\(\int \frac {-1+5 x+2 x^2}{-2 x+x^2+x^3} \, dx\) [109]

Optimal result

Integrand size = 23, antiderivative size = 23 \[ \int \frac {-1+5 x+2 x^2}{-2 x+x^2+x^3} \, dx=2 \log (1-x)+\frac {\log (x)}{2}-\frac {1}{2} \log (2+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1608, 1642} \[ \int \frac {-1+5 x+2 x^2}{-2 x+x^2+x^3} \, dx=2 \log (1-x)+\frac {\log (x)}{2}-\frac {1}{2} \log (x+2) \]

[In]

Int[(-1 + 5*x + 2*x^2)/(-2*x + x^2 + x^3),x]

[Out]

2*Log[1 - x] + Log[x]/2 - Log[2 + x]/2

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-1+5 x+2 x^2}{x \left (-2+x+x^2\right )} \, dx \\ & = \int \left (\frac {2}{-1+x}+\frac {1}{2 x}-\frac {1}{2 (2+x)}\right ) \, dx \\ & = 2 \log (1-x)+\frac {\log (x)}{2}-\frac {1}{2} \log (2+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-1+5 x+2 x^2}{-2 x+x^2+x^3} \, dx=2 \log (1-x)+\frac {\log (x)}{2}-\frac {1}{2} \log (2+x) \]

[In]

Integrate[(-1 + 5*x + 2*x^2)/(-2*x + x^2 + x^3),x]

[Out]

2*Log[1 - x] + Log[x]/2 - Log[2 + x]/2

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78

method	result	size
default	\(2 \ln \left (-1+x \right )-\frac {\ln \left (2+x \right )}{2}+\frac {\ln \left (x \right )}{2}\)	\(18\)
norman	\(2 \ln \left (-1+x \right )-\frac {\ln \left (2+x \right )}{2}+\frac {\ln \left (x \right )}{2}\)	\(18\)
risch	\(2 \ln \left (-1+x \right )-\frac {\ln \left (2+x \right )}{2}+\frac {\ln \left (x \right )}{2}\)	\(18\)
parallelrisch	\(2 \ln \left (-1+x \right )-\frac {\ln \left (2+x \right )}{2}+\frac {\ln \left (x \right )}{2}\)	\(18\)

[In]

int((2*x^2+5*x-1)/(x^3+x^2-2*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-1+5 x+2 x^2}{-2 x+x^2+x^3} \, dx=-\frac {1}{2} \, \log \left (x + 2\right ) + 2 \, \log \left (x - 1\right ) + \frac {1}{2} \, \log \left (x\right ) \]

[In]

integrate((2*x^2+5*x-1)/(x^3+x^2-2*x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-1+5 x+2 x^2}{-2 x+x^2+x^3} \, dx=\frac {\log {\left (x \right )}}{2} + 2 \log {\left (x - 1 \right )} - \frac {\log {\left (x + 2 \right )}}{2} \]

[In]

integrate((2*x**2+5*x-1)/(x**3+x**2-2*x),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-1+5 x+2 x^2}{-2 x+x^2+x^3} \, dx=-\frac {1}{2} \, \log \left (x + 2\right ) + 2 \, \log \left (x - 1\right ) + \frac {1}{2} \, \log \left (x\right ) \]

[In]

integrate((2*x^2+5*x-1)/(x^3+x^2-2*x),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-1+5 x+2 x^2}{-2 x+x^2+x^3} \, dx=-\frac {1}{2} \, \log \left ({\left | x + 2 \right |}\right ) + 2 \, \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate((2*x^2+5*x-1)/(x^3+x^2-2*x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-1+5 x+2 x^2}{-2 x+x^2+x^3} \, dx=2\,\ln \left (x-1\right )+\mathrm {atanh}\left (\frac {135}{11\,\left (11\,x-5\right )}+\frac {16}{11}\right ) \]

[In]

int((5*x + 2*x^2 - 1)/(x^2 - 2*x + x^3),x)

[Out]

2*log(x - 1) + atanh(135/(11*(11*x - 5)) + 16/11)