\(\int (a-a \cos (c+d x))^n \, dx\) [15]

Optimal result

Integrand size = 13, antiderivative size = 75 \[ \int (a-a \cos (c+d x))^n \, dx=-\frac {2^{\frac {1}{2}+n} (1-\cos (c+d x))^{-\frac {1}{2}-n} (a-a \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1+\cos (c+d x))\right ) \sin (c+d x)}{d} \]

[Out]

-2^(1/2+n)*(1-cos(d*x+c))^(-1/2-n)*(a-a*cos(d*x+c))^n*hypergeom([1/2, 1/2-n],[3/2],1/2+1/2*cos(d*x+c))*sin(d*x
+c)/d

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2731, 2730} \[ \int (a-a \cos (c+d x))^n \, dx=-\frac {2^{n+\frac {1}{2}} \sin (c+d x) (1-\cos (c+d x))^{-n-\frac {1}{2}} (a-a \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (\cos (c+d x)+1)\right )}{d} \]

[In]

Int[(a - a*Cos[c + d*x])^n,x]

[Out]

-((2^(1/2 + n)*(1 - Cos[c + d*x])^(-1/2 - n)*(a - a*Cos[c + d*x])^n*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1 +
Cos[c + d*x])/2]*Sin[c + d*x])/d)

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-\cos (c+d x))^{-n} (a-a \cos (c+d x))^n\right ) \int (1-\cos (c+d x))^n \, dx \\ & = -\frac {2^{\frac {1}{2}+n} (1-\cos (c+d x))^{-\frac {1}{2}-n} (a-a \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1+\cos (c+d x))\right ) \sin (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int (a-a \cos (c+d x))^n \, dx=\frac {\sqrt {2} \sqrt {1+\cos (c+d x)} (a-a \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}+n,\frac {3}{2}+n,\sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{d+2 d n} \]

[In]

Integrate[(a - a*Cos[c + d*x])^n,x]

[Out]

(Sqrt[2]*Sqrt[1 + Cos[c + d*x]]*(a - a*Cos[c + d*x])^n*Hypergeometric2F1[1/2, 1/2 + n, 3/2 + n, Sin[(c + d*x)/
2]^2]*Tan[(c + d*x)/2])/(d + 2*d*n)

Maple [F]

\[\int \left (a -\cos \left (d x +c \right ) a \right )^{n}d x\]

[In]

int((a-cos(d*x+c)*a)^n,x)

[Out]

int((a-cos(d*x+c)*a)^n,x)

Fricas [F]

\[ \int (a-a \cos (c+d x))^n \, dx=\int { {\left (-a \cos \left (d x + c\right ) + a\right )}^{n} \,d x } \]

[In]

integrate((a-a*cos(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((-a*cos(d*x + c) + a)^n, x)

Sympy [F]

\[ \int (a-a \cos (c+d x))^n \, dx=\int \left (- a \cos {\left (c + d x \right )} + a\right )^{n}\, dx \]

[In]

integrate((a-a*cos(d*x+c))**n,x)

[Out]

Integral((-a*cos(c + d*x) + a)**n, x)

Maxima [F]

\[ \int (a-a \cos (c+d x))^n \, dx=\int { {\left (-a \cos \left (d x + c\right ) + a\right )}^{n} \,d x } \]

[In]

integrate((a-a*cos(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((-a*cos(d*x + c) + a)^n, x)

Giac [F]

\[ \int (a-a \cos (c+d x))^n \, dx=\int { {\left (-a \cos \left (d x + c\right ) + a\right )}^{n} \,d x } \]

[In]

integrate((a-a*cos(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((-a*cos(d*x + c) + a)^n, x)

Mupad [F(-1)]

Timed out. \[ \int (a-a \cos (c+d x))^n \, dx=\int {\left (a-a\,\cos \left (c+d\,x\right )\right )}^n \,d x \]

[In]

int((a - a*cos(c + d*x))^n,x)

[Out]

int((a - a*cos(c + d*x))^n, x)