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

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

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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 = {} $$\begin{array}{c}\int \frac{e^{\frac{1}{e^2}}(50-10x)+e^x(10-2x)+(5e^{\frac{1}{e^2}}x+e^x(-4x+x^2))\log (x)}{5x{\log }^3(x)}dx\end{array}$$

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{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{5}\int \frac{e^{\frac{1}{e^2}}(50-10x)+e^x(10-2x)+(5e^{\frac{1}{e^2}}x+e^x(-4x+x^2))\log (x)}{x{\log }^3(x)}dx\\ & =\frac{1}{5}\int (\frac{5e^{\frac{1}{e^2}}(10-2x+x\log (x))}{x{\log }^3(x)}+\frac{e^x(10-2x-4x\log (x)+x^2\log (x))}{x{\log }^3(x)})dx\\ & =\frac{1}{5}\int \frac{e^x(10-2x-4x\log (x)+x^2\log (x))}{x{\log }^3(x)}dx+e^{\frac{1}{e^2}}\int \frac{10-2x+x\log (x)}{x{\log }^3(x)}dx\\ & =\frac{1}{5}\int (-\frac{2e^x(-5+x)}{x{\log }^3(x)}+\frac{e^x(-4+x)}{{\log }^2(x)})dx+e^{\frac{1}{e^2}}\int (-\frac{2(-5+x)}{x{\log }^3(x)}+\frac{1}{{\log }^2(x)})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(2e^{\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 (-\frac{4e^x}{{\log }^2(x)}+\frac{e^{x}x}{{\log }^2(x)})dx-\frac{2}{5}\int (\frac{e^x}{{\log }^3(x)}-\frac{5e^x}{x{\log }^3(x)})dx+e^{\frac{1}{e^2}}\int \frac{1}{\log (x)}dx-\left(2e^{\frac{1}{e^2}}\right)\int (\frac{1}{{\log }^3(x)}-\frac{5}{x{\log }^3(x)})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(2e^{\frac{1}{e^2}}\right)\int \frac{1}{{\log }^3(x)}dx+\left(10e^{\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(10e^{\frac{1}{e^2}}\right)\mathrm{Subst}(\int \frac{1}{x^3}dx,x,\log (x))\\ & =-\frac{5e^{\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{5e^{\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{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{(x-5)e^x+5(x-5)e^{\left(e^{(-2)}\right)}}{5\log (x{)}^2}\end{array}$$

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 $$\begin{array}{c}\frac{x{e}^x+5x{e}^{\left(e^{(-2)}\right)}-5e^x-25e^{\left(e^{(-2)}\right)}}{5\log (x{)}^2}\end{array}$$

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	$\frac{5x{\mathrm{e}}^{{\mathrm{e}}^{-2}}+{\mathrm{e}}^{x}x-25{\mathrm{e}}^{{\mathrm{e}}^{-2}}-5{\mathrm{e}}^x}{5\ln (x{)}^2}$	$27$

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 $$\begin{array}{c}2e^{\left(e^{(-2)}\right)}\mathrm{\Gamma }(-2,-\log (x))+e^{\left(e^{(-2)}\right)}\int \frac{1}{\log (x)}dx-\frac{5x{e}^{\left(e^{(-2)}\right)}\log (x)-(x-5)e^x}{5\log (x{)}^2}-\frac{5e^{\left(e^{(-2)}\right)}}{\log (x{)}^2}\end{array}$$

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 $$\begin{array}{c}\frac{(5{\mathrm{e}}^{{\mathrm{e}}^{-2}}+{\mathrm{e}}^x)(x-5)}{5{\ln (x)}^2}\end{array}$$

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 $$\begin{array}{c}\frac{(x-5)e^x}{5\log {(x)}^2}+\frac{x{e}^{e^{-2}}-5e^{e^{-2}}}{\log {(x)}^2}\end{array}$$

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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