Optimal. Leaf size=108 \[ \frac {2 a^2 (4 n+1) \tan (e+f x) \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};1-\sec (e+f x)\right )}{f (2 n+1) \sqrt {a \sec (e+f x)+a}}+\frac {2 a^2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) \sqrt {a \sec (e+f x)+a}} \]
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Rubi [A] time = 0.14, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3814, 21, 3806, 65} \[ \frac {2 a^2 (4 n+1) \tan (e+f x) \, _2F_1\left (\frac {1}{2},1-n;\frac {3}{2};1-\sec (e+f x)\right )}{f (2 n+1) \sqrt {a \sec (e+f x)+a}}+\frac {2 a^2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) \sqrt {a \sec (e+f x)+a}} \]
Antiderivative was successfully verified.
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Rule 21
Rule 65
Rule 3806
Rule 3814
Rubi steps
\begin {align*} \int \sec ^n(e+f x) (a+a \sec (e+f x))^{3/2} \, dx &=\frac {2 a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) \sqrt {a+a \sec (e+f x)}}+\frac {(2 a) \int \frac {\sec ^n(e+f x) \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 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) \sqrt {a+a \sec (e+f x)}}+\frac {(a (1+4 n)) \int \sec ^n(e+f x) \sqrt {a+a \sec (e+f x)} \, dx}{1+2 n}\\ &=\frac {2 a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) \sqrt {a+a \sec (e+f x)}}-\frac {\left (a^3 (1+4 n) \tan (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 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) \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 ) \tan (e+f x)}{f (1+2 n) \sqrt {a+a \sec (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 0.43, size = 86, normalized size = 0.80 \[ \frac {a \tan \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} \sec ^n(e+f x) \left ((4 n+1) \cos ^{n+\frac {1}{2}}(e+f x) \, _2F_1\left (\frac {1}{2},n+\frac {3}{2};\frac {3}{2};2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )-1\right )}{f n} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{n}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.09, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{n}\left (f x +e \right )\right ) \left (a +a \sec \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sec ^{n}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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