\(\int \frac {1}{x (1+x^2)^{3/4}} \, dx\) [252]

Optimal result

Integrand size = 13, antiderivative size = 25 \[ \int \frac {1}{x \left (1+x^2\right )^{3/4}} \, dx=-\arctan \left (\sqrt [4]{1+x^2}\right )-\text {arctanh}\left (\sqrt [4]{1+x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {272, 65, 218, 212, 209} \[ \int \frac {1}{x \left (1+x^2\right )^{3/4}} \, dx=-\arctan \left (\sqrt [4]{x^2+1}\right )-\text {arctanh}\left (\sqrt [4]{x^2+1}\right ) \]

[In]

Int[1/(x*(1 + x^2)^(3/4)),x]

[Out]

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

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 {1}{x (1+x)^{3/4}} \, dx,x,x^2\right ) \\ & = 2 \text {Subst}\left (\int \frac {1}{-1+x^4} \, 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 )-\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{1+x^2}\right ) \\ & = -\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 = 25, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+x^2\right )^{3/4}} \, dx=-\arctan \left (\sqrt [4]{1+x^2}\right )-\text {arctanh}\left (\sqrt [4]{1+x^2}\right ) \]

[In]

Integrate[1/(x*(1 + x^2)^(3/4)),x]

[Out]

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

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(-\arctan \left (\left (x^{2}+1\right )^{\frac {1}{4}}\right )-\operatorname {arctanh}\left (\left (x^{2}+1\right )^{\frac {1}{4}}\right )\)	\(22\)
meijerg	\(\frac {-\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{2} \operatorname {hypergeom}\left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], -x^{2}\right )}{4}+\left (-3 \ln \left (2\right )+\frac {\pi }{2}+2 \ln \left (x \right )\right ) \Gamma \left (\frac {3}{4}\right )}{2 \Gamma \left (\frac {3}{4}\right )}\)	\(43\)
trager	\(-\frac {\ln \left (-\frac {2 \left (x^{2}+1\right )^{\frac {3}{4}}+2 \sqrt {x^{2}+1}+x^{2}+2 \left (x^{2}+1\right )^{\frac {1}{4}}+2}{x^{2}}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \left (x^{2}+1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+2 \left (x^{2}+1\right )^{\frac {1}{4}}}{x^{2}}\right )}{2}\)	\(108\)

[In]

int(1/x/(x^2+1)^(3/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {1}{x \left (1+x^2\right )^{3/4}} \, dx=-\arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

integrate(1/x/(x^2+1)^(3/4),x, algorithm="fricas")

[Out]

-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 = 0.45 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {1}{x \left (1+x^2\right )^{3/4}} \, dx=- \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{2}}} \right )}}{2 x^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \]

[In]

integrate(1/x/(x**2+1)**(3/4),x)

[Out]

-gamma(3/4)*hyper((3/4, 3/4), (7/4,), exp_polar(I*pi)/x**2)/(2*x**(3/2)*gamma(7/4))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {1}{x \left (1+x^2\right )^{3/4}} \, dx=-\arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

integrate(1/x/(x^2+1)^(3/4),x, algorithm="maxima")

[Out]

-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 = 35, normalized size of antiderivative = 1.40 \[ \int \frac {1}{x \left (1+x^2\right )^{3/4}} \, dx=-\arctan \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

integrate(1/x/(x^2+1)^(3/4),x, algorithm="giac")

[Out]

-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 = 5.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x \left (1+x^2\right )^{3/4}} \, dx=-\mathrm {atan}\left ({\left (x^2+1\right )}^{1/4}\right )-\mathrm {atanh}\left ({\left (x^2+1\right )}^{1/4}\right ) \]

[In]

int(1/(x*(x^2 + 1)^(3/4)),x)

[Out]

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