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

Optimal. Leaf size=101 \[ \frac {3}{2} \sqrt [3]{a+b x^2}+\frac {3}{4} \sqrt [3]{a} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )-\frac {1}{2} \sqrt {3} \sqrt [3]{a} \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )-\frac {1}{2} \sqrt [3]{a} \log (x) \]

[Out]

3/2*(b*x^2+a)^(1/3)-1/2*a^(1/3)*ln(x)+3/4*a^(1/3)*ln(a^(1/3)-(b*x^2+a)^(1/3))-1/2*a^(1/3)*arctan(1/3*(a^(1/3)+
2*(b*x^2+a)^(1/3))/a^(1/3)*3^(1/2))*3^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {266, 50, 57, 617, 204, 31} \[ \frac {3}{2} \sqrt [3]{a+b x^2}+\frac {3}{4} \sqrt [3]{a} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )-\frac {1}{2} \sqrt {3} \sqrt [3]{a} \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^2}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )-\frac {1}{2} \sqrt [3]{a} \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/3)/x,x]

[Out]

(3*(a + b*x^2)^(1/3))/2 - (Sqrt[3]*a^(1/3)*ArcTan[(a^(1/3) + 2*(a + b*x^2)^(1/3))/(Sqrt[3]*a^(1/3))])/2 - (a^(
1/3)*Log[x])/2 + (3*a^(1/3)*Log[a^(1/3) - (a + b*x^2)^(1/3)])/4

Rule 31

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

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 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 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 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]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{a+b x^2}}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac {3}{2} \sqrt [3]{a+b x^2}+\frac {1}{2} a \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{2/3}} \, dx,x,x^2\right )\\ &=\frac {3}{2} \sqrt [3]{a+b x^2}-\frac {1}{2} \sqrt [3]{a} \log (x)-\frac {1}{4} \left (3 \sqrt [3]{a}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^2}\right )-\frac {1}{4} \left (3 a^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^2}\right )\\ &=\frac {3}{2} \sqrt [3]{a+b x^2}-\frac {1}{2} \sqrt [3]{a} \log (x)+\frac {3}{4} \sqrt [3]{a} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )+\frac {1}{2} \left (3 \sqrt [3]{a}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}\right )\\ &=\frac {3}{2} \sqrt [3]{a+b x^2}-\frac {1}{2} \sqrt {3} \sqrt [3]{a} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-\frac {1}{2} \sqrt [3]{a} \log (x)+\frac {3}{4} \sqrt [3]{a} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 126, normalized size = 1.25 \[ \frac {1}{4} \left (-\sqrt [3]{a} \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}\right )+6 \sqrt [3]{a+b x^2}+2 \sqrt [3]{a} \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )-2 \sqrt {3} \sqrt [3]{a} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a+b x^2}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(1/3)/x,x]

[Out]

(6*(a + b*x^2)^(1/3) - 2*Sqrt[3]*a^(1/3)*ArcTan[(1 + (2*(a + b*x^2)^(1/3))/a^(1/3))/Sqrt[3]] + 2*a^(1/3)*Log[a
^(1/3) - (a + b*x^2)^(1/3)] - a^(1/3)*Log[a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3)])/4

________________________________________________________________________________________

fricas [A]  time = 0.93, size = 102, normalized size = 1.01 \[ -\frac {1}{2} \, \sqrt {3} a^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {2}{3}} + \sqrt {3} a}{3 \, a}\right ) - \frac {1}{4} \, a^{\frac {1}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + \frac {1}{2} \, a^{\frac {1}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + \frac {3}{2} \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(3)*a^(1/3)*arctan(1/3*(2*sqrt(3)*(b*x^2 + a)^(1/3)*a^(2/3) + sqrt(3)*a)/a) - 1/4*a^(1/3)*log((b*x^2
+ a)^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3)) + 1/2*a^(1/3)*log((b*x^2 + a)^(1/3) - a^(1/3)) + 3/2*(b*x^2
+ a)^(1/3)

________________________________________________________________________________________

giac [A]  time = 2.41, size = 98, normalized size = 0.97 \[ -\frac {1}{2} \, \sqrt {3} a^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) - \frac {1}{4} \, a^{\frac {1}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + \frac {1}{2} \, a^{\frac {1}{3}} \log \left ({\left | {\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right ) + \frac {3}{2} \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(3)*a^(1/3)*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3)) - 1/4*a^(1/3)*log((b*x^2 + a)
^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3)) + 1/2*a^(1/3)*log(abs((b*x^2 + a)^(1/3) - a^(1/3))) + 3/2*(b*x^2
+ a)^(1/3)

________________________________________________________________________________________

maple [F]  time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{3}}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 3.03, size = 97, normalized size = 0.96 \[ -\frac {1}{2} \, \sqrt {3} a^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) - \frac {1}{4} \, a^{\frac {1}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {2}{3}} + {\left (b x^{2} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) + \frac {1}{2} \, a^{\frac {1}{3}} \log \left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right ) + \frac {3}{2} \, {\left (b x^{2} + a\right )}^{\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(3)*a^(1/3)*arctan(1/3*sqrt(3)*(2*(b*x^2 + a)^(1/3) + a^(1/3))/a^(1/3)) - 1/4*a^(1/3)*log((b*x^2 + a)
^(2/3) + (b*x^2 + a)^(1/3)*a^(1/3) + a^(2/3)) + 1/2*a^(1/3)*log((b*x^2 + a)^(1/3) - a^(1/3)) + 3/2*(b*x^2 + a)
^(1/3)

________________________________________________________________________________________

mupad [B]  time = 4.74, size = 115, normalized size = 1.14 \[ \frac {a^{1/3}\,\ln \left (\frac {9\,a\,{\left (b\,x^2+a\right )}^{1/3}}{4}-\frac {9\,a^{4/3}}{4}\right )}{2}+\frac {3\,{\left (b\,x^2+a\right )}^{1/3}}{2}-\frac {a^{1/3}\,\ln \left (\frac {9\,a^{4/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2}+\frac {9\,a\,{\left (b\,x^2+a\right )}^{1/3}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2}+a^{1/3}\,\ln \left (\frac {9\,a\,{\left (b\,x^2+a\right )}^{1/3}}{2}-9\,a^{4/3}\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/3)/x,x)

[Out]

(a^(1/3)*log((9*a*(a + b*x^2)^(1/3))/4 - (9*a^(4/3))/4))/2 + (3*(a + b*x^2)^(1/3))/2 - (a^(1/3)*log((9*a^(4/3)
*((3^(1/2)*1i)/2 + 1/2))/2 + (9*a*(a + b*x^2)^(1/3))/2)*((3^(1/2)*1i)/2 + 1/2))/2 + a^(1/3)*log((9*a*(a + b*x^
2)^(1/3))/2 - 9*a^(4/3)*((3^(1/2)*1i)/4 - 1/4))*((3^(1/2)*1i)/4 - 1/4)

________________________________________________________________________________________

sympy [C]  time = 1.07, size = 46, normalized size = 0.46 \[ - \frac {\sqrt [3]{b} x^{\frac {2}{3}} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{2 \Gamma \left (\frac {2}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________