3.900 \(\int \frac{5+2 x}{5+4 x+x^2} \, dx\)

Optimal. Leaf size=14 \[ \log \left (x^2+4 x+5\right )+\tan ^{-1}(x+2) \]

[Out]

ArcTan[2 + x] + Log[5 + 4*x + x^2]

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Rubi [A]  time = 0.0108742, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {634, 618, 204, 628} \[ \log \left (x^2+4 x+5\right )+\tan ^{-1}(x+2) \]

Antiderivative was successfully verified.

[In]

Int[(5 + 2*x)/(5 + 4*x + x^2),x]

[Out]

ArcTan[2 + x] + Log[5 + 4*x + x^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{5+2 x}{5+4 x+x^2} \, dx &=\int \frac{1}{5+4 x+x^2} \, dx+\int \frac{4+2 x}{5+4 x+x^2} \, dx\\ &=\log \left (5+4 x+x^2\right )-2 \operatorname{Subst}\left (\int \frac{1}{-4-x^2} \, dx,x,4+2 x\right )\\ &=\tan ^{-1}(2+x)+\log \left (5+4 x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0044674, size = 14, normalized size = 1. \[ \log \left (x^2+4 x+5\right )+\tan ^{-1}(x+2) \]

Antiderivative was successfully verified.

[In]

Integrate[(5 + 2*x)/(5 + 4*x + x^2),x]

[Out]

ArcTan[2 + x] + Log[5 + 4*x + x^2]

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Maple [A]  time = 0.001, size = 15, normalized size = 1.1 \begin{align*} \arctan \left ( 2+x \right ) +\ln \left ({x}^{2}+4\,x+5 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5+2*x)/(x^2+4*x+5),x)

[Out]

arctan(2+x)+ln(x^2+4*x+5)

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Maxima [A]  time = 1.65727, size = 19, normalized size = 1.36 \begin{align*} \arctan \left (x + 2\right ) + \log \left (x^{2} + 4 \, x + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5+2*x)/(x^2+4*x+5),x, algorithm="maxima")

[Out]

arctan(x + 2) + log(x^2 + 4*x + 5)

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Fricas [A]  time = 1.62262, size = 49, normalized size = 3.5 \begin{align*} \arctan \left (x + 2\right ) + \log \left (x^{2} + 4 \, x + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5+2*x)/(x^2+4*x+5),x, algorithm="fricas")

[Out]

arctan(x + 2) + log(x^2 + 4*x + 5)

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Sympy [A]  time = 0.139812, size = 14, normalized size = 1. \begin{align*} \log{\left (x^{2} + 4 x + 5 \right )} + \operatorname{atan}{\left (x + 2 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5+2*x)/(x**2+4*x+5),x)

[Out]

log(x**2 + 4*x + 5) + atan(x + 2)

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Giac [A]  time = 1.24733, size = 19, normalized size = 1.36 \begin{align*} \arctan \left (x + 2\right ) + \log \left (x^{2} + 4 \, x + 5\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5+2*x)/(x^2+4*x+5),x, algorithm="giac")

[Out]

arctan(x + 2) + log(x^2 + 4*x + 5)