\(\int \frac {\sqrt [4]{1+x^2}}{x} \, dx\) [449]

Optimal result

Integrand size = 13, antiderivative size = 36 \[ \int \frac {\sqrt [4]{1+x^2}}{x} \, dx=2 \sqrt [4]{1+x^2}-\arctan \left (\sqrt [4]{1+x^2}\right )-\text {arctanh}\left (\sqrt [4]{1+x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {272, 52, 65, 218, 212, 209} \[ \int \frac {\sqrt [4]{1+x^2}}{x} \, dx=-\arctan \left (\sqrt [4]{x^2+1}\right )-\text {arctanh}\left (\sqrt [4]{x^2+1}\right )+2 \sqrt [4]{x^2+1} \]

[In]

Int[(1 + x^2)^(1/4)/x,x]

[Out]

2*(1 + x^2)^(1/4) - ArcTan[(1 + x^2)^(1/4)] - ArcTanh[(1 + x^2)^(1/4)]

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 209

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

Rule 212

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

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/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}{2} \text {Subst}\left (\int \frac {\sqrt [4]{1+x}}{x} \, dx,x,x^2\right ) \\ & = 2 \sqrt [4]{1+x^2}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x (1+x)^{3/4}} \, dx,x,x^2\right ) \\ & = 2 \sqrt [4]{1+x^2}+2 \text {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\sqrt [4]{1+x^2}\right ) \\ & = 2 \sqrt [4]{1+x^2}-\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{1+x^2}\right )-\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{1+x^2}\right ) \\ & = 2 \sqrt [4]{1+x^2}-\arctan \left (\sqrt [4]{1+x^2}\right )-\text {arctanh}\left (\sqrt [4]{1+x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{1+x^2}}{x} \, dx=2 \sqrt [4]{1+x^2}-\arctan \left (\sqrt [4]{1+x^2}\right )-\text {arctanh}\left (\sqrt [4]{1+x^2}\right ) \]

[In]

Integrate[(1 + x^2)^(1/4)/x,x]

[Out]

2*(1 + x^2)^(1/4) - ArcTan[(1 + x^2)^(1/4)] - ArcTanh[(1 + x^2)^(1/4)]

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 1.31 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.25

method	result	size
meijerg	\(-\frac {-\Gamma \left (\frac {3}{4}\right ) x^{2} \operatorname {hypergeom}\left (\left [\frac {3}{4}, 1, 1\right ], \left [2, 2\right ], -x^{2}\right )-4 \left (4-3 \ln \left (2\right )+\frac {\pi }{2}+2 \ln \left (x \right )\right ) \Gamma \left (\frac {3}{4}\right )}{8 \Gamma \left (\frac {3}{4}\right )}\)	\(45\)
pseudoelliptic	\(2 \left (x^{2}+1\right )^{\frac {1}{4}}+\frac {\ln \left (\left (x^{2}+1\right )^{\frac {1}{4}}-1\right )}{2}-\frac {\ln \left (\left (x^{2}+1\right )^{\frac {1}{4}}+1\right )}{2}-\arctan \left (\left (x^{2}+1\right )^{\frac {1}{4}}\right )\)	\(45\)
trager	\(2 \left (x^{2}+1\right )^{\frac {1}{4}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \left (x^{2}+1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+2 \left (x^{2}+1\right )^{\frac {1}{4}}}{x^{2}}\right )}{2}-\frac {\ln \left (-\frac {2 \left (x^{2}+1\right )^{\frac {3}{4}}+2 \sqrt {x^{2}+1}+x^{2}+2 \left (x^{2}+1\right )^{\frac {1}{4}}+2}{x^{2}}\right )}{2}\)	\(117\)

[In]

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

[Out]

-1/8/GAMMA(3/4)*(-GAMMA(3/4)*x^2*hypergeom([3/4,1,1],[2,2],-x^2)-4*(4-3*ln(2)+1/2*Pi+2*ln(x))*GAMMA(3/4))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt [4]{1+x^2}}{x} \, dx=2 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}} - \arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.57 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt [4]{1+x^2}}{x} \, dx=- \frac {\sqrt {x} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{2}}} \right )}}{2 \Gamma \left (\frac {3}{4}\right )} \]

[In]

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

[Out]

-sqrt(x)*gamma(-1/4)*hyper((-1/4, -1/4), (3/4,), exp_polar(I*pi)/x**2)/(2*gamma(3/4))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt [4]{1+x^2}}{x} \, dx=2 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}} - \arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt [4]{1+x^2}}{x} \, dx=2 \, {\left (x^{2} + 1\right )}^{\frac {1}{4}} - \arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt [4]{1+x^2}}{x} \, dx=2\,{\left (x^2+1\right )}^{1/4}-\mathrm {atanh}\left ({\left (x^2+1\right )}^{1/4}\right )-\mathrm {atan}\left ({\left (x^2+1\right )}^{1/4}\right ) \]

[In]

int((x^2 + 1)^(1/4)/x,x)

[Out]

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