3.69.17 $\int \frac{e^{-2+e^{5e^x}}(-50x^2+50e^{-3+x}{x}^2)+e^{-4+2e^{5e^x}}(50e^{-3+x}x-50x^2+e^{5e^x+x}(250e^{-3+x}{x}^2-250x^3))\log (e^{-3+x}-x)\log (\log (e^{-3+x}-x))+(-50x^2+50e^{-3+x}{x}^2+e^{-2+e^{5e^x}}(100e^{-3+x}x-100x^2+e^{5e^x+x}(250e^{-3+x}{x}^2-250x^3))\log (e^{-3+x}-x)\log (\log (e^{-3+x}-x)))\log (\log (\log (e^{-3+x}-x)))+(50e^{-3+x}x-50x^2)\log (e^{-3+x}-x)\log (\log (e^{-3+x}-x)){\log }^2(\log (\log (e^{-3+x}-x)))}{(e^{-3+x}-x)\log (e^{-3+x}-x)\log (\log (e^{-3+x}-x))}dx$

Optimal. Leaf size=31 $$25x^2{(e^{-2+e^{5e^x}}+\log (\log (\log (e^{-3+x}-x))))}^2$$

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Rubi [A]  time = 3.74, antiderivative size = 36, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, integrand size = 293, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.010, Rules used = {6688, 12, 6687} $$\begin{array}{c}\frac{25x^2{(e^{e^{5e^x}}+e^2\log (\log (\log (e^{x-3}-x))))}^2}{e^4}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^(-2 + E^(5*E^x))*(-50*x^2 + 50*E^(-3 + x)*x^2) + E^(-4 + 2*E^(5*E^x))*(50*E^(-3 + x)*x - 50*x^2 + E^(5*
E^x + x)*(250*E^(-3 + x)*x^2 - 250*x^3))*Log[E^(-3 + x) - x]*Log[Log[E^(-3 + x) - x]] + (-50*x^2 + 50*E^(-3 +
x)*x^2 + E^(-2 + E^(5*E^x))*(100*E^(-3 + x)*x - 100*x^2 + E^(5*E^x + x)*(250*E^(-3 + x)*x^2 - 250*x^3))*Log[E^
(-3 + x) - x]*Log[Log[E^(-3 + x) - x]])*Log[Log[Log[E^(-3 + x) - x]]] + (50*E^(-3 + x)*x - 50*x^2)*Log[E^(-3 +
x) - x]*Log[Log[E^(-3 + x) - x]]*Log[Log[Log[E^(-3 + x) - x]]]^2)/((E^(-3 + x) - x)*Log[E^(-3 + x) - x]*Log[L
og[E^(-3 + x) - x]]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6687

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{50x(e^{e^{5e^x}}+e^2\log (\log (\log (e^{-3+x}-x))))(-e^{5}x+e^{2+x}x+(e^x-e^{3}x)\log (e^{-3+x}-x)\log (\log (e^{-3+x}-x))(e^{e^{5e^x}}(1+5e^{5e^x+x}x)+e^2\log (\log (\log (e^{-3+x}-x)))))}{e^4(e^x-e^{3}x)\log (e^{-3+x}-x)\log (\log (e^{-3+x}-x))}dx\\ & =\frac{50\int \frac{x(e^{e^{5e^x}}+e^2\log (\log (\log (e^{-3+x}-x))))(-e^{5}x+e^{2+x}x+(e^x-e^{3}x)\log (e^{-3+x}-x)\log (\log (e^{-3+x}-x))(e^{e^{5e^x}}(1+5e^{5e^x+x}x)+e^2\log (\log (\log (e^{-3+x}-x)))))}{(e^x-e^{3}x)\log (e^{-3+x}-x)\log (\log (e^{-3+x}-x))}dx}{e^4}\\ & =\frac{25x^2{(e^{e^{5e^x}}+e^2\log (\log (\log (e^{-3+x}-x))))}^2}{e^4}\end{array}\end{array}$$

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Mathematica [A]  time = 0.30, size = 36, normalized size = 1.16 $$\begin{array}{c}\frac{25x^2{(e^{e^{5e^x}}+e^2\log (\log (\log (e^{-3+x}-x))))}^2}{e^4}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2 + E^(5*E^x))*(-50*x^2 + 50*E^(-3 + x)*x^2) + E^(-4 + 2*E^(5*E^x))*(50*E^(-3 + x)*x - 50*x^2 +
E^(5*E^x + x)*(250*E^(-3 + x)*x^2 - 250*x^3))*Log[E^(-3 + x) - x]*Log[Log[E^(-3 + x) - x]] + (-50*x^2 + 50*E^
(-3 + x)*x^2 + E^(-2 + E^(5*E^x))*(100*E^(-3 + x)*x - 100*x^2 + E^(5*E^x + x)*(250*E^(-3 + x)*x^2 - 250*x^3))*
Log[E^(-3 + x) - x]*Log[Log[E^(-3 + x) - x]])*Log[Log[Log[E^(-3 + x) - x]]] + (50*E^(-3 + x)*x - 50*x^2)*Log[E
^(-3 + x) - x]*Log[Log[E^(-3 + x) - x]]*Log[Log[Log[E^(-3 + x) - x]]]^2)/((E^(-3 + x) - x)*Log[E^(-3 + x) - x]
*Log[Log[E^(-3 + x) - x]]),x]

[Out]

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

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fricas [B]  time = 0.62, size = 87, normalized size = 2.81 $$\begin{array}{c}50x^2{e}^{\left((e^{(x+5e^x)}-2e^x)e^{(-x)}\right)}\log (\log (\log (-(x{e}^3-e^x)e^{(-3)})))+25x^2\log {(\log (\log (-(x{e}^3-e^x)e^{(-3)})))}^2+25x^2{e}^{\left(2(e^{(x+5e^x)}-2e^x)e^{(-x)}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x*exp(x-3)-50*x^2)*log(exp(x-3)-x)*log(log(exp(x-3)-x))*log(log(log(exp(x-3)-x)))^2+(((250*x^2*
exp(x-3)-250*x^3)*exp(x)*exp(5*exp(x))+100*x*exp(x-3)-100*x^2)*log(exp(x-3)-x)*exp(exp(5*exp(x))-2)*log(log(ex
p(x-3)-x))+50*x^2*exp(x-3)-50*x^2)*log(log(log(exp(x-3)-x)))+((250*x^2*exp(x-3)-250*x^3)*exp(x)*exp(5*exp(x))+
50*x*exp(x-3)-50*x^2)*log(exp(x-3)-x)*exp(exp(5*exp(x))-2)^2*log(log(exp(x-3)-x))+(50*x^2*exp(x-3)-50*x^2)*exp
(exp(5*exp(x))-2))/(exp(x-3)-x)/log(exp(x-3)-x)/log(log(exp(x-3)-x)),x, algorithm="fricas")

[Out]

50*x^2*e^((e^(x + 5*e^x) - 2*e^x)*e^(-x))*log(log(log(-(x*e^3 - e^x)*e^(-3)))) + 25*x^2*log(log(log(-(x*e^3 -
e^x)*e^(-3))))^2 + 25*x^2*e^(2*(e^(x + 5*e^x) - 2*e^x)*e^(-x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{50((x^2-x{e}^{(x-3)})\log (-x+e^{(x-3)})\log (\log (-x+e^{(x-3)}))\log {(\log (\log (-x+e^{(x-3)})))}^2+(x^2+5(x^3-x^2{e}^{(x-3)})e^{(x+5e^x)}-x{e}^{(x-3)})e^{(2e^{\left(5e^x\right)}-4)}\log (-x+e^{(x-3)})\log (\log (-x+e^{(x-3)}))-(x^2{e}^{(x-3)}-x^2)e^{(e^{\left(5e^x\right)}-2)}+((2x^2+5(x^3-x^2{e}^{(x-3)})e^{(x+5e^x)}-2x{e}^{(x-3)})e^{(e^{\left(5e^x\right)}-2)}\log (-x+e^{(x-3)})\log (\log (-x+e^{(x-3)}))-x^2{e}^{(x-3)}+x^2)\log (\log (\log (-x+e^{(x-3)}))))}{(x-e^{(x-3)})\log (-x+e^{(x-3)})\log (\log (-x+e^{(x-3)}))}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x*exp(x-3)-50*x^2)*log(exp(x-3)-x)*log(log(exp(x-3)-x))*log(log(log(exp(x-3)-x)))^2+(((250*x^2*
exp(x-3)-250*x^3)*exp(x)*exp(5*exp(x))+100*x*exp(x-3)-100*x^2)*log(exp(x-3)-x)*exp(exp(5*exp(x))-2)*log(log(ex
p(x-3)-x))+50*x^2*exp(x-3)-50*x^2)*log(log(log(exp(x-3)-x)))+((250*x^2*exp(x-3)-250*x^3)*exp(x)*exp(5*exp(x))+
50*x*exp(x-3)-50*x^2)*log(exp(x-3)-x)*exp(exp(5*exp(x))-2)^2*log(log(exp(x-3)-x))+(50*x^2*exp(x-3)-50*x^2)*exp
(exp(5*exp(x))-2))/(exp(x-3)-x)/log(exp(x-3)-x)/log(log(exp(x-3)-x)),x, algorithm="giac")

[Out]

integrate(50*((x^2 - x*e^(x - 3))*log(-x + e^(x - 3))*log(log(-x + e^(x - 3)))*log(log(log(-x + e^(x - 3))))^2
+ (x^2 + 5*(x^3 - x^2*e^(x - 3))*e^(x + 5*e^x) - x*e^(x - 3))*e^(2*e^(5*e^x) - 4)*log(-x + e^(x - 3))*log(log
(-x + e^(x - 3))) - (x^2*e^(x - 3) - x^2)*e^(e^(5*e^x) - 2) + ((2*x^2 + 5*(x^3 - x^2*e^(x - 3))*e^(x + 5*e^x)
- 2*x*e^(x - 3))*e^(e^(5*e^x) - 2)*log(-x + e^(x - 3))*log(log(-x + e^(x - 3))) - x^2*e^(x - 3) + x^2)*log(log
(log(-x + e^(x - 3)))))/((x - e^(x - 3))*log(-x + e^(x - 3))*log(log(-x + e^(x - 3)))), x)

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maple [B]  time = 0.12, size = 59, normalized size = 1.90

method	result	size
risch	$25x^2{\mathrm{e}}^{2{\mathrm{e}}^{5{\mathrm{e}}^x}-4}+50x^2{\mathrm{e}}^{{\mathrm{e}}^{5{\mathrm{e}}^x}-2}\ln (\ln (\ln ({\mathrm{e}}^{x-3}-x)))+25\ln {(\ln (\ln ({\mathrm{e}}^{x-3}-x)))}^2{x}^2$	$59$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((50*x*exp(x-3)-50*x^2)*ln(exp(x-3)-x)*ln(ln(exp(x-3)-x))*ln(ln(ln(exp(x-3)-x)))^2+(((250*x^2*exp(x-3)-250
*x^3)*exp(x)*exp(5*exp(x))+100*x*exp(x-3)-100*x^2)*ln(exp(x-3)-x)*exp(exp(5*exp(x))-2)*ln(ln(exp(x-3)-x))+50*x
^2*exp(x-3)-50*x^2)*ln(ln(ln(exp(x-3)-x)))+((250*x^2*exp(x-3)-250*x^3)*exp(x)*exp(5*exp(x))+50*x*exp(x-3)-50*x
^2)*ln(exp(x-3)-x)*exp(exp(5*exp(x))-2)^2*ln(ln(exp(x-3)-x))+(50*x^2*exp(x-3)-50*x^2)*exp(exp(5*exp(x))-2))/(e
xp(x-3)-x)/ln(exp(x-3)-x)/ln(ln(exp(x-3)-x)),x,method=_RETURNVERBOSE)

[Out]

25*x^2*exp(2*exp(5*exp(x))-4)+50*x^2*exp(exp(5*exp(x))-2)*ln(ln(ln(exp(x-3)-x)))+25*ln(ln(ln(exp(x-3)-x)))^2*x
^2

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maxima [B]  time = 0.75, size = 64, normalized size = 2.06 $$\begin{array}{c}25(x^2{e}^4\log {(\log (\log (-x{e}^3+e^x)-3))}^2+2x^2{e}^{(e^{\left(5e^x\right)}+2)}\log (\log (\log (-x{e}^3+e^x)-3))+x^2{e}^{\left(2e^{\left(5e^x\right)}\right)})e^{(-4)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x*exp(x-3)-50*x^2)*log(exp(x-3)-x)*log(log(exp(x-3)-x))*log(log(log(exp(x-3)-x)))^2+(((250*x^2*
exp(x-3)-250*x^3)*exp(x)*exp(5*exp(x))+100*x*exp(x-3)-100*x^2)*log(exp(x-3)-x)*exp(exp(5*exp(x))-2)*log(log(ex
p(x-3)-x))+50*x^2*exp(x-3)-50*x^2)*log(log(log(exp(x-3)-x)))+((250*x^2*exp(x-3)-250*x^3)*exp(x)*exp(5*exp(x))+
50*x*exp(x-3)-50*x^2)*log(exp(x-3)-x)*exp(exp(5*exp(x))-2)^2*log(log(exp(x-3)-x))+(50*x^2*exp(x-3)-50*x^2)*exp
(exp(5*exp(x))-2))/(exp(x-3)-x)/log(exp(x-3)-x)/log(log(exp(x-3)-x)),x, algorithm="maxima")

[Out]

25*(x^2*e^4*log(log(log(-x*e^3 + e^x) - 3))^2 + 2*x^2*e^(e^(5*e^x) + 2)*log(log(log(-x*e^3 + e^x) - 3)) + x^2*
e^(2*e^(5*e^x)))*e^(-4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 $$\begin{array}{c}-\int \frac{\ln (\ln ({\mathrm{e}}^{x-3}-x))\ln ({\mathrm{e}}^{x-3}-x)(50x{\mathrm{e}}^{x-3}-50x^2){\ln (\ln (\ln ({\mathrm{e}}^{x-3}-x)))}^2+(50x^2{\mathrm{e}}^{x-3}-50x^2+\ln (\ln ({\mathrm{e}}^{x-3}-x)){\mathrm{e}}^{{\mathrm{e}}^{5{\mathrm{e}}^x}-2}\ln ({\mathrm{e}}^{x-3}-x)(100x{\mathrm{e}}^{x-3}-100x^2+{\mathrm{e}}^{x+5{\mathrm{e}}^x}(250x^2{\mathrm{e}}^{x-3}-250x^3)))\ln (\ln (\ln ({\mathrm{e}}^{x-3}-x)))+{\mathrm{e}}^{{\mathrm{e}}^{5{\mathrm{e}}^x}-2}(50x^2{\mathrm{e}}^{x-3}-50x^2)+\ln (\ln ({\mathrm{e}}^{x-3}-x)){\mathrm{e}}^{2{\mathrm{e}}^{5{\mathrm{e}}^x}-4}\ln ({\mathrm{e}}^{x-3}-x)(50x{\mathrm{e}}^{x-3}-50x^2+{\mathrm{e}}^{x+5{\mathrm{e}}^x}(250x^2{\mathrm{e}}^{x-3}-250x^3))}{\ln (\ln ({\mathrm{e}}^{x-3}-x))\ln ({\mathrm{e}}^{x-3}-x)(x-{\mathrm{e}}^{x-3})}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(log(exp(x - 3) - x)))*(50*x^2*exp(x - 3) - 50*x^2 + log(log(exp(x - 3) - x))*exp(exp(5*exp(x)) -
2)*log(exp(x - 3) - x)*(100*x*exp(x - 3) - 100*x^2 + exp(5*exp(x))*exp(x)*(250*x^2*exp(x - 3) - 250*x^3))) +
exp(exp(5*exp(x)) - 2)*(50*x^2*exp(x - 3) - 50*x^2) + log(log(exp(x - 3) - x))*log(log(log(exp(x - 3) - x)))^2
*log(exp(x - 3) - x)*(50*x*exp(x - 3) - 50*x^2) + log(log(exp(x - 3) - x))*exp(2*exp(5*exp(x)) - 4)*log(exp(x
- 3) - x)*(50*x*exp(x - 3) - 50*x^2 + exp(5*exp(x))*exp(x)*(250*x^2*exp(x - 3) - 250*x^3)))/(log(log(exp(x - 3
) - x))*log(exp(x - 3) - x)*(x - exp(x - 3))),x)

[Out]

-int((log(log(log(exp(x - 3) - x)))*(50*x^2*exp(x - 3) - 50*x^2 + log(log(exp(x - 3) - x))*exp(exp(5*exp(x)) -
2)*log(exp(x - 3) - x)*(100*x*exp(x - 3) - 100*x^2 + exp(x + 5*exp(x))*(250*x^2*exp(x - 3) - 250*x^3))) + exp
(exp(5*exp(x)) - 2)*(50*x^2*exp(x - 3) - 50*x^2) + log(log(exp(x - 3) - x))*log(log(log(exp(x - 3) - x)))^2*lo
g(exp(x - 3) - x)*(50*x*exp(x - 3) - 50*x^2) + log(log(exp(x - 3) - x))*exp(2*exp(5*exp(x)) - 4)*log(exp(x - 3
) - x)*(50*x*exp(x - 3) - 50*x^2 + exp(x + 5*exp(x))*(250*x^2*exp(x - 3) - 250*x^3)))/(log(log(exp(x - 3) - x)
)*log(exp(x - 3) - x)*(x - exp(x - 3))), 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(((50*x*exp(x-3)-50*x**2)*ln(exp(x-3)-x)*ln(ln(exp(x-3)-x))*ln(ln(ln(exp(x-3)-x)))**2+(((250*x**2*exp
(x-3)-250*x**3)*exp(x)*exp(5*exp(x))+100*x*exp(x-3)-100*x**2)*ln(exp(x-3)-x)*exp(exp(5*exp(x))-2)*ln(ln(exp(x-
3)-x))+50*x**2*exp(x-3)-50*x**2)*ln(ln(ln(exp(x-3)-x)))+((250*x**2*exp(x-3)-250*x**3)*exp(x)*exp(5*exp(x))+50*
x*exp(x-3)-50*x**2)*ln(exp(x-3)-x)*exp(exp(5*exp(x))-2)**2*ln(ln(exp(x-3)-x))+(50*x**2*exp(x-3)-50*x**2)*exp(e
xp(5*exp(x))-2))/(exp(x-3)-x)/ln(exp(x-3)-x)/ln(ln(exp(x-3)-x)),x)

[Out]

Timed out

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