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 = 0.16, antiderivative size = 141, normalized size of antiderivative = 1.00, 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 68
Rule 429
Rule 444
Rule 757
Rule 3512
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*}
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Mathematica [C] time = 15.05, 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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fricas [F] time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (d \sec \left (f x + e\right )\right )^{m}}{b \tan \left (f x + e\right ) + a}, x\right ) \]
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 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{m}}{b \tan \left (f x + e\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.26, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x +e \right )\right )^{m}}{a +b \tan \left (f x +e \right )}\, dx \]
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 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{m}}{b \tan \left (f x + e\right ) + a}\,{d x} \]
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 \[ \int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^m}{a+b\,\mathrm {tan}\left (e+f\,x\right )} \,d x \]
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 \[ \int \frac {\left (d \sec {\left (e + f x \right )}\right )^{m}}{a + b \tan {\left (e + f x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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