3.69.92 \(\int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+(5 e^{\frac {1}{e^2}} x+e^x (-4 x+x^2)) \log (x)}{5 x \log ^3(x)} \, dx\)

Optimal. Leaf size=24 \[ -\frac {\left (e^{\frac {1}{e^2}}+\frac {e^x}{5}\right ) (5-x)}{\log ^2(x)} \]

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Rubi [F]  time = 1.61, 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 {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^E^(-2)*(50 - 10*x) + E^x*(10 - 2*x) + (5*E^E^(-2)*x + E^x*(-4*x + x^2))*Log[x])/(5*x*Log[x]^3),x]

[Out]

(-5*E^E^(-2))/Log[x]^2 + (E^E^(-2)*x)/Log[x]^2 - (2*Defer[Int][E^x/Log[x]^3, x])/5 + 2*Defer[Int][E^x/(x*Log[x
]^3), x] - (4*Defer[Int][E^x/Log[x]^2, x])/5 + Defer[Int][(E^x*x)/Log[x]^2, x]/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{x \log ^3(x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {5 e^{\frac {1}{e^2}} (10-2 x+x \log (x))}{x \log ^3(x)}+\frac {e^x \left (10-2 x-4 x \log (x)+x^2 \log (x)\right )}{x \log ^3(x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^x \left (10-2 x-4 x \log (x)+x^2 \log (x)\right )}{x \log ^3(x)} \, dx+e^{\frac {1}{e^2}} \int \frac {10-2 x+x \log (x)}{x \log ^3(x)} \, dx\\ &=\frac {1}{5} \int \left (-\frac {2 e^x (-5+x)}{x \log ^3(x)}+\frac {e^x (-4+x)}{\log ^2(x)}\right ) \, dx+e^{\frac {1}{e^2}} \int \left (-\frac {2 (-5+x)}{x \log ^3(x)}+\frac {1}{\log ^2(x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^x (-4+x)}{\log ^2(x)} \, dx-\frac {2}{5} \int \frac {e^x (-5+x)}{x \log ^3(x)} \, dx+e^{\frac {1}{e^2}} \int \frac {1}{\log ^2(x)} \, dx-\left (2 e^{\frac {1}{e^2}}\right ) \int \frac {-5+x}{x \log ^3(x)} \, dx\\ &=-\frac {e^{\frac {1}{e^2}} x}{\log (x)}+\frac {1}{5} \int \left (-\frac {4 e^x}{\log ^2(x)}+\frac {e^x x}{\log ^2(x)}\right ) \, dx-\frac {2}{5} \int \left (\frac {e^x}{\log ^3(x)}-\frac {5 e^x}{x \log ^3(x)}\right ) \, dx+e^{\frac {1}{e^2}} \int \frac {1}{\log (x)} \, dx-\left (2 e^{\frac {1}{e^2}}\right ) \int \left (\frac {1}{\log ^3(x)}-\frac {5}{x \log ^3(x)}\right ) \, dx\\ &=-\frac {e^{\frac {1}{e^2}} x}{\log (x)}+e^{\frac {1}{e^2}} \text {li}(x)+\frac {1}{5} \int \frac {e^x x}{\log ^2(x)} \, dx-\frac {2}{5} \int \frac {e^x}{\log ^3(x)} \, dx-\frac {4}{5} \int \frac {e^x}{\log ^2(x)} \, dx+2 \int \frac {e^x}{x \log ^3(x)} \, dx-\left (2 e^{\frac {1}{e^2}}\right ) \int \frac {1}{\log ^3(x)} \, dx+\left (10 e^{\frac {1}{e^2}}\right ) \int \frac {1}{x \log ^3(x)} \, dx\\ &=\frac {e^{\frac {1}{e^2}} x}{\log ^2(x)}-\frac {e^{\frac {1}{e^2}} x}{\log (x)}+e^{\frac {1}{e^2}} \text {li}(x)+\frac {1}{5} \int \frac {e^x x}{\log ^2(x)} \, dx-\frac {2}{5} \int \frac {e^x}{\log ^3(x)} \, dx-\frac {4}{5} \int \frac {e^x}{\log ^2(x)} \, dx+2 \int \frac {e^x}{x \log ^3(x)} \, dx-e^{\frac {1}{e^2}} \int \frac {1}{\log ^2(x)} \, dx+\left (10 e^{\frac {1}{e^2}}\right ) \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right )\\ &=-\frac {5 e^{\frac {1}{e^2}}}{\log ^2(x)}+\frac {e^{\frac {1}{e^2}} x}{\log ^2(x)}+e^{\frac {1}{e^2}} \text {li}(x)+\frac {1}{5} \int \frac {e^x x}{\log ^2(x)} \, dx-\frac {2}{5} \int \frac {e^x}{\log ^3(x)} \, dx-\frac {4}{5} \int \frac {e^x}{\log ^2(x)} \, dx+2 \int \frac {e^x}{x \log ^3(x)} \, dx-e^{\frac {1}{e^2}} \int \frac {1}{\log (x)} \, dx\\ &=-\frac {5 e^{\frac {1}{e^2}}}{\log ^2(x)}+\frac {e^{\frac {1}{e^2}} x}{\log ^2(x)}+\frac {1}{5} \int \frac {e^x x}{\log ^2(x)} \, dx-\frac {2}{5} \int \frac {e^x}{\log ^3(x)} \, dx-\frac {4}{5} \int \frac {e^x}{\log ^2(x)} \, dx+2 \int \frac {e^x}{x \log ^3(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 22, normalized size = 0.92 \begin {gather*} \frac {\left (5 e^{\frac {1}{e^2}}+e^x\right ) (-5+x)}{5 \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^(-2)*(50 - 10*x) + E^x*(10 - 2*x) + (5*E^E^(-2)*x + E^x*(-4*x + x^2))*Log[x])/(5*x*Log[x]^3),x]

[Out]

((5*E^E^(-2) + E^x)*(-5 + x))/(5*Log[x]^2)

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fricas [A]  time = 0.76, size = 21, normalized size = 0.88 \begin {gather*} \frac {{\left (x - 5\right )} e^{x} + 5 \, {\left (x - 5\right )} e^{\left (e^{\left (-2\right )}\right )}}{5 \, \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((x^2-4*x)*exp(x)+5*x*exp(1/exp(2)))*log(x)+(-2*x+10)*exp(x)+(-10*x+50)*exp(1/exp(2)))/x/log(x)
^3,x, algorithm="fricas")

[Out]

1/5*((x - 5)*e^x + 5*(x - 5)*e^(e^(-2)))/log(x)^2

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giac [A]  time = 0.14, size = 26, normalized size = 1.08 \begin {gather*} \frac {x e^{x} + 5 \, x e^{\left (e^{\left (-2\right )}\right )} - 5 \, e^{x} - 25 \, e^{\left (e^{\left (-2\right )}\right )}}{5 \, \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((x^2-4*x)*exp(x)+5*x*exp(1/exp(2)))*log(x)+(-2*x+10)*exp(x)+(-10*x+50)*exp(1/exp(2)))/x/log(x)
^3,x, algorithm="giac")

[Out]

1/5*(x*e^x + 5*x*e^(e^(-2)) - 5*e^x - 25*e^(e^(-2)))/log(x)^2

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maple [A]  time = 0.03, size = 27, normalized size = 1.12

method	result	size
risch	\(\frac {5 x \,{\mathrm e}^{{\mathrm e}^{-2}}+{\mathrm e}^{x} x -25 \,{\mathrm e}^{{\mathrm e}^{-2}}-5 \,{\mathrm e}^{x}}{5 \ln \relax (x )^{2}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(((x^2-4*x)*exp(x)+5*x*exp(1/exp(2)))*ln(x)+(-2*x+10)*exp(x)+(-10*x+50)*exp(1/exp(2)))/x/ln(x)^3,x,met
hod=_RETURNVERBOSE)

[Out]

1/5*(5*x*exp(exp(-2))+exp(x)*x-25*exp(exp(-2))-5*exp(x))/ln(x)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, e^{\left (e^{\left (-2\right )}\right )} \Gamma \left (-2, -\log \relax (x)\right ) + e^{\left (e^{\left (-2\right )}\right )} \int \frac {1}{\log \relax (x)}\,{d x} - \frac {5 \, x e^{\left (e^{\left (-2\right )}\right )} \log \relax (x) - {\left (x - 5\right )} e^{x}}{5 \, \log \relax (x)^{2}} - \frac {5 \, e^{\left (e^{\left (-2\right )}\right )}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((x^2-4*x)*exp(x)+5*x*exp(1/exp(2)))*log(x)+(-2*x+10)*exp(x)+(-10*x+50)*exp(1/exp(2)))/x/log(x)
^3,x, algorithm="maxima")

[Out]

2*e^(e^(-2))*gamma(-2, -log(x)) + e^(e^(-2))*integrate(1/log(x), x) - 1/5*(5*x*e^(e^(-2))*log(x) - (x - 5)*e^x
)/log(x)^2 - 5*e^(e^(-2))/log(x)^2

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mupad [B]  time = 4.24, size = 17, normalized size = 0.71 \begin {gather*} \frac {\left (5\,{\mathrm {e}}^{{\mathrm {e}}^{-2}}+{\mathrm {e}}^x\right )\,\left (x-5\right )}{5\,{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(exp(-2))*(10*x - 50))/5 + (exp(x)*(2*x - 10))/5 - (log(x)*(5*x*exp(exp(-2)) - exp(x)*(4*x - x^2)))/
5)/(x*log(x)^3),x)

[Out]

((5*exp(exp(-2)) + exp(x))*(x - 5))/(5*log(x)^2)

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sympy [A]  time = 0.28, size = 34, normalized size = 1.42 \begin {gather*} \frac {\left (x - 5\right ) e^{x}}{5 \log {\relax (x )}^{2}} + \frac {x e^{e^{-2}} - 5 e^{e^{-2}}}{\log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((x**2-4*x)*exp(x)+5*x*exp(1/exp(2)))*ln(x)+(-2*x+10)*exp(x)+(-10*x+50)*exp(1/exp(2)))/x/ln(x)*
*3,x)

[Out]

(x - 5)*exp(x)/(5*log(x)**2) + (x*exp(exp(-2)) - 5*exp(exp(-2)))/log(x)**2

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