3.223 \(\int \sin ^2(c+d x) (a+b \sin (c+d x))^n \, dx\)

Optimal. Leaf size=274 \[ -\frac{\sqrt{2} \left (a^2+b^2 (n+1)\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right )}{b^2 d (n+2) \sqrt{\sin (c+d x)+1}}+\frac{\sqrt{2} a (a+b) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n-1;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right )}{b^2 d (n+2) \sqrt{\sin (c+d x)+1}}-\frac{\cos (c+d x) (a+b \sin (c+d x))^{n+1}}{b d (n+2)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.298298, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2791, 2756, 2665, 139, 138} \[ -\frac{\sqrt{2} \left (a^2+b^2 (n+1)\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right )}{b^2 d (n+2) \sqrt{\sin (c+d x)+1}}+\frac{\sqrt{2} a (a+b) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n-1;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right )}{b^2 d (n+2) \sqrt{\sin (c+d x)+1}}-\frac{\cos (c+d x) (a+b \sin (c+d x))^{n+1}}{b d (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^2*(a + b*Sin[c + d*x])^n,x]

[Out]

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

Rule 2791

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

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

Mathematica [F]  time = 6.78741, size = 0, normalized size = 0. \[ \int \sin ^2(c+d x) (a+b \sin (c+d x))^n \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sin[c + d*x]^2*(a + b*Sin[c + d*x])^n,x]

[Out]

Integrate[Sin[c + d*x]^2*(a + b*Sin[c + d*x])^n, x]

________________________________________________________________________________________

Maple [F]  time = 0.222, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( dx+c \right ) \right ) ^{2} \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)^2*(a+b*sin(d*x+c))^n,x)

[Out]

int(sin(d*x+c)^2*(a+b*sin(d*x+c))^n,x)

________________________________________________________________________________________

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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(a+b*sin(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^n*sin(d*x + c)^2, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (\cos \left (d x + c\right )^{2} - 1\right )}{\left (b \sin \left (d x + c\right ) + a\right )}^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(a+b*sin(d*x+c))^n,x, algorithm="fricas")

[Out]

integral(-(cos(d*x + c)^2 - 1)*(b*sin(d*x + c) + a)^n, x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(a+b*sin(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^n*sin(d*x + c)^2, x)