\(\int \frac {\sqrt {1+x^4}}{x} \, dx\) [320]

Optimal result

Integrand size = 13, antiderivative size = 28 \[ \int \frac {\sqrt {1+x^4}}{x} \, dx=\frac {\sqrt {1+x^4}}{2}-\frac {1}{2} \text {arctanh}\left (\sqrt {1+x^4}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {272, 52, 65, 213} \[ \int \frac {\sqrt {1+x^4}}{x} \, dx=\frac {\sqrt {x^4+1}}{2}-\frac {1}{2} \text {arctanh}\left (\sqrt {x^4+1}\right ) \]

[In]

Int[Sqrt[1 + x^4]/x,x]

[Out]

Sqrt[1 + x^4]/2 - ArcTanh[Sqrt[1 + x^4]]/2

Rule 52

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {1+x^4}}{x} \, dx=\frac {\sqrt {1+x^4}}{2}-\frac {1}{2} \text {arctanh}\left (\sqrt {1+x^4}\right ) \]

[In]

Integrate[Sqrt[1 + x^4]/x,x]

[Out]

Sqrt[1 + x^4]/2 - ArcTanh[Sqrt[1 + x^4]]/2

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {\sqrt {x^{4}+1}}{2}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{2}\)	\(21\)
elliptic	\(\frac {\sqrt {x^{4}+1}}{2}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{2}\)	\(21\)
pseudoelliptic	\(\frac {\sqrt {x^{4}+1}}{2}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{2}\)	\(21\)
trager	\(\frac {\sqrt {x^{4}+1}}{2}-\frac {\ln \left (\frac {1+\sqrt {x^{4}+1}}{x^{2}}\right )}{2}\)	\(27\)
meijerg	\(-\frac {4 \sqrt {\pi }-4 \sqrt {\pi }\, \sqrt {x^{4}+1}+4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{4}+1}}{2}\right )-2 \left (2-2 \ln \left (2\right )+4 \ln \left (x \right )\right ) \sqrt {\pi }}{8 \sqrt {\pi }}\)	\(56\)

[In]

int((x^4+1)^(1/2)/x,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {1+x^4}}{x} \, dx=\frac {1}{2} \, \sqrt {x^{4} + 1} - \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} + 1\right ) + \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} - 1\right ) \]

[In]

integrate((x^4+1)^(1/2)/x,x, algorithm="fricas")

[Out]

1/2*sqrt(x^4 + 1) - 1/4*log(sqrt(x^4 + 1) + 1) + 1/4*log(sqrt(x^4 + 1) - 1)

Sympy [A] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {\sqrt {1+x^4}}{x} \, dx=\frac {x^{2}}{2 \sqrt {1 + \frac {1}{x^{4}}}} - \frac {\operatorname {asinh}{\left (\frac {1}{x^{2}} \right )}}{2} + \frac {1}{2 x^{2} \sqrt {1 + \frac {1}{x^{4}}}} \]

[In]

integrate((x**4+1)**(1/2)/x,x)

[Out]

x**2/(2*sqrt(1 + x**(-4))) - asinh(x**(-2))/2 + 1/(2*x**2*sqrt(1 + x**(-4)))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {1+x^4}}{x} \, dx=\frac {1}{2} \, \sqrt {x^{4} + 1} - \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} + 1\right ) + \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} - 1\right ) \]

[In]

integrate((x^4+1)^(1/2)/x,x, algorithm="maxima")

[Out]

1/2*sqrt(x^4 + 1) - 1/4*log(sqrt(x^4 + 1) + 1) + 1/4*log(sqrt(x^4 + 1) - 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {1+x^4}}{x} \, dx=\frac {1}{2} \, \sqrt {x^{4} + 1} - \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} + 1\right ) + \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} - 1\right ) \]

[In]

integrate((x^4+1)^(1/2)/x,x, algorithm="giac")

[Out]

1/2*sqrt(x^4 + 1) - 1/4*log(sqrt(x^4 + 1) + 1) + 1/4*log(sqrt(x^4 + 1) - 1)

Mupad [B] (verification not implemented)

Time = 5.39 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {1+x^4}}{x} \, dx=\frac {\sqrt {x^4+1}}{2}-\frac {\mathrm {atanh}\left (\sqrt {x^4+1}\right )}{2} \]

[In]

int((x^4 + 1)^(1/2)/x,x)

[Out]

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