3.106 \(\int \frac{\sin (c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=93 \[ \frac{\cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{2^{5/6} d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}}-\frac{3 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.0626866, antiderivative size = 93, normalized size of antiderivative = 1., 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} \[ \frac{\cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{2^{5/6} d \sqrt [6]{\sin (c+d x)+1} \sqrt [3]{a \sin (c+d x)+a}}-\frac{3 \cos (c+d x)}{2 d \sqrt [3]{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]/(a + a*Sin[c + d*x])^(1/3),x]

[Out]

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

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)]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.148261, size = 84, normalized size = 0.9 \[ -\frac{3 \cos (c+d x) \left (2 \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};\sin ^2\left (\frac{1}{4} (2 c+2 d x+\pi )\right )\right )+\sqrt{2-2 \sin (c+d x)}\right )}{2 d \sqrt{2-2 \sin (c+d x)} \sqrt [3]{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]/(a + a*Sin[c + d*x])^(1/3),x]

[Out]

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

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Maple [F]  time = 0.132, size = 0, normalized size = 0. \begin{align*} \int{\sin \left ( dx+c \right ){\frac{1}{\sqrt [3]{a+a\sin \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)/(a+a*sin(d*x+c))^(1/3),x)

[Out]

int(sin(d*x+c)/(a+a*sin(d*x+c))^(1/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/(a+a*sin(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)/(a*sin(d*x + c) + a)^(1/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/(a+a*sin(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

integral(sin(d*x + c)/(a*sin(d*x + c) + a)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/(a+a*sin(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate(sin(d*x + c)/(a*sin(d*x + c) + a)^(1/3), x)