3.49.83 \(\int \frac {2+5 x^2+(-2+10 x) \log (x)+5 \log ^2(x)}{-2 x^2+5 x^3+(-2 x+10 x^2) \log (x)+5 x \log ^2(x)} \, dx\)

Optimal. Leaf size=20 \[ \log \left (\frac {5}{3} \left (-x+\frac {2 x}{5 (x+\log (x))}\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6742, 6684} \begin {gather*} \log (x)+\log (-5 x-5 \log (x)+2)-\log (x+\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 5*x^2 + (-2 + 10*x)*Log[x] + 5*Log[x]^2)/(-2*x^2 + 5*x^3 + (-2*x + 10*x^2)*Log[x] + 5*x*Log[x]^2),x]

[Out]

Log[x] + Log[2 - 5*x - 5*Log[x]] - Log[x + Log[x]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 20, normalized size = 1.00 \begin {gather*} \log (x)+\log (2-5 x-5 \log (x))-\log (x+\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 5*x^2 + (-2 + 10*x)*Log[x] + 5*Log[x]^2)/(-2*x^2 + 5*x^3 + (-2*x + 10*x^2)*Log[x] + 5*x*Log[x]^
2),x]

[Out]

Log[x] + Log[2 - 5*x - 5*Log[x]] - Log[x + Log[x]]

________________________________________________________________________________________

fricas [A]  time = 0.86, size = 20, normalized size = 1.00 \begin {gather*} \log \left (5 \, x + 5 \, \log \relax (x) - 2\right ) - \log \left (x + \log \relax (x)\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(x)^2+(10*x-2)*log(x)+5*x^2+2)/(5*x*log(x)^2+(10*x^2-2*x)*log(x)+5*x^3-2*x^2),x, algorithm="fr
icas")

[Out]

log(5*x + 5*log(x) - 2) - log(x + log(x)) + log(x)

________________________________________________________________________________________

giac [A]  time = 0.18, size = 20, normalized size = 1.00 \begin {gather*} \log \left (5 \, x + 5 \, \log \relax (x) - 2\right ) - \log \left (x + \log \relax (x)\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(x)^2+(10*x-2)*log(x)+5*x^2+2)/(5*x*log(x)^2+(10*x^2-2*x)*log(x)+5*x^3-2*x^2),x, algorithm="gi
ac")

[Out]

log(5*x + 5*log(x) - 2) - log(x + log(x)) + log(x)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 17, normalized size = 0.85




method result size



risch \(\ln \relax (x )+\ln \left (-\frac {2}{5}+x +\ln \relax (x )\right )-\ln \left (x +\ln \relax (x )\right )\) \(17\)
norman \(\ln \relax (x )-\ln \left (x +\ln \relax (x )\right )+\ln \left (5 x +5 \ln \relax (x )-2\right )\) \(21\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*ln(x)^2+(10*x-2)*ln(x)+5*x^2+2)/(5*x*ln(x)^2+(10*x^2-2*x)*ln(x)+5*x^3-2*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(x)+ln(-2/5+x+ln(x))-ln(x+ln(x))

________________________________________________________________________________________

maxima [A]  time = 0.40, size = 16, normalized size = 0.80 \begin {gather*} -\log \left (x + \log \relax (x)\right ) + \log \left (x + \log \relax (x) - \frac {2}{5}\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(x)^2+(10*x-2)*log(x)+5*x^2+2)/(5*x*log(x)^2+(10*x^2-2*x)*log(x)+5*x^3-2*x^2),x, algorithm="ma
xima")

[Out]

-log(x + log(x)) + log(x + log(x) - 2/5) + log(x)

________________________________________________________________________________________

mupad [B]  time = 3.60, size = 16, normalized size = 0.80 \begin {gather*} \ln \left (x+\ln \relax (x)-\frac {2}{5}\right )-\ln \left (x+\ln \relax (x)\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*log(x)^2 + log(x)*(10*x - 2) + 5*x^2 + 2)/(5*x*log(x)^2 - log(x)*(2*x - 10*x^2) - 2*x^2 + 5*x^3),x)

[Out]

log(x + log(x) - 2/5) - log(x + log(x)) + log(x)

________________________________________________________________________________________

sympy [A]  time = 0.28, size = 19, normalized size = 0.95 \begin {gather*} \log {\relax (x )} - \log {\left (x + \log {\relax (x )} \right )} + \log {\left (x + \log {\relax (x )} - \frac {2}{5} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*ln(x)**2+(10*x-2)*ln(x)+5*x**2+2)/(5*x*ln(x)**2+(10*x**2-2*x)*ln(x)+5*x**3-2*x**2),x)

[Out]

log(x) - log(x + log(x)) + log(x + log(x) - 2/5)

________________________________________________________________________________________