3.514 \(\int \frac{\sqrt [3]{a+b x^3}}{x} \, dx\)

Optimal. Leaf size=95 \[ \sqrt [3]{a+b x^3}+\frac{1}{2} \sqrt [3]{a} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )-\frac{\sqrt [3]{a} \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3}}-\frac{1}{2} \sqrt [3]{a} \log (x) \]

[Out]

(a + b*x^3)^(1/3) - (a^(1/3)*ArcTan[(a^(1/3) + 2*(a + b*x^3)^(1/3))/(Sqrt[3]*a^(1/3))])/Sqrt[3] - (a^(1/3)*Log
[x])/2 + (a^(1/3)*Log[a^(1/3) - (a + b*x^3)^(1/3)])/2

________________________________________________________________________________________

Rubi [A]  time = 0.0693723, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {266, 50, 57, 617, 204, 31} \[ \sqrt [3]{a+b x^3}+\frac{1}{2} \sqrt [3]{a} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )-\frac{\sqrt [3]{a} \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3}}-\frac{1}{2} \sqrt [3]{a} \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(1/3)/x,x]

[Out]

(a + b*x^3)^(1/3) - (a^(1/3)*ArcTan[(a^(1/3) + 2*(a + b*x^3)^(1/3))/(Sqrt[3]*a^(1/3))])/Sqrt[3] - (a^(1/3)*Log
[x])/2 + (a^(1/3)*Log[a^(1/3) - (a + b*x^3)^(1/3)])/2

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
 FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{\sqrt [3]{a+b x^3}}{x} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{\sqrt [3]{a+b x}}{x} \, dx,x,x^3\right )\\ &=\sqrt [3]{a+b x^3}+\frac{1}{3} a \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{2/3}} \, dx,x,x^3\right )\\ &=\sqrt [3]{a+b x^3}-\frac{1}{2} \sqrt [3]{a} \log (x)-\frac{1}{2} \sqrt [3]{a} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^3}\right )-\frac{1}{2} a^{2/3} \operatorname{Subst}\left (\int \frac{1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )\\ &=\sqrt [3]{a+b x^3}-\frac{1}{2} \sqrt [3]{a} \log (x)+\frac{1}{2} \sqrt [3]{a} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )+\sqrt [3]{a} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}\right )\\ &=\sqrt [3]{a+b x^3}-\frac{\sqrt [3]{a} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{1}{2} \sqrt [3]{a} \log (x)+\frac{1}{2} \sqrt [3]{a} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )\\ \end{align*}

Mathematica [A]  time = 0.0483776, size = 124, normalized size = 1.31 \[ -\frac{1}{6} \sqrt [3]{a} \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )+\sqrt [3]{a+b x^3}+\frac{1}{3} \sqrt [3]{a} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )-\frac{\sqrt [3]{a} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(1/3)/x,x]

[Out]

(a + b*x^3)^(1/3) - (a^(1/3)*ArcTan[(1 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]])/Sqrt[3] + (a^(1/3)*Log[a^(1/
3) - (a + b*x^3)^(1/3)])/3 - (a^(1/3)*Log[a^(2/3) + a^(1/3)*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/6

________________________________________________________________________________________

Maple [F]  time = 0.031, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x}\sqrt [3]{b{x}^{3}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.75128, size = 308, normalized size = 3.24 \begin{align*} -\frac{1}{3} \, \sqrt{3} a^{\frac{1}{3}} \arctan \left (\frac{2 \, \sqrt{3}{\left (b x^{3} + a\right )}^{\frac{1}{3}} a^{\frac{2}{3}} + \sqrt{3} a}{3 \, a}\right ) - \frac{1}{6} \, a^{\frac{1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac{2}{3}} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} a^{\frac{1}{3}} + a^{\frac{2}{3}}\right ) + \frac{1}{3} \, a^{\frac{1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}}\right ) +{\left (b x^{3} + a\right )}^{\frac{1}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(3)*a^(1/3)*arctan(1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3)*a^(2/3) + sqrt(3)*a)/a) - 1/6*a^(1/3)*log((b*x^3
+ a)^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3)) + 1/3*a^(1/3)*log((b*x^3 + a)^(1/3) - a^(1/3)) + (b*x^3 + a)
^(1/3)

________________________________________________________________________________________

Sympy [C]  time = 1.46582, size = 42, normalized size = 0.44 \begin{align*} - \frac{\sqrt [3]{b} x \Gamma \left (- \frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{3}, - \frac{1}{3} \\ \frac{2}{3} \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{3}}} \right )}}{3 \Gamma \left (\frac{2}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**(1/3)*x*gamma(-1/3)*hyper((-1/3, -1/3), (2/3,), a*exp_polar(I*pi)/(b*x**3))/(3*gamma(2/3))

________________________________________________________________________________________

Giac [A]  time = 3.17083, size = 130, normalized size = 1.37 \begin{align*} -\frac{1}{3} \, \sqrt{3} a^{\frac{1}{3}} \arctan \left (\frac{\sqrt{3}{\left (2 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} + a^{\frac{1}{3}}\right )}}{3 \, a^{\frac{1}{3}}}\right ) - \frac{1}{6} \, a^{\frac{1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac{2}{3}} +{\left (b x^{3} + a\right )}^{\frac{1}{3}} a^{\frac{1}{3}} + a^{\frac{2}{3}}\right ) + \frac{1}{3} \, a^{\frac{1}{3}} \log \left ({\left |{\left (b x^{3} + a\right )}^{\frac{1}{3}} - a^{\frac{1}{3}} \right |}\right ) +{\left (b x^{3} + a\right )}^{\frac{1}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(3)*a^(1/3)*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3)) - 1/6*a^(1/3)*log((b*x^3 + a)
^(2/3) + (b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3)) + 1/3*a^(1/3)*log(abs((b*x^3 + a)^(1/3) - a^(1/3))) + (b*x^3 + a
)^(1/3)