Optimal. Leaf size=101 \[ -\frac {c^2 2^{\frac {5}{2}-m} \tan (e+f x) (1-\sec (e+f x))^{m-\frac {1}{2}} (a \sec (e+f x)+a)^m (c-c \sec (e+f x))^{-m} \, _2F_1\left (m-\frac {3}{2},m+\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sec (e+f x)+1)\right )}{f (2 m+1)} \]
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Rubi [A] time = 0.15, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3961, 70, 69} \[ -\frac {c^2 2^{\frac {5}{2}-m} \tan (e+f x) (1-\sec (e+f x))^{m-\frac {1}{2}} (a \sec (e+f x)+a)^m (c-c \sec (e+f x))^{-m} \, _2F_1\left (m-\frac {3}{2},m+\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sec (e+f x)+1)\right )}{f (2 m+1)} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 3961
Rubi steps
\begin {align*} \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{2-m} \, dx &=-\frac {(a c \tan (e+f x)) \operatorname {Subst}\left (\int (a+a x)^{-\frac {1}{2}+m} (c-c x)^{\frac {3}{2}-m} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {\left (2^{\frac {3}{2}-m} a c^2 (c-c \sec (e+f x))^{-m} \left (\frac {c-c \sec (e+f x)}{c}\right )^{-\frac {1}{2}+m} \tan (e+f x)\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{\frac {3}{2}-m} (a+a x)^{-\frac {1}{2}+m} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)}}\\ &=-\frac {2^{\frac {5}{2}-m} c^2 \, _2F_1\left (-\frac {3}{2}+m,\frac {1}{2}+m;\frac {3}{2}+m;\frac {1}{2} (1+\sec (e+f x))\right ) (1-\sec (e+f x))^{-\frac {1}{2}+m} (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-m} \tan (e+f x)}{f (1+2 m)}\\ \end {align*}
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Mathematica [F] time = 2.71, size = 0, normalized size = 0.00 \[ \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{2-m} \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (f x + e\right ) + a\right )}^{m} {\left (-c \sec \left (f x + e\right ) + c\right )}^{-m + 2} \sec \left (f x + e\right ), x\right ) \]
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 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{m} {\left (-c \sec \left (f x + e\right ) + c\right )}^{-m + 2} \sec \left (f x + e\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 3.99, size = 0, normalized size = 0.00 \[ \int \sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m} \left (c -c \sec \left (f x +e \right )\right )^{2-m}\, dx \]
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 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{m} {\left (-c \sec \left (f x + e\right ) + c\right )}^{-m + 2} \sec \left (f x + e\right )\,{d x} \]
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 \[ \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{2-m}}{\cos \left (e+f\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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