3.94 \(\int \sin (c+d x) (a+a \sin (c+d x))^{2/3} \, dx\)

Optimal. Leaf size=96 \[ -\frac {4 \sqrt [6]{2} \cos (c+d x) (a \sin (c+d x)+a)^{2/3} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{5 d (\sin (c+d x)+1)^{7/6}}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d} \]

[Out]

-3/5*cos(d*x+c)*(a+a*sin(d*x+c))^(2/3)/d-4/5*cos(d*x+c)*hypergeom([-1/6, 1/2],[3/2],1/2-1/2*sin(d*x+c))*(a+a*s
in(d*x+c))^(2/3)*2^(1/6)/d/(1+sin(d*x+c))^(7/6)

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2751, 2652, 2651} \[ -\frac {4 \sqrt [6]{2} \cos (c+d x) (a \sin (c+d x)+a)^{2/3} \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{5 d (\sin (c+d x)+1)^{7/6}}-\frac {3 \cos (c+d x) (a \sin (c+d x)+a)^{2/3}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]*(a + a*Sin[c + d*x])^(2/3),x]

[Out]

(-3*Cos[c + d*x]*(a + a*Sin[c + d*x])^(2/3))/(5*d) - (4*2^(1/6)*Cos[c + d*x]*Hypergeometric2F1[-1/6, 1/2, 3/2,
(1 - Sin[c + d*x])/2]*(a + a*Sin[c + d*x])^(2/3))/(5*d*(1 + Sin[c + d*x])^(7/6))

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)]

Rubi steps

\begin {align*} \int \sin (c+d x) (a+a \sin (c+d x))^{2/3} \, dx &=-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 d}+\frac {2}{5} \int (a+a \sin (c+d x))^{2/3} \, dx\\ &=-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 d}+\frac {\left (2 (a+a \sin (c+d x))^{2/3}\right ) \int (1+\sin (c+d x))^{2/3} \, dx}{5 (1+\sin (c+d x))^{2/3}}\\ &=-\frac {3 \cos (c+d x) (a+a \sin (c+d x))^{2/3}}{5 d}-\frac {4 \sqrt [6]{2} \cos (c+d x) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x))\right ) (a+a \sin (c+d x))^{2/3}}{5 d (1+\sin (c+d x))^{7/6}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.23, size = 138, normalized size = 1.44 \[ -\frac {3 (a (\sin (c+d x)+1))^{2/3} \left (\sqrt {1-\sin (c+d x)} (\sin (c+d x)+2)-\sqrt {2} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\sin ^2\left (\frac {1}{4} (2 c+2 d x+\pi )\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{5 d \sqrt {1-\sin (c+d x)} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]*(a + a*Sin[c + d*x])^(2/3),x]

[Out]

(-3*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*(1 + Sin[c + d*x]))^(2/3)*(-(Sqrt[2]*Hypergeometric2F1[1/6, 1/2,
7/6, Sin[(2*c + Pi + 2*d*x)/4]^2]) + Sqrt[1 - Sin[c + d*x]]*(2 + Sin[c + d*x])))/(5*d*(Cos[(c + d*x)/2] + Sin[
(c + d*x)/2])*Sqrt[1 - Sin[c + d*x]])

________________________________________________________________________________________

fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sin \left (d x + c\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)*(a+a*sin(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

integral((a*sin(d*x + c) + a)^(2/3)*sin(d*x + c), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sin \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)*(a+a*sin(d*x+c))^(2/3),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^(2/3)*sin(d*x + c), x)

________________________________________________________________________________________

maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \sin \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right )^{\frac {2}{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)*(a+a*sin(d*x+c))^(2/3),x)

[Out]

int(sin(d*x+c)*(a+a*sin(d*x+c))^(2/3),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \sin \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)*(a+a*sin(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(2/3)*sin(d*x + c), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sin \left (c+d\,x\right )\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{2/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)*(a + a*sin(c + d*x))^(2/3),x)

[Out]

int(sin(c + d*x)*(a + a*sin(c + d*x))^(2/3), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {2}{3}} \sin {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)*(a+a*sin(d*x+c))**(2/3),x)

[Out]

Integral((a*(sin(c + d*x) + 1))**(2/3)*sin(c + d*x), x)

________________________________________________________________________________________