3.1.48 \(\int \frac {1}{x \sqrt {1+x^3}} \, dx\)

Optimal. Leaf size=14 \[ -\frac {2}{3} \tanh ^{-1}\left (\sqrt {x^3+1}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {266, 63, 207} \begin {gather*} -\frac {2}{3} \tanh ^{-1}\left (\sqrt {x^3+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[1 + x^3]),x]

[Out]

(-2*ArcTanh[Sqrt[1 + x^3]])/3

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[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]]

Rubi steps

\begin {align*} \int \frac {1}{x \sqrt {1+x^3}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,x^3\right )\\ &=\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+x^3}\right )\\ &=-\frac {2}{3} \tanh ^{-1}\left (\sqrt {1+x^3}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} -\frac {2}{3} \tanh ^{-1}\left (\sqrt {x^3+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[1 + x^3]),x]

[Out]

(-2*ArcTanh[Sqrt[1 + x^3]])/3

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.02, size = 14, normalized size = 1.00 \begin {gather*} -\frac {2}{3} \tanh ^{-1}\left (\sqrt {1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTanh[Sqrt[1 + x^3]])/3

________________________________________________________________________________________

fricas [B]  time = 0.48, size = 25, normalized size = 1.79 \begin {gather*} -\frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.34, size = 26, normalized size = 1.86 \begin {gather*} -\frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {1}{3} \, \log \left ({\left | \sqrt {x^{3} + 1} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.29, size = 11, normalized size = 0.79

method	result	size
default	\(-\frac {2 \arctanh \left (\sqrt {x^{3}+1}\right )}{3}\)	\(11\)
elliptic	\(-\frac {2 \arctanh \left (\sqrt {x^{3}+1}\right )}{3}\)	\(11\)
trager	\(-\frac {\ln \left (-\frac {x^{3}+2 \sqrt {x^{3}+1}+2}{x^{3}}\right )}{3}\)	\(23\)
meijerg	\(\frac {\left (-2 \ln \relax (2)+3 \ln \relax (x )\right ) \sqrt {\pi }-2 \ln \left (\frac {1}{2}+\frac {\sqrt {x^{3}+1}}{2}\right ) \sqrt {\pi }}{3 \sqrt {\pi }}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*arctanh((x^3+1)^(1/2))

________________________________________________________________________________________

maxima [B]  time = 0.49, size = 25, normalized size = 1.79 \begin {gather*} -\frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.25, size = 164, normalized size = 11.71 \begin {gather*} -\frac {\left (3+\sqrt {3}\,1{}\mathrm {i}\right )\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\Pi \left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(x^3 + 1)^(1/2)),x)

[Out]

-((3^(1/2)*1i + 3)*((x + (3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + 1)/((3^(1/2)*1i)/2 + 3/2))^
(1/2)*(((3^(1/2)*1i)/2 - x + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*ellipticPi((3^(1/2)*1i)/2 + 3/2, asin(((x + 1)
/((3^(1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(x^3 - x*(((3^(1/2)*1i)/2 -
1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) - ((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2)

________________________________________________________________________________________

sympy [A]  time = 0.78, size = 12, normalized size = 0.86 \begin {gather*} - \frac {2 \operatorname {asinh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*asinh(x**(-3/2))/3

________________________________________________________________________________________