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

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

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Rubi [A]  time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {1}{2} \log \left (x^2+2 x+5\right )-\frac {1}{2} \tan ^{-1}\left (\frac {x+1}{2}\right ) \end {gather*}

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 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 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 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]

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]

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*}

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Mathematica [A]  time = 0.00, size = 26, normalized size = 1.00 \begin {gather*} \frac {1}{2} \log \left (x^2+2 x+5\right )-\frac {1}{2} \tan ^{-1}\left (\frac {x+1}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(5 + 2*x + x^2),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{5+2 x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x/(5 + 2*x + x^2),x]

[Out]

IntegrateAlgebraic[x/(5 + 2*x + x^2), x]

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fricas [A]  time = 0.42, size = 20, normalized size = 0.77 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 0.16, size = 20, normalized size = 0.77 \begin {gather*} -\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 {gather*}

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)

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maple [A]  time = 0.07, size = 21, normalized size = 0.81 \begin {gather*} -\frac {\arctan \left (\frac {x}{2}+\frac {1}{2}\right )}{2}+\frac {\ln \left (x^{2}+2 x +5\right )}{2} \end {gather*}

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

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maxima [A]  time = 1.81, size = 20, normalized size = 0.77 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 0.04, size = 20, normalized size = 0.77 \begin {gather*} \frac {\ln \left (x^2+2\,x+5\right )}{2}-\frac {\mathrm {atan}\left (\frac {x}{2}+\frac {1}{2}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(2*x + x^2 + 5),x)

[Out]

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

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sympy [A]  time = 0.10, size = 20, normalized size = 0.77 \begin {gather*} \frac {\log {\left (x^{2} + 2 x + 5 \right )}}{2} - \frac {\operatorname {atan}{\left (\frac {x}{2} + \frac {1}{2} \right )}}{2} \end {gather*}

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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