3.315 \(\int \frac{\sqrt [3]{c \sin ^3(a+b x)}}{x} \, dx\)

Optimal. Leaf size=55 \[ \sin (a) \text{CosIntegral}(b x) \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)}+\cos (a) \text{Si}(b x) \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} \]

[Out]

CosIntegral[b*x]*Csc[a + b*x]*Sin[a]*(c*Sin[a + b*x]^3)^(1/3) + Cos[a]*Csc[a + b*x]*(c*Sin[a + b*x]^3)^(1/3)*S
inIntegral[b*x]

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Rubi [A]  time = 0.165793, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {6720, 3303, 3299, 3302} \[ \sin (a) \text{CosIntegral}(b x) \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)}+\cos (a) \text{Si}(b x) \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x]^3)^(1/3)/x,x]

[Out]

CosIntegral[b*x]*Csc[a + b*x]*Sin[a]*(c*Sin[a + b*x]^3)^(1/3) + Cos[a]*Csc[a + b*x]*(c*Sin[a + b*x]^3)^(1/3)*S
inIntegral[b*x]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt [3]{c \sin ^3(a+b x)}}{x} \, dx &=\left (\csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)}\right ) \int \frac{\sin (a+b x)}{x} \, dx\\ &=\left (\cos (a) \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)}\right ) \int \frac{\sin (b x)}{x} \, dx+\left (\csc (a+b x) \sin (a) \sqrt [3]{c \sin ^3(a+b x)}\right ) \int \frac{\cos (b x)}{x} \, dx\\ &=\text{Ci}(b x) \csc (a+b x) \sin (a) \sqrt [3]{c \sin ^3(a+b x)}+\cos (a) \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} \text{Si}(b x)\\ \end{align*}

Mathematica [A]  time = 0.0573214, size = 36, normalized size = 0.65 \[ \csc (a+b x) \sqrt [3]{c \sin ^3(a+b x)} (\sin (a) \text{CosIntegral}(b x)+\cos (a) \text{Si}(b x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Csc[a + b*x]*(c*Sin[a + b*x]^3)^(1/3)*(CosIntegral[b*x]*Sin[a] + Cos[a]*SinIntegral[b*x])

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Maple [C]  time = 0.078, size = 228, normalized size = 4.2 \begin{align*} -{\frac{{\it Ei} \left ( 1,-ibx \right ){{\rm e}^{i \left ( bx+2\,a \right ) }}}{2\,{{\rm e}^{2\,i \left ( bx+a \right ) }}-2}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }}}}-{\frac{{\frac{i}{2}}{{\rm e}^{ibx}}\pi \,{\it csgn} \left ( bx \right ) }{{{\rm e}^{2\,i \left ( bx+a \right ) }}-1}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }}}}+{\frac{i{{\rm e}^{ibx}}{\it Si} \left ( bx \right ) }{{{\rm e}^{2\,i \left ( bx+a \right ) }}-1}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }}}}+{\frac{{{\rm e}^{ibx}}{\it Ei} \left ( 1,-ibx \right ) }{2\,{{\rm e}^{2\,i \left ( bx+a \right ) }}-2}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(1/3)/(exp(2*I*(b*x+a))-1)*Ei(1,-I*b*x)*exp(I*(b*x+2*a))-1
/2*I*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(1/3)/(exp(2*I*(b*x+a))-1)*exp(I*b*x)*Pi*csgn(b*x)+I*(I*c*
(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(1/3)/(exp(2*I*(b*x+a))-1)*exp(I*b*x)*Si(b*x)+1/2*(I*c*(exp(2*I*(b*x
+a))-1)^3*exp(-3*I*(b*x+a)))^(1/3)/(exp(2*I*(b*x+a))-1)*exp(I*b*x)*Ei(1,-I*b*x)

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Maxima [C]  time = 1.64273, size = 57, normalized size = 1.04 \begin{align*} \frac{1}{4} \,{\left ({\left (i \, E_{1}\left (i \, b x\right ) - i \, E_{1}\left (-i \, b x\right )\right )} \cos \left (a\right ) +{\left (E_{1}\left (i \, b x\right ) + E_{1}\left (-i \, b x\right )\right )} \sin \left (a\right )\right )} c^{\frac{1}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((I*exp_integral_e(1, I*b*x) - I*exp_integral_e(1, -I*b*x))*cos(a) + (exp_integral_e(1, I*b*x) + exp_integ
ral_e(1, -I*b*x))*sin(a))*c^(1/3)

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Fricas [A]  time = 1.69393, size = 265, normalized size = 4.82 \begin{align*} -\frac{4^{\frac{1}{3}}{\left (2 \cdot 4^{\frac{2}{3}} \cos \left (a\right ) \operatorname{Si}\left (b x\right ) +{\left (4^{\frac{2}{3}} \operatorname{Ci}\left (b x\right ) + 4^{\frac{2}{3}} \operatorname{Ci}\left (-b x\right )\right )} \sin \left (a\right )\right )} \left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{\frac{1}{3}} \sin \left (b x + a\right )}{8 \,{\left (\cos \left (b x + a\right )^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*4^(1/3)*(2*4^(2/3)*cos(a)*sin_integral(b*x) + (4^(2/3)*cos_integral(b*x) + 4^(2/3)*cos_integral(-b*x))*si
n(a))*(-(c*cos(b*x + a)^2 - c)*sin(b*x + a))^(1/3)*sin(b*x + a)/(cos(b*x + a)^2 - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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