\(\int \frac {1}{x \sqrt {1+x^8}} \, dx\) [3]

Optimal result

Integrand size = 13, antiderivative size = 14 \[ \int \frac {1}{x \sqrt {1+x^8}} \, dx=-\frac {1}{4} \text {arctanh}\left (\sqrt {1+x^8}\right ) \]

[Out]

-1/4*arctanh((x^8+1)^(1/2))

Rubi [A] (verified)

Time = 0.00 (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} \[ \int \frac {1}{x \sqrt {1+x^8}} \, dx=-\frac {1}{4} \text {arctanh}\left (\sqrt {x^8+1}\right ) \]

[In]

Int[1/(x*Sqrt[1 + x^8]),x]

[Out]

-1/4*ArcTanh[Sqrt[1 + x^8]]

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}{8} \text {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,x^8\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+x^8}\right ) \\ & = -\frac {1}{4} \text {arctanh}\left (\sqrt {1+x^8}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \sqrt {1+x^8}} \, dx=-\frac {1}{4} \text {arctanh}\left (\sqrt {1+x^8}\right ) \]

[In]

Integrate[1/(x*Sqrt[1 + x^8]),x]

[Out]

-1/4*ArcTanh[Sqrt[1 + x^8]]

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{8}+1}}\right )}{4}\)	\(11\)
trager	\(-\frac {\ln \left (\frac {\sqrt {x^{8}+1}+1}{x^{4}}\right )}{4}\)	\(17\)
meijerg	\(\frac {\left (-2 \ln \left (2\right )+8 \ln \left (x \right )\right ) \sqrt {\pi }-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{8}+1}}{2}\right )}{8 \sqrt {\pi }}\)	\(37\)

[In]

int(1/(x^8+1)^(1/2)/x,x,method=_RETURNVERBOSE)

[Out]

-1/4*arctanh(1/(x^8+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 = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.79 \[ \int \frac {1}{x \sqrt {1+x^8}} \, dx=-\frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \]

[In]

integrate(1/(x^8+1)^(1/2)/x,x, algorithm="fricas")

[Out]

-1/8*log(sqrt(x^8 + 1) + 1) + 1/8*log(sqrt(x^8 + 1) - 1)

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.57 \[ \int \frac {1}{x \sqrt {1+x^8}} \, dx=- \frac {\operatorname {asinh}{\left (\frac {1}{x^{4}} \right )}}{4} \]

[In]

integrate(1/(x**8+1)**(1/2)/x,x)

[Out]

-asinh(x**(-4))/4

Maxima [B] (verification not implemented)

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

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.79 \[ \int \frac {1}{x \sqrt {1+x^8}} \, dx=-\frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \]

[In]

integrate(1/(x^8+1)^(1/2)/x,x, algorithm="maxima")

[Out]

-1/8*log(sqrt(x^8 + 1) + 1) + 1/8*log(sqrt(x^8 + 1) - 1)

Giac [B] (verification not implemented)

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

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.79 \[ \int \frac {1}{x \sqrt {1+x^8}} \, dx=-\frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \]

[In]

integrate(1/(x^8+1)^(1/2)/x,x, algorithm="giac")

[Out]

-1/8*log(sqrt(x^8 + 1) + 1) + 1/8*log(sqrt(x^8 + 1) - 1)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x \sqrt {1+x^8}} \, dx=-\frac {\mathrm {atanh}\left (\sqrt {x^8+1}\right )}{4} \]

[In]

int(1/(x*(x^8 + 1)^(1/2)),x)

[Out]

-atanh((x^8 + 1)^(1/2))/4