3.73.73 $\int \frac{-e^{e^{\frac{1}{x}}}{x}^2+e^{e^{\frac{1}{x}}}(-50x-2x^2+e^{\frac{1}{x}}(25+x)+(e^{\frac{1}{x}}(-25-x)+50x+2x^2)\log (25+x))\log (1-\log (25+x))}{(-50-2x+(50+2x)\log (25+x)){\log }^2(1-\log (25+x))}dx$

Optimal. Leaf size=25 $$\frac{e^{e^{\frac{1}{x}}}{x}^2}{2\log (1-\log (25+x))}$$

________________________________________________________________________________________

Rubi [F]  time = 4.94, 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 = {} $$\begin{array}{c}\int \frac{-e^{e^{\frac{1}{x}}}{x}^2+e^{e^{\frac{1}{x}}}(-50x-2x^2+e^{\frac{1}{x}}(25+x)+(e^{\frac{1}{x}}(-25-x)+50x+2x^2)\log (25+x))\log (1-\log (25+x))}{(-50-2x+(50+2x)\log (25+x)){\log }^2(1-\log (25+x))}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-(E^E^x^(-1)*x^2) + E^E^x^(-1)*(-50*x - 2*x^2 + E^x^(-1)*(25 + x) + (E^x^(-1)*(-25 - x) + 50*x + 2*x^2)*L
og[25 + x])*Log[1 - Log[25 + x]])/((-50 - 2*x + (50 + 2*x)*Log[25 + x])*Log[1 - Log[25 + x]]^2),x]

[Out]

(25*Defer[Int][E^E^x^(-1)/((-1 + Log[25 + x])*Log[1 - Log[25 + x]]^2), x])/2 - Defer[Int][(E^E^x^(-1)*x)/((-1
+ Log[25 + x])*Log[1 - Log[25 + x]]^2), x]/2 - (625*Defer[Int][E^E^x^(-1)/((25 + x)*(-1 + Log[25 + x])*Log[1 -
Log[25 + x]]^2), x])/2 - Defer[Int][E^(E^x^(-1) + x^(-1))/Log[1 - Log[25 + x]], x]/2 + Defer[Int][(E^E^x^(-1)
*x)/Log[1 - Log[25 + x]], x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^{e^{\frac{1}{x}}}(-\frac{x^2}{(25+x)(-1+\log (25+x))}-(e^{\frac{1}{x}}-2x)\log (1-\log (25+x)))}{2{\log }^2(1-\log (25+x))}dx\\ & =\frac{1}{2}\int \frac{e^{e^{\frac{1}{x}}}(-\frac{x^2}{(25+x)(-1+\log (25+x))}-(e^{\frac{1}{x}}-2x)\log (1-\log (25+x)))}{{\log }^2(1-\log (25+x))}dx\\ & =\frac{1}{2}\int (-\frac{e^{e^{\frac{1}{x}}+\frac{1}{x}}}{\log (1-\log (25+x))}+\frac{e^{e^{\frac{1}{x}}}x(-x-50\log (1-\log (25+x))-2x\log (1-\log (25+x))+50\log (25+x)\log (1-\log (25+x))+2x\log (25+x)\log (1-\log (25+x)))}{(25+x)(-1+\log (25+x)){\log }^2(1-\log (25+x))})dx\\ & =-(\frac{1}{2}\int \frac{e^{e^{\frac{1}{x}}+\frac{1}{x}}}{\log (1-\log (25+x))}dx)+\frac{1}{2}\int \frac{e^{e^{\frac{1}{x}}}x(-x-50\log (1-\log (25+x))-2x\log (1-\log (25+x))+50\log (25+x)\log (1-\log (25+x))+2x\log (25+x)\log (1-\log (25+x)))}{(25+x)(-1+\log (25+x)){\log }^2(1-\log (25+x))}dx\\ & =-(\frac{1}{2}\int \frac{e^{e^{\frac{1}{x}}+\frac{1}{x}}}{\log (1-\log (25+x))}dx)+\frac{1}{2}\int \frac{e^{e^{\frac{1}{x}}}x(x-2(25+x)(-1+\log (25+x))\log (1-\log (25+x)))}{(25+x)(1-\log (25+x)){\log }^2(1-\log (25+x))}dx\\ & =\frac{1}{2}\int (-\frac{e^{e^{\frac{1}{x}}}{x}^2}{(25+x)(-1+\log (25+x)){\log }^2(1-\log (25+x))}+\frac{2e^{e^{\frac{1}{x}}}x}{\log (1-\log (25+x))})dx-\frac{1}{2}\int \frac{e^{e^{\frac{1}{x}}+\frac{1}{x}}}{\log (1-\log (25+x))}dx\\ & =-(\frac{1}{2}\int \frac{e^{e^{\frac{1}{x}}}{x}^2}{(25+x)(-1+\log (25+x)){\log }^2(1-\log (25+x))}dx)-\frac{1}{2}\int \frac{e^{e^{\frac{1}{x}}+\frac{1}{x}}}{\log (1-\log (25+x))}dx+\int \frac{e^{e^{\frac{1}{x}}}x}{\log (1-\log (25+x))}dx\\ & =-(\frac{1}{2}\int (-\frac{25e^{e^{\frac{1}{x}}}}{(-1+\log (25+x)){\log }^2(1-\log (25+x))}+\frac{e^{e^{\frac{1}{x}}}x}{(-1+\log (25+x)){\log }^2(1-\log (25+x))}+\frac{625e^{e^{\frac{1}{x}}}}{(25+x)(-1+\log (25+x)){\log }^2(1-\log (25+x))})dx)-\frac{1}{2}\int \frac{e^{e^{\frac{1}{x}}+\frac{1}{x}}}{\log (1-\log (25+x))}dx+\int \frac{e^{e^{\frac{1}{x}}}x}{\log (1-\log (25+x))}dx\\ & =-(\frac{1}{2}\int \frac{e^{e^{\frac{1}{x}}}x}{(-1+\log (25+x)){\log }^2(1-\log (25+x))}dx)-\frac{1}{2}\int \frac{e^{e^{\frac{1}{x}}+\frac{1}{x}}}{\log (1-\log (25+x))}dx+\frac{25}{2}\int \frac{e^{e^{\frac{1}{x}}}}{(-1+\log (25+x)){\log }^2(1-\log (25+x))}dx-\frac{625}{2}\int \frac{e^{e^{\frac{1}{x}}}}{(25+x)(-1+\log (25+x)){\log }^2(1-\log (25+x))}dx+\int \frac{e^{e^{\frac{1}{x}}}x}{\log (1-\log (25+x))}dx\end{array}\end{array}$$

________________________________________________________________________________________

Mathematica [A]  time = 1.33, size = 25, normalized size = 1.00 $$\begin{array}{c}\frac{e^{e^{\frac{1}{x}}}{x}^2}{2\log (1-\log (25+x))}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-(E^E^x^(-1)*x^2) + E^E^x^(-1)*(-50*x - 2*x^2 + E^x^(-1)*(25 + x) + (E^x^(-1)*(-25 - x) + 50*x + 2*
x^2)*Log[25 + x])*Log[1 - Log[25 + x]])/((-50 - 2*x + (50 + 2*x)*Log[25 + x])*Log[1 - Log[25 + x]]^2),x]

[Out]

(E^E^x^(-1)*x^2)/(2*Log[1 - Log[25 + x]])

________________________________________________________________________________________

fricas [A]  time = 0.91, size = 21, normalized size = 0.84 $$\begin{array}{c}\frac{x^2{e}^{\left(e^{\frac{1}{x}}\right)}}{2\log (-\log (x+25)+1)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x-25)*exp(1/x)+2*x^2+50*x)*log(x+25)+(x+25)*exp(1/x)-2*x^2-50*x)*exp(exp(1/x))*log(-log(x+25)+1
)-x^2*exp(exp(1/x)))/((2*x+50)*log(x+25)-2*x-50)/log(-log(x+25)+1)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.27, size = 21, normalized size = 0.84 $$\begin{array}{c}\frac{x^2{e}^{\left(e^{\frac{1}{x}}\right)}}{2\log (-\log (x+25)+1)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x-25)*exp(1/x)+2*x^2+50*x)*log(x+25)+(x+25)*exp(1/x)-2*x^2-50*x)*exp(exp(1/x))*log(-log(x+25)+1
)-x^2*exp(exp(1/x)))/((2*x+50)*log(x+25)-2*x-50)/log(-log(x+25)+1)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.12, size = 22, normalized size = 0.88

method	result	size
risch	$\frac{x^2{\mathrm{e}}^{{\mathrm{e}}^{\frac{1}{x}}}}{2\ln (-\ln (x+25)+1)}$	$22$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-x-25)*exp(1/x)+2*x^2+50*x)*ln(x+25)+(x+25)*exp(1/x)-2*x^2-50*x)*exp(exp(1/x))*ln(-ln(x+25)+1)-x^2*exp
(exp(1/x)))/((2*x+50)*ln(x+25)-2*x-50)/ln(-ln(x+25)+1)^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 21, normalized size = 0.84 $$\begin{array}{c}\frac{x^2{e}^{\left(e^{\frac{1}{x}}\right)}}{2\log (-\log (x+25)+1)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x-25)*exp(1/x)+2*x^2+50*x)*log(x+25)+(x+25)*exp(1/x)-2*x^2-50*x)*exp(exp(1/x))*log(-log(x+25)+1
)-x^2*exp(exp(1/x)))/((2*x+50)*log(x+25)-2*x-50)/log(-log(x+25)+1)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 4.99, size = 21, normalized size = 0.84 $$\begin{array}{c}\frac{x^2{\mathrm{e}}^{{\mathrm{e}}^{1/x}}}{2\ln (1-\ln (x+25))}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*exp(exp(1/x)) + exp(exp(1/x))*log(1 - log(x + 25))*(50*x - exp(1/x)*(x + 25) - log(x + 25)*(50*x - ex
p(1/x)*(x + 25) + 2*x^2) + 2*x^2))/(log(1 - log(x + 25))^2*(2*x - log(x + 25)*(2*x + 50) + 50)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.47, size = 19, normalized size = 0.76 $$\begin{array}{c}\frac{x^2{e}^{e^{\frac{1}{x}}}}{2\log (1-\log (x+25))}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-x-25)*exp(1/x)+2*x**2+50*x)*ln(x+25)+(x+25)*exp(1/x)-2*x**2-50*x)*exp(exp(1/x))*ln(-ln(x+25)+1)
-x**2*exp(exp(1/x)))/((2*x+50)*ln(x+25)-2*x-50)/ln(-ln(x+25)+1)**2,x)

[Out]

x**2*exp(exp(1/x))/(2*log(1 - log(x + 25)))

________________________________________________________________________________________