3.17.30 $\int \frac{-1+e^{e^{e^{4x}}}(-20x+4e^{x}x+e^{e^{4x}+4x}(16x-4e^{x}x))+5x\log (x)}{e^{e^{e^{4x}}}(-20x+5e^{x}x)+5x\log (x)}dx$

Optimal. Leaf size=28 $$x-\frac{1}{5}\log (-e^{e^{e^{4x}}}(4-e^x)+\log (x))$$

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Rubi [F]  time = 44.36, 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 = {} $$\begin{array}{c}\int \frac{-1+e^{e^{e^{4x}}}(-20x+4e^{x}x+e^{e^{4x}+4x}(16x-4e^{x}x))+5x\log (x)}{e^{e^{e^{4x}}}(-20x+5e^{x}x)+5x\log (x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-1 + E^E^E^(4*x)*(-20*x + 4*E^x*x + E^(E^(4*x) + 4*x)*(16*x - 4*E^x*x)) + 5*x*Log[x])/(E^E^E^(4*x)*(-20*x
+ 5*E^x*x) + 5*x*Log[x]),x]

[Out]

-1/5*E^E^(4*x) + (4*x)/5 - (192*Defer[Int][E^(-2*E^E^(4*x) + E^(4*x))*Log[x]^2, x])/5 - (32*Defer[Int][E^(-2*E
^E^(4*x) + E^(4*x) + x)*Log[x]^2, x])/5 - (4*Defer[Int][E^(-2*E^E^(4*x) + E^(4*x) + 2*x)*Log[x]^2, x])/5 + (48
*Defer[Int][E^(-3*E^E^(4*x) + E^(4*x))*Log[x]^3, x])/5 + (4*Defer[Int][E^(-3*E^E^(4*x) + E^(4*x) + x)*Log[x]^3
, x])/5 - (4*Defer[Int][E^(-4*E^E^(4*x) + E^(4*x))*Log[x]^4, x])/5 - (4*Defer[Int][E^E^E^(4*x)/(E^E^E^(4*x)*(-
4 + E^x) + Log[x]), x])/5 - Defer[Int][1/(x*(E^E^E^(4*x)*(-4 + E^x) + Log[x])), x]/5 + Defer[Int][Log[x]/(E^E^
E^(4*x)*(-4 + E^x) + Log[x]), x]/5 + (1024*Defer[Int][(E^E^(4*x)*Log[x])/(E^E^E^(4*x)*(-4 + E^x) + Log[x]), x]
)/5 - (1024*Defer[Int][(E^(-E^E^(4*x) + E^(4*x))*Log[x]^2)/(E^E^E^(4*x)*(-4 + E^x) + Log[x]), x])/5 + (384*Def
er[Int][(E^(-2*E^E^(4*x) + E^(4*x))*Log[x]^3)/(E^E^E^(4*x)*(-4 + E^x) + Log[x]), x])/5 - (64*Defer[Int][(E^(-3
*E^E^(4*x) + E^(4*x))*Log[x]^4)/(E^E^E^(4*x)*(-4 + E^x) + Log[x]), x])/5 + (4*Defer[Int][(E^(-4*E^E^(4*x) + E^
(4*x))*Log[x]^5)/(E^E^E^(4*x)*(-4 + E^x) + Log[x]), x])/5 - (8*Defer[Int][Defer[Subst][Defer[Int][E^(-E^x^2 +
x^2), x], x, E^(2*x)]/x, x])/5 - (64*Defer[Int][Defer[Subst][Defer[Int][E^(-E^x^4 + x^4), x], x, E^x]/x, x])/5
- (64*Defer[Int][Defer[Subst][Defer[Int][E^(-E^x + x)/x, x], x, E^(4*x)]/x, x])/5 - (4*Defer[Int][Defer[Subst
][Defer[Int][E^(-E^x^4 + x^4)*x^2, x], x, E^x]/x, x])/5 + (8*Log[x]*Defer[Subst][Defer[Int][E^(-E^x^2 + x^2),
x], x, E^(2*x)])/5 + (64*Log[x]*Defer[Subst][Defer[Int][E^(-E^x^4 + x^4), x], x, E^x])/5 + (64*Log[x]*Defer[Su
bst][Defer[Int][E^(-E^x + x)/x, x], x, E^(4*x)])/5 + (4*Log[x]*Defer[Subst][Defer[Int][E^(-E^x^4 + x^4)*x^2, x
], x, E^x])/5

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{1-e^{e^{e^{4x}}}(-20x+4e^{x}x+e^{e^{4x}+4x}(16x-4e^{x}x))-5x\log (x)}{5x(4e^{e^{e^{4x}}}-e^{e^{e^{4x}}+x}-\log (x))}dx\\ & =\frac{1}{5}\int \frac{1-e^{e^{e^{4x}}}(-20x+4e^{x}x+e^{e^{4x}+4x}(16x-4e^{x}x))-5x\log (x)}{x(4e^{e^{e^{4x}}}-e^{e^{e^{4x}}+x}-\log (x))}dx\\ & =\frac{1}{5}\int (-4e^{e^{4x}+4x}+4e^{-e^{e^{4x}}+e^{4x}+3x}\log (x)+4e^{-2e^{e^{4x}}+e^{4x}+2x}(4e^{e^{e^{4x}}}-\log (x))\log (x)+4e^{-3e^{e^{4x}}+e^{4x}+x}{(4e^{e^{e^{4x}}}-\log (x))}^2\log (x)+4e^{-4e^{e^{4x}}}(e^{4e^{e^{4x}}}+64e^{3e^{e^{4x}}+e^{4x}}\log (x)-48e^{2e^{e^{4x}}+e^{4x}}{\log }^2(x)+12e^{e^{e^{4x}}+e^{4x}}{\log }^3(x)-e^{e^{4x}}{\log }^4(x))-\frac{e^{-4e^{e^{4x}}}(e^{4e^{e^{4x}}}+4e^{5e^{e^{4x}}}x-e^{4e^{e^{4x}}}x\log (x)-1024e^{4e^{e^{4x}}+e^{4x}}x\log (x)+1024e^{3e^{e^{4x}}+e^{4x}}x{\log }^2(x)-384e^{2e^{e^{4x}}+e^{4x}}x{\log }^3(x)+64e^{e^{e^{4x}}+e^{4x}}x{\log }^4(x)-4e^{e^{4x}}x{\log }^5(x))}{x(-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x))})dx\\ & =-(\frac{1}{5}\int \frac{e^{-4e^{e^{4x}}}(e^{4e^{e^{4x}}}+4e^{5e^{e^{4x}}}x-e^{4e^{e^{4x}}}x\log (x)-1024e^{4e^{e^{4x}}+e^{4x}}x\log (x)+1024e^{3e^{e^{4x}}+e^{4x}}x{\log }^2(x)-384e^{2e^{e^{4x}}+e^{4x}}x{\log }^3(x)+64e^{e^{e^{4x}}+e^{4x}}x{\log }^4(x)-4e^{e^{4x}}x{\log }^5(x))}{x(-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x))}dx)-\frac{4}{5}\int e^{e^{4x}+4x}dx+\frac{4}{5}\int e^{-e^{e^{4x}}+e^{4x}+3x}\log (x)dx+\frac{4}{5}\int e^{-2e^{e^{4x}}+e^{4x}+2x}(4e^{e^{e^{4x}}}-\log (x))\log (x)dx+\frac{4}{5}\int e^{-3e^{e^{4x}}+e^{4x}+x}{(4e^{e^{e^{4x}}}-\log (x))}^2\log (x)dx+\frac{4}{5}\int e^{-4e^{e^{4x}}}(e^{4e^{e^{4x}}}+64e^{3e^{e^{4x}}+e^{4x}}\log (x)-48e^{2e^{e^{4x}}+e^{4x}}{\log }^2(x)+12e^{e^{e^{4x}}+e^{4x}}{\log }^3(x)-e^{e^{4x}}{\log }^4(x))dx\\ & =-(\frac{1}{5}\int (\frac{4e^{e^{e^{4x}}}}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)}+\frac{1}{x(-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x))}-\frac{\log (x)}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)}-\frac{1024e^{e^{4x}}\log (x)}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)}+\frac{1024e^{-e^{e^{4x}}+e^{4x}}{\log }^2(x)}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)}-\frac{384e^{-2e^{e^{4x}}+e^{4x}}{\log }^3(x)}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)}+\frac{64e^{-3e^{e^{4x}}+e^{4x}}{\log }^4(x)}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)}-\frac{4e^{-4e^{e^{4x}}+e^{4x}}{\log }^5(x)}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)})dx)-\frac{1}{5}\mathrm{Subst}(\int e^{x}dx,x,e^{4x})+\frac{4}{5}\int (4e^{-e^{e^{4x}}+e^{4x}+2x}\log (x)-e^{-2e^{e^{4x}}+e^{4x}+2x}{\log }^2(x))dx+\frac{4}{5}\int (16e^{-e^{e^{4x}}+e^{4x}+x}\log (x)-8e^{-2e^{e^{4x}}+e^{4x}+x}{\log }^2(x)+e^{-3e^{e^{4x}}+e^{4x}+x}{\log }^3(x))dx+\frac{4}{5}\int (1+64e^{-e^{e^{4x}}+e^{4x}}\log (x)-48e^{-2e^{e^{4x}}+e^{4x}}{\log }^2(x)+12e^{-3e^{e^{4x}}+e^{4x}}{\log }^3(x)-e^{-4e^{e^{4x}}+e^{4x}}{\log }^4(x))dx-\frac{4}{5}\int \frac{\mathrm{Subst}(\int e^{-e^{x^4}+x^4}{x}^{2}dx,x,e^x)}{x}dx+\frac{1}{5}(4\log (x))\mathrm{Subst}(\int e^{-e^{x^4}+x^4}{x}^{2}dx,x,e^x)\\ & =-\frac{1}{5}{e}^{e^{4x}}+\frac{4x}{5}-\frac{1}{5}\int \frac{1}{x(-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x))}dx+\frac{1}{5}\int \frac{\log (x)}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)}dx-\frac{4}{5}\int e^{-2e^{e^{4x}}+e^{4x}+2x}{\log }^2(x)dx+\frac{4}{5}\int e^{-3e^{e^{4x}}+e^{4x}+x}{\log }^3(x)dx-\frac{4}{5}\int e^{-4e^{e^{4x}}+e^{4x}}{\log }^4(x)dx-\frac{4}{5}\int \frac{e^{e^{e^{4x}}}}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)}dx+\frac{4}{5}\int \frac{e^{-4e^{e^{4x}}+e^{4x}}{\log }^5(x)}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)}dx-\frac{4}{5}\int \frac{\mathrm{Subst}(\int e^{-e^{x^4}+x^4}{x}^{2}dx,x,e^x)}{x}dx+\frac{16}{5}\int e^{-e^{e^{4x}}+e^{4x}+2x}\log (x)dx-\frac{32}{5}\int e^{-2e^{e^{4x}}+e^{4x}+x}{\log }^2(x)dx+\frac{48}{5}\int e^{-3e^{e^{4x}}+e^{4x}}{\log }^3(x)dx+\frac{64}{5}\int e^{-e^{e^{4x}}+e^{4x}+x}\log (x)dx-\frac{64}{5}\int \frac{e^{-3e^{e^{4x}}+e^{4x}}{\log }^4(x)}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)}dx-\frac{192}{5}\int e^{-2e^{e^{4x}}+e^{4x}}{\log }^2(x)dx+\frac{256}{5}\int e^{-e^{e^{4x}}+e^{4x}}\log (x)dx+\frac{384}{5}\int \frac{e^{-2e^{e^{4x}}+e^{4x}}{\log }^3(x)}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)}dx+\frac{1024}{5}\int \frac{e^{e^{4x}}\log (x)}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)}dx-\frac{1024}{5}\int \frac{e^{-e^{e^{4x}}+e^{4x}}{\log }^2(x)}{-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)}dx+\frac{1}{5}(4\log (x))\mathrm{Subst}(\int e^{-e^{x^4}+x^4}{x}^{2}dx,x,e^x)\\ & =-\frac{1}{5}{e}^{e^{4x}}+\frac{4x}{5}-\frac{1}{5}\int \frac{1}{x(e^{e^{e^{4x}}}(-4+e^x)+\log (x))}dx+\frac{1}{5}\int \frac{\log (x)}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx-\frac{4}{5}\int e^{-2e^{e^{4x}}+e^{4x}+2x}{\log }^2(x)dx+\frac{4}{5}\int e^{-3e^{e^{4x}}+e^{4x}+x}{\log }^3(x)dx-\frac{4}{5}\int e^{-4e^{e^{4x}}+e^{4x}}{\log }^4(x)dx-\frac{4}{5}\int \frac{e^{e^{e^{4x}}}}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx+\frac{4}{5}\int \frac{e^{-4e^{e^{4x}}+e^{4x}}{\log }^5(x)}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx-\frac{4}{5}\int \frac{\mathrm{Subst}(\int e^{-e^{x^4}+x^4}{x}^{2}dx,x,e^x)}{x}dx-\frac{16}{5}\int \frac{\mathrm{Subst}(\int e^{-e^{x^2}+x^2}dx,x,e^{2x})}{2x}dx-\frac{32}{5}\int e^{-2e^{e^{4x}}+e^{4x}+x}{\log }^2(x)dx+\frac{48}{5}\int e^{-3e^{e^{4x}}+e^{4x}}{\log }^3(x)dx-\frac{64}{5}\int \frac{e^{-3e^{e^{4x}}+e^{4x}}{\log }^4(x)}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx-\frac{64}{5}\int \frac{\mathrm{Subst}(\int e^{-e^{x^4}+x^4}dx,x,e^x)}{x}dx-\frac{192}{5}\int e^{-2e^{e^{4x}}+e^{4x}}{\log }^2(x)dx-\frac{256}{5}\int \frac{\mathrm{Subst}(\int \frac{e^{-e^x+x}}{x}dx,x,e^{4x})}{4x}dx+\frac{384}{5}\int \frac{e^{-2e^{e^{4x}}+e^{4x}}{\log }^3(x)}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx+\frac{1024}{5}\int \frac{e^{e^{4x}}\log (x)}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx-\frac{1024}{5}\int \frac{e^{-e^{e^{4x}}+e^{4x}}{\log }^2(x)}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx+\frac{1}{5}(4\log (x))\mathrm{Subst}(\int e^{-e^{x^4}+x^4}{x}^{2}dx,x,e^x)+\frac{1}{5}(8\log (x))\mathrm{Subst}(\int e^{-e^{x^2}+x^2}dx,x,e^{2x})+\frac{1}{5}(64\log (x))\mathrm{Subst}(\int e^{-e^{x^4}+x^4}dx,x,e^x)+\frac{1}{5}(64\log (x))\mathrm{Subst}(\int \frac{e^{-e^x+x}}{x}dx,x,e^{4x})\\ & =-\frac{1}{5}{e}^{e^{4x}}+\frac{4x}{5}-\frac{1}{5}\int \frac{1}{x(e^{e^{e^{4x}}}(-4+e^x)+\log (x))}dx+\frac{1}{5}\int \frac{\log (x)}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx-\frac{4}{5}\int e^{-2e^{e^{4x}}+e^{4x}+2x}{\log }^2(x)dx+\frac{4}{5}\int e^{-3e^{e^{4x}}+e^{4x}+x}{\log }^3(x)dx-\frac{4}{5}\int e^{-4e^{e^{4x}}+e^{4x}}{\log }^4(x)dx-\frac{4}{5}\int \frac{e^{e^{e^{4x}}}}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx+\frac{4}{5}\int \frac{e^{-4e^{e^{4x}}+e^{4x}}{\log }^5(x)}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx-\frac{4}{5}\int \frac{\mathrm{Subst}(\int e^{-e^{x^4}+x^4}{x}^{2}dx,x,e^x)}{x}dx-\frac{8}{5}\int \frac{\mathrm{Subst}(\int e^{-e^{x^2}+x^2}dx,x,e^{2x})}{x}dx-\frac{32}{5}\int e^{-2e^{e^{4x}}+e^{4x}+x}{\log }^2(x)dx+\frac{48}{5}\int e^{-3e^{e^{4x}}+e^{4x}}{\log }^3(x)dx-\frac{64}{5}\int \frac{e^{-3e^{e^{4x}}+e^{4x}}{\log }^4(x)}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx-\frac{64}{5}\int \frac{\mathrm{Subst}(\int e^{-e^{x^4}+x^4}dx,x,e^x)}{x}dx-\frac{64}{5}\int \frac{\mathrm{Subst}(\int \frac{e^{-e^x+x}}{x}dx,x,e^{4x})}{x}dx-\frac{192}{5}\int e^{-2e^{e^{4x}}+e^{4x}}{\log }^2(x)dx+\frac{384}{5}\int \frac{e^{-2e^{e^{4x}}+e^{4x}}{\log }^3(x)}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx+\frac{1024}{5}\int \frac{e^{e^{4x}}\log (x)}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx-\frac{1024}{5}\int \frac{e^{-e^{e^{4x}}+e^{4x}}{\log }^2(x)}{e^{e^{e^{4x}}}(-4+e^x)+\log (x)}dx+\frac{1}{5}(4\log (x))\mathrm{Subst}(\int e^{-e^{x^4}+x^4}{x}^{2}dx,x,e^x)+\frac{1}{5}(8\log (x))\mathrm{Subst}(\int e^{-e^{x^2}+x^2}dx,x,e^{2x})+\frac{1}{5}(64\log (x))\mathrm{Subst}(\int e^{-e^{x^4}+x^4}dx,x,e^x)+\frac{1}{5}(64\log (x))\mathrm{Subst}(\int \frac{e^{-e^x+x}}{x}dx,x,e^{4x})\end{array}\end{array}$$

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Mathematica [A]  time = 0.99, size = 36, normalized size = 1.29 $$\begin{array}{c}\frac{1}{5}(5x-\log (-4e^{e^{e^{4x}}}+e^{e^{e^{4x}}+x}+\log (x)))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-1 + E^E^E^(4*x)*(-20*x + 4*E^x*x + E^(E^(4*x) + 4*x)*(16*x - 4*E^x*x)) + 5*x*Log[x])/(E^E^E^(4*x)*
(-20*x + 5*E^x*x) + 5*x*Log[x]),x]

[Out]

(5*x - Log[-4*E^E^E^(4*x) + E^(E^E^(4*x) + x) + Log[x]])/5

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fricas [A]  time = 0.88, size = 33, normalized size = 1.18 $$\begin{array}{c}x-\frac{1}{5}\log \left(\frac{(e^x-4)e^{\left(e^{\left(e^{\left(4x\right)}\right)}\right)}+\log (x)}{e^x-4}\right)-\frac{1}{5}\log (e^x-4)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(x)*x+16*x)*exp(4*x)*exp(exp(4*x))+4*exp(x)*x-20*x)*exp(exp(exp(4*x)))+5*x*log(x)-1)/((5*ex
p(x)*x-20*x)*exp(exp(exp(4*x)))+5*x*log(x)),x, algorithm="fricas")

[Out]

x - 1/5*log(((e^x - 4)*e^(e^(e^(4*x))) + log(x))/(e^x - 4)) - 1/5*log(e^x - 4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int -\frac{4((x{e}^x-4x)e^{(4x+e^{\left(4x\right)})}-x{e}^x+5x)e^{\left(e^{\left(e^{\left(4x\right)}\right)}\right)}-5x\log (x)+1}{5((x{e}^x-4x)e^{\left(e^{\left(e^{\left(4x\right)}\right)}\right)}+x\log (x))}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(x)*x+16*x)*exp(4*x)*exp(exp(4*x))+4*exp(x)*x-20*x)*exp(exp(exp(4*x)))+5*x*log(x)-1)/((5*ex
p(x)*x-20*x)*exp(exp(exp(4*x)))+5*x*log(x)),x, algorithm="giac")

[Out]

integrate(-1/5*(4*((x*e^x - 4*x)*e^(4*x + e^(4*x)) - x*e^x + 5*x)*e^(e^(e^(4*x))) - 5*x*log(x) + 1)/((x*e^x -
4*x)*e^(e^(e^(4*x))) + x*log(x)), x)

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maple [A]  time = 0.04, size = 29, normalized size = 1.04

method	result	size
risch	$x-\frac{\ln ({\mathrm{e}}^x-4)}{5}-\frac{\ln ({\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{4x}}}+\frac{\ln (x)}{{\mathrm{e}}^x-4})}{5}$	$29$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*exp(x)*x+16*x)*exp(4*x)*exp(exp(4*x))+4*exp(x)*x-20*x)*exp(exp(exp(4*x)))+5*x*ln(x)-1)/((5*exp(x)*x-
20*x)*exp(exp(exp(4*x)))+5*x*ln(x)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.63, size = 33, normalized size = 1.18 $$\begin{array}{c}x-\frac{1}{5}\log \left(\frac{(e^x-4)e^{\left(e^{\left(e^{\left(4x\right)}\right)}\right)}+\log (x)}{e^x-4}\right)-\frac{1}{5}\log (e^x-4)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(x)*x+16*x)*exp(4*x)*exp(exp(4*x))+4*exp(x)*x-20*x)*exp(exp(exp(4*x)))+5*x*log(x)-1)/((5*ex
p(x)*x-20*x)*exp(exp(exp(4*x)))+5*x*log(x)),x, algorithm="maxima")

[Out]

x - 1/5*log(((e^x - 4)*e^(e^(e^(4*x))) + log(x))/(e^x - 4)) - 1/5*log(e^x - 4)

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mupad [B]  time = 1.28, size = 39, normalized size = 1.39 $$\begin{array}{c}x-\frac{\ln \left(\frac{\ln (x)-4{\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{4x}}}+{\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{4x}}}{\mathrm{e}}^x}{{\mathrm{e}}^x-4}\right)}{5}-\frac{\ln ({\mathrm{e}}^x-4)}{5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(exp(4*x)))*(4*x*exp(x) - 20*x + exp(4*x)*exp(exp(4*x))*(16*x - 4*x*exp(x))) + 5*x*log(x) - 1)/(e
xp(exp(exp(4*x)))*(20*x - 5*x*exp(x)) - 5*x*log(x)),x)

[Out]

x - log((log(x) - 4*exp(exp(exp(4*x))) + exp(exp(exp(4*x)))*exp(x))/(exp(x) - 4))/5 - log(exp(x) - 4)/5

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sympy [A]  time = 1.14, size = 29, normalized size = 1.04 $$\begin{array}{c}x-\frac{\log (e^x-4)}{5}-\frac{\log (e^{e^{e^{4x}}}+\frac{\log (x)}{e^x-4})}{5}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(x)*x+16*x)*exp(4*x)*exp(exp(4*x))+4*exp(x)*x-20*x)*exp(exp(exp(4*x)))+5*x*ln(x)-1)/((5*exp
(x)*x-20*x)*exp(exp(exp(4*x)))+5*x*ln(x)),x)

[Out]

x - log(exp(x) - 4)/5 - log(exp(exp(exp(4*x))) + log(x)/(exp(x) - 4))/5

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