3.3.60 \(\int (a \cos (e+f x))^m (b \sec (e+f x))^n \, dx\) [260]

Optimal. Leaf size=88 \[ -\frac {(a \cos (e+f x))^{1+m} \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+m-n),\frac {1}{2} (3+m-n),\cos ^2(e+f x)\right ) (b \sec (e+f x))^n \sin (e+f x)}{a f (1+m-n) \sqrt {\sin ^2(e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2668, 2722} \begin {gather*} -\frac {\sin (e+f x) (a \cos (e+f x))^{m+1} (b \sec (e+f x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{2} (m-n+1);\frac {1}{2} (m-n+3);\cos ^2(e+f x)\right )}{a f (m-n+1) \sqrt {\sin ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a*Cos[e + f*x])^(1 + m)*Hypergeometric2F1[1/2, (1 + m - n)/2, (3 + m - n)/2, Cos[e + f*x]^2]*(b*Sec[e + f*
x])^n*Sin[e + f*x])/(a*f*(1 + m - n)*Sqrt[Sin[e + f*x]^2]))

Rule 2668

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[(a*b)^IntPar
t[n]*(a*Sin[e + f*x])^FracPart[n]*(b*Csc[e + f*x])^FracPart[n], Int[(a*Sin[e + f*x])^(m - n), x], x] /; FreeQ[
{a, b, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

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

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Mathematica [A]
time = 6.76, size = 89, normalized size = 1.01 \begin {gather*} -\frac {\cos (e+f x) (a \cos (e+f x))^m \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+m-n),\frac {1}{2} (3+m-n),\cos ^2(e+f x)\right ) (b \sec (e+f x))^n \sin (e+f x)}{f (1+m-n) \sqrt {\sin ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[e + f*x]*(a*Cos[e + f*x])^m*Hypergeometric2F1[1/2, (1 + m - n)/2, (3 + m - n)/2, Cos[e + f*x]^2]*(b*Sec
[e + f*x])^n*Sin[e + f*x])/(f*(1 + m - n)*Sqrt[Sin[e + f*x]^2]))

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \left (a \cos \left (f x +e \right )\right )^{m} \left (b \sec \left (f x +e \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \cos {\left (e + f x \right )}\right )^{m} \left (b \sec {\left (e + f x \right )}\right )^{n}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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