3.12.31 \(\int \frac {1+2 x^2}{x (1+x^2)^{2/3}} \, dx\)

Optimal. Leaf size=85 \[ 3 \sqrt [3]{x^2+1}+\frac {1}{2} \log \left (\sqrt [3]{x^2+1}-1\right )-\frac {1}{4} \log \left (\left (x^2+1\right )^{2/3}+\sqrt [3]{x^2+1}+1\right )-\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x^2+1}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 65, normalized size of antiderivative = 0.76, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {446, 80, 57, 618, 204, 31} \begin {gather*} 3 \sqrt [3]{x^2+1}+\frac {3}{4} \log \left (1-\sqrt [3]{x^2+1}\right )-\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x^2+1}+1}{\sqrt {3}}\right )-\frac {\log (x)}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 84, normalized size = 0.99 \begin {gather*} 3 \sqrt [3]{x^2+1}+\frac {1}{2} \log \left (1-\sqrt [3]{x^2+1}\right )-\frac {1}{4} \log \left (\left (x^2+1\right )^{2/3}+\sqrt [3]{x^2+1}+1\right )-\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x^2+1}+1}{\sqrt {3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.05, size = 85, normalized size = 1.00 \begin {gather*} 3 \sqrt [3]{1+x^2}-\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^2}}{\sqrt {3}}\right )+\frac {1}{2} \log \left (-1+\sqrt [3]{1+x^2}\right )-\frac {1}{4} \log \left (1+\sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 65, normalized size = 0.76 \begin {gather*} -\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{2} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + 3 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} - \frac {1}{4} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {2}{3}} + {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 63, normalized size = 0.74 \begin {gather*} -\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + 3 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} - \frac {1}{4} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {2}{3}} + {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.88, size = 62, normalized size = 0.73

method	result	size
meijerg	\(x^{2} \hypergeom \left (\left [\frac {2}{3}, 1\right ], \relax [2], -x^{2}\right )+\frac {-\frac {2 \Gamma \left (\frac {2}{3}\right ) x^{2} \hypergeom \left (\left [1, 1, \frac {5}{3}\right ], \left [2, 2\right ], -x^{2}\right )}{3}+\left (\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+2 \ln \relax (x )\right ) \Gamma \left (\frac {2}{3}\right )}{2 \Gamma \left (\frac {2}{3}\right )}\)	\(62\)
trager	\(3 \left (x^{2}+1\right )^{\frac {1}{3}}+\frac {\ln \left (\frac {-36 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}+90 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}-18 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{2}+1\right )^{\frac {2}{3}}-144 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{3}}+36 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2}+10 x^{2}-15 \left (x^{2}+1\right )^{\frac {1}{3}}+60 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )+25}{x^{2}}\right )}{2}-\frac {\ln \left (-\frac {-180 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}+90 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}-96 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}+24 \left (x^{2}+1\right )^{\frac {2}{3}}+54 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{3}}+180 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2}-3 x^{2}-15 \left (x^{2}+1\right )^{\frac {1}{3}}-114 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )-4}{x^{2}}\right )}{2}-3 \ln \left (-\frac {-180 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}+90 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}-96 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}+24 \left (x^{2}+1\right )^{\frac {2}{3}}+54 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{3}}+180 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2}-3 x^{2}-15 \left (x^{2}+1\right )^{\frac {1}{3}}-114 \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )-4}{x^{2}}\right ) \RootOf \left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )\)	\(426\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*hypergeom([2/3,1],[2],-x^2)+1/2/GAMMA(2/3)*(-2/3*GAMMA(2/3)*x^2*hypergeom([1,1,5/3],[2,2],-x^2)+(1/6*Pi*3^
(1/2)-3/2*ln(3)+2*ln(x))*GAMMA(2/3))

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maxima [A]  time = 0.43, size = 63, normalized size = 0.74 \begin {gather*} -\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + 3 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} - \frac {1}{4} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {2}{3}} + {\left (x^{2} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{3}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.00, size = 79, normalized size = 0.93 \begin {gather*} \frac {\ln \left (\frac {9\,{\left (x^2+1\right )}^{1/3}}{4}-\frac {9}{4}\right )}{2}+3\,{\left (x^2+1\right )}^{1/3}+\ln \left (\frac {9\,{\left (x^2+1\right )}^{1/3}}{2}+\frac {9}{4}-\frac {\sqrt {3}\,9{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )-\ln \left (\frac {9\,{\left (x^2+1\right )}^{1/3}}{2}+\frac {9}{4}+\frac {\sqrt {3}\,9{}\mathrm {i}}{4}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((9*(x^2 + 1)^(1/3))/4 - 9/4)/2 + 3*(x^2 + 1)^(1/3) + log((9*(x^2 + 1)^(1/3))/2 - (3^(1/2)*9i)/4 + 9/4)*((3
^(1/2)*1i)/4 - 1/4) - log((3^(1/2)*9i)/4 + (9*(x^2 + 1)^(1/3))/2 + 9/4)*((3^(1/2)*1i)/4 + 1/4)

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sympy [C]  time = 7.39, size = 41, normalized size = 0.48 \begin {gather*} 3 \sqrt [3]{x^{2} + 1} - \frac {\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 {e^{i \pi }}{x^{2}}} \right )}}{2 x^{\frac {4}{3}} \Gamma \left (\frac {5}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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