3.1.94 \(\int \frac {x^2}{5+2 x+x^2} \, dx\) [94]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, 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 = {717, 648, 632, 210, 642} \begin {gather*} -\frac {3}{2} \text {ArcTan}\left (\frac {x+1}{2}\right )-\log \left (x^2+2 x+5\right )+x \end {gather*}

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 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]

Rule 717

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]

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 \text {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*}

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Mathematica [A]
time = 0.00, size = 25, normalized size = 1.00 \begin {gather*} x-\frac {3}{2} \tan ^{-1}\left (\frac {1+x}{2}\right )-\log \left (5+2 x+x^2\right ) \end {gather*}

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 = 0.08, size = 22, normalized size = 0.88

method	result	size
default	\(x -\frac {3 \arctan \left (\frac {1}{2}+\frac {x}{2}\right )}{2}-\ln \left (x^{2}+2 x +5\right )\)	\(22\)
risch	\(x -\frac {3 \arctan \left (\frac {1}{2}+\frac {x}{2}\right )}{2}-\ln \left (x^{2}+2 x +5\right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x^2+2*x+5),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 2.37, size = 21, normalized size = 0.84 \begin {gather*} x - \frac {3}{2} \, \arctan \left (\frac {1}{2} \, x + \frac {1}{2}\right ) - \log \left (x^{2} + 2 \, x + 5\right ) \end {gather*}

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 = 0.40, size = 21, normalized size = 0.84 \begin {gather*} x - \frac {3}{2} \, \arctan \left (\frac {1}{2} \, x + \frac {1}{2}\right ) - \log \left (x^{2} + 2 \, x + 5\right ) \end {gather*}

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 = 0.03, size = 22, normalized size = 0.88 \begin {gather*} x - \log {\left (x^{2} + 2 x + 5 \right )} - \frac {3 \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**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 = 0.82, size = 21, normalized size = 0.84 \begin {gather*} x - \frac {3}{2} \, \arctan \left (\frac {1}{2} \, x + \frac {1}{2}\right ) - \log \left (x^{2} + 2 \, x + 5\right ) \end {gather*}

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)

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Mupad [B]
time = 0.04, size = 21, normalized size = 0.84 \begin {gather*} x-\ln \left (x^2+2\,x+5\right )-\frac {3\,\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/(2*x + x^2 + 5),x)

[Out]

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

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