3.97.22 \(\int \frac {e^{1-x} (1-x) x+e^{5-x} (1-5 x+e^4 x)-e^{5-x} x \log (x)}{e^{1-x} x^2+e^{5-x} (5 x-e^4 x)+e^{5-x} x \log (x)} \, dx\)

Optimal. Leaf size=28 \[ \log \left (e^{1-x} x+e^{5-x} \left (5-e^4+\log (x)\right )\right ) \]

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Rubi [A]  time = 0.48, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 89, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6688, 6742, 6684} \begin {gather*} \log \left (-x-e^4 \log (x)-e^4 \left (5-e^4\right )\right )-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1 - x)*(1 - x)*x + E^(5 - x)*(1 - 5*x + E^4*x) - E^(5 - x)*x*Log[x])/(E^(1 - x)*x^2 + E^(5 - x)*(5*x -
E^4*x) + E^(5 - x)*x*Log[x]),x]

[Out]

-x + Log[-(E^4*(5 - E^4)) - x - E^4*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 6688

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

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

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Mathematica [A]  time = 0.21, size = 24, normalized size = 0.86 \begin {gather*} -x+\log \left (-5 e^4+e^8-x-e^4 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 - x)*(1 - x)*x + E^(5 - x)*(1 - 5*x + E^4*x) - E^(5 - x)*x*Log[x])/(E^(1 - x)*x^2 + E^(5 - x)*
(5*x - E^4*x) + E^(5 - x)*x*Log[x]),x]

[Out]

-x + Log[-5*E^4 + E^8 - x - E^4*Log[x]]

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fricas [A]  time = 0.58, size = 20, normalized size = 0.71 \begin {gather*} -x + \log \left (e^{4} \log \relax (x) + x - e^{8} + 5 \, e^{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(1+log(x)-x)-x*exp(5-x)*log(x)+(x*exp(4)-5*x+1)*exp(5-x))/(x*exp(1+log(x)-x)+x*exp(5-x)*l
og(x)+(-x*exp(4)+5*x)*exp(5-x)),x, algorithm="fricas")

[Out]

-x + log(e^4*log(x) + x - e^8 + 5*e^4)

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giac [A]  time = 0.14, size = 21, normalized size = 0.75 \begin {gather*} -x + \log \left (-e^{4} \log \relax (x) - x + e^{8} - 5 \, e^{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(1+log(x)-x)-x*exp(5-x)*log(x)+(x*exp(4)-5*x+1)*exp(5-x))/(x*exp(1+log(x)-x)+x*exp(5-x)*l
og(x)+(-x*exp(4)+5*x)*exp(5-x)),x, algorithm="giac")

[Out]

-x + log(-e^4*log(x) - x + e^8 - 5*e^4)

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maple [A]  time = 0.06, size = 22, normalized size = 0.79

method	result	size
risch	\(-x +\ln \left (\ln \relax (x )+\left ({\mathrm e}^{-8} x +5 \,{\mathrm e}^{-4}-1\right ) {\mathrm e}^{4}\right )\)	\(22\)
norman	\(-x +\ln \left ({\mathrm e}^{8}-{\mathrm e}^{4} \ln \relax (x )-5 \,{\mathrm e}^{4}-x \right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1-x)*exp(1+ln(x)-x)-x*exp(5-x)*ln(x)+(x*exp(4)-5*x+1)*exp(5-x))/(x*exp(1+ln(x)-x)+x*exp(5-x)*ln(x)+(-x*e
xp(4)+5*x)*exp(5-x)),x,method=_RETURNVERBOSE)

[Out]

-x+ln(ln(x)+(exp(-8)*x+5*exp(-4)-1)*exp(4))

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maxima [A]  time = 0.39, size = 23, normalized size = 0.82 \begin {gather*} -x + \log \left ({\left (e^{4} \log \relax (x) + x - e^{8} + 5 \, e^{4}\right )} e^{\left (-4\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(1+log(x)-x)-x*exp(5-x)*log(x)+(x*exp(4)-5*x+1)*exp(5-x))/(x*exp(1+log(x)-x)+x*exp(5-x)*l
og(x)+(-x*exp(4)+5*x)*exp(5-x)),x, algorithm="maxima")

[Out]

-x + log((e^4*log(x) + x - e^8 + 5*e^4)*e^(-4))

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mupad [B]  time = 5.90, size = 17, normalized size = 0.61 \begin {gather*} \ln \left (\ln \relax (x)-{\mathrm {e}}^4+x\,{\mathrm {e}}^{-4}+5\right )-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(x) - x + 1)*(x - 1) - exp(5 - x)*(x*exp(4) - 5*x + 1) + x*exp(5 - x)*log(x))/(x*exp(log(x) - x +
1) + exp(5 - x)*(5*x - x*exp(4)) + x*exp(5 - x)*log(x)),x)

[Out]

log(log(x) - exp(4) + x*exp(-4) + 5) - x

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sympy [A]  time = 0.33, size = 19, normalized size = 0.68 \begin {gather*} - x + \log {\left (\frac {x - e^{8} + 5 e^{4}}{e^{4}} + \log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x+1)*exp(1+ln(x)-x)-x*exp(5-x)*ln(x)+(x*exp(4)-5*x+1)*exp(5-x))/(x*exp(1+ln(x)-x)+x*exp(5-x)*ln(x
)+(-x*exp(4)+5*x)*exp(5-x)),x)

[Out]

-x + log((x - exp(8) + 5*exp(4))*exp(-4) + log(x))

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