3.339 \(\int \frac{(c \sin ^3(a+b x))^{2/3}}{x} \, dx\)

Optimal. Leaf size=99 \[ -\frac{1}{2} \cos (2 a) \text{CosIntegral}(2 b x) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}+\frac{1}{2} \sin (2 a) \text{Si}(2 b x) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}+\frac{1}{2} \log (x) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} \]

[Out]

-(Cos[2*a]*CosIntegral[2*b*x]*Csc[a + b*x]^2*(c*Sin[a + b*x]^3)^(2/3))/2 + (Csc[a + b*x]^2*Log[x]*(c*Sin[a + b
*x]^3)^(2/3))/2 + (Csc[a + b*x]^2*Sin[2*a]*(c*Sin[a + b*x]^3)^(2/3)*SinIntegral[2*b*x])/2

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Rubi [A]  time = 0.208733, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {6720, 3312, 3303, 3299, 3302} \[ -\frac{1}{2} \cos (2 a) \text{CosIntegral}(2 b x) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}+\frac{1}{2} \sin (2 a) \text{Si}(2 b x) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}+\frac{1}{2} \log (x) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x]^3)^(2/3)/x,x]

[Out]

-(Cos[2*a]*CosIntegral[2*b*x]*Csc[a + b*x]^2*(c*Sin[a + b*x]^3)^(2/3))/2 + (Csc[a + b*x]^2*Log[x]*(c*Sin[a + b
*x]^3)^(2/3))/2 + (Csc[a + b*x]^2*Sin[2*a]*(c*Sin[a + b*x]^3)^(2/3)*SinIntegral[2*b*x])/2

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[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{\left (c \sin ^3(a+b x)\right )^{2/3}}{x} \, dx &=\left (\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int \frac{\sin ^2(a+b x)}{x} \, dx\\ &=\left (\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int \left (\frac{1}{2 x}-\frac{\cos (2 a+2 b x)}{2 x}\right ) \, dx\\ &=\frac{1}{2} \csc ^2(a+b x) \log (x) \left (c \sin ^3(a+b x)\right )^{2/3}-\frac{1}{2} \left (\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int \frac{\cos (2 a+2 b x)}{x} \, dx\\ &=\frac{1}{2} \csc ^2(a+b x) \log (x) \left (c \sin ^3(a+b x)\right )^{2/3}-\frac{1}{2} \left (\cos (2 a) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int \frac{\cos (2 b x)}{x} \, dx+\frac{1}{2} \left (\csc ^2(a+b x) \sin (2 a) \left (c \sin ^3(a+b x)\right )^{2/3}\right ) \int \frac{\sin (2 b x)}{x} \, dx\\ &=-\frac{1}{2} \cos (2 a) \text{Ci}(2 b x) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}+\frac{1}{2} \csc ^2(a+b x) \log (x) \left (c \sin ^3(a+b x)\right )^{2/3}+\frac{1}{2} \csc ^2(a+b x) \sin (2 a) \left (c \sin ^3(a+b x)\right )^{2/3} \text{Si}(2 b x)\\ \end{align*}

Mathematica [A]  time = 0.0870483, size = 50, normalized size = 0.51 \[ \frac{1}{2} \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} (-\cos (2 a) \text{CosIntegral}(2 b x)+\sin (2 a) \text{Si}(2 b x)+\log (x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[a + b*x]^2*(c*Sin[a + b*x]^3)^(2/3)*(-(Cos[2*a]*CosIntegral[2*b*x]) + Log[x] + Sin[2*a]*SinIntegral[2*b*x
]))/2

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Maple [C]  time = 0.079, size = 283, normalized size = 2.9 \begin{align*}{\frac{{\frac{i}{4}}{{\rm e}^{2\,ibx}}\pi \,{\it csgn} \left ( bx \right ) }{ \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{2}} \left ( ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }} \right ) ^{{\frac{2}{3}}}}-{\frac{{\frac{i}{2}}{{\rm e}^{2\,ibx}}{\it Si} \left ( 2\,bx \right ) }{ \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{2}} \left ( ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }} \right ) ^{{\frac{2}{3}}}}-{\frac{{{\rm e}^{2\,ibx}}{\it Ei} \left ( 1,-2\,ibx \right ) }{4\, \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{2}} \left ( ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }} \right ) ^{{\frac{2}{3}}}}-{\frac{{\it Ei} \left ( 1,-2\,ibx \right ){{\rm e}^{2\,i \left ( bx+2\,a \right ) }}}{4\, \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{2}} \left ( ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }} \right ) ^{{\frac{2}{3}}}}-{\frac{\ln \left ( x \right ){{\rm e}^{2\,i \left ( bx+a \right ) }}}{2\, \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{2}} \left ( ic \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( bx+a \right ) }} \right ) ^{{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(2/3)/(exp(2*I*(b*x+a))-1)^2*exp(2*I*b*x)*Pi*csgn(b*x)-1/
2*I*(I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(2/3)/(exp(2*I*(b*x+a))-1)^2*exp(2*I*b*x)*Si(2*b*x)-1/4*(I*
c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(2/3)/(exp(2*I*(b*x+a))-1)^2*exp(2*I*b*x)*Ei(1,-2*I*b*x)-1/4*(I*c*
(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(2/3)/(exp(2*I*(b*x+a))-1)^2*Ei(1,-2*I*b*x)*exp(2*I*(b*x+2*a))-1/2*(
I*c*(exp(2*I*(b*x+a))-1)^3*exp(-3*I*(b*x+a)))^(2/3)/(exp(2*I*(b*x+a))-1)^2*ln(x)*exp(2*I*(b*x+a))

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Maxima [C]  time = 1.57603, size = 70, normalized size = 0.71 \begin{align*} -\frac{1}{8} \,{\left ({\left (E_{1}\left (2 i \, b x\right ) + E_{1}\left (-2 i \, b x\right )\right )} \cos \left (2 \, a\right ) +{\left (-i \, E_{1}\left (2 i \, b x\right ) + i \, E_{1}\left (-2 i \, b x\right )\right )} \sin \left (2 \, a\right ) + 2 \, \log \left (b x\right )\right )} c^{\frac{2}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((exp_integral_e(1, 2*I*b*x) + exp_integral_e(1, -2*I*b*x))*cos(2*a) + (-I*exp_integral_e(1, 2*I*b*x) + I
*exp_integral_e(1, -2*I*b*x))*sin(2*a) + 2*log(b*x))*c^(2/3)

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Fricas [A]  time = 1.74951, size = 288, normalized size = 2.91 \begin{align*} -\frac{4^{\frac{2}{3}}{\left (2 \cdot 4^{\frac{1}{3}} \sin \left (2 \, a\right ) \operatorname{Si}\left (2 \, b x\right ) -{\left (4^{\frac{1}{3}} \operatorname{Ci}\left (2 \, b x\right ) + 4^{\frac{1}{3}} \operatorname{Ci}\left (-2 \, b x\right )\right )} \cos \left (2 \, a\right ) + 2 \cdot 4^{\frac{1}{3}} \log \left (x\right )\right )} \left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{\frac{2}{3}}}{16 \,{\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)^(2/3)/x,x, algorithm="fricas")

[Out]

-1/16*4^(2/3)*(2*4^(1/3)*sin(2*a)*sin_integral(2*b*x) - (4^(1/3)*cos_integral(2*b*x) + 4^(1/3)*cos_integral(-2
*b*x))*cos(2*a) + 2*4^(1/3)*log(x))*(-(c*cos(b*x + a)^2 - c)*sin(b*x + a))^(2/3)/(cos(b*x + a)^2 - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*sin(a + b*x)**3)**(2/3)/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{2}{3}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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