Optimal. Leaf size=130 \[ \frac{2 a^2 (4 n+1) \sin (e+f x) \sec ^{1-n}(e+f x) (d \sec (e+f x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},1-n,\frac{3}{2},1-\sec (e+f x)\right )}{f (2 n+1) \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt{a \sec (e+f x)+a}} \]
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Rubi [A] time = 0.161721, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3814, 21, 3806, 67, 65} \[ \frac{2 a^2 (4 n+1) \sin (e+f x) \sec ^{1-n}(e+f x) (d \sec (e+f x))^n \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};1-\sec (e+f x)\right )}{f (2 n+1) \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt{a \sec (e+f x)+a}} \]
Antiderivative was successfully verified.
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Rule 3814
Rule 21
Rule 3806
Rule 67
Rule 65
Rubi steps
\begin{align*} \int (d \sec (e+f x))^n (a+a \sec (e+f x))^{3/2} \, dx &=\frac{2 a^2 (d \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt{a+a \sec (e+f x)}}+\frac{(2 a) \int \frac{(d \sec (e+f x))^n \left (a \left (\frac{1}{2}+2 n\right )+a \left (\frac{1}{2}+2 n\right ) \sec (e+f x)\right )}{\sqrt{a+a \sec (e+f x)}} \, dx}{1+2 n}\\ &=\frac{2 a^2 (d \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt{a+a \sec (e+f x)}}+\frac{(a (1+4 n)) \int (d \sec (e+f x))^n \sqrt{a+a \sec (e+f x)} \, dx}{1+2 n}\\ &=\frac{2 a^2 (d \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt{a+a \sec (e+f x)}}-\frac{\left (a^3 d (1+4 n) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(d x)^{-1+n}}{\sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{f (1+2 n) \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a^2 (d \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt{a+a \sec (e+f x)}}-\frac{\left (a^3 (1+4 n) \sec ^{1-n}(e+f x) (d \sec (e+f x))^n \sin (e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^{-1+n}}{\sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{f (1+2 n) \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a^2 (1+4 n) \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};1-\sec (e+f x)\right ) \sec ^{1-n}(e+f x) (d \sec (e+f x))^n \sin (e+f x)}{f (1+2 n) \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 (d \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt{a+a \sec (e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.315141, size = 88, normalized size = 0.68 \[ \frac{a \tan \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} (d \sec (e+f x))^n \left ((4 n+1) \cos ^{n+\frac{1}{2}}(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},n+\frac{3}{2},\frac{3}{2},2 \sin ^2\left (\frac{1}{2} (e+f x)\right )\right )-1\right )}{f n} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.168, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{n} \left ( a+a\sec \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\, 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 (a \sec \left (f x + e\right ) + a\right )}^{\frac{3}{2}} \left (d \sec \left (f x + e\right )\right )^{n}\,{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 (a \sec \left (f x + e\right ) + a\right )}^{\frac{3}{2}} \left (d \sec \left (f x + e\right )\right )^{n}, 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(-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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