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

Optimal. Leaf size=87 \[ \frac {3}{4} \left (x^2+1\right )^{2/3}+\frac {1}{2} \log \left (\sqrt [3]{x^2+1}-1\right )-\frac {1}{4} \log \left (\left (x^2+1\right )^{2/3}+\sqrt [3]{x^2+1}+1\right )+\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x^2+1}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 0.77, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {266, 50, 55, 618, 204, 31} \begin {gather*} \frac {3}{4} \left (x^2+1\right )^{2/3}+\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 + x^2)^(2/3)/x,x]

[Out]

(3*(1 + x^2)^(2/3))/4 + (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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), 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 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 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 {\left (1+x^2\right )^{2/3}}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(1+x)^{2/3}}{x} \, dx,x,x^2\right )\\ &=\frac {3}{4} \left (1+x^2\right )^{2/3}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{1+x}} \, dx,x,x^2\right )\\ &=\frac {3}{4} \left (1+x^2\right )^{2/3}-\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 )\\ &=\frac {3}{4} \left (1+x^2\right )^{2/3}-\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 )\\ &=\frac {3}{4} \left (1+x^2\right )^{2/3}+\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*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 62, normalized size = 0.71 \begin {gather*} \frac {1}{4} \left (3 \left (\left (x^2+1\right )^{2/3}+\log \left (1-\sqrt [3]{x^2+1}\right )\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x^2+1}+1}{\sqrt {3}}\right )-2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^2)^(2/3)/x,x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.05, size = 87, normalized size = 1.00 \begin {gather*} \frac {3}{4} \left (1+x^2\right )^{2/3}+\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 + x^2)^(2/3)/x,x]

[Out]

(3*(1 + x^2)^(2/3))/4 + (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

________________________________________________________________________________________

fricas [A]  time = 0.65, size = 65, normalized size = 0.75 \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 ) + \frac {3}{4} \, {\left (x^{2} + 1\right )}^{\frac {2}{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((x^2+1)^(2/3)/x,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.22, size = 63, normalized size = 0.72 \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 ) + \frac {3}{4} \, {\left (x^{2} + 1\right )}^{\frac {2}{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((x^2+1)^(2/3)/x,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [C]  time = 1.51, size = 64, normalized size = 0.74

method	result	size
meijerg	\(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (-\frac {2 \pi \sqrt {3}\, x^{2} \hypergeom \left (\left [\frac {1}{3}, 1, 1\right ], \left [2, 2\right ], -x^{2}\right )}{3 \Gamma \left (\frac {2}{3}\right )}-\frac {\left (\frac {3}{2}-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+2 \ln \relax (x )\right ) \pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right )}\right )}{6 \pi }\)	\(64\)
trager	\(\frac {3 \left (x^{2}+1\right )^{\frac {2}{3}}}{4}+\frac {3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \ln \left (\frac {-36 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+45 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}+27 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{2}+1\right )^{\frac {2}{3}}+45 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{3}}+36 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}-2 x^{2}-9 \left (x^{2}+1\right )^{\frac {1}{3}}+57 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-5}{x^{2}}\right )}{2}-\frac {\ln \left (-\frac {36 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+45 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}+51 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+24 \left (x^{2}+1\right )^{\frac {2}{3}}+45 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{3}}-36 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}+15 x^{2}+24 \left (x^{2}+1\right )^{\frac {1}{3}}+33 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )+20}{x^{2}}\right )}{2}-\frac {3 \ln \left (-\frac {36 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{2}+45 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {2}{3}}+51 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{2}+24 \left (x^{2}+1\right )^{\frac {2}{3}}+45 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{2}+1\right )^{\frac {1}{3}}-36 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}+15 x^{2}+24 \left (x^{2}+1\right )^{\frac {1}{3}}+33 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )+20}{x^{2}}\right ) \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{2}\)	\(437\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 63, normalized size = 0.72 \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 ) + \frac {3}{4} \, {\left (x^{2} + 1\right )}^{\frac {2}{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((x^2+1)^(2/3)/x,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.87, size = 89, normalized size = 1.02 \begin {gather*} \frac {\ln \left (\frac {9\,{\left (x^2+1\right )}^{1/3}}{4}-\frac {9}{4}\right )}{2}+\ln \left (\frac {9\,{\left (x^2+1\right )}^{1/3}}{4}-9\,{\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}^2\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )-\ln \left (\frac {9\,{\left (x^2+1\right )}^{1/3}}{4}-9\,{\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}^2\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )+\frac {3\,{\left (x^2+1\right )}^{2/3}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + 1)^(2/3)/x,x)

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 0.81, size = 37, normalized size = 0.43 \begin {gather*} - \frac {x^{\frac {4}{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{2}}} \right )}}{2 \Gamma \left (\frac {1}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________