Optimal. Leaf size=117 \[ \frac {2 (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}-\frac {(1+4 n) \, _2F_1\left (\frac {1}{2},n;1+n;\sec (e+f x)\right ) (-\sec (e+f x))^n \tan (e+f x)}{f n (1+2 n) \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \]
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Rubi [A]
time = 0.09, antiderivative size = 117, 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 = {3899, 21, 3891,
66} \begin {gather*} \frac {2 \tan (e+f x) (-\sec (e+f x))^n}{f (2 n+1) \sqrt {\sec (e+f x)+1}}-\frac {(4 n+1) \tan (e+f x) (-\sec (e+f x))^n \, _2F_1\left (\frac {1}{2},n;n+1;\sec (e+f x)\right )}{f n (2 n+1) \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}} \end {gather*}
Antiderivative was successfully verified.
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Rule 21
Rule 66
Rule 3891
Rule 3899
Rubi steps
\begin {align*} \int (-\sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx &=\frac {2 (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}+\frac {2 \int \frac {(-\sec (e+f x))^n \left (\frac {1}{2}+2 n+\left (\frac {1}{2}+2 n\right ) \sec (e+f x)\right )}{\sqrt {1+\sec (e+f x)}} \, dx}{1+2 n}\\ &=\frac {2 (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}+\frac {(1+4 n) \int (-\sec (e+f x))^n \sqrt {1+\sec (e+f x)} \, dx}{1+2 n}\\ &=\frac {2 (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}+\frac {((1+4 n) \tan (e+f x)) \text {Subst}\left (\int \frac {(-x)^{-1+n}}{\sqrt {1-x}} \, dx,x,\sec (e+f x)\right )}{f (1+2 n) \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}\\ &=\frac {2 (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}-\frac {(1+4 n) \, _2F_1\left (\frac {1}{2},n;1+n;\sec (e+f x)\right ) (-\sec (e+f x))^n \tan (e+f x)}{f n (1+2 n) \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 0.42, size = 85, normalized size = 0.73 \begin {gather*} \frac {\left (-1+(1+4 n) \cos ^{\frac {1}{2}+n}(e+f x) \, _2F_1\left (\frac {1}{2},\frac {3}{2}+n;\frac {3}{2};2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) (-\sec (e+f x))^n \sqrt {1+\sec (e+f x)} \tan \left (\frac {1}{2} (e+f x)\right )}{f n} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \left (-\sec \left (f x +e \right )\right )^{n} \left (1+\sec \left (f x +e \right )\right )^{\frac {3}{2}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \left (- \sec {\left (e + f x \right )}\right )^{n} \left (\sec {\left (e + f x \right )} + 1\right )^{\frac {3}{2}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int {\left (\frac {1}{\cos \left (e+f\,x\right )}+1\right )}^{3/2}\,{\left (-\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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