\(\int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx\) [141]

Optimal result

Integrand size = 21, antiderivative size = 215 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=-\frac {(4+n) \cos (c+d x) (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n)}-\frac {\cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n)}-\frac {2^{\frac {1}{2}+n} n \left (5+3 n+n^2\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-\frac {1}{2}-n} (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n)}-\frac {n \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d \left (6+5 n+n^2\right )} \]

[Out]

-(4+n)*cos(d*x+c)*(a+a*sin(d*x+c))^n/d/(n^3+6*n^2+11*n+6)-cos(d*x+c)*sin(d*x+c)^2*(a+a*sin(d*x+c))^n/d/(3+n)-2
^(1/2+n)*n*(n^2+3*n+5)*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^3+6*n^2+11*n+6)-n*cos(d*x+c)*(a+a*sin(d*x+c))^(1+n)/a/d/(n^2+5*n+6)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2862, 3047, 3102, 2830, 2731, 2730} \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=-\frac {2^{n+\frac {1}{2}} n \left (n^2+3 n+5\right ) \cos (c+d x) (\sin (c+d x)+1)^{-n-\frac {1}{2}} (a \sin (c+d x)+a)^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{d (n+1) (n+2) (n+3)}-\frac {n \cos (c+d x) (a \sin (c+d x)+a)^{n+1}}{a d \left (n^2+5 n+6\right )}-\frac {\sin ^2(c+d x) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+3)}-\frac {(n+4) \cos (c+d x) (a \sin (c+d x)+a)^n}{d (n+1) (n+2) (n+3)} \]

[In]

Int[Sin[c + d*x]^3*(a + a*Sin[c + d*x])^n,x]

[Out]

-(((4 + n)*Cos[c + d*x]*(a + a*Sin[c + d*x])^n)/(d*(1 + n)*(2 + n)*(3 + n))) - (Cos[c + d*x]*Sin[c + d*x]^2*(a
+ a*Sin[c + d*x])^n)/(d*(3 + n)) - (2^(1/2 + n)*n*(5 + 3*n + n^2)*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))
- (n*Cos[c + d*x]*(a + a*Sin[c + d*x])^(1 + n))/(a*d*(6 + 5*n + n^2))

Rule 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

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

Rule 2830

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*S
in[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 2862

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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] && E
qQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 1] && IntegerQ[n]

Rule 3047

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 3102

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*} \text {integral}& = -\frac {\cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n)}+\frac {\int \sin (c+d x) (a+a \sin (c+d x))^n (2 a+a n \sin (c+d x)) \, dx}{a (3+n)} \\ & = -\frac {\cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n)}+\frac {\int (a+a \sin (c+d x))^n \left (2 a \sin (c+d x)+a n \sin ^2(c+d x)\right ) \, dx}{a (3+n)} \\ & = -\frac {\cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n)}-\frac {n \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d \left (6+5 n+n^2\right )}+\frac {\int (a+a \sin (c+d x))^n \left (a^2 n (1+n)+a^2 (4+n) \sin (c+d x)\right ) \, dx}{a^2 (2+n) (3+n)} \\ & = -\frac {(4+n) \cos (c+d x) (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n)}-\frac {\cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n)}-\frac {n \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d \left (6+5 n+n^2\right )}+\frac {\left (n \left (5+3 n+n^2\right )\right ) \int (a+a \sin (c+d x))^n \, dx}{(1+n) (2+n) (3+n)} \\ & = -\frac {(4+n) \cos (c+d x) (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n)}-\frac {\cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n)}-\frac {n \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d \left (6+5 n+n^2\right )}+\frac {\left (n \left (5+3 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) (3+n)} \\ & = -\frac {(4+n) \cos (c+d x) (a+a \sin (c+d x))^n}{d (1+n) (2+n) (3+n)}-\frac {\cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^n}{d (3+n)}-\frac {2^{\frac {1}{2}+n} n \left (5+3 n+n^2\right ) \cos (c+d x) \operatorname {Hypergeometric2F1}\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)}-\frac {n \cos (c+d x) (a+a \sin (c+d x))^{1+n}}{a d \left (6+5 n+n^2\right )} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2131\) vs. \(2(215)=430\).

Time = 23.74 (sec) , antiderivative size = 2131, normalized size of antiderivative = 9.91 \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=\text {Result too large to show} \]

[In]

Integrate[Sin[c + d*x]^3*(a + a*Sin[c + d*x])^n,x]

[Out]

(2*(Hypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] - 6*Hypergeometric2F1
[1/2 + n, 2 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] + 12*Hypergeometric2F1[1/2 + n, 3 + n, 3/2
+ n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] - 8*Hypergeometric2F1[1/2 + n, 4 + n, 3/2 + n, -(Sqrt[Sec[c +
d*x]^2] + Tan[c + d*x])^2])*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^n*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^
n*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c + d*x]^2)^2*(1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)
^n)/(d*(1 + 2*n)*(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^3*((4*n*(Hypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(S
qrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] - 6*Hypergeometric2F1[1/2 + n, 2 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] +
Tan[c + d*x])^2] + 12*Hypergeometric2F1[1/2 + n, 3 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] -
8*Hypergeometric2F1[1/2 + n, 4 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2])*(a + (a*Tan[c + d*x])/
Sqrt[Sec[c + d*x]^2])^n*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c
+ d*x] + 2*Tan[c + d*x]^2)^2*(1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^(-1 + n))/((1 + 2*n)*(Sqrt[Sec[c +
d*x]^2] + Tan[c + d*x])^2) - (6*(Hypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c +
d*x])^2] - 6*Hypergeometric2F1[1/2 + n, 2 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] + 12*Hyperge
ometric2F1[1/2 + n, 3 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] - 8*Hypergeometric2F1[1/2 + n, 4
+ n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2])*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^n*(Sec[c
+ d*x]^2 + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c + d*x]^2)^2*
(1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^n)/((1 + 2*n)*(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^4) + (2*n*(H
ypergeometric2F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] - 6*Hypergeometric2F1[1/2
+ n, 2 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] + 12*Hypergeometric2F1[1/2 + n, 3 + n, 3/2 + n,
-(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] - 8*Hypergeometric2F1[1/2 + n, 4 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^
2] + Tan[c + d*x])^2])*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^(-1 + n)*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c
+ d*x] + 2*Tan[c + d*x]^2)^2*(a*Sqrt[Sec[c + d*x]^2] - (a*Tan[c + d*x]^2)/Sqrt[Sec[c + d*x]^2])*(1 + (Sqrt[Sec
[c + d*x]^2] + Tan[c + d*x])^2)^n)/((1 + 2*n)*(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^3) + (4*(Hypergeometric2F1
[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] - 6*Hypergeometric2F1[1/2 + n, 2 + n, 3/2
+ n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] + 12*Hypergeometric2F1[1/2 + n, 3 + n, 3/2 + n, -(Sqrt[Sec[c +
d*x]^2] + Tan[c + d*x])^2] - 8*Hypergeometric2F1[1/2 + n, 4 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x
])^2])*(a + (a*Tan[c + d*x])/Sqrt[Sec[c + d*x]^2])^n*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c + d*x]
^2)*(2*(Sec[c + d*x]^2)^(3/2) + 4*Sec[c + d*x]^2*Tan[c + d*x] + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x]^2)*(1 + (S
qrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^n)/((1 + 2*n)*(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^3) + (2*(a + (a*Tan
[c + d*x])/Sqrt[Sec[c + d*x]^2])^n*(1 + 2*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x] + 2*Tan[c + d*x]^2)^2*(1 + (Sqrt[S
ec[c + d*x]^2] + Tan[c + d*x])^2)^n*((-16*(1/2 + n)*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])*(-Hyp
ergeometric2F1[1/2 + n, 4 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] + (1 + (Sqrt[Sec[c + d*x]^2]
+ Tan[c + d*x])^2)^(-4 - n)))/(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x]) + (24*(1/2 + n)*(Sec[c + d*x]^2 + Sqrt[Se
c[c + d*x]^2]*Tan[c + d*x])*(-Hypergeometric2F1[1/2 + n, 3 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x]
)^2] + (1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^(-3 - n)))/(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x]) - (12*(1
/2 + n)*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])*(-Hypergeometric2F1[1/2 + n, 2 + n, 3/2 + n, -(Sq
rt[Sec[c + d*x]^2] + Tan[c + d*x])^2] + (1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2)^(-2 - n)))/(Sqrt[Sec[c +
d*x]^2] + Tan[c + d*x]) + (2*(1/2 + n)*(Sec[c + d*x]^2 + Sqrt[Sec[c + d*x]^2]*Tan[c + d*x])*(-Hypergeometric2
F1[1/2 + n, 1 + n, 3/2 + n, -(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^2] + (1 + (Sqrt[Sec[c + d*x]^2] + Tan[c + d
*x])^2)^(-1 - n)))/(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])))/((1 + 2*n)*(Sqrt[Sec[c + d*x]^2] + Tan[c + d*x])^3)
))

Maple [F]

\[\int \left (\sin ^{3}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )^{n}d x\]

[In]

int(sin(d*x+c)^3*(a+a*sin(d*x+c))^n,x)

[Out]

int(sin(d*x+c)^3*(a+a*sin(d*x+c))^n,x)

Fricas [F]

\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sin(d*x+c)^3*(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*sin(d*x + c), x)

Sympy [F(-1)]

Timed out. \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=\text {Timed out} \]

[In]

integrate(sin(d*x+c)**3*(a+a*sin(d*x+c))**n,x)

[Out]

Timed out

Maxima [F]

\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^n*sin(d*x + c)^3, x)

Giac [F]

\[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sin(d*x+c)^3*(a+a*sin(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^n*sin(d*x + c)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \sin ^3(c+d x) (a+a \sin (c+d x))^n \, dx=\int {\sin \left (c+d\,x\right )}^3\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^n \,d x \]

[In]

int(sin(c + d*x)^3*(a + a*sin(c + d*x))^n,x)

[Out]

int(sin(c + d*x)^3*(a + a*sin(c + d*x))^n, x)