\(\int \frac {1}{x \sqrt {1+x^3}} \, dx\) [48]

Optimal result

Integrand size = 13, antiderivative size = 14 \[ \int \frac {1}{x \sqrt {1+x^3}} \, dx=-\frac {2}{3} \text {arctanh}\left (\sqrt {1+x^3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {272, 65, 213} \[ \int \frac {1}{x \sqrt {1+x^3}} \, dx=-\frac {2}{3} \text {arctanh}\left (\sqrt {x^3+1}\right ) \]

[In]

Int[1/(x*Sqrt[1 + x^3]),x]

[Out]

(-2*ArcTanh[Sqrt[1 + x^3]])/3

Rule 65

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 213

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

Rule 272

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \sqrt {1+x^3}} \, dx=-\frac {2}{3} \text {arctanh}\left (\sqrt {1+x^3}\right ) \]

[In]

Integrate[1/(x*Sqrt[1 + x^3]),x]

[Out]

(-2*ArcTanh[Sqrt[1 + x^3]])/3

Maple [A] (verified)

Time = 1.97 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 25 vs. \(2 (10) = 20\).

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.79 \[ \int \frac {1}{x \sqrt {1+x^3}} \, dx=-\frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} - 1\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x \sqrt {1+x^3}} \, dx=- \frac {2 \operatorname {asinh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 25 vs. \(2 (10) = 20\).

Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.79 \[ \int \frac {1}{x \sqrt {1+x^3}} \, dx=-\frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} - 1\right ) \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (10) = 20\).

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.86 \[ \int \frac {1}{x \sqrt {1+x^3}} \, dx=-\frac {1}{3} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {1}{3} \, \log \left ({\left | \sqrt {x^{3} + 1} - 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 5.25 (sec) , antiderivative size = 164, normalized size of antiderivative = 11.71 \[ \int \frac {1}{x \sqrt {1+x^3}} \, dx=-\frac {\left (3+\sqrt {3}\,1{}\mathrm {i}\right )\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\Pi \left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

[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)