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

Optimal. Leaf size=67 \[ 6 \sqrt [3]{\sqrt {x}+1}+3 \log \left (1-\sqrt [3]{\sqrt {x}+1}\right )-\frac {\log (x)}{2}-2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{\sqrt {x}+1}+1}{\sqrt {3}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {266, 50, 57, 618, 204, 31} \[ 6 \sqrt [3]{\sqrt {x}+1}+3 \log \left (1-\sqrt [3]{\sqrt {x}+1}\right )-\frac {\log (x)}{2}-2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{\sqrt {x}+1}+1}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Sqrt[x])^(1/3)/x,x]

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 88, normalized size = 1.31 \[ 6 \sqrt [3]{\sqrt {x}+1}+2 \log \left (1-\sqrt [3]{\sqrt {x}+1}\right )-\log \left (\left (\sqrt {x}+1\right )^{2/3}+\sqrt [3]{\sqrt {x}+1}+1\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{\sqrt {x}+1}+1}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Sqrt[x])^(1/3)/x,x]

[Out]

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

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fricas [A]  time = 0.98, size = 65, normalized size = 0.97 \[ -2 \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (\sqrt {x} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + 6 \, {\left (\sqrt {x} + 1\right )}^{\frac {1}{3}} - \log \left ({\left (\sqrt {x} + 1\right )}^{\frac {2}{3}} + {\left (\sqrt {x} + 1\right )}^{\frac {1}{3}} + 1\right ) + 2 \, \log \left ({\left (\sqrt {x} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.63, size = 64, normalized size = 0.96 \[ -2 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\sqrt {x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + 6 \, {\left (\sqrt {x} + 1\right )}^{\frac {1}{3}} - \log \left ({\left (\sqrt {x} + 1\right )}^{\frac {2}{3}} + {\left (\sqrt {x} + 1\right )}^{\frac {1}{3}} + 1\right ) + 2 \, \log \left ({\left | {\left (\sqrt {x} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 64, normalized size = 0.96 \[ -2 \sqrt {3}\, \arctan \left (\frac {\left (1+2 \left (\sqrt {x}+1\right )^{\frac {1}{3}}\right ) \sqrt {3}}{3}\right )+2 \ln \left (\left (\sqrt {x}+1\right )^{\frac {1}{3}}-1\right )-\ln \left (\left (\sqrt {x}+1\right )^{\frac {2}{3}}+\left (\sqrt {x}+1\right )^{\frac {1}{3}}+1\right )+6 \left (\sqrt {x}+1\right )^{\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.72, size = 63, normalized size = 0.94 \[ -2 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\sqrt {x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + 6 \, {\left (\sqrt {x} + 1\right )}^{\frac {1}{3}} - \log \left ({\left (\sqrt {x} + 1\right )}^{\frac {2}{3}} + {\left (\sqrt {x} + 1\right )}^{\frac {1}{3}} + 1\right ) + 2 \, \log \left ({\left (\sqrt {x} + 1\right )}^{\frac {1}{3}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.83, size = 73, normalized size = 1.09 \[ 2\,\ln \left ({\left (\sqrt {x}+1\right )}^{1/3}-1\right )+6\,{\left (\sqrt {x}+1\right )}^{1/3}+\ln \left ({\left (\sqrt {x}+1\right )}^{1/3}+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )-\ln \left ({\left (\sqrt {x}+1\right )}^{1/3}+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(1/2) + 1)^(1/3)/x,x)

[Out]

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

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sympy [C]  time = 1.05, size = 39, normalized size = 0.58 \[ - \frac {2 \sqrt [6]{x} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{\sqrt {x}}} \right )}}{\Gamma \left (\frac {2}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x**(1/6)*gamma(-1/3)*hyper((-1/3, -1/3), (2/3,), exp_polar(I*pi)/sqrt(x))/gamma(2/3)

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