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

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

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {72} \[ \frac {\log (x)}{3}-\log (x+1)+\frac {2}{3} \log (2 x+3) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(1 + x)*(3 + 2*x)),x]

[Out]

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{x (1+x) (3+2 x)} \, dx &=\int \left (\frac {1}{-1-x}+\frac {1}{3 x}+\frac {4}{3 (3+2 x)}\right ) \, dx\\ &=\frac {\log (x)}{3}-\log (1+x)+\frac {2}{3} \log (3+2 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 1.00 \[ \frac {\log (x)}{3}-\log (x+1)+\frac {2}{3} \log (2 x+3) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 + x)*(3 + 2*x)),x]

[Out]

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

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fricas [A]  time = 0.41, size = 19, normalized size = 0.83 \[ \frac {2}{3} \, \log \left (2 \, x + 3\right ) - \log \left (x + 1\right ) + \frac {1}{3} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.86, size = 22, normalized size = 0.96 \[ \frac {2}{3} \, \log \left ({\left | 2 \, x + 3 \right |}\right ) - \log \left ({\left | x + 1 \right |}\right ) + \frac {1}{3} \, \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 20, normalized size = 0.87 \[ \frac {\ln \relax (x )}{3}-\ln \left (x +1\right )+\frac {2 \ln \left (2 x +3\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 19, normalized size = 0.83 \[ \frac {2}{3} \, \log \left (2 \, x + 3\right ) - \log \left (x + 1\right ) + \frac {1}{3} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.08, size = 17, normalized size = 0.74 \[ \frac {2\,\ln \left (x+\frac {3}{2}\right )}{3}-\ln \left (x+1\right )+\frac {\ln \relax (x)}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(2*x + 3)*(x + 1)),x)

[Out]

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

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sympy [A]  time = 0.14, size = 19, normalized size = 0.83 \[ \frac {\log {\relax (x )}}{3} - \log {\left (x + 1 \right )} + \frac {2 \log {\left (x + \frac {3}{2} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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