3.15.8 \(\int \frac {(-1+x^3) \sqrt [3]{1+x^3}}{x^2} \, dx\)

Optimal. Leaf size=101 \[ \frac {\sqrt [3]{x^3+1} \left (x^3+3\right )}{3 x}+\frac {2}{9} \log \left (\sqrt [3]{x^3+1}-x\right )+\frac {2 \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+1}+x}\right )}{3 \sqrt {3}}-\frac {1}{9} \log \left (\sqrt [3]{x^3+1} x+\left (x^3+1\right )^{2/3}+x^2\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 110, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 9, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {453, 279, 331, 292, 31, 634, 618, 204, 628} \begin {gather*} \frac {\left (x^3+1\right )^{4/3}}{x}+\frac {2}{9} \log \left (1-\frac {x}{\sqrt [3]{x^3+1}}\right )+\frac {2 \tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {2}{3} \sqrt [3]{x^3+1} x^2-\frac {1}{9} \log \left (\frac {x}{\sqrt [3]{x^3+1}}+\frac {x^2}{\left (x^3+1\right )^{2/3}}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 + x^3)*(1 + x^3)^(1/3))/x^2,x]

[Out]

(-2*x^2*(1 + x^3)^(1/3))/3 + (1 + x^3)^(4/3)/x + (2*ArcTan[(1 + (2*x)/(1 + x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[3]) +
(2*Log[1 - x/(1 + x^3)^(1/3)])/9 - Log[1 + x^2/(1 + x^3)^(2/3) + x/(1 + x^3)^(1/3)]/9

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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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 453

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 {\left (-1+x^3\right ) \sqrt [3]{1+x^3}}{x^2} \, dx &=\frac {\left (1+x^3\right )^{4/3}}{x}-2 \int x \sqrt [3]{1+x^3} \, dx\\ &=-\frac {2}{3} x^2 \sqrt [3]{1+x^3}+\frac {\left (1+x^3\right )^{4/3}}{x}-\frac {2}{3} \int \frac {x}{\left (1+x^3\right )^{2/3}} \, dx\\ &=-\frac {2}{3} x^2 \sqrt [3]{1+x^3}+\frac {\left (1+x^3\right )^{4/3}}{x}-\frac {2}{3} \operatorname {Subst}\left (\int \frac {x}{1-x^3} \, dx,x,\frac {x}{\sqrt [3]{1+x^3}}\right )\\ &=-\frac {2}{3} x^2 \sqrt [3]{1+x^3}+\frac {\left (1+x^3\right )^{4/3}}{x}-\frac {2}{9} \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\frac {x}{\sqrt [3]{1+x^3}}\right )+\frac {2}{9} \operatorname {Subst}\left (\int \frac {1-x}{1+x+x^2} \, dx,x,\frac {x}{\sqrt [3]{1+x^3}}\right )\\ &=-\frac {2}{3} x^2 \sqrt [3]{1+x^3}+\frac {\left (1+x^3\right )^{4/3}}{x}+\frac {2}{9} \log \left (1-\frac {x}{\sqrt [3]{1+x^3}}\right )-\frac {1}{9} \operatorname {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\frac {x}{\sqrt [3]{1+x^3}}\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {x}{\sqrt [3]{1+x^3}}\right )\\ &=-\frac {2}{3} x^2 \sqrt [3]{1+x^3}+\frac {\left (1+x^3\right )^{4/3}}{x}+\frac {2}{9} \log \left (1-\frac {x}{\sqrt [3]{1+x^3}}\right )-\frac {1}{9} \log \left (1+\frac {x^2}{\left (1+x^3\right )^{2/3}}+\frac {x}{\sqrt [3]{1+x^3}}\right )-\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 x}{\sqrt [3]{1+x^3}}\right )\\ &=-\frac {2}{3} x^2 \sqrt [3]{1+x^3}+\frac {\left (1+x^3\right )^{4/3}}{x}+\frac {2 \tan ^{-1}\left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {2}{9} \log \left (1-\frac {x}{\sqrt [3]{1+x^3}}\right )-\frac {1}{9} \log \left (1+\frac {x^2}{\left (1+x^3\right )^{2/3}}+\frac {x}{\sqrt [3]{1+x^3}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 34, normalized size = 0.34 \begin {gather*} \frac {\left (x^3+1\right )^{4/3}}{x}-x^2 \, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};-x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1 + x^3)*(1 + x^3)^(1/3))/x^2,x]

[Out]

(1 + x^3)^(4/3)/x - x^2*Hypergeometric2F1[-1/3, 2/3, 5/3, -x^3]

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IntegrateAlgebraic [A]  time = 0.18, size = 101, normalized size = 1.00 \begin {gather*} \frac {\sqrt [3]{1+x^3} \left (3+x^3\right )}{3 x}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{1+x^3}}\right )}{3 \sqrt {3}}+\frac {2}{9} \log \left (-x+\sqrt [3]{1+x^3}\right )-\frac {1}{9} \log \left (x^2+x \sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + x^3)*(1 + x^3)^(1/3))/x^2,x]

[Out]

((1 + x^3)^(1/3)*(3 + x^3))/(3*x) + (2*ArcTan[(Sqrt[3]*x)/(x + 2*(1 + x^3)^(1/3))])/(3*Sqrt[3]) + (2*Log[-x +
(1 + x^3)^(1/3)])/9 - Log[x^2 + x*(1 + x^3)^(1/3) + (1 + x^3)^(2/3)]/9

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fricas [A]  time = 0.78, size = 105, normalized size = 1.04 \begin {gather*} \frac {2 \, \sqrt {3} x \arctan \left (-\frac {25382 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} - 13720 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {2}{3}} x + \sqrt {3} {\left (5831 \, x^{3} + 7200\right )}}{58653 \, x^{3} + 8000}\right ) + x \log \left (3 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{2} - 3 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} x + 1\right ) + 3 \, {\left (x^{3} + 3\right )} {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{9 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(2*sqrt(3)*x*arctan(-(25382*sqrt(3)*(x^3 + 1)^(1/3)*x^2 - 13720*sqrt(3)*(x^3 + 1)^(2/3)*x + sqrt(3)*(5831*
x^3 + 7200))/(58653*x^3 + 8000)) + x*log(3*(x^3 + 1)^(1/3)*x^2 - 3*(x^3 + 1)^(2/3)*x + 1) + 3*(x^3 + 3)*(x^3 +
1)^(1/3))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}} {\left (x^{3} - 1\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^3 + 1)^(1/3)*(x^3 - 1)/x^2, x)

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maple [C]  time = 2.17, size = 33, normalized size = 0.33

method	result	size
meijerg	\(\frac {\hypergeom \left (\left [-\frac {1}{3}, -\frac {1}{3}\right ], \left [\frac {2}{3}\right ], -x^{3}\right )}{x}+\frac {x^{2} \hypergeom \left (\left [-\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -x^{3}\right )}{2}\)	\(33\)
risch	\(\frac {x^{6}+4 x^{3}+3}{3 x \left (x^{3}+1\right )^{\frac {2}{3}}}-\frac {x^{2} \hypergeom \left (\left [\frac {2}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -x^{3}\right )}{3}\)	\(40\)
trager	\(\frac {\left (x^{3}+1\right )^{\frac {1}{3}} \left (x^{3}+3\right )}{3 x}+\frac {2 \ln \left (317 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-555 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x +2358 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}} x^{2}-1486 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}-2358 x \left (x^{3}+1\right )^{\frac {2}{3}}+1803 x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}+872 x^{3}-317 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-733 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+654\right )}{9}+\frac {2 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (-535 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+555 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x +1803 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}} x^{2}-2893 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}-1803 x \left (x^{3}+1\right )^{\frac {2}{3}}+2358 x^{2} \left (x^{3}+1\right )^{\frac {1}{3}}-1090 x^{3}+535 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-852 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-436\right )}{9}\)	\(254\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3-1)*(x^3+1)^(1/3)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/x*hypergeom([-1/3,-1/3],[2/3],-x^3)+1/2*x^2*hypergeom([-1/3,2/3],[5/3],-x^3)

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maxima [A]  time = 0.46, size = 105, normalized size = 1.04 \begin {gather*} -\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} + 1\right )}\right ) + \frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{3 \, x {\left (\frac {x^{3} + 1}{x^{3}} - 1\right )}} - \frac {1}{9} \, \log \left (\frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} + \frac {{\left (x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}} + 1\right ) + \frac {2}{9} \, \log \left (\frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 1)^(1/3)/x + 1)) + (x^3 + 1)^(1/3)/x + 1/3*(x^3 + 1)^(1/3)/(x*((x^3
+ 1)/x^3 - 1)) - 1/9*log((x^3 + 1)^(1/3)/x + (x^3 + 1)^(2/3)/x^2 + 1) + 2/9*log((x^3 + 1)^(1/3)/x - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 2.16, size = 65, normalized size = 0.64 \begin {gather*} \frac {x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {5}{3}\right )} - \frac {\Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 x \Gamma \left (\frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3-1)*(x**3+1)**(1/3)/x**2,x)

[Out]

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

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