\(\int \frac {-5+2 x}{2+3 x^2} \, dx\) [28]

Optimal result

Integrand size = 15, antiderivative size = 30 \[ \int \frac {-5+2 x}{2+3 x^2} \, dx=-\frac {5 \arctan \left (\sqrt {\frac {3}{2}} x\right )}{\sqrt {6}}+\frac {1}{3} \log \left (2+3 x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {649, 209, 266} \[ \int \frac {-5+2 x}{2+3 x^2} \, dx=\frac {1}{3} \log \left (3 x^2+2\right )-\frac {5 \arctan \left (\sqrt {\frac {3}{2}} x\right )}{\sqrt {6}} \]

[In]

Int[(-5 + 2*x)/(2 + 3*x^2),x]

[Out]

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

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]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {-5+2 x}{2+3 x^2} \, dx=-\frac {5 \arctan \left (\sqrt {\frac {3}{2}} x\right )}{\sqrt {6}}+\frac {1}{3} \log \left (2+3 x^2\right ) \]

[In]

Integrate[(-5 + 2*x)/(2 + 3*x^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {\ln \left (3 x^{2}+2\right )}{3}-\frac {5 \arctan \left (\frac {x \sqrt {6}}{2}\right ) \sqrt {6}}{6}\)	\(24\)
risch	\(\frac {\ln \left (9 x^{2}+6\right )}{3}-\frac {5 \arctan \left (\frac {x \sqrt {6}}{2}\right ) \sqrt {6}}{6}\)	\(24\)
meijerg	\(-\frac {5 \sqrt {6}\, \arctan \left (\frac {x \sqrt {2}\, \sqrt {3}}{2}\right )}{6}+\frac {\ln \left (1+\frac {3 x^{2}}{2}\right )}{3}\)	\(27\)

[In]

int((-5+2*x)/(3*x^2+2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {-5+2 x}{2+3 x^2} \, dx=-\frac {5}{6} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x\right ) + \frac {1}{3} \, \log \left (3 \, x^{2} + 2\right ) \]

[In]

integrate((-5+2*x)/(3*x^2+2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-5+2 x}{2+3 x^2} \, dx=\frac {\log {\left (x^{2} + \frac {2}{3} \right )}}{3} - \frac {5 \sqrt {6} \operatorname {atan}{\left (\frac {\sqrt {6} x}{2} \right )}}{6} \]

[In]

integrate((-5+2*x)/(3*x**2+2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {-5+2 x}{2+3 x^2} \, dx=-\frac {5}{6} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x\right ) + \frac {1}{3} \, \log \left (3 \, x^{2} + 2\right ) \]

[In]

integrate((-5+2*x)/(3*x^2+2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {-5+2 x}{2+3 x^2} \, dx=-\frac {5}{6} \, \sqrt {6} \arctan \left (\frac {1}{2} \, \sqrt {6} x\right ) + \frac {1}{3} \, \log \left (x^{2} + \frac {2}{3}\right ) \]

[In]

integrate((-5+2*x)/(3*x^2+2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {-5+2 x}{2+3 x^2} \, dx=\frac {\ln \left (x^2+\frac {2}{3}\right )}{3}-\frac {5\,\sqrt {6}\,\mathrm {atan}\left (\frac {\sqrt {6}\,x}{2}\right )}{6} \]

[In]

int((2*x - 5)/(3*x^2 + 2),x)

[Out]

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