3.30.24 \(\int \frac {e^{\frac {e^{4-x} x+x \log (x)}{\log (x)}} (-e^{4-x}+e^{4-x} (1-x) \log (x)+\log ^2(x))}{e^{\frac {e^{4-x} x+x \log (x)}{\log (x)}} \log ^2(x)+(-4+4 e^{10}-e^{20}) \log ^2(x)} \, dx\)

Optimal. Leaf size=30 \[ \log \left (-e^{x+\frac {e^{4-x} x}{\log (x)}}+\left (2-e^{10}\right )^2\right ) \]

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Rubi [F]  time = 6.39, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{\frac {e^{4-x} x+x \log (x)}{\log (x)}} \left (-e^{4-x}+e^{4-x} (1-x) \log (x)+\log ^2(x)\right )}{e^{\frac {e^{4-x} x+x \log (x)}{\log (x)}} \log ^2(x)+\left (-4+4 e^{10}-e^{20}\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((E^(4 - x)*x + x*Log[x])/Log[x])*(-E^(4 - x) + E^(4 - x)*(1 - x)*Log[x] + Log[x]^2))/(E^((E^(4 - x)*x
+ x*Log[x])/Log[x])*Log[x]^2 + (-4 + 4*E^10 - E^20)*Log[x]^2),x]

[Out]

x - (2 - E^10)^2*Defer[Int][(-E^(x + (E^(4 - x)*x)/Log[x]) + 4*(1 + (E^10*(-4 + E^10))/4))^(-1), x] + Defer[In
t][E^(4 + (E^(4 - x)*x)/Log[x])/((-E^(x + (E^(4 - x)*x)/Log[x]) + 4*(1 + (E^10*(-4 + E^10))/4))*Log[x]^2), x]
+ Defer[Int][E^(4 + (E^(4 - x)*x)/Log[x])/((E^(x + (E^(4 - x)*x)/Log[x]) - 4*(1 + (E^10*(-4 + E^10))/4))*Log[x
]), x] + Defer[Int][(E^(4 + (E^(4 - x)*x)/Log[x])*x)/((-E^(x + (E^(4 - x)*x)/Log[x]) + 4*(1 + (E^10*(-4 + E^10
))/4))*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {e^{4-x} x+x \log (x)}{\log (x)}} \left (-e^{4-x}+e^{4-x} (1-x) \log (x)+\log ^2(x)\right )}{\left (e^{x+\frac {e^{4-x} x}{\log (x)}}-4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log ^2(x)} \, dx\\ &=\int \left (\frac {e^{\frac {e^{4-x} x+x \log (x)}{\log (x)}}}{e^{x+\frac {e^{4-x} x}{\log (x)}}-4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )}+\frac {e^{4-x+\frac {e^{4-x} x+x \log (x)}{\log (x)}}}{\left (-e^{x+\frac {e^{4-x} x}{\log (x)}}+4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log ^2(x)}+\frac {e^{4-x+\frac {e^{4-x} x+x \log (x)}{\log (x)}}}{\left (e^{x+\frac {e^{4-x} x}{\log (x)}}-4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log (x)}+\frac {e^{4-x+\frac {e^{4-x} x+x \log (x)}{\log (x)}} x}{\left (-e^{x+\frac {e^{4-x} x}{\log (x)}}+4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log (x)}\right ) \, dx\\ &=\int \frac {e^{\frac {e^{4-x} x+x \log (x)}{\log (x)}}}{e^{x+\frac {e^{4-x} x}{\log (x)}}-4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )} \, dx+\int \frac {e^{4-x+\frac {e^{4-x} x+x \log (x)}{\log (x)}}}{\left (-e^{x+\frac {e^{4-x} x}{\log (x)}}+4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log ^2(x)} \, dx+\int \frac {e^{4-x+\frac {e^{4-x} x+x \log (x)}{\log (x)}}}{\left (e^{x+\frac {e^{4-x} x}{\log (x)}}-4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log (x)} \, dx+\int \frac {e^{4-x+\frac {e^{4-x} x+x \log (x)}{\log (x)}} x}{\left (-e^{x+\frac {e^{4-x} x}{\log (x)}}+4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log (x)} \, dx\\ &=\int \left (1-\frac {\left (-2+e^{10}\right )^2}{4-4 e^{10}+e^{20}-e^{x+\frac {e^{4-x} x}{\log (x)}}}\right ) \, dx+\int \frac {e^{4+\frac {e^{4-x} x}{\log (x)}}}{\left (-e^{x+\frac {e^{4-x} x}{\log (x)}}+4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log ^2(x)} \, dx+\int \frac {e^{4+\frac {e^{4-x} x}{\log (x)}}}{\left (e^{x+\frac {e^{4-x} x}{\log (x)}}-4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log (x)} \, dx+\int \frac {e^{4+\frac {e^{4-x} x}{\log (x)}} x}{\left (-e^{x+\frac {e^{4-x} x}{\log (x)}}+4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log (x)} \, dx\\ &=x-\left (-2+e^{10}\right )^2 \int \frac {1}{4-4 e^{10}+e^{20}-e^{x+\frac {e^{4-x} x}{\log (x)}}} \, dx+\int \frac {e^{4+\frac {e^{4-x} x}{\log (x)}}}{\left (-e^{x+\frac {e^{4-x} x}{\log (x)}}+4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log ^2(x)} \, dx+\int \frac {e^{4+\frac {e^{4-x} x}{\log (x)}}}{\left (e^{x+\frac {e^{4-x} x}{\log (x)}}-4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log (x)} \, dx+\int \frac {e^{4+\frac {e^{4-x} x}{\log (x)}} x}{\left (-e^{x+\frac {e^{4-x} x}{\log (x)}}+4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log (x)} \, dx\\ &=x-\left (-2+e^{10}\right )^2 \int \frac {1}{-e^{x+\frac {e^{4-x} x}{\log (x)}}+4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )} \, dx+\int \frac {e^{4+\frac {e^{4-x} x}{\log (x)}}}{\left (-e^{x+\frac {e^{4-x} x}{\log (x)}}+4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log ^2(x)} \, dx+\int \frac {e^{4+\frac {e^{4-x} x}{\log (x)}}}{\left (e^{x+\frac {e^{4-x} x}{\log (x)}}-4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log (x)} \, dx+\int \frac {e^{4+\frac {e^{4-x} x}{\log (x)}} x}{\left (-e^{x+\frac {e^{4-x} x}{\log (x)}}+4 \left (1+\frac {1}{4} e^{10} \left (-4+e^{10}\right )\right )\right ) \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.92, size = 30, normalized size = 1.00 \begin {gather*} \log \left (-4+4 e^{10}-e^{20}+e^{x+\frac {e^{4-x} x}{\log (x)}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((E^(4 - x)*x + x*Log[x])/Log[x])*(-E^(4 - x) + E^(4 - x)*(1 - x)*Log[x] + Log[x]^2))/(E^((E^(4 -
x)*x + x*Log[x])/Log[x])*Log[x]^2 + (-4 + 4*E^10 - E^20)*Log[x]^2),x]

[Out]

Log[-4 + 4*E^10 - E^20 + E^(x + (E^(4 - x)*x)/Log[x])]

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fricas [A]  time = 0.75, size = 30, normalized size = 1.00 \begin {gather*} \log \left (-e^{20} + 4 \, e^{10} + e^{\left (\frac {x e^{\left (-x + 4\right )} + x \log \relax (x)}{\log \relax (x)}\right )} - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)^2+(-x+1)*exp(-x+4)*log(x)-exp(-x+4))*exp((x*log(x)+x*exp(-x+4))/log(x))/(log(x)^2*exp((x*log
(x)+x*exp(-x+4))/log(x))+(-exp(5)^4+4*exp(5)^2-4)*log(x)^2),x, algorithm="fricas")

[Out]

log(-e^20 + 4*e^10 + e^((x*e^(-x + 4) + x*log(x))/log(x)) - 4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (x - 1\right )} e^{\left (-x + 4\right )} \log \relax (x) - \log \relax (x)^{2} + e^{\left (-x + 4\right )}\right )} e^{\left (\frac {x e^{\left (-x + 4\right )} + x \log \relax (x)}{\log \relax (x)}\right )}}{{\left (e^{20} - 4 \, e^{10} + 4\right )} \log \relax (x)^{2} - e^{\left (\frac {x e^{\left (-x + 4\right )} + x \log \relax (x)}{\log \relax (x)}\right )} \log \relax (x)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)^2+(-x+1)*exp(-x+4)*log(x)-exp(-x+4))*exp((x*log(x)+x*exp(-x+4))/log(x))/(log(x)^2*exp((x*log
(x)+x*exp(-x+4))/log(x))+(-exp(5)^4+4*exp(5)^2-4)*log(x)^2),x, algorithm="giac")

[Out]

integrate(((x - 1)*e^(-x + 4)*log(x) - log(x)^2 + e^(-x + 4))*e^((x*e^(-x + 4) + x*log(x))/log(x))/((e^20 - 4*
e^10 + 4)*log(x)^2 - e^((x*e^(-x + 4) + x*log(x))/log(x))*log(x)^2), x)

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maple [B]  time = 0.05, size = 61, normalized size = 2.03

method	result	size
risch	\(x +\frac {x \,{\mathrm e}^{-x +4}}{\ln \relax (x )}-\frac {x \ln \relax (x )+x \,{\mathrm e}^{-x +4}}{\ln \relax (x )}+\ln \left ({\mathrm e}^{\frac {x \left (\ln \relax (x )+{\mathrm e}^{-x +4}\right )}{\ln \relax (x )}}-{\mathrm e}^{20}+4 \,{\mathrm e}^{10}-4\right )\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(x)^2+(1-x)*exp(-x+4)*ln(x)-exp(-x+4))*exp((x*ln(x)+x*exp(-x+4))/ln(x))/(ln(x)^2*exp((x*ln(x)+x*exp(-x+
4))/ln(x))+(-exp(5)^4+4*exp(5)^2-4)*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

x+x*exp(-x+4)/ln(x)-(x*ln(x)+x*exp(-x+4))/ln(x)+ln(exp(x*(ln(x)+exp(-x+4))/ln(x))-exp(20)+4*exp(10)-4)

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maxima [A]  time = 0.64, size = 34, normalized size = 1.13 \begin {gather*} x + \log \left (-{\left (e^{20} - 4 \, e^{10} - e^{\left (x + \frac {x e^{\left (-x + 4\right )}}{\log \relax (x)}\right )} + 4\right )} e^{\left (-x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)^2+(-x+1)*exp(-x+4)*log(x)-exp(-x+4))*exp((x*log(x)+x*exp(-x+4))/log(x))/(log(x)^2*exp((x*log
(x)+x*exp(-x+4))/log(x))+(-exp(5)^4+4*exp(5)^2-4)*log(x)^2),x, algorithm="maxima")

[Out]

x + log(-(e^20 - 4*e^10 - e^(x + x*e^(-x + 4)/log(x)) + 4)*e^(-x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{4-x}+x\,\ln \relax (x)}{\ln \relax (x)}}\,\left (-{\ln \relax (x)}^2+{\mathrm {e}}^{4-x}\,\left (x-1\right )\,\ln \relax (x)+{\mathrm {e}}^{4-x}\right )}{{\ln \relax (x)}^2\,\left ({\mathrm {e}}^{20}-4\,{\mathrm {e}}^{10}+4\right )-{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{4-x}+x\,\ln \relax (x)}{\ln \relax (x)}}\,{\ln \relax (x)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x*exp(4 - x) + x*log(x))/log(x))*(exp(4 - x) - log(x)^2 + exp(4 - x)*log(x)*(x - 1)))/(log(x)^2*(exp
(20) - 4*exp(10) + 4) - exp((x*exp(4 - x) + x*log(x))/log(x))*log(x)^2),x)

[Out]

int((exp((x*exp(4 - x) + x*log(x))/log(x))*(exp(4 - x) - log(x)^2 + exp(4 - x)*log(x)*(x - 1)))/(log(x)^2*(exp
(20) - 4*exp(10) + 4) - exp((x*exp(4 - x) + x*log(x))/log(x))*log(x)^2), x)

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sympy [A]  time = 0.68, size = 27, normalized size = 0.90 \begin {gather*} \log {\left (e^{\frac {x e^{4 - x} + x \log {\relax (x )}}{\log {\relax (x )}}} - e^{20} - 4 + 4 e^{10} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(x)**2+(-x+1)*exp(-x+4)*ln(x)-exp(-x+4))*exp((x*ln(x)+x*exp(-x+4))/ln(x))/(ln(x)**2*exp((x*ln(x)+
x*exp(-x+4))/ln(x))+(-exp(5)**4+4*exp(5)**2-4)*ln(x)**2),x)

[Out]

log(exp((x*exp(4 - x) + x*log(x))/log(x)) - exp(20) - 4 + 4*exp(10))

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