\(\int \sin (c+d x) (a+a \sin (c+d x))^n \, dx\) [143]

Optimal result

Integrand size = 19, antiderivative size = 109 \[ \int \sin (c+d x) (a+a \sin (c+d x))^n \, dx=-\frac {\cos (c+d x) (a+a \sin (c+d x))^n}{d (1+n)}-\frac {2^{\frac {1}{2}+n} n \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-\frac {1}{2}-n} (a+a \sin (c+d x))^n}{d (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2830, 2731, 2730} \[ \int \sin (c+d x) (a+a \sin (c+d x))^n \, dx=-\frac {2^{n+\frac {1}{2}} n \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac {1}{2}} (a \sin (c+d x)+a)^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{d (n+1)}-\frac {\cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1)} \]

[In]

Int[Sin[c + d*x]*(a + a*Sin[c + d*x])^n,x]

[Out]

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

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]

Rule 2830

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m
, -2^(-1)]

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

Mathematica [C] (verified)

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

Time = 0.39 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.87 \[ \int \sin (c+d x) (a+a \sin (c+d x))^n \, dx=-\frac {2^n \left (B_{\frac {1}{2} (1+\sin (c+d x))}\left (\frac {1}{2}+n,\frac {1}{2}\right )-2 B_{\frac {1}{2} (1+\sin (c+d x))}\left (\frac {3}{2}+n,\frac {1}{2}\right )\right ) \sqrt {\cos ^2(c+d x)} \sec (c+d x) (1+\sin (c+d x))^{-n} (a (1+\sin (c+d x)))^n}{d} \]

[In]

Integrate[Sin[c + d*x]*(a + a*Sin[c + d*x])^n,x]

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((a*sin(d*x + c) + a)^n*sin(d*x + c), x)

Sympy [F]

\[ \int \sin (c+d x) (a+a \sin (c+d x))^n \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{n} \sin {\left (c + d x \right )}\, dx \]

[In]

integrate(sin(d*x+c)*(a+a*sin(d*x+c))**n,x)

[Out]

Integral((a*(sin(c + d*x) + 1))**n*sin(c + d*x), x)

Maxima [F]

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

[In]

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

[Out]

integrate((a*sin(d*x + c) + a)^n*sin(d*x + c), x)

Giac [F]

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

[In]

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

[Out]

integrate((a*sin(d*x + c) + a)^n*sin(d*x + c), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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