3.1.94 \(\int \sin (c+d x) (a+a \sin (c+d x))^{2/3} \, dx\) [94]

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

[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)

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Rubi [A]
time = 0.05, 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 = {2830, 2731, 2730} \begin {gather*} -\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} \end {gather*}

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 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)]

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*}

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Mathematica [A]
time = 0.15, size = 138, normalized size = 1.44 \begin {gather*} -\frac {3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) (a (1+\sin (c+d x)))^{2/3} \left (-\sqrt {2} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\sin ^2\left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right )+\sqrt {1-\sin (c+d x)} (2+\sin (c+d x))\right )}{5 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {1-\sin (c+d x)}} \end {gather*}

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]])

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Maple [F]
time = 0.05, 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)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

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)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

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)

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