3.4.27 \(\int \frac {2+x}{2+x+x^2} \, dx\) [327]

Optimal. Leaf size=31 \[ \frac {3 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {7}}\right )}{\sqrt {7}}+\frac {1}{2} \log \left (2+x+x^2\right ) \]

[Out]

1/2*ln(x^2+x+2)+3/7*arctan(1/7*(1+2*x)*7^(1/2))*7^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 31, 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 = {648, 632, 210, 642} \begin {gather*} \frac {1}{2} \log \left (x^2+x+2\right )+\frac {3 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {7}}\right )}{\sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + x)/(2 + x + x^2),x]

[Out]

(3*ArcTan[(1 + 2*x)/Sqrt[7]])/Sqrt[7] + Log[2 + x + x^2]/2

Rule 210

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

Rule 632

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 642

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 648

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 {2+x}{2+x+x^2} \, dx &=\frac {1}{2} \int \frac {1+2 x}{2+x+x^2} \, dx+\frac {3}{2} \int \frac {1}{2+x+x^2} \, dx\\ &=\frac {1}{2} \log \left (2+x+x^2\right )-3 \text {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,1+2 x\right )\\ &=\frac {3 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {7}}\right )}{\sqrt {7}}+\frac {1}{2} \log \left (2+x+x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 31, normalized size = 1.00 \begin {gather*} \frac {3 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {7}}\right )}{\sqrt {7}}+\frac {1}{2} \log \left (2+x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + x)/(2 + x + x^2),x]

[Out]

(3*ArcTan[(1 + 2*x)/Sqrt[7]])/Sqrt[7] + Log[2 + x + x^2]/2

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Mathics [A]
time = 1.85, size = 26, normalized size = 0.84 \begin {gather*} \frac {3 \sqrt {7} \text {ArcTan}\left [\frac {\sqrt {7} \left (1+2 x\right )}{7}\right ]}{7}+\frac {\text {Log}\left [2+x+x^2\right ]}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(2 + x)/(2 + x + x^2),x]')

[Out]

3 Sqrt[7] ArcTan[Sqrt[7] (1 + 2 x) / 7] / 7 + Log[2 + x + x ^ 2] / 2

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Maple [A]
time = 0.18, size = 27, normalized size = 0.87

method	result	size
default	\(\frac {\ln \left (x^{2}+x +2\right )}{2}+\frac {3 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {7}}{7}\right ) \sqrt {7}}{7}\)	\(27\)
risch	\(\frac {\ln \left (4 x^{2}+4 x +8\right )}{2}+\frac {3 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {7}}{7}\right ) \sqrt {7}}{7}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2+x)/(x^2+x+2),x,method=_RETURNVERBOSE)

[Out]

1/2*ln(x^2+x+2)+3/7*arctan(1/7*(1+2*x)*7^(1/2))*7^(1/2)

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Maxima [A]
time = 0.35, size = 26, normalized size = 0.84 \begin {gather*} \frac {3}{7} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x + 1\right )}\right ) + \frac {1}{2} \, \log \left (x^{2} + x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/7*sqrt(7)*arctan(1/7*sqrt(7)*(2*x + 1)) + 1/2*log(x^2 + x + 2)

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Fricas [A]
time = 0.33, size = 26, normalized size = 0.84 \begin {gather*} \frac {3}{7} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x + 1\right )}\right ) + \frac {1}{2} \, \log \left (x^{2} + x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/7*sqrt(7)*arctan(1/7*sqrt(7)*(2*x + 1)) + 1/2*log(x^2 + x + 2)

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Sympy [A]
time = 0.06, size = 36, normalized size = 1.16 \begin {gather*} \frac {\log {\left (x^{2} + x + 2 \right )}}{2} + \frac {3 \sqrt {7} \operatorname {atan}{\left (\frac {2 \sqrt {7} x}{7} + \frac {\sqrt {7}}{7} \right )}}{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x**2 + x + 2)/2 + 3*sqrt(7)*atan(2*sqrt(7)*x/7 + sqrt(7)/7)/7

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Giac [A]
time = 0.00, size = 32, normalized size = 1.03 \begin {gather*} \frac {\ln \left (x^{2}+x+2\right )}{2}+\frac {3 \arctan \left (\frac {2 x+1}{\sqrt {7}}\right )}{\sqrt {7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)/(x^2+x+2),x)

[Out]

3/7*sqrt(7)*arctan(1/7*sqrt(7)*(2*x + 1)) + 1/2*log(x^2 + x + 2)

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Mupad [B]
time = 0.04, size = 28, normalized size = 0.90 \begin {gather*} \frac {\ln \left (x^2+x+2\right )}{2}+\frac {3\,\sqrt {7}\,\mathrm {atan}\left (\frac {2\,\sqrt {7}\,x}{7}+\frac {\sqrt {7}}{7}\right )}{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 2)/(x + x^2 + 2),x)

[Out]

log(x + x^2 + 2)/2 + (3*7^(1/2)*atan((2*7^(1/2)*x)/7 + 7^(1/2)/7))/7

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