3.140 \(\int \sin ^4(c+d x) (a+a \sin (c+d x))^n \, dx\)

Optimal. Leaf size=294 \[ -\frac {2^{n+\frac {1}{2}} \left (n^4+6 n^3+17 n^2+12 n+9\right ) \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac {1}{2}} (a \sin (c+d x)+a)^n \, _2F_1\left (\frac {1}{2},\frac {1}{2}-n;\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d (n+1) (n+2) (n+3) (n+4)}+\frac {\left (-n^2-n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1) (n+2) (n+3) (n+4)}-\frac {\left (n^2+3 n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{a d (n+2) (n+3) (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}-\frac {n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3) (n+4)} \]

[Out]

(-n^2-n+9)*cos(d*x+c)*(a+a*sin(d*x+c))^n/d/(n^4+10*n^3+35*n^2+50*n+24)-n*cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+
c))^n/d/(3+n)/(4+n)-cos(d*x+c)*sin(d*x+c)^3*(a+a*sin(d*x+c))^n/d/(4+n)-2^(1/2+n)*(n^4+6*n^3+17*n^2+12*n+9)*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/(4+n)/(n
^3+6*n^2+11*n+6)-(n^2+3*n+9)*cos(d*x+c)*(a+a*sin(d*x+c))^(1+n)/a/d/(4+n)/(n^2+5*n+6)

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Rubi [A]  time = 0.51, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2783, 2983, 2968, 3023, 2751, 2652, 2651} \[ -\frac {2^{n+\frac {1}{2}} \left (n^4+6 n^3+17 n^2+12 n+9\right ) \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac {1}{2}} (a \sin (c+d x)+a)^n \, _2F_1\left (\frac {1}{2},\frac {1}{2}-n;\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d (n+1) (n+2) (n+3) (n+4)}+\frac {\left (-n^2-n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1) (n+2) (n+3) (n+4)}-\frac {\left (n^2+3 n+9\right ) \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{a d (n+2) (n+3) (n+4)}-\frac {\sin ^3(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+4)}-\frac {n \sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3) (n+4)} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^4*(a + a*Sin[c + d*x])^n,x]

[Out]

((9 - n - n^2)*Cos[c + d*x]*(a + a*Sin[c + d*x])^n)/(d*(1 + n)*(2 + n)*(3 + n)*(4 + n)) - (n*Cos[c + d*x]*Sin[
c + d*x]^2*(a + a*Sin[c + d*x])^n)/(d*(3 + n)*(4 + n)) - (Cos[c + d*x]*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^n)/
(d*(4 + n)) - (2^(1/2 + n)*(9 + 12*n + 17*n^2 + 6*n^3 + n^4)*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)*(2 + n)*(3 + n)*(4 + n
)) - ((9 + 3*n + n^2)*Cos[c + d*x]*(a + a*Sin[c + d*x])^(1 + n))/(a*d*(2 + n)*(3 + n)*(4 + 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 2783

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

Rule 2968

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

Rule 2983

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(B*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c*(
m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (I
ntegerQ[n] || EqQ[m + 1/2, 0])

Rule 3023

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

Rubi steps

\begin {align*} \int \sin ^4(c+d x) (a+a \sin (c+d x))^n \, dx &=-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}+\frac {\int \sin ^2(c+d x) (a+a \sin (c+d x))^n (3 a+a n \sin (c+d x)) \, dx}{a (4+n)}\\ &=-\frac {n \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n) (4+n)}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}+\frac {\int \sin (c+d x) (a+a \sin (c+d x))^n \left (2 a^2 n+a^2 \left (9+3 n+n^2\right ) \sin (c+d x)\right ) \, dx}{a^2 (3+n) (4+n)}\\ &=-\frac {n \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n) (4+n)}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}+\frac {\int (a+a \sin (c+d x))^n \left (2 a^2 n \sin (c+d x)+a^2 \left (9+3 n+n^2\right ) \sin ^2(c+d x)\right ) \, dx}{a^2 (3+n) (4+n)}\\ &=-\frac {n \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n) (4+n)}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}-\frac {\left (9+3 n+n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n) (3+n) (4+n)}+\frac {\int (a+a \sin (c+d x))^n \left (a^3 (1+n) \left (9+3 n+n^2\right )-a^3 \left (9-n-n^2\right ) \sin (c+d x)\right ) \, dx}{a^3 (2+n) (3+n) (4+n)}\\ &=\frac {\left (9-n-n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n) (4+n)}-\frac {n \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n) (4+n)}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}-\frac {\left (9+3 n+n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n) (3+n) (4+n)}+\frac {\left (9+12 n+17 n^2+6 n^3+n^4\right ) \int (a+a \sin (c+d x))^n \, dx}{(1+n) (2+n) (3+n) (4+n)}\\ &=\frac {\left (9-n-n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n) (4+n)}-\frac {n \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n) (4+n)}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}-\frac {\left (9+3 n+n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n) (3+n) (4+n)}+\frac {\left (\left (9+12 n+17 n^2+6 n^3+n^4\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) (3+n) (4+n)}\\ &=\frac {\left (9-n-n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n) (4+n)}-\frac {n \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n) (4+n)}-\frac {\cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^n}{d (4+n)}-\frac {2^{\frac {1}{2}+n} \left (9+12 n+17 n^2+6 n^3+n^4\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) (3+n) (4+n)}-\frac {\left (9+3 n+n^2\right ) \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d (2+n) (3+n) (4+n)}\\ \end {align*}

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Mathematica [F]  time = 180.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[Sin[c + d*x]^4*(a + a*Sin[c + d*x])^n,x]

[Out]

$Aborted

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4*(a+a*sin(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((cos(d*x + c)^4 - 2*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 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4*(a+a*sin(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^n*sin(d*x + c)^4, x)

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maple [F]  time = 3.04, size = 0, normalized size = 0.00 \[ \int \left (\sin ^{4}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^4*(a+a*sin(d*x+c))^n,x)

[Out]

int(sin(d*x+c)^4*(a+a*sin(d*x+c))^n,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4*(a+a*sin(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^n*sin(d*x + c)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\sin \left (c+d\,x\right )}^4\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^4*(a + a*sin(c + d*x))^n,x)

[Out]

int(sin(c + d*x)^4*(a + a*sin(c + d*x))^n, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{n} \sin ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**4*(a+a*sin(d*x+c))**n,x)

[Out]

Integral((a*(sin(c + d*x) + 1))**n*sin(c + d*x)**4, x)

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