3.19.58 \(\int \frac {1}{x \sqrt [4]{-b+a x^3}} \, dx\)

Optimal. Leaf size=128 \[ -\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}{\sqrt {a x^3-b}-\sqrt {b}}\right )}{3 \sqrt [4]{b}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\frac {\sqrt {a x^3-b}}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{b}}{\sqrt {2}}}{\sqrt [4]{a x^3-b}}\right )}{3 \sqrt [4]{b}} \]

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Rubi [A]  time = 0.19, antiderivative size = 201, normalized size of antiderivative = 1.57, number of steps used = 11, number of rules used = 8, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {266, 63, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{3 \sqrt {2} \sqrt [4]{b}}-\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {a x^3-b}+\sqrt {b}\right )}{3 \sqrt {2} \sqrt [4]{b}}-\frac {\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{3 \sqrt [4]{b}}+\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{3 \sqrt [4]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(-b + a*x^3)^(1/4)),x]

[Out]

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

Rule 63

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 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 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
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 628

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 1162

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 1165

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

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Mathematica [A]  time = 0.02, size = 57, normalized size = 0.45 \begin {gather*} \frac {2 \left (\tan ^{-1}\left (\frac {\sqrt [4]{a x^3-b}}{\sqrt [4]{-b}}\right )+\tanh ^{-1}\left (\frac {b \sqrt [4]{a x^3-b}}{(-b)^{5/4}}\right )\right )}{3 \sqrt [4]{-b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(-b + a*x^3)^(1/4)),x]

[Out]

(2*(ArcTan[(-b + a*x^3)^(1/4)/(-b)^(1/4)] + ArcTanh[(b*(-b + a*x^3)^(1/4))/(-b)^(5/4)]))/(3*(-b)^(1/4))

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IntegrateAlgebraic [A]  time = 0.16, size = 127, normalized size = 0.99 \begin {gather*} \frac {\sqrt {2} \tan ^{-1}\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 )}{3 \sqrt [4]{b}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{\sqrt {b}+\sqrt {-b+a x^3}}\right )}{3 \sqrt [4]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x*(-b + a*x^3)^(1/4)),x]

[Out]

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

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fricas [A]  time = 0.46, size = 127, normalized size = 0.99 \begin {gather*} -\frac {4}{3} \, \left (-\frac {1}{b}\right )^{\frac {1}{4}} \arctan \left (\sqrt {-b \sqrt {-\frac {1}{b}} + \sqrt {a x^{3} - b}} \left (-\frac {1}{b}\right )^{\frac {1}{4}} - {\left (a x^{3} - b\right )}^{\frac {1}{4}} \left (-\frac {1}{b}\right )^{\frac {1}{4}}\right ) + \frac {1}{3} \, \left (-\frac {1}{b}\right )^{\frac {1}{4}} \log \left (b \left (-\frac {1}{b}\right )^{\frac {3}{4}} + {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right ) - \frac {1}{3} \, \left (-\frac {1}{b}\right )^{\frac {1}{4}} \log \left (-b \left (-\frac {1}{b}\right )^{\frac {3}{4}} + {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.33, size = 162, normalized size = 1.27 \begin {gather*} \frac {\sqrt {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 )}{3 \, b^{\frac {1}{4}}} + \frac {\sqrt {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 )}{3 \, b^{\frac {1}{4}}} - \frac {\sqrt {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 )}{6 \, b^{\frac {1}{4}}} + \frac {\sqrt {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 )}{6 \, b^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{x \left (a \,x^{3}-b \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 162, normalized size = 1.27 \begin {gather*} \frac {\sqrt {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 )}{3 \, b^{\frac {1}{4}}} + \frac {\sqrt {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 )}{3 \, b^{\frac {1}{4}}} - \frac {\sqrt {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 )}{6 \, b^{\frac {1}{4}}} + \frac {\sqrt {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 )}{6 \, b^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.96, size = 51, normalized size = 0.40 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{3\,{\left (-b\right )}^{1/4}}-\frac {2\,\mathrm {atanh}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{3\,{\left (-b\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a*x^3 - b)^(1/4)),x)

[Out]

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

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sympy [C]  time = 1.02, size = 42, normalized size = 0.33 \begin {gather*} - \frac {\Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{3}}} \right )}}{3 \sqrt [4]{a} x^{\frac {3}{4}} \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(1/4)*hyper((1/4, 1/4), (5/4,), b*exp_polar(2*I*pi)/(a*x**3))/(3*a**(1/4)*x**(3/4)*gamma(5/4))

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