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

Optimal. Leaf size=351 \[ \frac{\sqrt{2} a \left (2 a^2+b^2 \left (n^2+5 n+4\right )\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right )}{b^3 d (n+2) (n+3) \sqrt{\sin (c+d x)+1}}-\frac{\sqrt{2} (a+b) \left (2 a^2+b^2 (n+2)^2\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n-1;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right )}{b^3 d (n+2) (n+3) \sqrt{\sin (c+d x)+1}}+\frac{2 a \cos (c+d x) (a+b \sin (c+d x))^{n+1}}{b^2 d (n+2) (n+3)}-\frac{\sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^{n+1}}{b d (n+3)} \]

[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)

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Rubi [A]  time = 0.485529, antiderivative size = 351, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2793, 3023, 2756, 2665, 139, 138} \[ \frac{\sqrt{2} a \left (2 a^2+b^2 \left (n^2+5 n+4\right )\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right )}{b^3 d (n+2) (n+3) \sqrt{\sin (c+d x)+1}}-\frac{\sqrt{2} (a+b) \left (2 a^2+b^2 (n+2)^2\right ) \cos (c+d x) (a+b \sin (c+d x))^n \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{-n} F_1\left (\frac{1}{2};\frac{1}{2},-n-1;\frac{3}{2};\frac{1}{2} (1-\sin (c+d x)),\frac{b (1-\sin (c+d x))}{a+b}\right )}{b^3 d (n+2) (n+3) \sqrt{\sin (c+d x)+1}}+\frac{2 a \cos (c+d x) (a+b \sin (c+d x))^{n+1}}{b^2 d (n+2) (n+3)}-\frac{\sin (c+d x) \cos (c+d x) (a+b \sin (c+d x))^{n+1}}{b d (n+3)} \]

Antiderivative was successfully verified.

[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 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(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 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]

Rule 2756

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 2665

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 139

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 138

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

Rubi steps

\begin{align*} \int \sin ^3(c+d x) (a+b \sin (c+d x))^n \, dx &=-\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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) F_1\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 ) F_1\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]  time = 4.27136, size = 0, normalized size = 0. \[ \int \sin ^3(c+d x) (a+b \sin (c+d x))^n \, dx \]

Verification is Not applicable to the result.

[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]

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Maple [F]  time = 0.282, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( a+b\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)^3*(a+b*sin(d*x+c))^n,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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 (b \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)