Optimal. Leaf size=117 \[ \frac {2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt {\sec (e+f x)+1}}-\frac {(4 n+1) \tan (e+f x) (d \sec (e+f x))^n \, _2F_1\left (\frac {1}{2},n;n+1;\sec (e+f x)\right )}{f n (2 n+1) \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}} \]
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Rubi [A] time = 0.13, 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 = {3814, 21, 3806, 64} \[ \frac {2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt {\sec (e+f x)+1}}-\frac {(4 n+1) \tan (e+f x) (d \sec (e+f x))^n \, _2F_1\left (\frac {1}{2},n;n+1;\sec (e+f x)\right )}{f n (2 n+1) \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}} \]
Antiderivative was successfully verified.
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Rule 21
Rule 64
Rule 3806
Rule 3814
Rubi steps
\begin {align*} \int (d \sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx &=\frac {2 (d \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}+\frac {2 \int \frac {(d \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 (d \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}+\frac {(1+4 n) \int (d \sec (e+f x))^n \sqrt {1+\sec (e+f x)} \, dx}{1+2 n}\\ &=\frac {2 (d \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}-\frac {(d (1+4 n) \tan (e+f x)) \operatorname {Subst}\left (\int \frac {(d 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 (d \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 ) (d \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 = 0.41, size = 85, normalized size = 0.73 \[ \frac {\tan \left (\frac {1}{2} (e+f x)\right ) \sqrt {\sec (e+f x)+1} (d \sec (e+f x))^n \left ((4 n+1) \cos ^{n+\frac {1}{2}}(e+f x) \, _2F_1\left (\frac {1}{2},n+\frac {3}{2};\frac {3}{2};2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )-1\right )}{f n} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (d \sec \left (f x + e\right )\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}, x\right ) \]
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 \[ \int \left (d \sec \left (f x + e\right )\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.07, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{n} \left (1+\sec \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]
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 \[ \int \left (d \sec \left (f x + e\right )\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}\,{d x} \]
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 \[ \int {\left (\frac {1}{\cos \left (e+f\,x\right )}+1\right )}^{3/2}\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]
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 \[ \int \left (d \sec {\left (e + f x \right )}\right )^{n} \left (\sec {\left (e + f x \right )} + 1\right )^{\frac {3}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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