3.85.11 \(\int \frac {1+(5-x) \log (5-x)+e^3 (-100+20 x) \log ^2(5-x)}{(5-x) \log (5-x)+e^3 (-100+20 x) \log ^2(5-x)} \, dx\)

Optimal. Leaf size=19 \[ x+\log \left (e^3-\frac {1}{20 \log (5-x)}\right ) \]

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Rubi [A]  time = 0.45, antiderivative size = 25, normalized size of antiderivative = 1.32, number of steps used = 8, number of rules used = 4, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {6741, 6742, 2302, 29} \begin {gather*} x-\log (\log (5-x))+\log \left (1-20 e^3 \log (5-x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + (5 - x)*Log[5 - x] + E^3*(-100 + 20*x)*Log[5 - x]^2)/((5 - x)*Log[5 - x] + E^3*(-100 + 20*x)*Log[5 -
x]^2),x]

[Out]

x - Log[Log[5 - x]] + Log[1 - 20*E^3*Log[5 - x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

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} &=\operatorname {Subst}\left (\int \frac {1-x \log (-x)+20 e^3 x \log ^2(-x)}{-x \log (-x)+20 e^3 x \log ^2(-x)} \, dx,x,-5+x\right )\\ &=\operatorname {Subst}\left (\int \frac {-1+x \log (-x)-20 e^3 x \log ^2(-x)}{x \log (-x) \left (1-20 e^3 \log (-x)\right )} \, dx,x,-5+x\right )\\ &=\operatorname {Subst}\left (\int \left (1-\frac {1}{x \log (-x)}+\frac {20 e^3}{x \left (-1+20 e^3 \log (-x)\right )}\right ) \, dx,x,-5+x\right )\\ &=x+\left (20 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (-1+20 e^3 \log (-x)\right )} \, dx,x,-5+x\right )-\operatorname {Subst}\left (\int \frac {1}{x \log (-x)} \, dx,x,-5+x\right )\\ &=x-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (5-x)\right )+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,-1+20 e^3 \log (5-x)\right )\\ &=x-\log (\log (5-x))+\log \left (1-20 e^3 \log (5-x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 25, normalized size = 1.32 \begin {gather*} x-\log (\log (5-x))+\log \left (1-20 e^3 \log (5-x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + (5 - x)*Log[5 - x] + E^3*(-100 + 20*x)*Log[5 - x]^2)/((5 - x)*Log[5 - x] + E^3*(-100 + 20*x)*Lo
g[5 - x]^2),x]

[Out]

x - Log[Log[5 - x]] + Log[1 - 20*E^3*Log[5 - x]]

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fricas [A]  time = 0.81, size = 24, normalized size = 1.26 \begin {gather*} x + \log \left (20 \, e^{3} \log \left (-x + 5\right ) - 1\right ) - \log \left (\log \left (-x + 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x-100)*exp(3)*log(5-x)^2+(5-x)*log(5-x)+1)/((20*x-100)*exp(3)*log(5-x)^2+(5-x)*log(5-x)),x, alg
orithm="fricas")

[Out]

x + log(20*e^3*log(-x + 5) - 1) - log(log(-x + 5))

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giac [A]  time = 0.21, size = 24, normalized size = 1.26 \begin {gather*} x + \log \left (20 \, e^{3} \log \left (-x + 5\right ) - 1\right ) - \log \left (\log \left (-x + 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x-100)*exp(3)*log(5-x)^2+(5-x)*log(5-x)+1)/((20*x-100)*exp(3)*log(5-x)^2+(5-x)*log(5-x)),x, alg
orithm="giac")

[Out]

x + log(20*e^3*log(-x + 5) - 1) - log(log(-x + 5))

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maple [A]  time = 0.09, size = 24, normalized size = 1.26

method	result	size
risch	\(x -\ln \left (\ln \left (5-x \right )\right )+\ln \left (\ln \left (5-x \right )-\frac {{\mathrm e}^{-3}}{20}\right )\)	\(24\)
norman	\(x -\ln \left (\ln \left (5-x \right )\right )+\ln \left (20 \,{\mathrm e}^{3} \ln \left (5-x \right )-1\right )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((20*x-100)*exp(3)*ln(5-x)^2+(5-x)*ln(5-x)+1)/((20*x-100)*exp(3)*ln(5-x)^2+(5-x)*ln(5-x)),x,method=_RETURN
VERBOSE)

[Out]

x-ln(ln(5-x))+ln(ln(5-x)-1/20*exp(-3))

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maxima [B]  time = 0.52, size = 263, normalized size = 13.84 \begin {gather*} -100 \, {\left (\log \left (\frac {1}{20} \, {\left (20 \, e^{3} \log \left (-x + 5\right ) - 1\right )} e^{\left (-3\right )}\right ) - \log \left (\log \left (-x + 5\right )\right )\right )} e^{3} \log \left (-x + 5\right )^{2} - \frac {1}{4} \, {\left (20 \, e^{3} \log \left (-x + 5\right ) - 1\right )} e^{\left (-3\right )} \log \left (\frac {1}{20} \, {\left (20 \, e^{3} \log \left (-x + 5\right ) - 1\right )} e^{\left (-3\right )}\right ) - \frac {1}{4} \, {\left (400 \, {\left (\log \relax (5) + 2 \, \log \relax (2) + 3\right )} e^{6} \log \left (-x + 5\right )^{2} + 400 \, e^{6} \log \left (-x + 5\right )^{2} \log \left (\log \left (-x + 5\right )\right ) - {\left (400 \, e^{6} \log \left (-x + 5\right )^{2} - 1\right )} \log \left (20 \, e^{3} \log \left (-x + 5\right ) - 1\right ) + 20 \, e^{3} \log \left (-x + 5\right )\right )} e^{\left (-3\right )} + \frac {1}{4} \, {\left (20 \, e^{3} \log \left (-x + 5\right ) - 1\right )} e^{\left (-3\right )} + 5 \, {\left (\log \left (\frac {1}{20} \, {\left (20 \, e^{3} \log \left (-x + 5\right ) - 1\right )} e^{\left (-3\right )}\right ) - \log \left (\log \left (-x + 5\right )\right )\right )} \log \left (-x + 5\right ) + 5 \, \log \left (-x + 5\right ) \log \left (\log \left (-x + 5\right )\right ) + x + \log \left (\frac {1}{20} \, {\left (20 \, e^{3} \log \left (-x + 5\right ) - 1\right )} e^{\left (-3\right )}\right ) + 5 \, \log \left (x - 5\right ) - 5 \, \log \left (-x + 5\right ) - \log \left (\log \left (-x + 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x-100)*exp(3)*log(5-x)^2+(5-x)*log(5-x)+1)/((20*x-100)*exp(3)*log(5-x)^2+(5-x)*log(5-x)),x, alg
orithm="maxima")

[Out]

-100*(log(1/20*(20*e^3*log(-x + 5) - 1)*e^(-3)) - log(log(-x + 5)))*e^3*log(-x + 5)^2 - 1/4*(20*e^3*log(-x + 5
) - 1)*e^(-3)*log(1/20*(20*e^3*log(-x + 5) - 1)*e^(-3)) - 1/4*(400*(log(5) + 2*log(2) + 3)*e^6*log(-x + 5)^2 +
400*e^6*log(-x + 5)^2*log(log(-x + 5)) - (400*e^6*log(-x + 5)^2 - 1)*log(20*e^3*log(-x + 5) - 1) + 20*e^3*log
(-x + 5))*e^(-3) + 1/4*(20*e^3*log(-x + 5) - 1)*e^(-3) + 5*(log(1/20*(20*e^3*log(-x + 5) - 1)*e^(-3)) - log(lo
g(-x + 5)))*log(-x + 5) + 5*log(-x + 5)*log(log(-x + 5)) + x + log(1/20*(20*e^3*log(-x + 5) - 1)*e^(-3)) + 5*l
og(x - 5) - 5*log(-x + 5) - log(log(-x + 5))

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mupad [B]  time = 0.36, size = 36, normalized size = 1.89 \begin {gather*} x-\ln \left (\frac {\ln \left (5-x\right )}{x-5}\right )+\ln \left (\frac {20\,{\mathrm {e}}^3\,\ln \left (5-x\right )-1}{x-5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3)*log(5 - x)^2*(20*x - 100) - log(5 - x)*(x - 5) + 1)/(log(5 - x)*(x - 5) - exp(3)*log(5 - x)^2*(20
*x - 100)),x)

[Out]

x - log(log(5 - x)/(x - 5)) + log((20*exp(3)*log(5 - x) - 1)/(x - 5))

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((20*x-100)*exp(3)*ln(5-x)**2+(5-x)*ln(5-x)+1)/((20*x-100)*exp(3)*ln(5-x)**2+(5-x)*ln(5-x)),x)

[Out]

Exception raised: PolynomialError

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