\(\int \frac {1+x^2}{x \sqrt {1+x^4}} \, dx\) [322]

Optimal result

Integrand size = 18, antiderivative size = 16 \[ \int \frac {1+x^2}{x \sqrt {1+x^4}} \, dx=\text {arctanh}\left (\frac {-1+x^2}{\sqrt {1+x^4}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1266, 858, 221, 272, 65, 213} \[ \int \frac {1+x^2}{x \sqrt {1+x^4}} \, dx=\frac {\text {arcsinh}\left (x^2\right )}{2}-\frac {1}{2} \text {arctanh}\left (\sqrt {x^4+1}\right ) \]

[In]

Int[(1 + x^2)/(x*Sqrt[1 + x^4]),x]

[Out]

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

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 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

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]]

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1266

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(41\) vs. \(2(16)=32\).

Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.56 \[ \int \frac {1+x^2}{x \sqrt {1+x^4}} \, dx=\text {arctanh}\left (1+2 x^2-2 \sqrt {1+x^4}\right )-\frac {1}{2} \log \left (1-x^2+\sqrt {1+x^4}\right ) \]

[In]

Integrate[(1 + x^2)/(x*Sqrt[1 + x^4]),x]

[Out]

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12

method	result	size
default	\(\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{2}\)	\(18\)
trager	\(\ln \left (\frac {x^{2}+\sqrt {x^{4}+1}-1}{x}\right )\)	\(18\)
elliptic	\(\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{2}\)	\(18\)
pseudoelliptic	\(\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{4}+1}}\right )}{2}\)	\(18\)
meijerg	\(\frac {\operatorname {arcsinh}\left (x^{2}\right )}{2}+\frac {\left (-2 \ln \left (2\right )+4 \ln \left (x \right )\right ) \sqrt {\pi }-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{4}+1}}{2}\right )}{4 \sqrt {\pi }}\)	\(44\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (14) = 28\).

Time = 0.24 (sec) , antiderivative size = 49, normalized size of antiderivative = 3.06 \[ \int \frac {1+x^2}{x \sqrt {1+x^4}} \, dx=-\frac {1}{2} \, \log \left (2 \, x^{4} - x^{2} - \sqrt {x^{4} + 1} {\left (2 \, x^{2} - 1\right )} + 1\right ) + \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 1} - 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 3.63 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {1+x^2}{x \sqrt {1+x^4}} \, dx=- \frac {\operatorname {asinh}{\left (\frac {1}{x^{2}} \right )}}{2} + \frac {\operatorname {asinh}{\left (x^{2} \right )}}{2} \]

[In]

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

[Out]

-asinh(x**(-2))/2 + asinh(x**2)/2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (14) = 28\).

Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 3.56 \[ \int \frac {1+x^2}{x \sqrt {1+x^4}} \, dx=-\frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} + 1\right ) + \frac {1}{4} \, \log \left (\sqrt {x^{4} + 1} - 1\right ) + \frac {1}{4} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} + 1\right ) - \frac {1}{4} \, \log \left (\frac {\sqrt {x^{4} + 1}}{x^{2}} - 1\right ) \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (14) = 28\).

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 3.19 \[ \int \frac {1+x^2}{x \sqrt {1+x^4}} \, dx=\frac {1}{2} \, \log \left (x^{2} - \sqrt {x^{4} + 1} + 1\right ) - \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 1} + 1\right ) - \frac {1}{2} \, \log \left (-x^{2} + \sqrt {x^{4} + 1}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {1+x^2}{x \sqrt {1+x^4}} \, dx=\frac {\mathrm {asinh}\left (x^2\right )}{2}-\frac {\mathrm {atanh}\left (\sqrt {x^4+1}\right )}{2} \]

[In]

int((x^2 + 1)/(x*(x^4 + 1)^(1/2)),x)

[Out]

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