3.45.95 \(\int \frac {2-x+(-x-2 x^2) \log (\frac {x}{3+6 x})}{x+2 x^2} \, dx\)

Optimal. Leaf size=34 \[ -2+\log (4)+\frac {\left (2 x-x^2\right ) \log \left (\frac {x^2}{3 \left (x+2 x^2\right )}\right )}{x} \]

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Rubi [A]  time = 0.15, antiderivative size = 29, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1593, 6742, 72, 2486, 31} \begin {gather*} 2 \log (x)-x \log \left (\frac {x}{3 (2 x+1)}\right )-2 \log (2 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - x + (-x - 2*x^2)*Log[x/(3 + 6*x)])/(x + 2*x^2),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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]

Rule 1593

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]

Rule 2486

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.), x_Symbol] :> Simp[((
a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/b, x] + Dist[(q*r*s*(b*c - a*d))/b, Int[Log[e*(f*(a + b*x)^p*
(c + d*x)^q)^r]^(s - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] &&
EqQ[p + q, 0] && IGtQ[s, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2-x+\left (-x-2 x^2\right ) \log \left (\frac {x}{3+6 x}\right )}{x (1+2 x)} \, dx\\ &=\int \left (\frac {2-x}{x (1+2 x)}-\log \left (\frac {x}{3+6 x}\right )\right ) \, dx\\ &=\int \frac {2-x}{x (1+2 x)} \, dx-\int \log \left (\frac {x}{3+6 x}\right ) \, dx\\ &=-x \log \left (\frac {x}{3 (1+2 x)}\right )+3 \int \frac {1}{3+6 x} \, dx+\int \left (\frac {2}{x}-\frac {5}{1+2 x}\right ) \, dx\\ &=2 \log (x)-x \log \left (\frac {x}{3 (1+2 x)}\right )-2 \log (1+2 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 26, normalized size = 0.76 \begin {gather*} 2 \log (x)-2 \log (1+2 x)-x \log \left (\frac {x}{3+6 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*Log[x] - 2*Log[1 + 2*x] - x*Log[x/(3 + 6*x)]

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fricas [A]  time = 0.52, size = 16, normalized size = 0.47 \begin {gather*} -{\left (x - 2\right )} \log \left (\frac {x}{3 \, {\left (2 \, x + 1\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x - 2)*log(1/3*x/(2*x + 1))

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giac [A]  time = 0.24, size = 41, normalized size = 1.21 \begin {gather*} \frac {\log \left (\frac {x}{3 \, {\left (2 \, x + 1\right )}}\right )}{2 \, {\left (\frac {2 \, x}{2 \, x + 1} - 1\right )}} + \frac {5}{2} \, \log \left (\frac {x}{3 \, {\left (2 \, x + 1\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(1/3*x/(2*x + 1))/(2*x/(2*x + 1) - 1) + 5/2*log(1/3*x/(2*x + 1))

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maple [A]  time = 0.14, size = 27, normalized size = 0.79

method	result	size
norman	\(2 \ln \left (\frac {x}{6 x +3}\right )-x \ln \left (\frac {x}{6 x +3}\right )\)	\(27\)
risch	\(-x \ln \left (\frac {x}{6 x +3}\right )+2 \ln \relax (x )-2 \ln \left (2 x +1\right )\)	\(27\)
derivativedivides	\(-3 \ln \left (\frac {1}{6}-\frac {1}{6 \left (2 x +1\right )}\right ) \left (\frac {1}{6}-\frac {1}{6 \left (2 x +1\right )}\right ) \left (2 x +1\right )+2 \ln \left (\frac {1}{6}-\frac {1}{6 \left (2 x +1\right )}\right )\)	\(46\)
default	\(-3 \ln \left (\frac {1}{6}-\frac {1}{6 \left (2 x +1\right )}\right ) \left (\frac {1}{6}-\frac {1}{6 \left (2 x +1\right )}\right ) \left (2 x +1\right )+2 \ln \left (\frac {1}{6}-\frac {1}{6 \left (2 x +1\right )}\right )\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*ln(x/(6*x+3))-x*ln(x/(6*x+3))

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maxima [A]  time = 0.50, size = 35, normalized size = 1.03 \begin {gather*} x \log \relax (3) + \frac {1}{2} \, {\left (2 \, x + 1\right )} \log \left (2 \, x + 1\right ) - x \log \relax (x) - \frac {5}{2} \, \log \left (2 \, x + 1\right ) + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(3) + 1/2*(2*x + 1)*log(2*x + 1) - x*log(x) - 5/2*log(2*x + 1) + 2*log(x)

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mupad [B]  time = 3.39, size = 15, normalized size = 0.44 \begin {gather*} -\ln \left (\frac {x}{6\,x+3}\right )\,\left (x-2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + log(x/(6*x + 3))*(x + 2*x^2) - 2)/(x + 2*x^2),x)

[Out]

-log(x/(6*x + 3))*(x - 2)

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sympy [A]  time = 0.15, size = 22, normalized size = 0.65 \begin {gather*} - x \log {\left (\frac {x}{6 x + 3} \right )} + 2 \log {\relax (x )} - 2 \log {\left (x + \frac {1}{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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