3.4.23 \(\int \frac {\sqrt [3]{c \sin ^3(a+b x^2)}}{x} \, dx\) [323]

Optimal. Leaf size=73 \[ \frac {1}{2} \text {Ci}\left (b x^2\right ) \csc \left (a+b x^2\right ) \sin (a) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}+\frac {1}{2} \cos (a) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \text {Si}\left (b x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6852, 3458, 3457, 3456} \begin {gather*} \frac {1}{2} \sin (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} \cos (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 )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x^2]^3)^(1/3)/x,x]

[Out]

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

Rule 3456

Int[Sin[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[SinIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 3457

Int[Cos[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[CosIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 3458

Int[Sin[(c_) + (d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Dist[Sin[c], Int[Cos[d*x^n]/x, x], x] + Dist[Cos[c], Int[Si
n[d*x^n]/x, x], x] /; FreeQ[{c, d, n}, x]

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} \, 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} \, dx\\ &=\left (\cos (a) \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \int \frac {\sin \left (b x^2\right )}{x} \, dx+\left (\csc \left (a+b x^2\right ) \sin (a) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}\right ) \int \frac {\cos \left (b x^2\right )}{x} \, dx\\ &=\frac {1}{2} \text {Ci}\left (b x^2\right ) \csc \left (a+b x^2\right ) \sin (a) \sqrt [3]{c \sin ^3\left (a+b x^2\right )}+\frac {1}{2} \cos (a) \csc \left (a+b x^2\right ) \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 = 0.04, size = 47, normalized size = 0.64 \begin {gather*} \frac {1}{2} \csc \left (a+b x^2\right ) \sqrt [3]{c \sin ^3\left (a+b x^2\right )} \left (\text {Ci}\left (b x^2\right ) \sin (a)+\cos (a) \text {Si}\left (b x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.16, size = 268, normalized size = 3.67

method	result	size
risch	\(-\frac {\expIntegral \left (1, -i x^{2} b \right ) \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{i \left (b \,x^{2}+2 a \right )}}{4 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )}-\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{i x^{2} b} \pi \,\mathrm {csgn}\left (b \,x^{2}\right )}{4 \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )}+\frac {i \left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{i x^{2} b} \sinIntegral \left (b \,x^{2}\right )}{2 \,{\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-2}+\frac {\left (i c \left ({\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-1\right )^{3} {\mathrm e}^{-3 i \left (b \,x^{2}+a \right )}\right )^{\frac {1}{3}} {\mathrm e}^{i x^{2} b} \expIntegral \left (1, -i x^{2} b \right )}{4 \,{\mathrm e}^{2 i \left (b \,x^{2}+a \right )}-4}\)	\(268\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.60, size = 47, normalized size = 0.64 \begin {gather*} \frac {1}{8} \, {\left ({\left (i \, {\rm Ei}\left (i \, b x^{2}\right ) - i \, {\rm Ei}\left (-i \, b x^{2}\right )\right )} \cos \left (a\right ) - {\left ({\rm Ei}\left (i \, b x^{2}\right ) + {\rm Ei}\left (-i \, b x^{2}\right )\right )} \sin \left (a\right )\right )} 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,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 94, normalized size = 1.29 \begin {gather*} -\frac {4^{\frac {1}{3}} {\left (2 \cdot 4^{\frac {2}{3}} \cos \left (a\right ) \operatorname {Si}\left (b x^{2}\right ) + {\left (4^{\frac {2}{3}} \operatorname {Ci}\left (b x^{2}\right ) + 4^{\frac {2}{3}} \operatorname {Ci}\left (-b x^{2}\right )\right )} \sin \left (a\right )\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}} \sin \left (b x^{2} + a\right )}{16 \, {\left (\cos \left (b x^{2} + a\right )^{2} - 1\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,x, algorithm="fricas")

[Out]

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

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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}\, 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,x)

[Out]

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

[Out]

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

[Out]

int((c*sin(a + b*x^2)^3)^(1/3)/x, x)

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