3.100.83 \(\int \frac {-1+4 x+x^2-x \log (x)}{-9 x^3-3 x^4+3 x^3 \log (x)+(-18 x^2-6 x^3+6 x^2 \log (x)) \log (\frac {1}{5} (3+x-\log (x)))+(-9 x-3 x^2+3 x \log (x)) \log ^2(\frac {1}{5} (3+x-\log (x)))} \, dx\)

Optimal. Leaf size=22 \[ 25+\frac {1}{3 \left (x+\log \left (\frac {1}{5} (3+x-\log (x))\right )\right )} \]

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Rubi [A]  time = 0.24, antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 3, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6688, 12, 6686} \begin {gather*} \frac {1}{3 \left (x+\log \left (\frac {1}{5} (x-\log (x)+3)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + 4*x + x^2 - x*Log[x])/(-9*x^3 - 3*x^4 + 3*x^3*Log[x] + (-18*x^2 - 6*x^3 + 6*x^2*Log[x])*Log[(3 + x -
Log[x])/5] + (-9*x - 3*x^2 + 3*x*Log[x])*Log[(3 + x - Log[x])/5]^2),x]

[Out]

1/(3*(x + Log[(3 + x - Log[x])/5]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1-4 x-x^2+x \log (x)}{3 x (3+x-\log (x)) \left (x+\log \left (\frac {1}{5} (3+x-\log (x))\right )\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {1-4 x-x^2+x \log (x)}{x (3+x-\log (x)) \left (x+\log \left (\frac {1}{5} (3+x-\log (x))\right )\right )^2} \, dx\\ &=\frac {1}{3 \left (x+\log \left (\frac {1}{5} (3+x-\log (x))\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 20, normalized size = 0.91 \begin {gather*} \frac {1}{3 \left (x+\log \left (\frac {1}{5} (3+x-\log (x))\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 4*x + x^2 - x*Log[x])/(-9*x^3 - 3*x^4 + 3*x^3*Log[x] + (-18*x^2 - 6*x^3 + 6*x^2*Log[x])*Log[(3
+ x - Log[x])/5] + (-9*x - 3*x^2 + 3*x*Log[x])*Log[(3 + x - Log[x])/5]^2),x]

[Out]

1/(3*(x + Log[(3 + x - Log[x])/5]))

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fricas [A]  time = 0.55, size = 16, normalized size = 0.73 \begin {gather*} \frac {1}{3 \, {\left (x + \log \left (\frac {1}{5} \, x - \frac {1}{5} \, \log \relax (x) + \frac {3}{5}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)+x^2+4*x-1)/((3*x*log(x)-3*x^2-9*x)*log(-1/5*log(x)+3/5+1/5*x)^2+(6*x^2*log(x)-6*x^3-18*x^
2)*log(-1/5*log(x)+3/5+1/5*x)+3*x^3*log(x)-3*x^4-9*x^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 2.37, size = 18, normalized size = 0.82 \begin {gather*} \frac {1}{3 \, {\left (x - \log \relax (5) + \log \left (x - \log \relax (x) + 3\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)+x^2+4*x-1)/((3*x*log(x)-3*x^2-9*x)*log(-1/5*log(x)+3/5+1/5*x)^2+(6*x^2*log(x)-6*x^3-18*x^
2)*log(-1/5*log(x)+3/5+1/5*x)+3*x^3*log(x)-3*x^4-9*x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 17, normalized size = 0.77

method	result	size
risch	\(\frac {1}{3 \ln \left (-\frac {\ln \relax (x )}{5}+\frac {3}{5}+\frac {x}{5}\right )+3 x}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(x)+x^2+4*x-1)/((3*x*ln(x)-3*x^2-9*x)*ln(-1/5*ln(x)+3/5+1/5*x)^2+(6*x^2*ln(x)-6*x^3-18*x^2)*ln(-1/5*
ln(x)+3/5+1/5*x)+3*x^3*ln(x)-3*x^4-9*x^3),x,method=_RETURNVERBOSE)

[Out]

1/3/(ln(-1/5*ln(x)+3/5+1/5*x)+x)

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maxima [A]  time = 0.47, size = 18, normalized size = 0.82 \begin {gather*} \frac {1}{3 \, {\left (x - \log \relax (5) + \log \left (x - \log \relax (x) + 3\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)+x^2+4*x-1)/((3*x*log(x)-3*x^2-9*x)*log(-1/5*log(x)+3/5+1/5*x)^2+(6*x^2*log(x)-6*x^3-18*x^
2)*log(-1/5*log(x)+3/5+1/5*x)+3*x^3*log(x)-3*x^4-9*x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.37, size = 18, normalized size = 0.82 \begin {gather*} \frac {1}{3\,\left (x+\ln \left (\frac {x}{5}-\frac {\ln \relax (x)}{5}+\frac {3}{5}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x - x*log(x) + x^2 - 1)/(log(x/5 - log(x)/5 + 3/5)*(18*x^2 - 6*x^2*log(x) + 6*x^3) - 3*x^3*log(x) + lo
g(x/5 - log(x)/5 + 3/5)^2*(9*x - 3*x*log(x) + 3*x^2) + 9*x^3 + 3*x^4),x)

[Out]

1/(3*(x + log(x/5 - log(x)/5 + 3/5)))

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sympy [A]  time = 0.32, size = 19, normalized size = 0.86 \begin {gather*} \frac {1}{3 x + 3 \log {\left (\frac {x}{5} - \frac {\log {\relax (x )}}{5} + \frac {3}{5} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*ln(x)+x**2+4*x-1)/((3*x*ln(x)-3*x**2-9*x)*ln(-1/5*ln(x)+3/5+1/5*x)**2+(6*x**2*ln(x)-6*x**3-18*x*
*2)*ln(-1/5*ln(x)+3/5+1/5*x)+3*x**3*ln(x)-3*x**4-9*x**3),x)

[Out]

1/(3*x + 3*log(x/5 - log(x)/5 + 3/5))

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