3.2.6 \(\int \frac {\sin (c+d x)}{\sqrt [3]{a+a \sin (c+d x)}} \, dx\) [106]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 93, 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 {\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}} \end {gather*}

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 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 \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*}

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Mathematica [A]
time = 0.10, size = 84, normalized size = 0.90 \begin {gather*} -\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+\pi +2 d x)\right )\right )+\sqrt {2-2 \sin (c+d x)}\right )}{2 d \sqrt {2-2 \sin (c+d x)} \sqrt [3]{a (1+\sin (c+d x))}} \end {gather*}

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 = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\sin \left (d x +c \right )}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx\]

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 = 0.00, 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))^(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 = 0.00, 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))^(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin {\left (c + d x \right )}}{\sqrt [3]{a \left (\sin {\left (c + d x \right )} + 1\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))**(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 = 0.00, 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))^(1/3),x, algorithm="giac")

[Out]

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

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

[Out]

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

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