\(\int \frac {2 x}{(-1+x) (5+x^2)} \, dx\) [447]

Optimal result

Integrand size = 15, antiderivative size = 38 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {1}{3} \sqrt {5} \arctan \left (\frac {x}{\sqrt {5}}\right )+\frac {1}{3} \log (1-x)-\frac {1}{6} \log \left (5+x^2\right ) \]

[Out]

1/3*ln(1-x)-1/6*ln(x^2+5)+1/3*arctan(1/5*x*5^(1/2))*5^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {12, 815, 649, 209, 266} \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {1}{3} \sqrt {5} \arctan \left (\frac {x}{\sqrt {5}}\right )-\frac {1}{6} \log \left (x^2+5\right )+\frac {1}{3} \log (1-x) \]

[In]

Int[(2*x)/((-1 + x)*(5 + x^2)),x]

[Out]

(Sqrt[5]*ArcTan[x/Sqrt[5]])/3 + Log[1 - x]/3 - Log[5 + x^2]/6

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {x}{(-1+x) \left (5+x^2\right )} \, dx \\ & = 2 \int \left (\frac {1}{6 (-1+x)}+\frac {5-x}{6 \left (5+x^2\right )}\right ) \, dx \\ & = \frac {1}{3} \log (1-x)+\frac {1}{3} \int \frac {5-x}{5+x^2} \, dx \\ & = \frac {1}{3} \log (1-x)-\frac {1}{3} \int \frac {x}{5+x^2} \, dx+\frac {5}{3} \int \frac {1}{5+x^2} \, dx \\ & = \frac {1}{3} \sqrt {5} \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )+\frac {1}{3} \log (1-x)-\frac {1}{6} \log \left (5+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=2 \left (\frac {1}{6} \sqrt {5} \arctan \left (\frac {x}{\sqrt {5}}\right )+\frac {1}{6} \log (1-x)-\frac {1}{12} \log \left (5+x^2\right )\right ) \]

[In]

Integrate[(2*x)/((-1 + x)*(5 + x^2)),x]

[Out]

2*((Sqrt[5]*ArcTan[x/Sqrt[5]])/6 + Log[1 - x]/6 - Log[5 + x^2]/12)

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74

method	result	size
default	\(-\frac {\ln \left (x^{2}+5\right )}{6}+\frac {\arctan \left (\frac {x \sqrt {5}}{5}\right ) \sqrt {5}}{3}+\frac {\ln \left (x -1\right )}{3}\)	\(28\)
risch	\(-\frac {\ln \left (x^{2}+5\right )}{6}+\frac {\arctan \left (\frac {x \sqrt {5}}{5}\right ) \sqrt {5}}{3}+\frac {\ln \left (x -1\right )}{3}\)	\(28\)

[In]

int(2*x/(x-1)/(x^2+5),x,method=_RETURNVERBOSE)

[Out]

-1/6*ln(x^2+5)+1/3*arctan(1/5*x*5^(1/2))*5^(1/2)+1/3*ln(x-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {1}{3} \, \sqrt {5} \arctan \left (\frac {1}{5} \, \sqrt {5} x\right ) - \frac {1}{6} \, \log \left (x^{2} + 5\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]

[In]

integrate(2*x/(-1+x)/(x^2+5),x, algorithm="fricas")

[Out]

1/3*sqrt(5)*arctan(1/5*sqrt(5)*x) - 1/6*log(x^2 + 5) + 1/3*log(x - 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {\log {\left (x - 1 \right )}}{3} - \frac {\log {\left (x^{2} + 5 \right )}}{6} + \frac {\sqrt {5} \operatorname {atan}{\left (\frac {\sqrt {5} x}{5} \right )}}{3} \]

[In]

integrate(2*x/(-1+x)/(x**2+5),x)

[Out]

log(x - 1)/3 - log(x**2 + 5)/6 + sqrt(5)*atan(sqrt(5)*x/5)/3

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {1}{3} \, \sqrt {5} \arctan \left (\frac {1}{5} \, \sqrt {5} x\right ) - \frac {1}{6} \, \log \left (x^{2} + 5\right ) + \frac {1}{3} \, \log \left (x - 1\right ) \]

[In]

integrate(2*x/(-1+x)/(x^2+5),x, algorithm="maxima")

[Out]

1/3*sqrt(5)*arctan(1/5*sqrt(5)*x) - 1/6*log(x^2 + 5) + 1/3*log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {1}{3} \, \sqrt {5} \arctan \left (\frac {1}{5} \, \sqrt {5} x\right ) - \frac {1}{6} \, \log \left (x^{2} + 5\right ) + \frac {1}{3} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

integrate(2*x/(-1+x)/(x^2+5),x, algorithm="giac")

[Out]

1/3*sqrt(5)*arctan(1/5*sqrt(5)*x) - 1/6*log(x^2 + 5) + 1/3*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int \frac {2 x}{(-1+x) \left (5+x^2\right )} \, dx=\frac {\ln \left (x-1\right )}{3}-\ln \left (x-\sqrt {5}\,1{}\mathrm {i}\right )\,\left (\frac {1}{6}+\frac {\sqrt {5}\,1{}\mathrm {i}}{6}\right )+\ln \left (x+\sqrt {5}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {5}\,1{}\mathrm {i}}{6}\right ) \]

[In]

int((2*x)/((x^2 + 5)*(x - 1)),x)

[Out]

log(x - 1)/3 - log(x - 5^(1/2)*1i)*((5^(1/2)*1i)/6 + 1/6) + log(x + 5^(1/2)*1i)*((5^(1/2)*1i)/6 - 1/6)