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

Optimal. Leaf size=25 \[ -\log \left (x^2+2 x+5\right )+x-\frac{3}{2} \tan ^{-1}\left (\frac{x+1}{2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0159191, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {703, 634, 618, 204, 628} \[ -\log \left (x^2+2 x+5\right )+x-\frac{3}{2} \tan ^{-1}\left (\frac{x+1}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2/(5 + 2*x + x^2),x]

[Out]

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

Rule 703

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

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^2}{5+2 x+x^2} \, dx &=x+\int \frac{-5-2 x}{5+2 x+x^2} \, dx\\ &=x-3 \int \frac{1}{5+2 x+x^2} \, dx-\int \frac{2+2 x}{5+2 x+x^2} \, dx\\ &=x-\log \left (5+2 x+x^2\right )+6 \operatorname{Subst}\left (\int \frac{1}{-16-x^2} \, dx,x,2+2 x\right )\\ &=x-\frac{3}{2} \tan ^{-1}\left (\frac{1+x}{2}\right )-\log \left (5+2 x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0036703, size = 25, normalized size = 1. \[ -\log \left (x^2+2 x+5\right )+x-\frac{3}{2} \tan ^{-1}\left (\frac{x+1}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(5 + 2*x + x^2),x]

[Out]

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

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Maple [A]  time = 0.003, size = 22, normalized size = 0.9 \begin{align*} x-{\frac{3}{2}\arctan \left ({\frac{1}{2}}+{\frac{x}{2}} \right ) }-\ln \left ({x}^{2}+2\,x+5 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.40652, size = 28, normalized size = 1.12 \begin{align*} x - \frac{3}{2} \, \arctan \left (\frac{1}{2} \, x + \frac{1}{2}\right ) - \log \left (x^{2} + 2 \, x + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.75202, size = 68, normalized size = 2.72 \begin{align*} x - \frac{3}{2} \, \arctan \left (\frac{1}{2} \, x + \frac{1}{2}\right ) - \log \left (x^{2} + 2 \, x + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.096622, size = 22, normalized size = 0.88 \begin{align*} x - \log{\left (x^{2} + 2 x + 5 \right )} - \frac{3 \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**2/(x**2+2*x+5),x)

[Out]

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

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Giac [A]  time = 1.04478, size = 28, normalized size = 1.12 \begin{align*} x - \frac{3}{2} \, \arctan \left (\frac{1}{2} \, x + \frac{1}{2}\right ) - \log \left (x^{2} + 2 \, x + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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