sec (e+f x))^n}{a}+1\right )}{a f n (p+1)} \]
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Rubi [A] time = 0.5245, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {4139, 6742, 367, 12, 266, 65, 365, 364} \[ \frac{\sec ^4(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac{b (c \sec (e+f x))^n}{a}+1\right )^{-p} \, _2F_1\left (\frac{4}{n},-p;\frac{n+4}{n};-\frac{b (c \sec (e+f x))^n}{a}\right )}{4 f}-\frac{\sec ^2(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (\frac{b (c \sec (e+f x))^n}{a}+1\right )^{-p} \, _2F_1\left (\frac{2}{n},-p;\frac{n+2}{n};-\frac{b (c \sec (e+f x))^n}{a}\right )}{f}-\frac{\left (a+b (c \sec (e+f x))^n\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b (c \sec (e+f x))^n}{a}+1\right )}{a f n (p+1)} \]
Antiderivative was successfully verified.
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Rule 4139
Rule 6742
Rule 367
Rule 12
Rule 266
Rule 65
Rule 365
Rule 364
Rubi steps
\begin{align*} \int \left (a+b (c \sec (e+f x))^n\right )^p \tan ^5(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2 \left (a+b (c x)^n\right )^p}{x} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{\left (a+b (c x)^n\right )^p}{x}-2 x \left (a+b (c x)^n\right )^p+x^3 \left (a+b (c x)^n\right )^p\right ) \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b (c x)^n\right )^p}{x} \, dx,x,\sec (e+f x)\right )}{f}+\frac{\operatorname{Subst}\left (\int x^3 \left (a+b (c x)^n\right )^p \, dx,x,\sec (e+f x)\right )}{f}-\frac{2 \operatorname{Subst}\left (\int x \left (a+b (c x)^n\right )^p \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{c \left (a+b x^n\right )^p}{x} \, dx,x,c \sec (e+f x)\right )}{c f}+\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (a+b x^n\right )^p}{c^3} \, dx,x,c \sec (e+f x)\right )}{c f}-\frac{2 \operatorname{Subst}\left (\int \frac{x \left (a+b x^n\right )^p}{c} \, dx,x,c \sec (e+f x)\right )}{c f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^n\right )^p}{x} \, dx,x,c \sec (e+f x)\right )}{f}+\frac{\operatorname{Subst}\left (\int x^3 \left (a+b x^n\right )^p \, dx,x,c \sec (e+f x)\right )}{c^4 f}-\frac{2 \operatorname{Subst}\left (\int x \left (a+b x^n\right )^p \, dx,x,c \sec (e+f x)\right )}{c^2 f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p}{x} \, dx,x,(c \sec (e+f x))^n\right )}{f n}+\frac{\left (\left (a+b (c \sec (e+f x))^n\right )^p \left (1+\frac{b (c \sec (e+f x))^n}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int x^3 \left (1+\frac{b x^n}{a}\right )^p \, dx,x,c \sec (e+f x)\right )}{c^4 f}-\frac{\left (2 \left (a+b (c \sec (e+f x))^n\right )^p \left (1+\frac{b (c \sec (e+f x))^n}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int x \left (1+\frac{b x^n}{a}\right )^p \, dx,x,c \sec (e+f x)\right )}{c^2 f}\\ &=-\frac{\, _2F_1\left (1,1+p;2+p;1+\frac{b (c \sec (e+f x))^n}{a}\right ) \left (a+b (c \sec (e+f x))^n\right )^{1+p}}{a f n (1+p)}-\frac{\, _2F_1\left (\frac{2}{n},-p;\frac{2+n}{n};-\frac{b (c \sec (e+f x))^n}{a}\right ) \sec ^2(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (1+\frac{b (c \sec (e+f x))^n}{a}\right )^{-p}}{f}+\frac{\, _2F_1\left (\frac{4}{n},-p;\frac{4+n}{n};-\frac{b (c \sec (e+f x))^n}{a}\right ) \sec ^4(e+f x) \left (a+b (c \sec (e+f x))^n\right )^p \left (1+\frac{b (c \sec (e+f x))^n}{a}\right )^{-p}}{4 f}\\ \end{align*}
Mathematica [A] time = 6.89216, size = 221, normalized size = 0.98 \[ \frac{\left (a+b (c \sec (e+f x))^n\right )^p \left (\frac{4 \left (\frac{a \left (c \sqrt{\sec ^2(e+f x)}\right )^{-n}}{b}+1\right )^{-p} \text{Hypergeometric2F1}\left (-p,-p,1-p,-\frac{a \left (c \sqrt{\sec ^2(e+f x)}\right )^{-n}}{b}\right )}{n p}+\sec ^2(e+f x) \left (\frac{b \left (c \sqrt{\sec ^2(e+f x)}\right )^n}{a}+1\right )^{-p} \left (\sec ^2(e+f x) \text{Hypergeometric2F1}\left (\frac{4}{n},-p,\frac{n+4}{n},-\frac{b \left (c \sqrt{\sec ^2(e+f x)}\right )^n}{a}\right )-4 \text{Hypergeometric2F1}\left (\frac{2}{n},-p,\frac{n+2}{n},-\frac{b \left (c \sqrt{\sec ^2(e+f x)}\right )^n}{a}\right )\right )\right )}{4 f} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.573, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ( c\sec \left ( fx+e \right ) \right ) ^{n} \right ) ^{p} \left ( \tan \left ( fx+e \right ) \right ) ^{5}\, 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 (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right )^{5}\,{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 (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \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 (\left (c \sec \left (f x + e\right )\right )^{n} b + a\right )}^{p} \tan \left (f x + e\right )^{5}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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