3.362 \(\int (a \sec (e+f x))^m (b \tan (e+f x))^n \, dx\)

Optimal. Leaf size=82 \[ \frac{(a \sec (e+f x))^m (b \tan (e+f x))^{n+1} \cos ^2(e+f x)^{\frac{1}{2} (m+n+1)} \, _2F_1\left (\frac{n+1}{2},\frac{1}{2} (m+n+1);\frac{n+3}{2};\sin ^2(e+f x)\right )}{b f (n+1)} \]

[Out]

((Cos[e + f*x]^2)^((1 + m + n)/2)*Hypergeometric2F1[(1 + n)/2, (1 + m + n)/2, (3 + n)/2, Sin[e + f*x]^2]*(a*Se
c[e + f*x])^m*(b*Tan[e + f*x])^(1 + n))/(b*f*(1 + n))

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Rubi [A]  time = 0.045036, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {2617} \[ \frac{(a \sec (e+f x))^m (b \tan (e+f x))^{n+1} \cos ^2(e+f x)^{\frac{1}{2} (m+n+1)} \, _2F_1\left (\frac{n+1}{2},\frac{1}{2} (m+n+1);\frac{n+3}{2};\sin ^2(e+f x)\right )}{b f (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sec[e + f*x])^m*(b*Tan[e + f*x])^n,x]

[Out]

((Cos[e + f*x]^2)^((1 + m + n)/2)*Hypergeometric2F1[(1 + n)/2, (1 + m + n)/2, (3 + n)/2, Sin[e + f*x]^2]*(a*Se
c[e + f*x])^m*(b*Tan[e + f*x])^(1 + n))/(b*f*(1 + n))

Rule 2617

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1)*(Cos[e + f*x]^2)^((m + n + 1)/2)*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2,
(n + 3)/2, Sin[e + f*x]^2])/(b*f*(n + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rubi steps

\begin{align*} \int (a \sec (e+f x))^m (b \tan (e+f x))^n \, dx &=\frac{\cos ^2(e+f x)^{\frac{1}{2} (1+m+n)} \, _2F_1\left (\frac{1+n}{2},\frac{1}{2} (1+m+n);\frac{3+n}{2};\sin ^2(e+f x)\right ) (a \sec (e+f x))^m (b \tan (e+f x))^{1+n}}{b f (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.144983, size = 80, normalized size = 0.98 \[ \frac{b \left (-\tan ^2(e+f x)\right )^{\frac{1-n}{2}} (a \sec (e+f x))^m (b \tan (e+f x))^{n-1} \, _2F_1\left (\frac{m}{2},\frac{1-n}{2};\frac{m+2}{2};\sec ^2(e+f x)\right )}{f m} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sec[e + f*x])^m*(b*Tan[e + f*x])^n,x]

[Out]

(b*Hypergeometric2F1[m/2, (1 - n)/2, (2 + m)/2, Sec[e + f*x]^2]*(a*Sec[e + f*x])^m*(b*Tan[e + f*x])^(-1 + n)*(
-Tan[e + f*x]^2)^((1 - n)/2))/(f*m)

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Maple [F]  time = 0.748, size = 0, normalized size = 0. \begin{align*} \int \left ( a\sec \left ( fx+e \right ) \right ) ^{m} \left ( b\tan \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*sec(f*x+e))^m*(b*tan(f*x+e))^n,x)

[Out]

int((a*sec(f*x+e))^m*(b*tan(f*x+e))^n,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \sec \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(f*x+e))^m*(b*tan(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((a*sec(f*x + e))^m*(b*tan(f*x + e))^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (a \sec \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(f*x+e))^m*(b*tan(f*x+e))^n,x, algorithm="fricas")

[Out]

integral((a*sec(f*x + e))^m*(b*tan(f*x + e))^n, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \sec{\left (e + f x \right )}\right )^{m} \left (b \tan{\left (e + f x \right )}\right )^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(f*x+e))**m*(b*tan(f*x+e))**n,x)

[Out]

Integral((a*sec(e + f*x))**m*(b*tan(e + f*x))**n, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \sec \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*sec(f*x+e))^m*(b*tan(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((a*sec(f*x + e))^m*(b*tan(f*x + e))^n, x)