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(f*x+e)^2)^(1/2+1/2*m+1/2*n)*hypergeom([1/2+1/2*n, 1/2+1/2*m+1/2*n],[3/2+1/2*n],sin(f*x+e)^2)*(a*sec(f*x+e
))^m*(b*tan(f*x+e))^(1+n)/b/f/(1+n)

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Rubi [A]  time = 0.05, antiderivative size = 82, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.13, 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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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (a \sec \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sec \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n}\,{d x} \]

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)

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maple [F]  time = 1.46, size = 0, normalized size = 0.00 \[ \int \left (a \sec \left (f x +e \right )\right )^{m} \left (b \tan \left (f x +e \right )\right )^{n}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sec \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (\frac {a}{\cos \left (e+f\,x\right )}\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sec {\left (e + f x \right )}\right )^{m} \left (b \tan {\left (e + f x \right )}\right )^{n}\, dx \]

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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