3.85.35 \(\int \frac {75+90 x+3 x^2+(120 x-24 x^2+(75-30 x+3 x^2) \log (\frac {x}{e})) \log (\frac {-8 x+(-5+x) \log (\frac {x}{e})}{-5+x}) \log (-\log (\frac {-8 x+(-5+x) \log (\frac {x}{e})}{-5+x}))}{(40 x-8 x^2+(25-10 x+x^2) \log (\frac {x}{e})) \log (\frac {-8 x+(-5+x) \log (\frac {x}{e})}{-5+x})} \, dx\)

Optimal. Leaf size=24 \[ 3 x \log \left (-\log \left (\frac {8 x}{5-x}+\log \left (\frac {x}{e}\right )\right )\right ) \]

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Rubi [A]  time = 1.95, antiderivative size = 30, normalized size of antiderivative = 1.25, number of steps used = 8, number of rules used = 3, integrand size = 132, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6741, 6742, 2549} \begin {gather*} 3 x \log \left (-\log \left (-\frac {-9 x-(5-x) \log (x)+5}{5-x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(75 + 90*x + 3*x^2 + (120*x - 24*x^2 + (75 - 30*x + 3*x^2)*Log[x/E])*Log[(-8*x + (-5 + x)*Log[x/E])/(-5 +
x)]*Log[-Log[(-8*x + (-5 + x)*Log[x/E])/(-5 + x)]])/((40*x - 8*x^2 + (25 - 10*x + x^2)*Log[x/E])*Log[(-8*x + (
-5 + x)*Log[x/E])/(-5 + x)]),x]

[Out]

3*x*Log[-Log[-((5 - 9*x - (5 - x)*Log[x])/(5 - x))]]

Rule 2549

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*Simplify[D[u, x]/u], x], x] /; ProductQ[
u]

Rule 6741

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

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 {-75-90 x-3 x^2-\left (120 x-24 x^2+\left (75-30 x+3 x^2\right ) \log \left (\frac {x}{e}\right )\right ) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right ) \log \left (-\log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )\right )}{(5-x) (5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {-8 x+(-5+x) \log \left (\frac {x}{e}\right )}{-5+x}\right )} \, dx\\ &=\int \left (\frac {3 \left (25+30 x+x^2\right )}{(-5+x) (5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}+3 \log \left (-\log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )\right )\right ) \, dx\\ &=3 \int \frac {25+30 x+x^2}{(-5+x) (5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )} \, dx+3 \int \log \left (-\log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )\right ) \, dx\\ &=3 x \log \left (-\log \left (-\frac {5-9 x-(5-x) \log (x)}{5-x}\right )\right )+3 \int \left (\frac {35}{(5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}+\frac {200}{(-5+x) (5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}+\frac {x}{(5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}\right ) \, dx-3 \int \frac {25+30 x+x^2}{(-5+x) (5-9 x+(-5+x) \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )} \, dx\\ &=3 x \log \left (-\log \left (-\frac {5-9 x-(5-x) \log (x)}{5-x}\right )\right )-3 \int \left (\frac {35}{(5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}+\frac {200}{(-5+x) (5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}+\frac {x}{(5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )}\right ) \, dx+3 \int \frac {x}{(5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )} \, dx+105 \int \frac {1}{(5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )} \, dx+600 \int \frac {1}{(-5+x) (5-9 x-5 \log (x)+x \log (x)) \log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )} \, dx\\ &=3 x \log \left (-\log \left (-\frac {5-9 x-(5-x) \log (x)}{5-x}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 24, normalized size = 1.00 \begin {gather*} 3 x \log \left (-\log \left (\frac {5-9 x+(-5+x) \log (x)}{-5+x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(75 + 90*x + 3*x^2 + (120*x - 24*x^2 + (75 - 30*x + 3*x^2)*Log[x/E])*Log[(-8*x + (-5 + x)*Log[x/E])/
(-5 + x)]*Log[-Log[(-8*x + (-5 + x)*Log[x/E])/(-5 + x)]])/((40*x - 8*x^2 + (25 - 10*x + x^2)*Log[x/E])*Log[(-8
*x + (-5 + x)*Log[x/E])/(-5 + x)]),x]

[Out]

3*x*Log[-Log[(5 - 9*x + (-5 + x)*Log[x])/(-5 + x)]]

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fricas [A]  time = 2.82, size = 26, normalized size = 1.08 \begin {gather*} 3 \, x \log \left (-\log \left (\frac {{\left (x - 5\right )} \log \left (x e^{\left (-1\right )}\right ) - 8 \, x}{x - 5}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-30*x+75)*log(x/exp(1))-24*x^2+120*x)*log(((x-5)*log(x/exp(1))-8*x)/(x-5))*log(-log(((x-5)*l
og(x/exp(1))-8*x)/(x-5)))+3*x^2+90*x+75)/((x^2-10*x+25)*log(x/exp(1))-8*x^2+40*x)/log(((x-5)*log(x/exp(1))-8*x
)/(x-5)),x, algorithm="fricas")

[Out]

3*x*log(-log(((x - 5)*log(x*e^(-1)) - 8*x)/(x - 5)))

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giac [A]  time = 1.03, size = 25, normalized size = 1.04 \begin {gather*} 3 \, x \log \left (-\log \left (x \log \relax (x) - 9 \, x - 5 \, \log \relax (x) + 5\right ) + \log \left (x - 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-30*x+75)*log(x/exp(1))-24*x^2+120*x)*log(((x-5)*log(x/exp(1))-8*x)/(x-5))*log(-log(((x-5)*l
og(x/exp(1))-8*x)/(x-5)))+3*x^2+90*x+75)/((x^2-10*x+25)*log(x/exp(1))-8*x^2+40*x)/log(((x-5)*log(x/exp(1))-8*x
)/(x-5)),x, algorithm="giac")

[Out]

3*x*log(-log(x*log(x) - 9*x - 5*log(x) + 5) + log(x - 5))

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maple [C]  time = 0.24, size = 148, normalized size = 6.17

method	result	size
risch	\(3 x \ln \left (\ln \left (x -5\right )-\ln \left (\left (\ln \left ({\mathrm e}^{-1} x \right )-8\right ) x -5 \ln \left ({\mathrm e}^{-1} x \right )\right )+\frac {i \pi \,\mathrm {csgn}\left (\frac {i \left (\left (\ln \left ({\mathrm e}^{-1} x \right )-8\right ) x -5 \ln \left ({\mathrm e}^{-1} x \right )\right )}{x -5}\right ) \left (-\mathrm {csgn}\left (\frac {i \left (\left (\ln \left ({\mathrm e}^{-1} x \right )-8\right ) x -5 \ln \left ({\mathrm e}^{-1} x \right )\right )}{x -5}\right )+\mathrm {csgn}\left (\frac {i}{x -5}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \left (\left (\ln \left ({\mathrm e}^{-1} x \right )-8\right ) x -5 \ln \left ({\mathrm e}^{-1} x \right )\right )}{x -5}\right )+\mathrm {csgn}\left (i \left (\left (\ln \left ({\mathrm e}^{-1} x \right )-8\right ) x -5 \ln \left ({\mathrm e}^{-1} x \right )\right )\right )\right )}{2}\right )\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3*x^2-30*x+75)*ln(x/exp(1))-24*x^2+120*x)*ln(((x-5)*ln(x/exp(1))-8*x)/(x-5))*ln(-ln(((x-5)*ln(x/exp(1))
-8*x)/(x-5)))+3*x^2+90*x+75)/((x^2-10*x+25)*ln(x/exp(1))-8*x^2+40*x)/ln(((x-5)*ln(x/exp(1))-8*x)/(x-5)),x,meth
od=_RETURNVERBOSE)

[Out]

3*x*ln(ln(x-5)-ln((ln(exp(-1)*x)-8)*x-5*ln(exp(-1)*x))+1/2*I*Pi*csgn(I*((ln(exp(-1)*x)-8)*x-5*ln(exp(-1)*x))/(
x-5))*(-csgn(I*((ln(exp(-1)*x)-8)*x-5*ln(exp(-1)*x))/(x-5))+csgn(I/(x-5)))*(-csgn(I*((ln(exp(-1)*x)-8)*x-5*ln(
exp(-1)*x))/(x-5))+csgn(I*((ln(exp(-1)*x)-8)*x-5*ln(exp(-1)*x)))))

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maxima [A]  time = 0.49, size = 23, normalized size = 0.96 \begin {gather*} 3 \, x \log \left (-\log \left ({\left (x - 5\right )} \log \relax (x) - 9 \, x + 5\right ) + \log \left (x - 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2-30*x+75)*log(x/exp(1))-24*x^2+120*x)*log(((x-5)*log(x/exp(1))-8*x)/(x-5))*log(-log(((x-5)*l
og(x/exp(1))-8*x)/(x-5)))+3*x^2+90*x+75)/((x^2-10*x+25)*log(x/exp(1))-8*x^2+40*x)/log(((x-5)*log(x/exp(1))-8*x
)/(x-5)),x, algorithm="maxima")

[Out]

3*x*log(-log((x - 5)*log(x) - 9*x + 5) + log(x - 5))

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mupad [B]  time = 7.52, size = 28, normalized size = 1.17 \begin {gather*} 3\,x\,\ln \left (-\ln \left (-\frac {8\,x-\ln \left (x\,{\mathrm {e}}^{-1}\right )\,\left (x-5\right )}{x-5}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((90*x + 3*x^2 + log(-(8*x - log(x*exp(-1))*(x - 5))/(x - 5))*log(-log(-(8*x - log(x*exp(-1))*(x - 5))/(x -
5)))*(120*x + log(x*exp(-1))*(3*x^2 - 30*x + 75) - 24*x^2) + 75)/(log(-(8*x - log(x*exp(-1))*(x - 5))/(x - 5)
)*(40*x + log(x*exp(-1))*(x^2 - 10*x + 25) - 8*x^2)),x)

[Out]

3*x*log(-log(-(8*x - log(x*exp(-1))*(x - 5))/(x - 5)))

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sympy [B]  time = 7.45, size = 51, normalized size = 2.12 \begin {gather*} \left (3 x - \frac {5}{2}\right ) \log {\left (- \log {\left (\frac {- 8 x + \left (x - 5\right ) \log {\left (\frac {x}{e} \right )}}{x - 5} \right )} \right )} + \frac {5 \log {\left (\log {\left (\frac {- 8 x + \left (x - 5\right ) \log {\left (\frac {x}{e} \right )}}{x - 5} \right )} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x**2-30*x+75)*ln(x/exp(1))-24*x**2+120*x)*ln(((x-5)*ln(x/exp(1))-8*x)/(x-5))*ln(-ln(((x-5)*ln(x
/exp(1))-8*x)/(x-5)))+3*x**2+90*x+75)/((x**2-10*x+25)*ln(x/exp(1))-8*x**2+40*x)/ln(((x-5)*ln(x/exp(1))-8*x)/(x
-5)),x)

[Out]

(3*x - 5/2)*log(-log((-8*x + (x - 5)*log(x*exp(-1)))/(x - 5))) + 5*log(log((-8*x + (x - 5)*log(x*exp(-1)))/(x
- 5)))/2

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