\(\int \frac {\sqrt [4]{-b+a x^3}}{x^4} \, dx\) [2073]

Optimal result

Integrand size = 17, antiderivative size = 150 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^4} \, dx=-\frac {\sqrt [4]{-b+a x^3}}{3 x^3}-\frac {a \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{-\sqrt {b}+\sqrt {-b+a x^3}}\right )}{6 \sqrt {2} b^{3/4}}+\frac {a \text {arctanh}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^3}}\right )}{6 \sqrt {2} b^{3/4}} \]

[Out]

-1/3*(a*x^3-b)^(1/4)/x^3-1/12*a*arctan(2^(1/2)*b^(1/4)*(a*x^3-b)^(1/4)/(-b^(1/2)+(a*x^3-b)^(1/2)))*2^(1/2)/b^(
3/4)+1/12*a*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))*2^(1/2)/b^(3/4)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 225, 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, 43, 65, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^4} \, dx=-\frac {a \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{6 \sqrt {2} b^{3/4}}+\frac {a \arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{6 \sqrt {2} b^{3/4}}-\frac {a \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{12 \sqrt {2} b^{3/4}}+\frac {a \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{12 \sqrt {2} b^{3/4}}-\frac {\sqrt [4]{a x^3-b}}{3 x^3} \]

[In]

Int[(-b + a*x^3)^(1/4)/x^4,x]

[Out]

-1/3*(-b + a*x^3)^(1/4)/x^3 - (a*ArcTan[1 - (Sqrt[2]*(-b + a*x^3)^(1/4))/b^(1/4)])/(6*Sqrt[2]*b^(3/4)) + (a*Ar
cTan[1 + (Sqrt[2]*(-b + a*x^3)^(1/4))/b^(1/4)])/(6*Sqrt[2]*b^(3/4)) - (a*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*(-b + a
*x^3)^(1/4) + Sqrt[-b + a*x^3]])/(12*Sqrt[2]*b^(3/4)) + (a*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*(-b + a*x^3)^(1/4) +
Sqrt[-b + a*x^3]])/(12*Sqrt[2]*b^(3/4))

Rule 43

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

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^2} \, dx,x,x^3\right ) \\ & = -\frac {\sqrt [4]{-b+a x^3}}{3 x^3}+\frac {1}{12} a \text {Subst}\left (\int \frac {1}{x (-b+a x)^{3/4}} \, dx,x,x^3\right ) \\ & = -\frac {\sqrt [4]{-b+a x^3}}{3 x^3}+\frac {1}{3} \text {Subst}\left (\int \frac {1}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^3}\right ) \\ & = -\frac {\sqrt [4]{-b+a x^3}}{3 x^3}+\frac {\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 )}{6 \sqrt {b}}+\frac {\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 )}{6 \sqrt {b}} \\ & = -\frac {\sqrt [4]{-b+a x^3}}{3 x^3}-\frac {a \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 )}{12 \sqrt {2} b^{3/4}}-\frac {a \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 )}{12 \sqrt {2} b^{3/4}}+\frac {a \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 )}{12 \sqrt {b}}+\frac {a \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 )}{12 \sqrt {b}} \\ & = -\frac {\sqrt [4]{-b+a x^3}}{3 x^3}-\frac {a \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}+\sqrt {-b+a x^3}\right )}{12 \sqrt {2} b^{3/4}}+\frac {a \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}+\sqrt {-b+a x^3}\right )}{12 \sqrt {2} b^{3/4}}+\frac {a \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 )}{6 \sqrt {2} b^{3/4}}-\frac {a \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 )}{6 \sqrt {2} b^{3/4}} \\ & = -\frac {\sqrt [4]{-b+a x^3}}{3 x^3}-\frac {a \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-b+a x^3}}{\sqrt [4]{b}}\right )}{6 \sqrt {2} b^{3/4}}+\frac {a \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-b+a x^3}}{\sqrt [4]{b}}\right )}{6 \sqrt {2} b^{3/4}}-\frac {a \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}+\sqrt {-b+a x^3}\right )}{12 \sqrt {2} b^{3/4}}+\frac {a \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}+\sqrt {-b+a x^3}\right )}{12 \sqrt {2} b^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^4} \, dx=\frac {-4 b^{3/4} \sqrt [4]{-b+a x^3}+\sqrt {2} a x^3 \arctan \left (\frac {-\sqrt {b}+\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}\right )+\sqrt {2} a x^3 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{\sqrt {b}+\sqrt {-b+a x^3}}\right )}{12 b^{3/4} x^3} \]

[In]

Integrate[(-b + a*x^3)^(1/4)/x^4,x]

[Out]

(-4*b^(3/4)*(-b + a*x^3)^(1/4) + Sqrt[2]*a*x^3*ArcTan[(-Sqrt[b] + Sqrt[-b + a*x^3])/(Sqrt[2]*b^(1/4)*(-b + a*x
^3)^(1/4))] + Sqrt[2]*a*x^3*ArcTanh[(Sqrt[2]*b^(1/4)*(-b + a*x^3)^(1/4))/(Sqrt[b] + Sqrt[-b + a*x^3])])/(12*b^
(3/4)*x^3)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.20

method	result	size
pseudoelliptic	\(\frac {\ln \left (\frac {-b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{3}-b}-\sqrt {b}}{b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{3}-b}-\sqrt {b}}\right ) \sqrt {2}\, a \,x^{3}+2 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a \,x^{3}-2 \arctan \left (\frac {-\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a \,x^{3}-8 \left (a \,x^{3}-b \right )^{\frac {1}{4}} b^{\frac {3}{4}}}{24 x^{3} b^{\frac {3}{4}}}\)	\(180\)

[In]

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

[Out]

1/24*(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)))*2^(1/2)*a*x^3+2*arctan((2^(1/2)*(a*x^3-b)^(1/4)+b^(1/4))/b^(1/4))*2^(1/2)*a*x^3-2*arctan((-2^
(1/2)*(a*x^3-b)^(1/4)+b^(1/4))/b^(1/4))*2^(1/2)*a*x^3-8*(a*x^3-b)^(1/4)*b^(3/4))/x^3/b^(3/4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.26 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^4} \, dx=\frac {\left (-\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} x^{3} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} a + \left (-\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} b\right ) + i \, \left (-\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} x^{3} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} a + i \, \left (-\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} b\right ) - i \, \left (-\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} x^{3} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} a - i \, \left (-\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} b\right ) - \left (-\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} x^{3} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} a - \left (-\frac {a^{4}}{b^{3}}\right )^{\frac {1}{4}} b\right ) - 4 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}}{12 \, x^{3}} \]

[In]

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

[Out]

1/12*((-a^4/b^3)^(1/4)*x^3*log((a*x^3 - b)^(1/4)*a + (-a^4/b^3)^(1/4)*b) + I*(-a^4/b^3)^(1/4)*x^3*log((a*x^3 -
b)^(1/4)*a + I*(-a^4/b^3)^(1/4)*b) - I*(-a^4/b^3)^(1/4)*x^3*log((a*x^3 - b)^(1/4)*a - I*(-a^4/b^3)^(1/4)*b) -
(-a^4/b^3)^(1/4)*x^3*log((a*x^3 - b)^(1/4)*a - (-a^4/b^3)^(1/4)*b) - 4*(a*x^3 - b)^(1/4))/x^3

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.82 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.29 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^4} \, dx=- \frac {\sqrt [4]{a} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{3}}} \right )}}{3 x^{\frac {9}{4}} \Gamma \left (\frac {7}{4}\right )} \]

[In]

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

[Out]

-a**(1/4)*gamma(3/4)*hyper((-1/4, 3/4), (7/4,), b*exp_polar(2*I*pi)/(a*x**3))/(3*x**(9/4)*gamma(7/4))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^4} \, dx=\frac {\sqrt {2} a \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{12 \, b^{\frac {3}{4}}} + \frac {\sqrt {2} a \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{12 \, b^{\frac {3}{4}}} + \frac {\sqrt {2} a \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{24 \, b^{\frac {3}{4}}} - \frac {\sqrt {2} a \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{24 \, b^{\frac {3}{4}}} - \frac {{\left (a x^{3} - b\right )}^{\frac {1}{4}}}{3 \, x^{3}} \]

[In]

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

[Out]

1/12*sqrt(2)*a*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^3 - b)^(1/4))/b^(1/4))/b^(3/4) + 1/12*sqrt(2)*a*ar
ctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^3 - b)^(1/4))/b^(1/4))/b^(3/4) + 1/24*sqrt(2)*a*log(sqrt(2)*(a*x^3
- b)^(1/4)*b^(1/4) + sqrt(a*x^3 - b) + sqrt(b))/b^(3/4) - 1/24*sqrt(2)*a*log(-sqrt(2)*(a*x^3 - b)^(1/4)*b^(1/
4) + sqrt(a*x^3 - b) + sqrt(b))/b^(3/4) - 1/3*(a*x^3 - b)^(1/4)/x^3

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.30 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^4} \, dx=\frac {\frac {2 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} a^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {\sqrt {2} a^{2} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} a^{2} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {8 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} a}{x^{3}}}{24 \, a} \]

[In]

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

[Out]

1/24*(2*sqrt(2)*a^2*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^3 - b)^(1/4))/b^(1/4))/b^(3/4) + 2*sqrt(2)*a^
2*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^3 - b)^(1/4))/b^(1/4))/b^(3/4) + sqrt(2)*a^2*log(sqrt(2)*(a*x^
3 - b)^(1/4)*b^(1/4) + sqrt(a*x^3 - b) + sqrt(b))/b^(3/4) - sqrt(2)*a^2*log(-sqrt(2)*(a*x^3 - b)^(1/4)*b^(1/4)
+ sqrt(a*x^3 - b) + sqrt(b))/b^(3/4) - 8*(a*x^3 - b)^(1/4)*a/x^3)/a

Mupad [B] (verification not implemented)

Time = 6.18 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.46 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x^4} \, dx=-\frac {{\left (a\,x^3-b\right )}^{1/4}}{3\,x^3}-\frac {a\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{6\,{\left (-b\right )}^{3/4}}-\frac {a\,\mathrm {atanh}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{6\,{\left (-b\right )}^{3/4}} \]

[In]

int((a*x^3 - b)^(1/4)/x^4,x)

[Out]

- (a*x^3 - b)^(1/4)/(3*x^3) - (a*atan((a*x^3 - b)^(1/4)/(-b)^(1/4)))/(6*(-b)^(3/4)) - (a*atanh((a*x^3 - b)^(1/
4)/(-b)^(1/4)))/(6*(-b)^(3/4))