3.2255 \(\int \frac{x}{5+2 x+x^2} \, dx\)

Optimal. Leaf size=26 \[ \frac{1}{2} \log \left (x^2+2 x+5\right )-\frac{1}{2} \tan ^{-1}\left (\frac{x+1}{2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0134394, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {634, 618, 204, 628} \[ \frac{1}{2} \log \left (x^2+2 x+5\right )-\frac{1}{2} \tan ^{-1}\left (\frac{x+1}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x/(5 + 2*x + x^2),x]

[Out]

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

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{x}{5+2 x+x^2} \, dx &=\frac{1}{2} \int \frac{2+2 x}{5+2 x+x^2} \, dx-\int \frac{1}{5+2 x+x^2} \, dx\\ &=\frac{1}{2} \log \left (5+2 x+x^2\right )+2 \operatorname{Subst}\left (\int \frac{1}{-16-x^2} \, dx,x,2+2 x\right )\\ &=-\frac{1}{2} \tan ^{-1}\left (\frac{1+x}{2}\right )+\frac{1}{2} \log \left (5+2 x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0035023, size = 26, normalized size = 1. \[ \frac{1}{2} \log \left (x^2+2 x+5\right )-\frac{1}{2} \tan ^{-1}\left (\frac{x+1}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x/(5 + 2*x + x^2),x]

[Out]

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

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Maple [A]  time = 0.04, size = 21, normalized size = 0.8 \begin{align*} -{\frac{1}{2}\arctan \left ({\frac{1}{2}}+{\frac{x}{2}} \right ) }+{\frac{\ln \left ({x}^{2}+2\,x+5 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^2+2*x+5),x)

[Out]

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

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Maxima [A]  time = 1.50598, size = 27, normalized size = 1.04 \begin{align*} -\frac{1}{2} \, \arctan \left (\frac{1}{2} \, x + \frac{1}{2}\right ) + \frac{1}{2} \, \log \left (x^{2} + 2 \, x + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.44496, size = 69, normalized size = 2.65 \begin{align*} -\frac{1}{2} \, \arctan \left (\frac{1}{2} \, x + \frac{1}{2}\right ) + \frac{1}{2} \, \log \left (x^{2} + 2 \, x + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.103024, size = 20, normalized size = 0.77 \begin{align*} \frac{\log{\left (x^{2} + 2 x + 5 \right )}}{2} - \frac{\operatorname{atan}{\left (\frac{x}{2} + \frac{1}{2} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.09128, size = 27, normalized size = 1.04 \begin{align*} -\frac{1}{2} \, \arctan \left (\frac{1}{2} \, x + \frac{1}{2}\right ) + \frac{1}{2} \, \log \left (x^{2} + 2 \, x + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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