\(\int \frac {x^2}{2-3 x+x^2} \, dx\) [2260]

Optimal result

Integrand size = 14, antiderivative size = 18 \[ \int \frac {x^2}{2-3 x+x^2} \, dx=x-\log (1-x)+4 \log (2-x) \]

[Out]

x-ln(1-x)+4*ln(2-x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {717, 646, 31} \[ \int \frac {x^2}{2-3 x+x^2} \, dx=x-\log (1-x)+4 \log (2-x) \]

[In]

Int[x^2/(2 - 3*x + x^2),x]

[Out]

x - Log[1 - x] + 4*Log[2 - x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 646

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), 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*} \text {integral}& = x+\int \frac {-2+3 x}{2-3 x+x^2} \, dx \\ & = x+4 \int \frac {1}{-2+x} \, dx-\int \frac {1}{-1+x} \, dx \\ & = x-\log (1-x)+4 \log (2-x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{2-3 x+x^2} \, dx=x-\log (1-x)+4 \log (2-x) \]

[In]

Integrate[x^2/(2 - 3*x + x^2),x]

[Out]

x - Log[1 - x] + 4*Log[2 - x]

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
default	\(x -\ln \left (-1+x \right )+4 \ln \left (-2+x \right )\)	\(15\)
norman	\(x -\ln \left (-1+x \right )+4 \ln \left (-2+x \right )\)	\(15\)
risch	\(x -\ln \left (-1+x \right )+4 \ln \left (-2+x \right )\)	\(15\)
parallelrisch	\(x -\ln \left (-1+x \right )+4 \ln \left (-2+x \right )\)	\(15\)

[In]

int(x^2/(x^2-3*x+2),x,method=_RETURNVERBOSE)

[Out]

x-ln(-1+x)+4*ln(-2+x)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {x^2}{2-3 x+x^2} \, dx=x - \log \left (x - 1\right ) + 4 \, \log \left (x - 2\right ) \]

[In]

integrate(x^2/(x^2-3*x+2),x, algorithm="fricas")

[Out]

x - log(x - 1) + 4*log(x - 2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {x^2}{2-3 x+x^2} \, dx=x + 4 \log {\left (x - 2 \right )} - \log {\left (x - 1 \right )} \]

[In]

integrate(x**2/(x**2-3*x+2),x)

[Out]

x + 4*log(x - 2) - log(x - 1)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {x^2}{2-3 x+x^2} \, dx=x - \log \left (x - 1\right ) + 4 \, \log \left (x - 2\right ) \]

[In]

integrate(x^2/(x^2-3*x+2),x, algorithm="maxima")

[Out]

x - log(x - 1) + 4*log(x - 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {x^2}{2-3 x+x^2} \, dx=x - \log \left ({\left | x - 1 \right |}\right ) + 4 \, \log \left ({\left | x - 2 \right |}\right ) \]

[In]

integrate(x^2/(x^2-3*x+2),x, algorithm="giac")

[Out]

x - log(abs(x - 1)) + 4*log(abs(x - 2))

Mupad [B] (verification not implemented)

Time = 9.76 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {x^2}{2-3 x+x^2} \, dx=x-\ln \left (x-1\right )+4\,\ln \left (x-2\right ) \]

[In]

int(x^2/(x^2 - 3*x + 2),x)

[Out]

x - log(x - 1) + 4*log(x - 2)