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

Optimal. Leaf size=156 \[ -\frac{2^{n+\frac{1}{2}} \left (n^2+n+1\right ) \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac{1}{2}} (a \sin (c+d x)+a)^n \, _2F_1\left (\frac{1}{2},\frac{1}{2}-n;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d (n+1) (n+2)}+\frac{\cos (c+d x) (a \sin (c+d x)+a)^n}{d \left (n^2+3 n+2\right )}-\frac{\cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{a d (n+2)} \]

[Out]

(Cos[c + d*x]*(a + a*Sin[c + d*x])^n)/(d*(2 + 3*n + n^2)) - (2^(1/2 + n)*(1 + n + n^2)*Cos[c + d*x]*Hypergeome
tric2F1[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)*(2 + n)) - (Cos[c + d*x]*(a + a*Sin[c + d*x])^(1 + n))/(a*d*(2 + n))

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Rubi [A]  time = 0.142928, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2759, 2751, 2652, 2651} \[ -\frac{2^{n+\frac{1}{2}} \left (n^2+n+1\right ) \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac{1}{2}} (a \sin (c+d x)+a)^n \, _2F_1\left (\frac{1}{2},\frac{1}{2}-n;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d (n+1) (n+2)}+\frac{\cos (c+d x) (a \sin (c+d x)+a)^n}{d \left (n^2+3 n+2\right )}-\frac{\cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{a d (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]*(a + a*Sin[c + d*x])^n)/(d*(2 + 3*n + n^2)) - (2^(1/2 + n)*(1 + n + n^2)*Cos[c + d*x]*Hypergeome
tric2F1[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)*(2 + n)) - (Cos[c + d*x]*(a + a*Sin[c + d*x])^(1 + n))/(a*d*(2 + n))

Rule 2759

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

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

Mathematica [C]  time = 54.1764, size = 28439, normalized size = 182.3 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]  time = 0.907, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( dx+c \right ) \right ) ^{2} \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)^2*(a+a*sin(d*x+c))^n,x)

[Out]

int(sin(d*x+c)^2*(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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integral(-(cos(d*x + c)^2 - 1)*(a*sin(d*x + c) + a)^n, 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 ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sin(c + d*x) + 1))**n*sin(c + d*x)**2, 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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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