\(\int \sin ^3(c+d x) (a+b \sin (c+d x))^n \, dx\) [222]

Optimal result

Integrand size = 21, antiderivative size = 351 \[ \int \sin ^3(c+d x) (a+b \sin (c+d x))^n \, dx=\frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}-\frac {\sqrt {2} (a+b) \left (2 a^2+b^2 (2+n)^2\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n}}{b^3 d (2+n) (3+n) \sqrt {1+\sin (c+d x)}}+\frac {\sqrt {2} a \left (2 a^2+b^2 \left (4+5 n+n^2\right )\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n}}{b^3 d (2+n) (3+n) \sqrt {1+\sin (c+d x)}} \]

[Out]

2*a*cos(d*x+c)*(a+b*sin(d*x+c))^(1+n)/b^2/d/(2+n)/(3+n)-cos(d*x+c)*sin(d*x+c)*(a+b*sin(d*x+c))^(1+n)/b/d/(3+n)
-(a+b)*(2*a^2+b^2*(2+n)^2)*AppellF1(1/2,-1-n,1/2,3/2,b*(1-sin(d*x+c))/(a+b),1/2-1/2*sin(d*x+c))*cos(d*x+c)*(a+
b*sin(d*x+c))^n*2^(1/2)/b^3/d/(2+n)/(3+n)/(((a+b*sin(d*x+c))/(a+b))^n)/(1+sin(d*x+c))^(1/2)+a*(2*a^2+b^2*(n^2+
5*n+4))*AppellF1(1/2,-n,1/2,3/2,b*(1-sin(d*x+c))/(a+b),1/2-1/2*sin(d*x+c))*cos(d*x+c)*(a+b*sin(d*x+c))^n*2^(1/
2)/b^3/d/(2+n)/(3+n)/(((a+b*sin(d*x+c))/(a+b))^n)/(1+sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2872, 3102, 2835, 2744, 144, 143} \[ \int \sin ^3(c+d x) (a+b \sin (c+d x))^n \, dx=\frac {\sqrt {2} a \left (2 a^2+b^2 \left (n^2+5 n+4\right )\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right )}{b^3 d (n+2) (n+3) \sqrt {\sin (c+d x)+1}}-\frac {\sqrt {2} (a+b) \left (2 a^2+b^2 (n+2)^2\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n-1,\frac {3}{2},\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right )}{b^3 d (n+2) (n+3) \sqrt {\sin (c+d x)+1}}+\frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{n+1}}{b^2 d (n+2) (n+3)}-\frac {\sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^{n+1}}{b d (n+3)} \]

[In]

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

[Out]

(2*a*Cos[c + d*x]*(a + b*Sin[c + d*x])^(1 + n))/(b^2*d*(2 + n)*(3 + n)) - (Cos[c + d*x]*Sin[c + d*x]*(a + b*Si
n[c + d*x])^(1 + n))/(b*d*(3 + n)) - (Sqrt[2]*(a + b)*(2*a^2 + b^2*(2 + n)^2)*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^3*d*(2 + n)*(3 +
n)*Sqrt[1 + Sin[c + d*x]]*((a + b*Sin[c + d*x])/(a + b))^n) + (Sqrt[2]*a*(2*a^2 + b^2*(4 + 5*n + n^2))*Appell
F1[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^3*d*(2 + n)*(3 + n)*Sqrt[1 + Sin[c + d*x]]*((a + b*Sin[c + d*x])/(a + b))^n)

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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])

Rule 144

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 2744

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 2835

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 2872

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Dist[1/
(d*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a
*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n
- 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m]
|| (EqQ[a, 0] && NeQ[c, 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 (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}+\frac {\int (a+b \sin (c+d x))^n \left (a+b (2+n) \sin (c+d x)-2 a \sin ^2(c+d x)\right ) \, dx}{b (3+n)} \\ & = \frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}+\frac {\int (a+b \sin (c+d x))^n \left (-a b n+\left (2 a^2+b^2 (2+n)^2\right ) \sin (c+d x)\right ) \, dx}{b^2 (2+n) (3+n)} \\ & = \frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}+\frac {\left (2 a^2+b^2 (2+n)^2\right ) \int (a+b \sin (c+d x))^{1+n} \, dx}{b^3 (2+n) (3+n)}-\frac {\left (a \left (2 a^2+b^2 \left (4+5 n+n^2\right )\right )\right ) \int (a+b \sin (c+d x))^n \, dx}{b^3 (2+n) (3+n)} \\ & = \frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}+\frac {\left (\left (2 a^2+b^2 (2+n)^2\right ) \cos (c+d x)\right ) \text {Subst}\left (\int \frac {(a+b x)^{1+n}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (c+d x)\right )}{b^3 d (2+n) (3+n) \sqrt {1-\sin (c+d x)} \sqrt {1+\sin (c+d x)}}-\frac {\left (a \left (2 a^2+b^2 \left (4+5 n+n^2\right )\right ) \cos (c+d x)\right ) \text {Subst}\left (\int \frac {(a+b x)^n}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (c+d x)\right )}{b^3 d (2+n) (3+n) \sqrt {1-\sin (c+d x)} \sqrt {1+\sin (c+d x)}} \\ & = \frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}-\frac {\left ((-a-b) \left (2 a^2+b^2 (2+n)^2\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 ) \text {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^3 d (2+n) (3+n) \sqrt {1-\sin (c+d x)} \sqrt {1+\sin (c+d x)}}-\frac {\left (a \left (2 a^2+b^2 \left (4+5 n+n^2\right )\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 ) \text {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^3 d (2+n) (3+n) \sqrt {1-\sin (c+d x)} \sqrt {1+\sin (c+d x)}} \\ & = \frac {2 a \cos (c+d x) (a+b \sin (c+d x))^{1+n}}{b^2 d (2+n) (3+n)}-\frac {\cos (c+d x) \sin (c+d x) (a+b \sin (c+d x))^{1+n}}{b d (3+n)}-\frac {\sqrt {2} (a+b) \left (2 a^2+b^2 (2+n)^2\right ) \operatorname {AppellF1}\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^3 d (2+n) (3+n) \sqrt {1+\sin (c+d x)}}+\frac {\sqrt {2} a \left (2 a^2+b^2 \left (4+5 n+n^2\right )\right ) \operatorname {AppellF1}\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^3 d (2+n) (3+n) \sqrt {1+\sin (c+d x)}} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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