\(\int \frac {1+x}{-6 x+x^2+x^3} \, dx\) [441]

Optimal result

Integrand size = 16, antiderivative size = 25 \[ \int \frac {1+x}{-6 x+x^2+x^3} \, dx=\frac {3}{10} \log (2-x)-\frac {\log (x)}{6}-\frac {2}{15} \log (3+x) \]

[Out]

3/10*ln(2-x)-1/6*ln(x)-2/15*ln(3+x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1608, 814} \[ \int \frac {1+x}{-6 x+x^2+x^3} \, dx=\frac {3}{10} \log (2-x)-\frac {\log (x)}{6}-\frac {2}{15} \log (x+3) \]

[In]

Int[(1 + x)/(-6*x + x^2 + x^3),x]

[Out]

(3*Log[2 - x])/10 - Log[x]/6 - (2*Log[3 + x])/15

Rule 814

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m*((f + g*x)/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1+x}{x \left (-6+x+x^2\right )} \, dx \\ & = \int \left (\frac {3}{10 (-2+x)}-\frac {1}{6 x}-\frac {2}{15 (3+x)}\right ) \, dx \\ & = \frac {3}{10} \log (2-x)-\frac {\log (x)}{6}-\frac {2}{15} \log (3+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {1+x}{-6 x+x^2+x^3} \, dx=\frac {3}{10} \log (2-x)-\frac {\log (x)}{6}-\frac {2}{15} \log (3+x) \]

[In]

Integrate[(1 + x)/(-6*x + x^2 + x^3),x]

[Out]

(3*Log[2 - x])/10 - Log[x]/6 - (2*Log[3 + x])/15

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72

method	result	size
default	\(-\frac {\ln \left (x \right )}{6}-\frac {2 \ln \left (3+x \right )}{15}+\frac {3 \ln \left (x -2\right )}{10}\)	\(18\)
norman	\(-\frac {\ln \left (x \right )}{6}-\frac {2 \ln \left (3+x \right )}{15}+\frac {3 \ln \left (x -2\right )}{10}\)	\(18\)
risch	\(-\frac {\ln \left (x \right )}{6}-\frac {2 \ln \left (3+x \right )}{15}+\frac {3 \ln \left (x -2\right )}{10}\)	\(18\)
parallelrisch	\(-\frac {\ln \left (x \right )}{6}-\frac {2 \ln \left (3+x \right )}{15}+\frac {3 \ln \left (x -2\right )}{10}\)	\(18\)

[In]

int((x+1)/(x^3+x^2-6*x),x,method=_RETURNVERBOSE)

[Out]

-1/6*ln(x)-2/15*ln(3+x)+3/10*ln(x-2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {1+x}{-6 x+x^2+x^3} \, dx=-\frac {2}{15} \, \log \left (x + 3\right ) + \frac {3}{10} \, \log \left (x - 2\right ) - \frac {1}{6} \, \log \left (x\right ) \]

[In]

integrate((1+x)/(x^3+x^2-6*x),x, algorithm="fricas")

[Out]

-2/15*log(x + 3) + 3/10*log(x - 2) - 1/6*log(x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {1+x}{-6 x+x^2+x^3} \, dx=- \frac {\log {\left (x \right )}}{6} + \frac {3 \log {\left (x - 2 \right )}}{10} - \frac {2 \log {\left (x + 3 \right )}}{15} \]

[In]

integrate((1+x)/(x**3+x**2-6*x),x)

[Out]

-log(x)/6 + 3*log(x - 2)/10 - 2*log(x + 3)/15

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {1+x}{-6 x+x^2+x^3} \, dx=-\frac {2}{15} \, \log \left (x + 3\right ) + \frac {3}{10} \, \log \left (x - 2\right ) - \frac {1}{6} \, \log \left (x\right ) \]

[In]

integrate((1+x)/(x^3+x^2-6*x),x, algorithm="maxima")

[Out]

-2/15*log(x + 3) + 3/10*log(x - 2) - 1/6*log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {1+x}{-6 x+x^2+x^3} \, dx=-\frac {2}{15} \, \log \left ({\left | x + 3 \right |}\right ) + \frac {3}{10} \, \log \left ({\left | x - 2 \right |}\right ) - \frac {1}{6} \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate((1+x)/(x^3+x^2-6*x),x, algorithm="giac")

[Out]

-2/15*log(abs(x + 3)) + 3/10*log(abs(x - 2)) - 1/6*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {1+x}{-6 x+x^2+x^3} \, dx=\frac {3\,\ln \left (x-2\right )}{10}-\frac {2\,\ln \left (x+3\right )}{15}-\frac {\ln \left (x\right )}{6} \]

[In]

int((x + 1)/(x^2 - 6*x + x^3),x)

[Out]

(3*log(x - 2))/10 - (2*log(x + 3))/15 - log(x)/6