3.1511 \(\int \frac {\sqrt {1+x^8}}{x} \, dx\)

Optimal. Leaf size=28 \[ \frac {\sqrt {x^8+1}}{4}-\frac {1}{4} \tanh ^{-1}\left (\sqrt {x^8+1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {266, 50, 63, 207} \[ \frac {\sqrt {x^8+1}}{4}-\frac {1}{4} \tanh ^{-1}\left (\sqrt {x^8+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x^8]/x,x]

[Out]

Sqrt[1 + x^8]/4 - ArcTanh[Sqrt[1 + x^8]]/4

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

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Mathematica [A]  time = 0.00, size = 28, normalized size = 1.00 \[ \frac {\sqrt {x^8+1}}{4}-\frac {1}{4} \tanh ^{-1}\left (\sqrt {x^8+1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + x^8]/x,x]

[Out]

Sqrt[1 + x^8]/4 - ArcTanh[Sqrt[1 + x^8]]/4

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fricas [A]  time = 0.78, size = 34, normalized size = 1.21 \[ \frac {1}{4} \, \sqrt {x^{8} + 1} - \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(x^8 + 1) - 1/8*log(sqrt(x^8 + 1) + 1) + 1/8*log(sqrt(x^8 + 1) - 1)

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giac [A]  time = 0.15, size = 34, normalized size = 1.21 \[ \frac {1}{4} \, \sqrt {x^{8} + 1} - \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(x^8 + 1) - 1/8*log(sqrt(x^8 + 1) + 1) + 1/8*log(sqrt(x^8 + 1) - 1)

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maple [B]  time = 0.17, size = 56, normalized size = 2.00 \[ -\frac {4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{8}+1}}{2}\right )+4 \sqrt {\pi }-4 \sqrt {\pi }\, \sqrt {x^{8}+1}-2 \left (8 \ln \relax (x )+2-2 \ln \relax (2)\right ) \sqrt {\pi }}{16 \sqrt {\pi }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8+1)^(1/2)/x,x)

[Out]

-1/16/Pi^(1/2)*(4*Pi^(1/2)-4*Pi^(1/2)*(x^8+1)^(1/2)+4*Pi^(1/2)*ln(1/2+1/2*(x^8+1)^(1/2))-2*(2-2*ln(2)+8*ln(x))
*Pi^(1/2))

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maxima [A]  time = 1.05, size = 34, normalized size = 1.21 \[ \frac {1}{4} \, \sqrt {x^{8} + 1} - \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(x^8 + 1) - 1/8*log(sqrt(x^8 + 1) + 1) + 1/8*log(sqrt(x^8 + 1) - 1)

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mupad [B]  time = 1.14, size = 20, normalized size = 0.71 \[ \frac {\sqrt {x^8+1}}{4}-\frac {\mathrm {atanh}\left (\sqrt {x^8+1}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8 + 1)^(1/2)/x,x)

[Out]

(x^8 + 1)^(1/2)/4 - atanh((x^8 + 1)^(1/2))/4

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sympy [A]  time = 1.30, size = 39, normalized size = 1.39 \[ \frac {x^{4}}{4 \sqrt {1 + \frac {1}{x^{8}}}} - \frac {\operatorname {asinh}{\left (\frac {1}{x^{4}} \right )}}{4} + \frac {1}{4 x^{4} \sqrt {1 + \frac {1}{x^{8}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4/(4*sqrt(1 + x**(-8))) - asinh(x**(-4))/4 + 1/(4*x**4*sqrt(1 + x**(-8)))

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