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

Optimal. Leaf size=220 \[ \frac{\sqrt{2} a \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right )}{b d \sqrt{\sin (c+d x)+1}}-\frac{\sqrt{2} (a+b) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n-1;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right )}{b d \sqrt{\sin (c+d x)+1}} \]

[Out]

-((Sqrt[2]*(a + b)*AppellF1[1/2, 1/2, -1 - n, 3/2, (1 - Sin[c + d*x])/2, (b*(1 - Sin[c + d*x]))/(a + b)]*Cos[c
+ d*x]*(a + b*Sin[c + d*x])^n)/(b*d*Sqrt[1 + Sin[c + d*x]]*((a + b*Sin[c + d*x])/(a + b))^n)) + (Sqrt[2]*a*Ap
pellF1[1/2, 1/2, -n, 3/2, (1 - Sin[c + d*x])/2, (b*(1 - Sin[c + d*x]))/(a + b)]*Cos[c + d*x]*(a + b*Sin[c + d*
x])^n)/(b*d*Sqrt[1 + Sin[c + d*x]]*((a + b*Sin[c + d*x])/(a + b))^n)

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Rubi [A]  time = 0.188408, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2756, 2665, 139, 138} \[ \frac{\sqrt{2} a \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right )}{b d \sqrt{\sin (c+d x)+1}}-\frac{\sqrt{2} (a+b) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n-1;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right )}{b d \sqrt{\sin (c+d x)+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(a + b)*AppellF1[1/2, 1/2, -1 - n, 3/2, (1 - Sin[c + d*x])/2, (b*(1 - Sin[c + d*x]))/(a + b)]*Cos[c
+ d*x]*(a + b*Sin[c + d*x])^n)/(b*d*Sqrt[1 + Sin[c + d*x]]*((a + b*Sin[c + d*x])/(a + b))^n)) + (Sqrt[2]*a*Ap
pellF1[1/2, 1/2, -n, 3/2, (1 - Sin[c + d*x])/2, (b*(1 - Sin[c + d*x]))/(a + b)]*Cos[c + d*x]*(a + b*Sin[c + d*
x])^n)/(b*d*Sqrt[1 + Sin[c + d*x]]*((a + b*Sin[c + d*x])/(a + b))^n)

Rule 2756

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

Rule 2665

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt
[1 - Sin[c + d*x]]), Subst[Int[(a + b*x)^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b,
c, d, n}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*n]

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rubi steps

\begin{align*} \int \sin (c+d x) (a+b \sin (c+d x))^n \, dx &=\frac{\int (a+b \sin (c+d x))^{1+n} \, dx}{b}-\frac{a \int (a+b \sin (c+d x))^n \, dx}{b}\\ &=\frac{\cos (c+d x) \operatorname{Subst}\left (\int \frac{(a+b x)^{1+n}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sin (c+d x)\right )}{b d \sqrt{1-\sin (c+d x)} \sqrt{1+\sin (c+d x)}}-\frac{(a \cos (c+d x)) \operatorname{Subst}\left (\int \frac{(a+b x)^n}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sin (c+d x)\right )}{b d \sqrt{1-\sin (c+d x)} \sqrt{1+\sin (c+d x)}}\\ &=-\frac{\left (a \cos (c+d x) (a+b \sin (c+d x))^n \left (-\frac{a+b \sin (c+d x)}{-a-b}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^n}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sin (c+d x)\right )}{b d \sqrt{1-\sin (c+d x)} \sqrt{1+\sin (c+d x)}}-\frac{\left ((-a-b) \cos (c+d x) (a+b \sin (c+d x))^n \left (-\frac{a+b \sin (c+d x)}{-a-b}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^{1+n}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\sin (c+d x)\right )}{b d \sqrt{1-\sin (c+d x)} \sqrt{1+\sin (c+d x)}}\\ &=-\frac{\sqrt{2} (a+b) F_1\left (\frac{1}{2};\frac{1}{2},-1-n;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n}}{b d \sqrt{1+\sin (c+d x)}}+\frac{\sqrt{2} a F_1\left (\frac{1}{2};\frac{1}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n}}{b d \sqrt{1+\sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.470251, size = 193, normalized size = 0.88 \[ \frac{\sec (c+d x) \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}} \sqrt{\frac{b (\sin (c+d x)+1)}{b-a}} (a+b \sin (c+d x))^{n+1} \left ((n+1) (a+b \sin (c+d x)) F_1\left (n+2;\frac{1}{2},\frac{1}{2};n+3;\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right )-a (n+2) F_1\left (n+1;\frac{1}{2},\frac{1}{2};n+2;\frac{a+b \sin (c+d x)}{a-b},\frac{a+b \sin (c+d x)}{a+b}\right )\right )}{b^2 d (n+1) (n+2)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sec[c + d*x]*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]*Sqrt[(b*(1 + Sin[c + d*x]))/(-a + b)]*(a + b*Sin[c + d*
x])^(1 + n)*(-(a*(2 + n)*AppellF1[1 + n, 1/2, 1/2, 2 + n, (a + b*Sin[c + d*x])/(a - b), (a + b*Sin[c + d*x])/(
a + b)]) + (1 + n)*AppellF1[2 + n, 1/2, 1/2, 3 + n, (a + b*Sin[c + d*x])/(a - b), (a + b*Sin[c + d*x])/(a + b)
]*(a + b*Sin[c + d*x])))/(b^2*d*(1 + n)*(2 + n))

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Maple [F]  time = 0.145, size = 0, normalized size = 0. \begin{align*} \int \sin \left ( dx+c \right ) \left ( a+b\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+b*sin(d*x+c))^n,x)

[Out]

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

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

[Out]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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