\(\int \frac {x^2}{(-1+x) (1+2 x+x^2)} \, dx\) [356]

Optimal result

Integrand size = 19, antiderivative size = 28 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {1}{2 (1+x)}+\frac {1}{4} \log (1-x)+\frac {3}{4} \log (1+x) \]

[Out]

1/2/(1+x)+1/4*ln(1-x)+3/4*ln(1+x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {27, 90} \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {1}{2 (x+1)}+\frac {1}{4} \log (1-x)+\frac {3}{4} \log (x+1) \]

[In]

Int[x^2/((-1 + x)*(1 + 2*x + x^2)),x]

[Out]

1/(2*(1 + x)) + Log[1 - x]/4 + (3*Log[1 + x])/4

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2}{(-1+x) (1+x)^2} \, dx \\ & = \int \left (\frac {1}{4 (-1+x)}-\frac {1}{2 (1+x)^2}+\frac {3}{4 (1+x)}\right ) \, dx \\ & = \frac {1}{2 (1+x)}+\frac {1}{4} \log (1-x)+\frac {3}{4} \log (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {1}{4} \left (\frac {2}{1+x}+\log (-1+x)+3 \log (1+x)\right ) \]

[In]

Integrate[x^2/((-1 + x)*(1 + 2*x + x^2)),x]

[Out]

(2/(1 + x) + Log[-1 + x] + 3*Log[1 + x])/4

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {1}{2 x +2}+\frac {3 \ln \left (x +1\right )}{4}+\frac {\ln \left (x -1\right )}{4}\)	\(21\)
norman	\(\frac {1}{2 x +2}+\frac {3 \ln \left (x +1\right )}{4}+\frac {\ln \left (x -1\right )}{4}\)	\(21\)
risch	\(\frac {1}{2 x +2}+\frac {3 \ln \left (x +1\right )}{4}+\frac {\ln \left (x -1\right )}{4}\)	\(21\)
parallelrisch	\(\frac {\ln \left (x -1\right ) x +3 \ln \left (x +1\right ) x +2+\ln \left (x -1\right )+3 \ln \left (x +1\right )}{4+4 x}\)	\(33\)

[In]

int(x^2/(x-1)/(x^2+2*x+1),x,method=_RETURNVERBOSE)

[Out]

1/2/(x+1)+3/4*ln(x+1)+1/4*ln(x-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {3 \, {\left (x + 1\right )} \log \left (x + 1\right ) + {\left (x + 1\right )} \log \left (x - 1\right ) + 2}{4 \, {\left (x + 1\right )}} \]

[In]

integrate(x^2/(-1+x)/(x^2+2*x+1),x, algorithm="fricas")

[Out]

1/4*(3*(x + 1)*log(x + 1) + (x + 1)*log(x - 1) + 2)/(x + 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {\log {\left (x - 1 \right )}}{4} + \frac {3 \log {\left (x + 1 \right )}}{4} + \frac {1}{2 x + 2} \]

[In]

integrate(x**2/(-1+x)/(x**2+2*x+1),x)

[Out]

log(x - 1)/4 + 3*log(x + 1)/4 + 1/(2*x + 2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {1}{2 \, {\left (x + 1\right )}} + \frac {3}{4} \, \log \left (x + 1\right ) + \frac {1}{4} \, \log \left (x - 1\right ) \]

[In]

integrate(x^2/(-1+x)/(x^2+2*x+1),x, algorithm="maxima")

[Out]

1/2/(x + 1) + 3/4*log(x + 1) + 1/4*log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {1}{2 \, {\left (x + 1\right )}} + \frac {3}{4} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{4} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

integrate(x^2/(-1+x)/(x^2+2*x+1),x, algorithm="giac")

[Out]

1/2/(x + 1) + 3/4*log(abs(x + 1)) + 1/4*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {x^2}{(-1+x) \left (1+2 x+x^2\right )} \, dx=\frac {\ln \left (x-1\right )}{4}+\frac {3\,\ln \left (x+1\right )}{4}+\frac {1}{2\,\left (x+1\right )} \]

[In]

int(x^2/((x - 1)*(2*x + x^2 + 1)),x)

[Out]

log(x - 1)/4 + (3*log(x + 1))/4 + 1/(2*(x + 1))