Optimal. Leaf size=227 \[ \frac{b^2 \tan ^3(e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m F_1\left (\frac{3}{2};2,1-\frac{m}{2};\frac{5}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4 f}+\frac{\tan (e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m F_1\left (\frac{1}{2};2,1-\frac{m}{2};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2 f}-\frac{2 a b (d \sec (e+f x))^m \, _2F_1\left (2,\frac{m}{2};\frac{m+2}{2};\frac{b^2 \sec ^2(e+f x)}{a^2+b^2}\right )}{f m \left (a^2+b^2\right )^2} \]
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Rubi [A] time = 0.197627, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3512, 757, 429, 444, 68, 510} \[ \frac{b^2 \tan ^3(e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m F_1\left (\frac{3}{2};2,1-\frac{m}{2};\frac{5}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4 f}+\frac{\tan (e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m F_1\left (\frac{1}{2};2,1-\frac{m}{2};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2 f}-\frac{2 a b (d \sec (e+f x))^m \, _2F_1\left (2,\frac{m}{2};\frac{m+2}{2};\frac{b^2 \sec ^2(e+f x)}{a^2+b^2}\right )}{f m \left (a^2+b^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 3512
Rule 757
Rule 429
Rule 444
Rule 68
Rule 510
Rubi steps
\begin{align*} \int \frac{(d \sec (e+f x))^m}{(a+b \tan (e+f x))^2} \, 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)^2} \, 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^2 \left (1+\frac{x^2}{b^2}\right )^{-1+\frac{m}{2}}}{\left (a^2-x^2\right )^2}-\frac{2 a x \left (1+\frac{x^2}{b^2}\right )^{-1+\frac{m}{2}}}{\left (a^2-x^2\right )^2}+\frac{x^2 \left (1+\frac{x^2}{b^2}\right )^{-1+\frac{m}{2}}}{\left (-a^2+x^2\right )^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^2 \left (1+\frac{x^2}{b^2}\right )^{-1+\frac{m}{2}}}{\left (-a^2+x^2\right )^2} \, dx,x,b \tan (e+f x)\right )}{b f}-\frac{\left (2 a (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}}}{\left (a^2-x^2\right )^2} \, dx,x,b \tan (e+f x)\right )}{b f}+\frac{\left (a^2 (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}}}{\left (a^2-x^2\right )^2} \, dx,x,b \tan (e+f x)\right )}{b f}\\ &=\frac{F_1\left (\frac{1}{2};2,1-\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^2 f}+\frac{b^2 F_1\left (\frac{3}{2};2,1-\frac{m}{2};\frac{5}{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 ^3(e+f x)}{3 a^4 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}{b^2}\right )^{-1+\frac{m}{2}}}{\left (a^2-x\right )^2} \, dx,x,b^2 \tan ^2(e+f x)\right )}{b f}\\ &=-\frac{2 a b \, _2F_1\left (2,\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 )^2 f m}+\frac{F_1\left (\frac{1}{2};2,1-\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^2 f}+\frac{b^2 F_1\left (\frac{3}{2};2,1-\frac{m}{2};\frac{5}{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 ^3(e+f x)}{3 a^4 f}\\ \end{align*}
Mathematica [C] time = 3.24919, size = 356, normalized size = 1.57 \[ \frac{2 (m-4) (d \sec (e+f x))^m (a \cos (e+f x)+b \sin (e+f x))^2 F_1\left (3-m;1-\frac{m}{2},1-\frac{m}{2};4-m;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right )}{b f (m-3) (a+b \tan (e+f x))^2 \left ((m-2) \left ((a+i b) F_1\left (4-m;1-\frac{m}{2},2-\frac{m}{2};5-m;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right )+(a-i b) F_1\left (4-m;2-\frac{m}{2},1-\frac{m}{2};5-m;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right )\right )+2 (m-4) (a+b \tan (e+f x)) F_1\left (3-m;1-\frac{m}{2},1-\frac{m}{2};4-m;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right )\right )} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.267, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\sec \left ( fx+e \right ) \right ) ^{m}}{ \left ( a+b\tan \left ( fx+e \right ) \right ) ^{2}}}\, 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}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}\,{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^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}}, 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}}{\left (a + b \tan{\left (e + f x \right )}\right )^{2}}\, 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}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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