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

Optimal result

Integrand size = 17, antiderivative size = 23 \[ \int \frac {x}{(1+x) (2+x) (3+x)} \, dx=-\frac {1}{2} \log (1+x)+2 \log (2+x)-\frac {3}{2} \log (3+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {153} \[ \int \frac {x}{(1+x) (2+x) (3+x)} \, dx=-\frac {1}{2} \log (x+1)+2 \log (x+2)-\frac {3}{2} \log (x+3) \]

[In]

Int[x/((1 + x)*(2 + x)*(3 + x)),x]

[Out]

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

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x), x], x] /; FreeQ[{a, b, c, d, e, f, g
, h, m}, x] && (IntegersQ[m, n, p] || (IGtQ[n, 0] && IGtQ[p, 0]))

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {x}{(1+x) (2+x) (3+x)} \, dx=-\frac {1}{2} \log (1+x)+2 \log (2+x)-\frac {3}{2} \log (3+x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {\ln \left (1+x \right )}{2}+2 \ln \left (2+x \right )-\frac {3 \ln \left (3+x \right )}{2}\)	\(20\)
norman	\(-\frac {\ln \left (1+x \right )}{2}+2 \ln \left (2+x \right )-\frac {3 \ln \left (3+x \right )}{2}\)	\(20\)
risch	\(-\frac {\ln \left (1+x \right )}{2}+2 \ln \left (2+x \right )-\frac {3 \ln \left (3+x \right )}{2}\)	\(20\)
parallelrisch	\(-\frac {\ln \left (1+x \right )}{2}+2 \ln \left (2+x \right )-\frac {3 \ln \left (3+x \right )}{2}\)	\(20\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x}{(1+x) (2+x) (3+x)} \, dx=-\frac {3}{2} \, \log \left (x + 3\right ) + 2 \, \log \left (x + 2\right ) - \frac {1}{2} \, \log \left (x + 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {x}{(1+x) (2+x) (3+x)} \, dx=- \frac {\log {\left (x + 1 \right )}}{2} + 2 \log {\left (x + 2 \right )} - \frac {3 \log {\left (x + 3 \right )}}{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x}{(1+x) (2+x) (3+x)} \, dx=-\frac {3}{2} \, \log \left (x + 3\right ) + 2 \, \log \left (x + 2\right ) - \frac {1}{2} \, \log \left (x + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {x}{(1+x) (2+x) (3+x)} \, dx=-\frac {3}{2} \, \log \left ({\left | x + 3 \right |}\right ) + 2 \, \log \left ({\left | x + 2 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | x + 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {x}{(1+x) (2+x) (3+x)} \, dx=2\,\ln \left (x+2\right )-\frac {\ln \left (x+1\right )}{2}-\frac {3\,\ln \left (x+3\right )}{2} \]

[In]

int(x/((x + 1)*(x + 2)*(x + 3)),x)

[Out]

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