3.325 \(\int \frac{\sqrt [3]{c \sin ^3(a+b x^2)}}{x^3} \, dx\)

Optimal. Leaf size=98 \[ \frac{1}{2} b \cos (a) \text{CosIntegral}\left (b x^2\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\frac{1}{2} b \sin (a) \text{Si}\left (b x^2\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\frac{\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 x^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.20267, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {6720, 3379, 3297, 3303, 3299, 3302} \[ \frac{1}{2} b \cos (a) \text{CosIntegral}\left (b x^2\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\frac{1}{2} b \sin (a) \text{Si}\left (b x^2\right ) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}-\frac{\sqrt [3]{c \sin ^3\left (a+b x^2\right )}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x^2]^3)^(1/3)/x^3,x]

[Out]

-(c*Sin[a + b*x^2]^3)^(1/3)/(2*x^2) + (b*Cos[a]*CosIntegral[b*x^2]*Csc[a + b*x^2]*(c*Sin[a + b*x^2]^3)^(1/3))/
2 - (b*Csc[a + b*x^2]*Sin[a]*(c*Sin[a + b*x^2]^3)^(1/3)*SinIntegral[b*x^2])/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 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3297

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

Mathematica [A]  time = 0.127633, size = 67, normalized size = 0.68 \[ -\frac{\csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \left (-b x^2 \cos (a) \text{CosIntegral}\left (b x^2\right )+b x^2 \sin (a) \text{Si}\left (b x^2\right )+\sin \left (a+b x^2\right )\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x^2]^3)^(1/3)/x^3,x]

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.087, size = 214, normalized size = 2.2 \begin{align*}{\frac{1}{2\,{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-2}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( b{x}^{2}+a \right ) }}} \left ( -{\frac{{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}}{2\,{x}^{2}}}-{\frac{i}{2}}b{\it Ei} \left ( 1,-ib{x}^{2} \right ){{\rm e}^{i \left ( b{x}^{2}+2\,a \right ) }} \right ) }+{\frac{1}{ \left ( 4\,{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-4 \right ){x}^{2}}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( b{x}^{2}+a \right ) }}}}-{\frac{{\frac{i}{4}}{{\rm e}^{ib{x}^{2}}}b{\it Ei} \left ( 1,ib{x}^{2} \right ) }{{{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1}\sqrt [3]{ic \left ({{\rm e}^{2\,i \left ( b{x}^{2}+a \right ) }}-1 \right ) ^{3}{{\rm e}^{-3\,i \left ( b{x}^{2}+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(b*x^2+a)^3)^(1/3)/x^3,x)

[Out]

1/2*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(1/3)/(exp(2*I*(b*x^2+a))-1)*(-1/2/x^2*exp(2*I*(b*x^2+a
))-1/2*I*b*Ei(1,-I*b*x^2)*exp(I*(b*x^2+2*a)))+1/4*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(1/3)/(ex
p(2*I*(b*x^2+a))-1)/x^2-1/4*I*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(1/3)/(exp(2*I*(b*x^2+a))-1)*
exp(I*b*x^2)*b*Ei(1,I*b*x^2)

________________________________________________________________________________________

Maxima [C]  time = 1.69097, size = 70, normalized size = 0.71 \begin{align*} -\frac{1}{8} \,{\left ({\left (\Gamma \left (-1, i \, b x^{2}\right ) + \Gamma \left (-1, -i \, b x^{2}\right )\right )} \cos \left (a\right ) -{\left (i \, \Gamma \left (-1, i \, b x^{2}\right ) - i \, \Gamma \left (-1, -i \, b x^{2}\right )\right )} \sin \left (a\right )\right )} b c^{\frac{1}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x^2+a)^3)^(1/3)/x^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.66378, size = 375, normalized size = 3.83 \begin{align*} -\frac{4^{\frac{1}{3}}{\left (2 \cdot 4^{\frac{2}{3}} \cos \left (b x^{2} + a\right )^{2} -{\left (2 \cdot 4^{\frac{2}{3}} b x^{2} \sin \left (a\right ) \operatorname{Si}\left (b x^{2}\right ) -{\left (4^{\frac{2}{3}} b x^{2} \operatorname{Ci}\left (b x^{2}\right ) + 4^{\frac{2}{3}} b x^{2} \operatorname{Ci}\left (-b x^{2}\right )\right )} \cos \left (a\right )\right )} \sin \left (b x^{2} + a\right ) - 2 \cdot 4^{\frac{2}{3}}\right )} \left (-{\left (c \cos \left (b x^{2} + a\right )^{2} - c\right )} \sin \left (b x^{2} + a\right )\right )^{\frac{1}{3}}}{16 \,{\left (x^{2} \cos \left (b x^{2} + a\right )^{2} - x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x^2+a)^3)^(1/3)/x^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x**2+a)**3)**(1/3)/x**3,x)

[Out]

Integral((c*sin(a + b*x**2)**3)**(1/3)/x**3, x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c \sin \left (b x^{2} + a\right )^{3}\right )^{\frac{1}{3}}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x^2+a)^3)^(1/3)/x^3,x, algorithm="giac")

[Out]

integrate((c*sin(b*x^2 + a)^3)^(1/3)/x^3, x)