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

Optimal. Leaf size=86 \[ b \sin (2 a) \text {Ci}(2 b x) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}+b \cos (2 a) \text {Si}(2 b x) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}-\frac {\left (c \sin ^3(a+b x)\right )^{2/3}}{x} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6720, 3313, 12, 3303, 3299, 3302} \[ b \sin (2 a) \text {CosIntegral}(2 b x) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}+b \cos (2 a) \text {Si}(2 b x) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}-\frac {\left (c \sin ^3(a+b x)\right )^{2/3}}{x} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x]^3)^(2/3)/x^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3313

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

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Mathematica [A]  time = 0.15, size = 65, normalized size = 0.76 \[ \frac {\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} (2 b x \sin (2 a) \text {Ci}(2 b x)+2 b x \cos (2 a) \text {Si}(2 b x)+\cos (2 (a+b x))-1)}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[a + b*x]^2*(c*Sin[a + b*x]^3)^(2/3)*(-1 + Cos[2*(a + b*x)] + 2*b*x*CosIntegral[2*b*x]*Sin[2*a] + 2*b*x*Co
s[2*a]*SinIntegral[2*b*x]))/(2*x)

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fricas [A]  time = 0.67, size = 108, normalized size = 1.26 \[ -\frac {4^{\frac {2}{3}} {\left (2 \cdot 4^{\frac {1}{3}} b x \cos \left (2 \, a\right ) \operatorname {Si}\left (2 \, b x\right ) + 2 \cdot 4^{\frac {1}{3}} \cos \left (b x + a\right )^{2} + {\left (4^{\frac {1}{3}} b x \operatorname {Ci}\left (2 \, b x\right ) + 4^{\frac {1}{3}} b x \operatorname {Ci}\left (-2 \, b x\right )\right )} \sin \left (2 \, a\right ) - 2 \cdot 4^{\frac {1}{3}}\right )} \left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{\frac {2}{3}}}{8 \, {\left (x \cos \left (b x + a\right )^{2} - x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.20, size = 211, normalized size = 2.45 \[ \frac {i \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} b \left (\frac {i}{b x}+2 \,{\mathrm e}^{2 i b x} \Ei \left (1, 2 i b x \right )\right )}{4 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {i b \left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} \left (\frac {i {\mathrm e}^{4 i \left (b x +a \right )}}{x b}-2 \Ei \left (1, -2 i b x \right ) {\mathrm e}^{2 i \left (b x +2 a \right )}\right )}{4 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {\left (i c \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b x +a \right )}\right )^{\frac {2}{3}} {\mathrm e}^{2 i \left (b x +a \right )}}{2 x \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(b*x+a)^3)^(2/3)/x^2,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*b*(I/b/x+2*exp(2*I*b*x)*Ei(1
,2*I*b*x))+1/4*I*b/(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)*(I/x/b*exp(4*I*
(b*x+a))-2*Ei(1,-2*I*b*x)*exp(2*I*(b*x+2*a)))+1/2/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 = 1.05, size = 280, normalized size = 3.26 \[ \frac {{\left (64 \, {\left ({\left (-i \, \sqrt {3} + 1\right )} E_{2}\left (2 i \, b x\right ) + {\left (i \, \sqrt {3} + 1\right )} E_{2}\left (-2 i \, b x\right )\right )} \cos \left (2 \, a\right )^{3} - {\left ({\left (64 \, \sqrt {3} + 64 i\right )} E_{2}\left (2 i \, b x\right ) + {\left (64 \, \sqrt {3} - 64 i\right )} E_{2}\left (-2 i \, b x\right )\right )} \sin \left (2 \, a\right )^{3} + 64 \, {\left ({\left ({\left (-i \, \sqrt {3} + 1\right )} E_{2}\left (2 i \, b x\right ) + {\left (i \, \sqrt {3} + 1\right )} E_{2}\left (-2 i \, b x\right )\right )} \cos \left (2 \, a\right ) - 4\right )} \sin \left (2 \, a\right )^{2} + 64 \, {\left ({\left (i \, \sqrt {3} + 1\right )} E_{2}\left (2 i \, b x\right ) + {\left (-i \, \sqrt {3} + 1\right )} E_{2}\left (-2 i \, b x\right )\right )} \cos \left (2 \, a\right ) - 256 \, \cos \left (2 \, a\right )^{2} - {\left ({\left ({\left (64 \, \sqrt {3} + 64 i\right )} E_{2}\left (2 i \, b x\right ) + {\left (64 \, \sqrt {3} - 64 i\right )} E_{2}\left (-2 i \, b x\right )\right )} \cos \left (2 \, a\right )^{2} - {\left (64 \, \sqrt {3} - 64 i\right )} E_{2}\left (2 i \, b x\right ) - {\left (64 \, \sqrt {3} + 64 i\right )} E_{2}\left (-2 i \, b x\right )\right )} \sin \left (2 \, a\right )\right )} b c^{\frac {2}{3}}}{1024 \, {\left (a \cos \left (2 \, a\right )^{2} + a \sin \left (2 \, a\right )^{2} - {\left (b x + a\right )} {\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1024*(64*((-I*sqrt(3) + 1)*exp_integral_e(2, 2*I*b*x) + (I*sqrt(3) + 1)*exp_integral_e(2, -2*I*b*x))*cos(2*a
)^3 - ((64*sqrt(3) + 64*I)*exp_integral_e(2, 2*I*b*x) + (64*sqrt(3) - 64*I)*exp_integral_e(2, -2*I*b*x))*sin(2
*a)^3 + 64*(((-I*sqrt(3) + 1)*exp_integral_e(2, 2*I*b*x) + (I*sqrt(3) + 1)*exp_integral_e(2, -2*I*b*x))*cos(2*
a) - 4)*sin(2*a)^2 + 64*((I*sqrt(3) + 1)*exp_integral_e(2, 2*I*b*x) + (-I*sqrt(3) + 1)*exp_integral_e(2, -2*I*
b*x))*cos(2*a) - 256*cos(2*a)^2 - (((64*sqrt(3) + 64*I)*exp_integral_e(2, 2*I*b*x) + (64*sqrt(3) - 64*I)*exp_i
ntegral_e(2, -2*I*b*x))*cos(2*a)^2 - (64*sqrt(3) - 64*I)*exp_integral_e(2, 2*I*b*x) - (64*sqrt(3) + 64*I)*exp_
integral_e(2, -2*I*b*x))*sin(2*a))*b*c^(2/3)/(a*cos(2*a)^2 + a*sin(2*a)^2 - (b*x + a)*(cos(2*a)^2 + sin(2*a)^2
))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a + b*x)^3)^(2/3)/x^2,x)

[Out]

int((c*sin(a + b*x)^3)^(2/3)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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