\(\int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx\) [2029]

Optimal result

Integrand size = 17, antiderivative size = 145 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=\frac {4}{3} \sqrt [4]{-b+a x^3}+\frac {1}{3} \sqrt {2} \sqrt [4]{b} \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{-\sqrt {b}+\sqrt {-b+a x^3}}\right )-\frac {1}{3} \sqrt {2} \sqrt [4]{b} \text {arctanh}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^3}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.50, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {272, 52, 65, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=\frac {1}{3} \sqrt {2} \sqrt [4]{b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )-\frac {1}{3} \sqrt {2} \sqrt [4]{b} \arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )+\frac {4}{3} \sqrt [4]{a x^3-b}+\frac {\sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{3 \sqrt {2}}-\frac {\sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{3 \sqrt {2}} \]

[In]

Int[(-b + a*x^3)^(1/4)/x,x]

[Out]

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

Rule 52

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 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 210

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

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 272

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 631

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 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=\frac {1}{3} \left (4 \sqrt [4]{-b+a x^3}-\sqrt {2} \sqrt [4]{b} \arctan \left (\frac {-\sqrt {b}+\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}\right )-\sqrt {2} \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{\sqrt {b}+\sqrt {-b+a x^3}}\right )\right ) \]

[In]

Integrate[(-b + a*x^3)^(1/4)/x,x]

[Out]

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

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.15

method	result	size
pseudoelliptic	\(\frac {4 \left (a \,x^{3}-b \right )^{\frac {1}{4}}}{3}-\frac {\ln \left (\frac {-b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{3}-b}-\sqrt {b}}{b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{3}-b}-\sqrt {b}}\right ) b^{\frac {1}{4}} \sqrt {2}}{6}-\frac {\arctan \left (\frac {\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) b^{\frac {1}{4}} \sqrt {2}}{3}+\frac {\arctan \left (\frac {-\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) b^{\frac {1}{4}} \sqrt {2}}{3}\)	\(167\)

[In]

int((a*x^3-b)^(1/4)/x,x,method=_RETURNVERBOSE)

[Out]

4/3*(a*x^3-b)^(1/4)-1/6*ln((-b^(1/4)*(a*x^3-b)^(1/4)*2^(1/2)-(a*x^3-b)^(1/2)-b^(1/2))/(b^(1/4)*(a*x^3-b)^(1/4)
*2^(1/2)-(a*x^3-b)^(1/2)-b^(1/2)))*b^(1/4)*2^(1/2)-1/3*arctan((2^(1/2)*(a*x^3-b)^(1/4)+b^(1/4))/b^(1/4))*b^(1/
4)*2^(1/2)+1/3*arctan((-2^(1/2)*(a*x^3-b)^(1/4)+b^(1/4))/b^(1/4))*b^(1/4)*2^(1/2)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=-\frac {1}{3} \, \left (-b\right )^{\frac {1}{4}} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} + \left (-b\right )^{\frac {1}{4}}\right ) - \frac {1}{3} i \, \left (-b\right )^{\frac {1}{4}} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} + i \, \left (-b\right )^{\frac {1}{4}}\right ) + \frac {1}{3} i \, \left (-b\right )^{\frac {1}{4}} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} - i \, \left (-b\right )^{\frac {1}{4}}\right ) + \frac {1}{3} \, \left (-b\right )^{\frac {1}{4}} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} - \left (-b\right )^{\frac {1}{4}}\right ) + \frac {4}{3} \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} \]

[In]

integrate((a*x^3-b)^(1/4)/x,x, algorithm="fricas")

[Out]

-1/3*(-b)^(1/4)*log((a*x^3 - b)^(1/4) + (-b)^(1/4)) - 1/3*I*(-b)^(1/4)*log((a*x^3 - b)^(1/4) + I*(-b)^(1/4)) +
1/3*I*(-b)^(1/4)*log((a*x^3 - b)^(1/4) - I*(-b)^(1/4)) + 1/3*(-b)^(1/4)*log((a*x^3 - b)^(1/4) - (-b)^(1/4)) +
4/3*(a*x^3 - b)^(1/4)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.74 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.33 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=- \frac {\sqrt [4]{a} x^{\frac {3}{4}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{3}}} \right )}}{3 \Gamma \left (\frac {3}{4}\right )} \]

[In]

integrate((a*x**3-b)**(1/4)/x,x)

[Out]

-a**(1/4)*x**(3/4)*gamma(-1/4)*hyper((-1/4, -1/4), (3/4,), b*exp_polar(2*I*pi)/(a*x**3))/(3*gamma(3/4))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=-\frac {1}{3} \, \sqrt {2} b^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) - \frac {1}{3} \, \sqrt {2} b^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) - \frac {1}{6} \, \sqrt {2} b^{\frac {1}{4}} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right ) + \frac {1}{6} \, \sqrt {2} b^{\frac {1}{4}} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right ) + \frac {4}{3} \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} \]

[In]

integrate((a*x^3-b)^(1/4)/x,x, algorithm="maxima")

[Out]

-1/3*sqrt(2)*b^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^3 - b)^(1/4))/b^(1/4)) - 1/3*sqrt(2)*b^(1/4)
*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^3 - b)^(1/4))/b^(1/4)) - 1/6*sqrt(2)*b^(1/4)*log(sqrt(2)*(a*x^3
- b)^(1/4)*b^(1/4) + sqrt(a*x^3 - b) + sqrt(b)) + 1/6*sqrt(2)*b^(1/4)*log(-sqrt(2)*(a*x^3 - b)^(1/4)*b^(1/4)
+ sqrt(a*x^3 - b) + sqrt(b)) + 4/3*(a*x^3 - b)^(1/4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=-\frac {1}{3} \, \sqrt {2} b^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) - \frac {1}{3} \, \sqrt {2} b^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) - \frac {1}{6} \, \sqrt {2} b^{\frac {1}{4}} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right ) + \frac {1}{6} \, \sqrt {2} b^{\frac {1}{4}} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right ) + \frac {4}{3} \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} \]

[In]

integrate((a*x^3-b)^(1/4)/x,x, algorithm="giac")

[Out]

-1/3*sqrt(2)*b^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^3 - b)^(1/4))/b^(1/4)) - 1/3*sqrt(2)*b^(1/4)
*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^3 - b)^(1/4))/b^(1/4)) - 1/6*sqrt(2)*b^(1/4)*log(sqrt(2)*(a*x^3
- b)^(1/4)*b^(1/4) + sqrt(a*x^3 - b) + sqrt(b)) + 1/6*sqrt(2)*b^(1/4)*log(-sqrt(2)*(a*x^3 - b)^(1/4)*b^(1/4)
+ sqrt(a*x^3 - b) + sqrt(b)) + 4/3*(a*x^3 - b)^(1/4)

Mupad [B] (verification not implemented)

Time = 6.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=\frac {4\,{\left (a\,x^3-b\right )}^{1/4}}{3}-\frac {2\,{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{3}-\frac {2\,{\left (-b\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{3} \]

[In]

int((a*x^3 - b)^(1/4)/x,x)

[Out]

(4*(a*x^3 - b)^(1/4))/3 - (2*(-b)^(1/4)*atan((a*x^3 - b)^(1/4)/(-b)^(1/4)))/3 - (2*(-b)^(1/4)*atanh((a*x^3 - b
)^(1/4)/(-b)^(1/4)))/3