3.569 \(\int \frac {x}{(a+b x^3)^{2/3}} \, dx\)

Optimal. Leaf size=72 \[ -\frac {\log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}}-\frac {\tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 122, normalized size of antiderivative = 1.69, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {331, 292, 31, 634, 617, 204, 628} \[ -\frac {\log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac {\log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )}{6 b^{2/3}}-\frac {\tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*x^3)^(2/3),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (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]

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 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 40, normalized size = 0.56 \[ \frac {x^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {b x^3}{b x^3+a}\right )}{2 \left (a+b x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*x^3)^(2/3),x]

[Out]

(x^2*Hypergeometric2F1[2/3, 1, 5/3, (b*x^3)/(a + b*x^3)])/(2*(a + b*x^3)^(2/3))

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fricas [B]  time = 0.69, size = 173, normalized size = 2.40 \[ \frac {2 \, \sqrt {3} b \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\left (\sqrt {3} \left (-b^{2}\right )^{\frac {1}{3}} b x - 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}}}{3 \, b^{2} x}\right ) - 2 \, \left (-b^{2}\right )^{\frac {2}{3}} \log \left (-\frac {\left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + \left (-b^{2}\right )^{\frac {2}{3}} \log \left (-\frac {\left (-b^{2}\right )^{\frac {1}{3}} b x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right )}{6 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(b*x^3 + a)^(2/3), x)

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.88, size = 100, normalized size = 1.39 \[ \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{3 \, b^{\frac {2}{3}}} + \frac {\log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{6 \, b^{\frac {2}{3}}} - \frac {\log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{3 \, b^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*x^3)^(2/3),x)

[Out]

int(x/(a + b*x^3)^(2/3), x)

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sympy [C]  time = 2.52, size = 37, normalized size = 0.51 \[ \frac {x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {5}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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