Optimal. Leaf size=141 \[ \frac{\tan (e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m F_1\left (\frac{1}{2};1,1-\frac{m}{2};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a f}-\frac{b (d \sec (e+f x))^m \, _2F_1\left (1,\frac{m}{2};\frac{m+2}{2};\frac{b^2 \sec ^2(e+f x)}{a^2+b^2}\right )}{f m \left (a^2+b^2\right )} \]
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Rubi [A] time = 0.157447, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3512, 757, 429, 444, 68} \[ \frac{\tan (e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m F_1\left (\frac{1}{2};1,1-\frac{m}{2};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a f}-\frac{b (d \sec (e+f x))^m \, _2F_1\left (1,\frac{m}{2};\frac{m+2}{2};\frac{b^2 \sec ^2(e+f x)}{a^2+b^2}\right )}{f m \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
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Rule 3512
Rule 757
Rule 429
Rule 444
Rule 68
Rubi steps
\begin{align*} \int \frac{(d \sec (e+f x))^m}{a+b \tan (e+f x)} \, dx &=\frac{\left ((d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{x^2}{b^2}\right )^{-1+\frac{m}{2}}}{a+x} \, dx,x,b \tan (e+f x)\right )}{b f}\\ &=\frac{\left ((d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}\right ) \operatorname{Subst}\left (\int \left (\frac{a \left (1+\frac{x^2}{b^2}\right )^{-1+\frac{m}{2}}}{a^2-x^2}+\frac{x \left (1+\frac{x^2}{b^2}\right )^{-1+\frac{m}{2}}}{-a^2+x^2}\right ) \, dx,x,b \tan (e+f x)\right )}{b f}\\ &=\frac{\left ((d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}\right ) \operatorname{Subst}\left (\int \frac{x \left (1+\frac{x^2}{b^2}\right )^{-1+\frac{m}{2}}}{-a^2+x^2} \, dx,x,b \tan (e+f x)\right )}{b f}+\frac{\left (a (d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{x^2}{b^2}\right )^{-1+\frac{m}{2}}}{a^2-x^2} \, dx,x,b \tan (e+f x)\right )}{b f}\\ &=\frac{F_1\left (\frac{1}{2};1,1-\frac{m}{2};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2} \tan (e+f x)}{a f}+\frac{\left ((d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{x}{b^2}\right )^{-1+\frac{m}{2}}}{-a^2+x} \, dx,x,b^2 \tan ^2(e+f x)\right )}{2 b f}\\ &=-\frac{b \, _2F_1\left (1,\frac{m}{2};\frac{2+m}{2};\frac{b^2 \sec ^2(e+f x)}{a^2+b^2}\right ) (d \sec (e+f x))^m}{\left (a^2+b^2\right ) f m}+\frac{F_1\left (\frac{1}{2};1,1-\frac{m}{2};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2} \tan (e+f x)}{a f}\\ \end{align*}
Mathematica [C] time = 15.053, size = 1158, normalized size = 8.21 \[ \frac{(d \sec (e+f x))^m \left (b F_1\left (-m;-\frac{m}{2},-\frac{m}{2};1-m;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right ) \sec ^2(e+f x)^{m/2} \left (\frac{b (\tan (e+f x)-i)}{a+b \tan (e+f x)}\right )^{-m/2} \left (\frac{b (\tan (e+f x)+i)}{a+b \tan (e+f x)}\right )^{-m/2}-b \sec ^2(e+f x)^{m/2}+b+a m \, _2F_1\left (\frac{1}{2},1-\frac{m}{2};\frac{3}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)\right )}{f (a+b \tan (e+f x)) \left (-\frac{1}{2} b m F_1\left (-m;-\frac{m}{2},-\frac{m}{2};1-m;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right ) \sec ^2(e+f x)^{m/2} \left (\frac{b (\tan (e+f x)+i)}{a+b \tan (e+f x)}\right )^{-m/2} \left (\frac{b \sec ^2(e+f x)}{a+b \tan (e+f x)}-\frac{b^2 \sec ^2(e+f x) (\tan (e+f x)-i)}{(a+b \tan (e+f x))^2}\right ) \left (\frac{b (\tan (e+f x)-i)}{a+b \tan (e+f x)}\right )^{-\frac{m}{2}-1}+b m F_1\left (-m;-\frac{m}{2},-\frac{m}{2};1-m;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right ) \sec ^2(e+f x)^{m/2} \tan (e+f x) \left (\frac{b (\tan (e+f x)+i)}{a+b \tan (e+f x)}\right )^{-m/2} \left (\frac{b (\tan (e+f x)-i)}{a+b \tan (e+f x)}\right )^{-m/2}+b \sec ^2(e+f x)^{m/2} \left (\frac{b (\tan (e+f x)+i)}{a+b \tan (e+f x)}\right )^{-m/2} \left (-\frac{(a-i b) b m^2 F_1\left (1-m;1-\frac{m}{2},-\frac{m}{2};2-m;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right ) \sec ^2(e+f x)}{2 (1-m) (a+b \tan (e+f x))^2}-\frac{(a+i b) b m^2 F_1\left (1-m;-\frac{m}{2},1-\frac{m}{2};2-m;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right ) \sec ^2(e+f x)}{2 (1-m) (a+b \tan (e+f x))^2}\right ) \left (\frac{b (\tan (e+f x)-i)}{a+b \tan (e+f x)}\right )^{-m/2}-\frac{1}{2} b m F_1\left (-m;-\frac{m}{2},-\frac{m}{2};1-m;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right ) \sec ^2(e+f x)^{m/2} \left (\frac{b (\tan (e+f x)+i)}{a+b \tan (e+f x)}\right )^{-\frac{m}{2}-1} \left (\frac{b \sec ^2(e+f x)}{a+b \tan (e+f x)}-\frac{b^2 \sec ^2(e+f x) (\tan (e+f x)+i)}{(a+b \tan (e+f x))^2}\right ) \left (\frac{b (\tan (e+f x)-i)}{a+b \tan (e+f x)}\right )^{-m/2}+a m \, _2F_1\left (\frac{1}{2},1-\frac{m}{2};\frac{3}{2};-\tan ^2(e+f x)\right ) \sec ^2(e+f x)-b m \sec ^2(e+f x)^{m/2} \tan (e+f x)+a m \sec ^2(e+f x) \left (\left (\tan ^2(e+f x)+1\right )^{\frac{m}{2}-1}-\, _2F_1\left (\frac{1}{2},1-\frac{m}{2};\frac{3}{2};-\tan ^2(e+f x)\right )\right )\right )} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.228, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\sec \left ( fx+e \right ) \right ) ^{m}}{a+b\tan \left ( fx+e \right ) }}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \sec \left (f x + e\right )\right )^{m}}{b \tan \left (f x + e\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (d \sec \left (f x + e\right )\right )^{m}}{b \tan \left (f x + e\right ) + a}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \sec{\left (e + f x \right )}\right )^{m}}{a + b \tan{\left (e + f x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \sec \left (f x + e\right )\right )^{m}}{b \tan \left (f x + e\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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