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

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Rubi [A]  time = 0.14, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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]

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 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 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)]

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*}

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Mathematica [C]  time = 54.58, size = 28439, normalized size = 182.30 \[ \text {Result too large to show} \]

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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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (\cos \left (d x + c\right )^{2} - 1\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{n}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{2}\,{d x} \]

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)

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maple [F]  time = 1.82, size = 0, normalized size = 0.00 \[ \int \left (\sin ^{2}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{n}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{2}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{n} \sin ^{2}{\left (c + d x \right )}\, dx \]

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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