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

Optimal. Leaf size=88 \[ -\frac{\sin (e+f x) (a \cos (e+f x))^{m+1} (b \sec (e+f x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{2} (m-n+1);\frac{1}{2} (m-n+3);\cos ^2(e+f x)\right )}{a f (m-n+1) \sqrt{\sin ^2(e+f x)}} \]

[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]))

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Rubi [A]  time = 0.070436, antiderivative size = 88, normalized size of antiderivative = 1., 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 = {2588, 2643} \[ -\frac{\sin (e+f x) (a \cos (e+f x))^{m+1} (b \sec (e+f x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{2} (m-n+1);\frac{1}{2} (m-n+3);\cos ^2(e+f x)\right )}{a f (m-n+1) \sqrt{\sin ^2(e+f x)}} \]

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 2588

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 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[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*}

Mathematica [A]  time = 10.4935, size = 89, normalized size = 1.01 \[ -\frac{\sin (e+f x) \cos (e+f x) (a \cos (e+f x))^m (b \sec (e+f x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{2} (m-n+1);\frac{1}{2} (m-n+3);\cos ^2(e+f x)\right )}{f (m-n+1) \sqrt{\sin ^2(e+f x)}} \]

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 = 1.065, size = 0, normalized size = 0. \begin{align*} \int \left ( a\cos \left ( fx+e \right ) \right ) ^{m} \left ( b\sec \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

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

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

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

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

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)