3.116 \(\int \frac{x}{(1+x) (2+x) (3+x)} \, dx\)

Optimal. Leaf size=23 \[ -\frac{1}{2} \log (x+1)+2 \log (x+2)-\frac{3}{2} \log (x+3) \]

[Out]

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

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Rubi [A]  time = 0.0076509, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {148} \[ -\frac{1}{2} \log (x+1)+2 \log (x+2)-\frac{3}{2} \log (x+3) \]

Antiderivative was successfully verified.

[In]

Int[x/((1 + x)*(2 + x)*(3 + x)),x]

[Out]

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

Rule 148

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*} \int \frac{x}{(1+x) (2+x) (3+x)} \, dx &=\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]  time = 0.0052298, size = 23, normalized size = 1. \[ -\frac{1}{2} \log (x+1)+2 \log (x+2)-\frac{3}{2} \log (x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 20, normalized size = 0.9 \begin{align*} -{\frac{\ln \left ( 1+x \right ) }{2}}+2\,\ln \left ( 2+x \right ) -{\frac{3\,\ln \left ( 3+x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.96862, size = 26, normalized size = 1.13 \begin{align*} -\frac{3}{2} \, \log \left (x + 3\right ) + 2 \, \log \left (x + 2\right ) - \frac{1}{2} \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 0.876651, size = 66, normalized size = 2.87 \begin{align*} -\frac{3}{2} \, \log \left (x + 3\right ) + 2 \, \log \left (x + 2\right ) - \frac{1}{2} \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.114764, size = 20, normalized size = 0.87 \begin{align*} - \frac{\log{\left (x + 1 \right )}}{2} + 2 \log{\left (x + 2 \right )} - \frac{3 \log{\left (x + 3 \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.07683, size = 30, normalized size = 1.3 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))