3.4.25 \(\int \frac {\sqrt [3]{c \sin ^3(a+b x^2)}}{x^3} \, dx\) [325]

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

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6852, 3460, 3378, 3384, 3380, 3383} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x^2]^3)^(1/3)/x^3,x]

[Out]

-1/2*(c*Sin[a + b*x^2]^3)^(1/3)/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 3378

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 3380

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 3383

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 3384

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 3460

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 6852

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 {\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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.09, size = 67, normalized size = 0.68 \begin {gather*} -\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 {Ci}\left (b x^2\right )+\sin \left (a+b x^2\right )+b x^2 \sin (a) \text {Si}\left (b x^2\right )\right )}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(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*S
in[a]*SinIntegral[b*x^2]))/x^2

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Maple [C] Result contains complex when optimal does not.
time = 0.16, size = 214, normalized size = 2.18

method	result	size
risch	\(\frac {\left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} \left (-\frac {{\mathrm e}^{2 i \left (b \,x^{2}+a \right )}}{2 x^{2}}-\frac {i b \expIntegral \left (1, -i x^{2} b \right ) {\mathrm e}^{i \left (b \,x^{2}+2 a \right )}}{2}\right )}{2 \,{\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-2}+\frac {\left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}}}{4 x^{2} \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )}-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{i x^{2} b} b \expIntegral \left (1, i x^{2} b \right )}{4 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )}\)	\(214\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(exp(2*I*(b*x^2+a))-1)*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp(-3*I*(b*x^2+a)))^(1/3)*(-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/x^2/(exp(2*I*(b*x^2+a))-1)*(I*c*(exp(2*I*(b*x^2+a))-1)^3*exp
(-3*I*(b*x^2+a)))^(1/3)-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)

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Maxima [C] Result contains complex when optimal does not.
time = 0.61, size = 52, normalized size = 0.53 \begin {gather*} -\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 {gather*}

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)

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Fricas [A]
time = 0.37, size = 138, normalized size = 1.41 \begin {gather*} -\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 {gather*}

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)

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

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)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a + b*x^2)^3)^(1/3)/x^3,x)

[Out]

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

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