\(\int (3+3 \sin (e+f x))^m \, dx\) [610]

Optimal result

Integrand size = 12, antiderivative size = 74 \[ \int (3+3 \sin (e+f x))^m \, dx=-\frac {2^{\frac {1}{2}+m} \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (3+3 \sin (e+f x))^m}{f} \]

[Out]

-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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2731, 2730} \[ \int (3+3 \sin (e+f x))^m \, dx=-\frac {2^{m+\frac {1}{2}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{f} \]

[In]

Int[(a + a*Sin[e + f*x])^m,x]

[Out]

-((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 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*} \text {integral}& = \left ((1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^m \, dx \\ & = -\frac {2^{\frac {1}{2}+m} \cos (e+f x) \operatorname {Hypergeometric2F1}\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 [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.

Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.62 \[ \int (3+3 \sin (e+f x))^m \, dx=\frac {6^m B_{\frac {1}{2} (1+\sin (e+f x))}\left (\frac {1}{2}+m,\frac {1}{2}\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x)}{f} \]

[In]

Integrate[(3 + 3*Sin[e + f*x])^m,x]

[Out]

(6^m*Beta[(1 + Sin[e + f*x])/2, 1/2 + m, 1/2]*Sqrt[Cos[e + f*x]^2]*Sec[e + f*x])/f

Maple [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

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

[Out]

integral((a*sin(f*x + e) + a)^m, x)

Sympy [F]

\[ \int (3+3 \sin (e+f x))^m \, dx=\int \left (a \sin {\left (e + f x \right )} + a\right )^{m}\, dx \]

[In]

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

[Out]

Integral((a*sin(e + f*x) + a)**m, x)

Maxima [F]

\[ \int (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

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

[Out]

integrate((a*sin(f*x + e) + a)^m, x)

Giac [F]

\[ \int (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

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

[Out]

integrate((a*sin(f*x + e) + a)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (3+3 \sin (e+f x))^m \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]

[In]

int((a + a*sin(e + f*x))^m,x)

[Out]

int((a + a*sin(e + f*x))^m, x)