3.41.62 $\int \frac{e^{9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+\frac{5x}{\log (e^{e^x}-\log (x))}}(45-45e^{e^x+x}x+(45e^{e^x}-45\log (x))\log (e^{e^x}-\log (x)))}{(e^{e^x}-\log (x)){\log }^2(e^{e^x}-\log (x))}dx$

Optimal. Leaf size=22 $$e^{9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}}$$

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Rubi [F]  time = 4.90, 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 = {} $$\begin{array}{c}\int \frac{\exp (9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+\frac{5x}{\log (e^{e^x}-\log (x))})(45-45e^{e^x+x}x+(45e^{e^x}-45\log (x))\log (e^{e^x}-\log (x)))}{(e^{e^x}-\log (x)){\log }^2(e^{e^x}-\log (x))}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(E^(9*E^((5*x)/Log[E^E^x - Log[x]]) + (5*x)/Log[E^E^x - Log[x]])*(45 - 45*E^(E^x + x)*x + (45*E^E^x - 45*L
og[x])*Log[E^E^x - Log[x]]))/((E^E^x - Log[x])*Log[E^E^x - Log[x]]^2),x]

[Out]

45*Defer[Int][E^(9*E^((5*x)/Log[E^E^x - Log[x]]) + (5*x)/Log[E^E^x - Log[x]])/((E^E^x - Log[x])*Log[E^E^x - Lo
g[x]]^2), x] - 45*Defer[Int][(E^(E^x + 9*E^((5*x)/Log[E^E^x - Log[x]]) + x + (5*x)/Log[E^E^x - Log[x]])*x)/((E
^E^x - Log[x])*Log[E^E^x - Log[x]]^2), x] + 45*Defer[Int][E^(9*E^((5*x)/Log[E^E^x - Log[x]]) + (5*x)/Log[E^E^x
- Log[x]])/Log[E^E^x - Log[x]], x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-\frac{45\exp (e^x+9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+x+\frac{5x}{\log (e^{e^x}-\log (x))})x}{(e^{e^x}-\log (x)){\log }^2(e^{e^x}-\log (x))}+\frac{45\exp (9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+\frac{5x}{\log (e^{e^x}-\log (x))})(1+e^{e^x}\log (e^{e^x}-\log (x))-\log (x)\log (e^{e^x}-\log (x)))}{(e^{e^x}-\log (x)){\log }^2(e^{e^x}-\log (x))})dx\\ & =-(45\int \frac{\exp (e^x+9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+x+\frac{5x}{\log (e^{e^x}-\log (x))})x}{(e^{e^x}-\log (x)){\log }^2(e^{e^x}-\log (x))}dx)+45\int \frac{\exp (9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+\frac{5x}{\log (e^{e^x}-\log (x))})(1+e^{e^x}\log (e^{e^x}-\log (x))-\log (x)\log (e^{e^x}-\log (x)))}{(e^{e^x}-\log (x)){\log }^2(e^{e^x}-\log (x))}dx\\ & =-(45\int \frac{\exp (e^x+9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+x+\frac{5x}{\log (e^{e^x}-\log (x))})x}{(e^{e^x}-\log (x)){\log }^2(e^{e^x}-\log (x))}dx)+45\int \frac{\exp (9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+\frac{5x}{\log (e^{e^x}-\log (x))})(1+(e^{e^x}-\log (x))\log (e^{e^x}-\log (x)))}{(e^{e^x}-\log (x)){\log }^2(e^{e^x}-\log (x))}dx\\ & =45\int (\frac{\exp (9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+\frac{5x}{\log (e^{e^x}-\log (x))})}{(e^{e^x}-\log (x)){\log }^2(e^{e^x}-\log (x))}+\frac{\exp (9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+\frac{5x}{\log (e^{e^x}-\log (x))})}{\log (e^{e^x}-\log (x))})dx-45\int \frac{\exp (e^x+9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+x+\frac{5x}{\log (e^{e^x}-\log (x))})x}{(e^{e^x}-\log (x)){\log }^2(e^{e^x}-\log (x))}dx\\ & =45\int \frac{\exp (9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+\frac{5x}{\log (e^{e^x}-\log (x))})}{(e^{e^x}-\log (x)){\log }^2(e^{e^x}-\log (x))}dx-45\int \frac{\exp (e^x+9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+x+\frac{5x}{\log (e^{e^x}-\log (x))})x}{(e^{e^x}-\log (x)){\log }^2(e^{e^x}-\log (x))}dx+45\int \frac{\exp (9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}+\frac{5x}{\log (e^{e^x}-\log (x))})}{\log (e^{e^x}-\log (x))}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.11, size = 22, normalized size = 1.00 $$\begin{array}{c}{e}^{9e^{\frac{5x}{\log (e^{e^x}-\log (x))}}}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(9*E^((5*x)/Log[E^E^x - Log[x]]) + (5*x)/Log[E^E^x - Log[x]])*(45 - 45*E^(E^x + x)*x + (45*E^E^x
- 45*Log[x])*Log[E^E^x - Log[x]]))/((E^E^x - Log[x])*Log[E^E^x - Log[x]]^2),x]

[Out]

E^(9*E^((5*x)/Log[E^E^x - Log[x]]))

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fricas [B]  time = 0.66, size = 102, normalized size = 4.64 $$\begin{array}{c}{e}^{(\frac{9e^{\left(\frac{5x}{\log (-(e^x\log (x)-e^{(x+e^x)})e^{(-x)})}\right)}\log (-(e^x\log (x)-e^{(x+e^x)})e^{(-x)})+5x}{\log (-(e^x\log (x)-e^{(x+e^x)})e^{(-x)})}-\frac{5x}{\log (-(e^x\log (x)-e^{(x+e^x)})e^{(-x)})})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((45*exp(exp(x))-45*log(x))*log(exp(exp(x))-log(x))-45*x*exp(x)*exp(exp(x))+45)*exp(5*x/log(exp(exp(
x))-log(x)))*exp(9*exp(5*x/log(exp(exp(x))-log(x))))/(exp(exp(x))-log(x))/log(exp(exp(x))-log(x))^2,x, algorit
hm="fricas")

[Out]

e^((9*e^(5*x/log(-(e^x*log(x) - e^(x + e^x))*e^(-x)))*log(-(e^x*log(x) - e^(x + e^x))*e^(-x)) + 5*x)/log(-(e^x
*log(x) - e^(x + e^x))*e^(-x)) - 5*x/log(-(e^x*log(x) - e^(x + e^x))*e^(-x)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\mathit{undef}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((45*exp(exp(x))-45*log(x))*log(exp(exp(x))-log(x))-45*x*exp(x)*exp(exp(x))+45)*exp(5*x/log(exp(exp(
x))-log(x)))*exp(9*exp(5*x/log(exp(exp(x))-log(x))))/(exp(exp(x))-log(x))/log(exp(exp(x))-log(x))^2,x, algorit
hm="giac")

[Out]

undef

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maple [A]  time = 0.08, size = 19, normalized size = 0.86

method	result	size
risch	${\mathrm{e}}^{9{\mathrm{e}}^{\frac{5x}{\ln ({\mathrm{e}}^{{\mathrm{e}}^x}-\ln (x))}}}$	$19$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((45*exp(exp(x))-45*ln(x))*ln(exp(exp(x))-ln(x))-45*x*exp(x)*exp(exp(x))+45)*exp(5*x/ln(exp(exp(x))-ln(x))
)*exp(9*exp(5*x/ln(exp(exp(x))-ln(x))))/(exp(exp(x))-ln(x))/ln(exp(exp(x))-ln(x))^2,x,method=_RETURNVERBOSE)

[Out]

exp(9*exp(5*x/ln(exp(exp(x))-ln(x))))

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maxima [A]  time = 0.78, size = 18, normalized size = 0.82 $$\begin{array}{c}{e}^{\left(9e^{\left(\frac{5x}{\log (e^{\left(e^x\right)}-\log (x))}\right)}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((45*exp(exp(x))-45*log(x))*log(exp(exp(x))-log(x))-45*x*exp(x)*exp(exp(x))+45)*exp(5*x/log(exp(exp(
x))-log(x)))*exp(9*exp(5*x/log(exp(exp(x))-log(x))))/(exp(exp(x))-log(x))/log(exp(exp(x))-log(x))^2,x, algorit
hm="maxima")

[Out]

e^(9*e^(5*x/log(e^(e^x) - log(x))))

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mupad [B]  time = 3.37, size = 18, normalized size = 0.82 $$\begin{array}{c}{\mathrm{e}}^{9{\mathrm{e}}^{\frac{5x}{\ln ({\mathrm{e}}^{{\mathrm{e}}^x}-\ln (x))}}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(9*exp((5*x)/log(exp(exp(x)) - log(x))))*exp((5*x)/log(exp(exp(x)) - log(x)))*(log(exp(exp(x)) - log(x
))*(45*exp(exp(x)) - 45*log(x)) - 45*x*exp(exp(x))*exp(x) + 45))/(log(exp(exp(x)) - log(x))^2*(exp(exp(x)) - l
og(x))),x)

[Out]

exp(9*exp((5*x)/log(exp(exp(x)) - log(x))))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\text{Timed out}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((45*exp(exp(x))-45*ln(x))*ln(exp(exp(x))-ln(x))-45*x*exp(x)*exp(exp(x))+45)*exp(5*x/ln(exp(exp(x))-
ln(x)))*exp(9*exp(5*x/ln(exp(exp(x))-ln(x))))/(exp(exp(x))-ln(x))/ln(exp(exp(x))-ln(x))**2,x)

[Out]

Timed out

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