3.143 \(\int \sin (c+d x) (a+a \sin (c+d x))^n \, dx\)

Optimal. Leaf size=109 \[ -\frac{2^{n+\frac{1}{2}} n \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac{1}{2}} (a \sin (c+d x)+a)^n \, _2F_1\left (\frac{1}{2},\frac{1}{2}-n;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d (n+1)}-\frac{\cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1)} \]

[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))

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Rubi [A]  time = 0.0638162, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2751, 2652, 2651} \[ -\frac{2^{n+\frac{1}{2}} n \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac{1}{2}} (a \sin (c+d x)+a)^n \, _2F_1\left (\frac{1}{2},\frac{1}{2}-n;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d (n+1)}-\frac{\cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1)} \]

Antiderivative was successfully verified.

[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 2751

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*Sin
[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)]

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 \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{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) \, _2F_1\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]  time = 0.443593, size = 178, normalized size = 1.63 \[ -\frac{\sqrt [4]{-1} 2^{-2 n-1} e^{-\frac{3}{2} i (c+d x)} \left (-(-1)^{3/4} e^{-\frac{1}{2} i (c+d x)} \left (e^{i (c+d x)}+i\right )\right )^{2 n+1} \left ((n-1) e^{2 i (c+d x)} \, _2F_1\left (1,n;-n;-i e^{-i (c+d x)}\right )-(n+1) \, _2F_1\left (1,n+2;2-n;-i e^{-i (c+d x)}\right )\right ) \sin ^{-2 n}\left (\frac{1}{4} (2 c+2 d x+\pi )\right ) (a (\sin (c+d x)+1))^n}{d (n-1) (n+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(((-1)^(1/4)*2^(-1 - 2*n)*(-(((-1)^(3/4)*(I + E^(I*(c + d*x))))/E^((I/2)*(c + d*x))))^(1 + 2*n)*(E^((2*I)*(c
+ d*x))*(-1 + n)*Hypergeometric2F1[1, n, -n, (-I)/E^(I*(c + d*x))] - (1 + n)*Hypergeometric2F1[1, 2 + n, 2 - n
, (-I)/E^(I*(c + d*x))])*(a*(1 + Sin[c + d*x]))^n)/(d*E^(((3*I)/2)*(c + d*x))*(-1 + n)*(1 + n)*Sin[(2*c + Pi +
2*d*x)/4]^(2*n)))

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Maple [F]  time = 0.875, size = 0, normalized size = 0. \begin{align*} \int \sin \left ( dx+c \right ) \left ( a+a\sin \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sin{\left (c + d x \right )} + 1\right )\right )^{n} \sin{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)