3.2256 \(\int \frac {\sqrt [3]{x}}{1+\sqrt {x}} \, dx\)

Optimal. Leaf size=58 \[ \frac {6 x^{5/6}}{5}-3 \sqrt [3]{x}-3 \log \left (\sqrt [6]{x}+1\right )+\log \left (\sqrt {x}+1\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {1-2 \sqrt [6]{x}}{\sqrt {3}}\right ) \]

[Out]

-3*x^(1/3)+6/5*x^(5/6)-3*ln(1+x^(1/6))+ln(1+x^(1/2))-2*arctan(1/3*(1-2*x^(1/6))*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {341, 50, 56, 618, 204, 31} \[ \frac {6 x^{5/6}}{5}-3 \sqrt [3]{x}-3 \log \left (\sqrt [6]{x}+1\right )+\log \left (\sqrt {x}+1\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {1-2 \sqrt [6]{x}}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^(1/3)/(1 + Sqrt[x]),x]

[Out]

-3*x^(1/3) + (6*x^(5/6))/5 - 2*Sqrt[3]*ArcTan[(1 - 2*x^(1/6))/Sqrt[3]] - 3*Log[1 + x^(1/6)] + Log[1 + Sqrt[x]]

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 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[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] && Ne
gQ[(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 341

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k*(
m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[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 {\sqrt [3]{x}}{1+\sqrt {x}} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^{5/3}}{1+x} \, dx,x,\sqrt {x}\right )\\ &=\frac {6 x^{5/6}}{5}-2 \operatorname {Subst}\left (\int \frac {x^{2/3}}{1+x} \, dx,x,\sqrt {x}\right )\\ &=-3 \sqrt [3]{x}+\frac {6 x^{5/6}}{5}+2 \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{x} (1+x)} \, dx,x,\sqrt {x}\right )\\ &=-3 \sqrt [3]{x}+\frac {6 x^{5/6}}{5}+\log \left (1+\sqrt {x}\right )-3 \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\sqrt [6]{x}\right )+3 \operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\sqrt [6]{x}\right )\\ &=-3 \sqrt [3]{x}+\frac {6 x^{5/6}}{5}-3 \log \left (1+\sqrt [6]{x}\right )+\log \left (1+\sqrt {x}\right )-6 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \sqrt [6]{x}\right )\\ &=-3 \sqrt [3]{x}+\frac {6 x^{5/6}}{5}-2 \sqrt {3} \tan ^{-1}\left (\frac {1-2 \sqrt [6]{x}}{\sqrt {3}}\right )-3 \log \left (1+\sqrt [6]{x}\right )+\log \left (1+\sqrt {x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.60 \[ \frac {3}{5} \sqrt [3]{x} \left (5 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\sqrt {x}\right )+2 \sqrt {x}-5\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(1/3)/(1 + Sqrt[x]),x]

[Out]

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

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fricas [A]  time = 0.89, size = 50, normalized size = 0.86 \[ 2 \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} x^{\frac {1}{6}} - \frac {1}{3} \, \sqrt {3}\right ) + \frac {6}{5} \, x^{\frac {5}{6}} - 3 \, x^{\frac {1}{3}} + \log \left (x^{\frac {1}{3}} - x^{\frac {1}{6}} + 1\right ) - 2 \, \log \left (x^{\frac {1}{6}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(3)*arctan(2/3*sqrt(3)*x^(1/6) - 1/3*sqrt(3)) + 6/5*x^(5/6) - 3*x^(1/3) + log(x^(1/3) - x^(1/6) + 1) - 2
*log(x^(1/6) + 1)

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giac [A]  time = 0.16, size = 48, normalized size = 0.83 \[ 2 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{\frac {1}{6}} - 1\right )}\right ) + \frac {6}{5} \, x^{\frac {5}{6}} - 3 \, x^{\frac {1}{3}} + \log \left (x^{\frac {1}{3}} - x^{\frac {1}{6}} + 1\right ) - 2 \, \log \left (x^{\frac {1}{6}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^(1/6) - 1)) + 6/5*x^(5/6) - 3*x^(1/3) + log(x^(1/3) - x^(1/6) + 1) - 2*log(x
^(1/6) + 1)

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maple [A]  time = 0.01, size = 49, normalized size = 0.84 \[ 2 \sqrt {3}\, \arctan \left (\frac {\left (2 x^{\frac {1}{6}}-1\right ) \sqrt {3}}{3}\right )-2 \ln \left (x^{\frac {1}{6}}+1\right )+\ln \left (x^{\frac {1}{3}}-x^{\frac {1}{6}}+1\right )+\frac {6 x^{\frac {5}{6}}}{5}-3 x^{\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6/5*x^(5/6)-3*x^(1/3)+ln(x^(1/3)-x^(1/6)+1)+2*3^(1/2)*arctan(1/3*(2*x^(1/6)-1)*3^(1/2))-2*ln(1+x^(1/6))

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maxima [A]  time = 1.98, size = 48, normalized size = 0.83 \[ 2 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{\frac {1}{6}} - 1\right )}\right ) + \frac {6}{5} \, x^{\frac {5}{6}} - 3 \, x^{\frac {1}{3}} + \log \left (x^{\frac {1}{3}} - x^{\frac {1}{6}} + 1\right ) - 2 \, \log \left (x^{\frac {1}{6}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^(1/6) - 1)) + 6/5*x^(5/6) - 3*x^(1/3) + log(x^(1/3) - x^(1/6) + 1) - 2*log(x
^(1/6) + 1)

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mupad [B]  time = 1.12, size = 78, normalized size = 1.34 \[ \frac {6\,x^{5/6}}{5}-\ln \left (9\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2+36\,x^{1/6}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )+\ln \left (9\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2+36\,x^{1/6}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )-3\,x^{1/3}-2\,\ln \left (36\,x^{1/6}+36\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/3)/(x^(1/2) + 1),x)

[Out]

log(9*(3^(1/2)*1i + 1)^2 + 36*x^(1/6))*(3^(1/2)*1i + 1) - log(9*(3^(1/2)*1i - 1)^2 + 36*x^(1/6))*(3^(1/2)*1i -
1) - 2*log(36*x^(1/6) + 36) - 3*x^(1/3) + (6*x^(5/6))/5

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sympy [C]  time = 1.21, size = 138, normalized size = 2.38 \[ \frac {16 x^{\frac {5}{6}} \Gamma \left (\frac {8}{3}\right )}{5 \Gamma \left (\frac {11}{3}\right )} - \frac {8 \sqrt [3]{x} \Gamma \left (\frac {8}{3}\right )}{\Gamma \left (\frac {11}{3}\right )} - \frac {16 e^{- \frac {2 i \pi }{3}} \log {\left (- \sqrt [6]{x} e^{\frac {i \pi }{3}} + 1 \right )} \Gamma \left (\frac {8}{3}\right )}{3 \Gamma \left (\frac {11}{3}\right )} - \frac {16 \log {\left (- \sqrt [6]{x} e^{i \pi } + 1 \right )} \Gamma \left (\frac {8}{3}\right )}{3 \Gamma \left (\frac {11}{3}\right )} - \frac {16 e^{\frac {2 i \pi }{3}} \log {\left (- \sqrt [6]{x} e^{\frac {5 i \pi }{3}} + 1 \right )} \Gamma \left (\frac {8}{3}\right )}{3 \Gamma \left (\frac {11}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16*x**(5/6)*gamma(8/3)/(5*gamma(11/3)) - 8*x**(1/3)*gamma(8/3)/gamma(11/3) - 16*exp(-2*I*pi/3)*log(-x**(1/6)*e
xp_polar(I*pi/3) + 1)*gamma(8/3)/(3*gamma(11/3)) - 16*log(-x**(1/6)*exp_polar(I*pi) + 1)*gamma(8/3)/(3*gamma(1
1/3)) - 16*exp(2*I*pi/3)*log(-x**(1/6)*exp_polar(5*I*pi/3) + 1)*gamma(8/3)/(3*gamma(11/3))

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