3.7.47 \(\int (-3)^{-1-m} (a+a \sin (e+f x))^m \, dx\) [647]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {12, 2731, 2730} \begin {gather*} -\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} \end {gather*}

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 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[a^IntPart[n]*((a + b*Sin[c + d*x])^FracPart
[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n]), Int[(1 + (b/a)*Sin[c + d*x])^n, 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*}

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Mathematica [A]
time = 0.17, size = 97, normalized size = 1.20 \begin {gather*} \frac {(-3)^{-1-m} \sqrt {2} \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}+m;\frac {3}{2}+m;\frac {1}{4} \cos ^2(e+f x) \csc ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )\right ) (a (1+\sin (e+f x)))^m}{(f+2 f m) \sqrt {1-\sin (e+f x)}} \end {gather*}

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 = 0.04, size = 0, normalized size = 0.00 \[\int \left (-3\right )^{-1-m} \left (a +a \sin \left (f x +e \right )\right )^{m}\, dx\]

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 = 0.00, 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((-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 = 0.00, 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((-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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \left (-3\right )^{- m - 1} \int \left (a \sin {\left (e + f x \right )} + a\right )^{m}\, dx \end {gather*}

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 = 0.00, 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((-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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (-3\right )}^{m+1}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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