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

Optimal. Leaf size=88 \[ -\frac {\sqrt [3]{a+b x^3}}{x}-\frac {1}{2} \sqrt [3]{b} \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )-\frac {\sqrt [3]{b} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 138, normalized size of antiderivative = 1.57, number of steps used = 8, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {277, 331, 292, 31, 634, 617, 204, 628} \[ \frac {1}{6} \sqrt [3]{b} \log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )-\frac {\sqrt [3]{a+b x^3}}{x}-\frac {1}{3} \sqrt [3]{b} \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )-\frac {\sqrt [3]{b} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(1/3)/x^2,x]

[Out]

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

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 277

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

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

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Mathematica [C]  time = 0.01, size = 49, normalized size = 0.56 \[ -\frac {\sqrt [3]{a+b x^3} \, _2F_1\left (-\frac {1}{3},-\frac {1}{3};\frac {2}{3};-\frac {b x^3}{a}\right )}{x \sqrt [3]{\frac {b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(1/3)/x^2,x]

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(1/3)/x^2, x)

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maple [F]  time = 0.13, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.87, size = 114, normalized size = 1.30 \[ \frac {1}{3} \, \sqrt {3} b^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right ) + \frac {1}{6} \, b^{\frac {1}{3}} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - \frac {1}{3} \, b^{\frac {1}{3}} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.21, size = 40, normalized size = 0.45 \[ -\frac {{\left (b\,x^3+a\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},-\frac {1}{3};\ \frac {2}{3};\ -\frac {b\,x^3}{a}\right )}{x\,{\left (\frac {b\,x^3}{a}+1\right )}^{1/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(1/3)/x^2,x)

[Out]

-((a + b*x^3)^(1/3)*hypergeom([-1/3, -1/3], 2/3, -(b*x^3)/a))/(x*((b*x^3)/a + 1)^(1/3))

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sympy [C]  time = 2.21, size = 41, normalized size = 0.47 \[ \frac {\sqrt [3]{a} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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