3.647 \(\int (-3)^{-1-m} (a+a \sin (e+f x))^m \, dx\)

Optimal. Leaf size=81 \[ -\frac{(-3)^{-m-1} 2^{m+\frac{1}{2}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.0470816, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {12, 2652, 2651} \[ -\frac{(-3)^{-m-1} 2^{m+\frac{1}{2}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(-3)^(-1 - m)*(a + a*Sin[e + f*x])^m,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2652

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Sin[c + d*x])^FracPart
[n])/(1 + (b*Sin[c + d*x])/a)^FracPart[n], Int[(1 + (b*Sin[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rubi steps

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

Mathematica [A]  time = 0.156529, size = 97, normalized size = 1.2 \[ \frac{\sqrt{2} (-3)^{-m-1} \cos (e+f x) (a (\sin (e+f x)+1))^m \, _2F_1\left (\frac{1}{2},m+\frac{1}{2};m+\frac{3}{2};\frac{1}{4} \cos ^2(e+f x) \csc ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )}{(2 f m+f) \sqrt{1-\sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-3)^(-1 - m)*(a + a*Sin[e + f*x])^m,x]

[Out]

((-3)^(-1 - m)*Sqrt[2]*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 + m, 3/2 + m, (Cos[e + f*x]^2*Csc[(2*e - Pi + 2
*f*x)/4]^2)/4]*(a*(1 + Sin[e + f*x]))^m)/((f + 2*f*m)*Sqrt[1 - Sin[e + f*x]])

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Maple [F]  time = 0.332, size = 0, normalized size = 0. \begin{align*} \int \left ( -3 \right ) ^{-1-m} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3)^(-1-m)*(a+a*sin(f*x+e))^m,x)

[Out]

int((-3)^(-1-m)*(a+a*sin(f*x+e))^m,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left (-3\right )^{-m - 1} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3)^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="maxima")

[Out]

(-3)^(-m - 1)*integrate((a*sin(f*x + e) + a)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (-3\right )^{-m - 1}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3)^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((-3)^(-m - 1)*(a*sin(f*x + e) + a)^m, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left (-3\right )^{- m - 1} \int \left (a \sin{\left (e + f x \right )} + a\right )^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3)**(-1-m)*(a+a*sin(f*x+e))**m,x)

[Out]

(-3)**(-m - 1)*Integral((a*sin(e + f*x) + a)**m, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (-3\right )^{-m - 1}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3)^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((-3)^(-m - 1)*(a*sin(f*x + e) + a)^m, x)