Optimal. Leaf size=130 \[ \frac {e (e \tan (c+d x))^{m-1} \, _2F_1\left (1,\frac {m-1}{2};\frac {m+1}{2};-\tan ^2(c+d x)\right )}{a d (1-m)}-\frac {e \sec (c+d x) \cos ^2(c+d x)^{m/2} (e \tan (c+d x))^{m-1} \, _2F_1\left (\frac {m-1}{2},\frac {m}{2};\frac {m+1}{2};\sin ^2(c+d x)\right )}{a d (1-m)} \]
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Rubi [A] time = 0.16, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3888, 3884, 3476, 364, 2617} \[ \frac {e (e \tan (c+d x))^{m-1} \, _2F_1\left (1,\frac {m-1}{2};\frac {m+1}{2};-\tan ^2(c+d x)\right )}{a d (1-m)}-\frac {e \sec (c+d x) \cos ^2(c+d x)^{m/2} (e \tan (c+d x))^{m-1} \, _2F_1\left (\frac {m-1}{2},\frac {m}{2};\frac {m+1}{2};\sin ^2(c+d x)\right )}{a d (1-m)} \]
Antiderivative was successfully verified.
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Rule 364
Rule 2617
Rule 3476
Rule 3884
Rule 3888
Rubi steps
\begin {align*} \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx &=\frac {e^2 \int (-a+a \sec (c+d x)) (e \tan (c+d x))^{-2+m} \, dx}{a^2}\\ &=-\frac {e^2 \int (e \tan (c+d x))^{-2+m} \, dx}{a}+\frac {e^2 \int \sec (c+d x) (e \tan (c+d x))^{-2+m} \, dx}{a}\\ &=-\frac {e \cos ^2(c+d x)^{m/2} \, _2F_1\left (\frac {1}{2} (-1+m),\frac {m}{2};\frac {1+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-1+m}}{a d (1-m)}-\frac {e^3 \operatorname {Subst}\left (\int \frac {x^{-2+m}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a d}\\ &=\frac {e \, _2F_1\left (1,\frac {1}{2} (-1+m);\frac {1+m}{2};-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-1+m}}{a d (1-m)}-\frac {e \cos ^2(c+d x)^{m/2} \, _2F_1\left (\frac {1}{2} (-1+m),\frac {m}{2};\frac {1+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-1+m}}{a d (1-m)}\\ \end {align*}
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Mathematica [F] time = 0.53, size = 0, normalized size = 0.00 \[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (e \tan \left (d x + c\right )\right )^{m}}{a \sec \left (d x + c\right ) + a}, x\right ) \]
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 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{a \sec \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 2.45, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x +c \right )\right )^{m}}{a +a \sec \left (d x +c \right )}\, dx \]
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 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{a \sec \left (d x + c\right ) + a}\,{d x} \]
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 \[ \int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]
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 \[ \frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{m}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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