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

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

[Out]

-1/2*Ci(2*b*x)*cos(2*a)*csc(b*x+a)^2*(c*sin(b*x+a)^3)^(2/3)+1/2*csc(b*x+a)^2*ln(x)*(c*sin(b*x+a)^3)^(2/3)+1/2*
csc(b*x+a)^2*Si(2*b*x)*sin(2*a)*(c*sin(b*x+a)^3)^(2/3)

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Rubi [A]  time = 0.21, antiderivative size = 99, normalized size of antiderivative = 1.00, 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 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]

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 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 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])

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*}

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Mathematica [A]  time = 0.08, size = 50, normalized size = 0.51 \[ \frac {1}{2} \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} (-\cos (2 a) \text {Ci}(2 b x)+\sin (2 a) \text {Si}(2 b x)+\log (x)) \]

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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fricas [A]  time = 0.78, size = 88, normalized size = 0.89 \[ -\frac {4^{\frac {2}{3}} {\left (2 \cdot 4^{\frac {1}{3}} \sin \left (2 \, a\right ) \operatorname {Si}\left (2 \, b x\right ) - {\left (4^{\frac {1}{3}} \operatorname {Ci}\left (2 \, b x\right ) + 4^{\frac {1}{3}} \operatorname {Ci}\left (-2 \, b x\right )\right )} \cos \left (2 \, a\right ) + 2 \cdot 4^{\frac {1}{3}} \log \relax (x)\right )} \left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{\frac {2}{3}}}{16 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \sin \left (b x + a\right )^{3}\right )^{\frac {2}{3}}}{x}\,{d x} \]

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)

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maple [C]  time = 0.19, size = 283, normalized size = 2.86 \[ \frac {i \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i b x} \pi \,\mathrm {csgn}\left (b x \right )}{4 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i b x} \Si \left (2 b x \right )}{2 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {\left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i b x} \Ei \left (1, -2 i b x \right )}{4 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {\Ei \left (1, -2 i b x \right ) \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b x +2 a \right )}}{4 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {\ln \relax (x ) \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b x +a \right )}}{2 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}} \]

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*Ei(1,
-2*I*b*x)/(exp(2*I*(b*x+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+2*a))-1/2*l
n(x)/(exp(2*I*(b*x+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))

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maxima [C]  time = 1.11, size = 52, normalized size = 0.53 \[ -\frac {1}{8} \, {\left ({\left (E_{1}\left (2 i \, b x\right ) + E_{1}\left (-2 i \, b x\right )\right )} \cos \left (2 \, a\right ) + {\left (-i \, E_{1}\left (2 i \, b x\right ) + i \, E_{1}\left (-2 i \, b x\right )\right )} \sin \left (2 \, a\right ) + 2 \, \log \left (b x\right )\right )} c^{\frac {2}{3}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,{\sin \left (a+b\,x\right )}^3\right )}^{2/3}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a + b*x)^3)^(2/3)/x,x)

[Out]

int((c*sin(a + b*x)^3)^(2/3)/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \sin ^{3}{\left (a + b x \right )}\right )^{\frac {2}{3}}}{x}\, dx \]

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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