\(\int \frac {1-x+3 x^2}{-x^2+x^3} \, dx\) [352]

Optimal result

Integrand size = 22, antiderivative size = 12 \[ \int \frac {1-x+3 x^2}{-x^2+x^3} \, dx=\frac {1}{x}+3 \log (1-x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1607, 907} \[ \int \frac {1-x+3 x^2}{-x^2+x^3} \, dx=\frac {1}{x}+3 \log (1-x) \]

[In]

Int[(1 - x + 3*x^2)/(-x^2 + x^3),x]

[Out]

x^(-1) + 3*Log[1 - x]

Rule 907

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1607

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {1-x+3 x^2}{-x^2+x^3} \, dx=\frac {1}{x}+3 \log (1-x) \]

[In]

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

[Out]

x^(-1) + 3*Log[1 - x]

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92

method	result	size
default	\(\frac {1}{x}+3 \ln \left (x -1\right )\)	\(11\)
norman	\(\frac {1}{x}+3 \ln \left (x -1\right )\)	\(11\)
risch	\(\frac {1}{x}+3 \ln \left (x -1\right )\)	\(11\)
meijerg	\(\frac {1}{x}+3 \ln \left (1-x \right )\)	\(13\)
parallelrisch	\(\frac {3 \ln \left (x -1\right ) x +1}{x}\)	\(14\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {1-x+3 x^2}{-x^2+x^3} \, dx=\frac {3 \, x \log \left (x - 1\right ) + 1}{x} \]

[In]

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

[Out]

(3*x*log(x - 1) + 1)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {1-x+3 x^2}{-x^2+x^3} \, dx=3 \log {\left (x - 1 \right )} + \frac {1}{x} \]

[In]

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

[Out]

3*log(x - 1) + 1/x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {1-x+3 x^2}{-x^2+x^3} \, dx=\frac {1}{x} + 3 \, \log \left (x - 1\right ) \]

[In]

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

[Out]

1/x + 3*log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {1-x+3 x^2}{-x^2+x^3} \, dx=\frac {1}{x} + 3 \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

1/x + 3*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {1-x+3 x^2}{-x^2+x^3} \, dx=3\,\ln \left (x-1\right )+\frac {1}{x} \]

[In]

int(-(3*x^2 - x + 1)/(x^2 - x^3),x)

[Out]

3*log(x - 1) + 1/x