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

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

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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 {gather*} \frac {25 x^2 \left (e^{e^{5 e^x}}+e^2 \log \left (\log \left (\log \left (e^{x-3}-x\right )\right )\right )\right )^2}{e^4} \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {50 x \left (e^{e^{5 e^x}}+e^2 \log \left (\log \left (\log \left (e^{-3+x}-x\right )\right )\right )\right ) \left (-e^5 x+e^{2+x} x+\left (e^x-e^3 x\right ) \log \left (e^{-3+x}-x\right ) \log \left (\log \left (e^{-3+x}-x\right )\right ) \left (e^{e^{5 e^x}} \left (1+5 e^{5 e^x+x} x\right )+e^2 \log \left (\log \left (\log \left (e^{-3+x}-x\right )\right )\right )\right )\right )}{e^4 \left (e^x-e^3 x\right ) \log \left (e^{-3+x}-x\right ) \log \left (\log \left (e^{-3+x}-x\right )\right )} \, dx\\ &=\frac {50 \int \frac {x \left (e^{e^{5 e^x}}+e^2 \log \left (\log \left (\log \left (e^{-3+x}-x\right )\right )\right )\right ) \left (-e^5 x+e^{2+x} x+\left (e^x-e^3 x\right ) \log \left (e^{-3+x}-x\right ) \log \left (\log \left (e^{-3+x}-x\right )\right ) \left (e^{e^{5 e^x}} \left (1+5 e^{5 e^x+x} x\right )+e^2 \log \left (\log \left (\log \left (e^{-3+x}-x\right )\right )\right )\right )\right )}{\left (e^x-e^3 x\right ) \log \left (e^{-3+x}-x\right ) \log \left (\log \left (e^{-3+x}-x\right )\right )} \, dx}{e^4}\\ &=\frac {25 x^2 \left (e^{e^{5 e^x}}+e^2 \log \left (\log \left (\log \left (e^{-3+x}-x\right )\right )\right )\right )^2}{e^4}\\ \end {aligned} \end {gather*}

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

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 {gather*} 50 \, x^{2} e^{\left ({\left (e^{\left (x + 5 \, e^{x}\right )} - 2 \, e^{x}\right )} e^{\left (-x\right )}\right )} \log \left (\log \left (\log \left (-{\left (x e^{3} - e^{x}\right )} e^{\left (-3\right )}\right )\right )\right ) + 25 \, x^{2} \log \left (\log \left (\log \left (-{\left (x e^{3} - e^{x}\right )} e^{\left (-3\right )}\right )\right )\right )^{2} + 25 \, x^{2} e^{\left (2 \, {\left (e^{\left (x + 5 \, e^{x}\right )} - 2 \, e^{x}\right )} e^{\left (-x\right )}\right )} \end {gather*}

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 {gather*} \int \frac {50 \, {\left ({\left (x^{2} - x e^{\left (x - 3\right )}\right )} \log \left (-x + e^{\left (x - 3\right )}\right ) \log \left (\log \left (-x + e^{\left (x - 3\right )}\right )\right ) \log \left (\log \left (\log \left (-x + e^{\left (x - 3\right )}\right )\right )\right )^{2} + {\left (x^{2} + 5 \, {\left (x^{3} - x^{2} e^{\left (x - 3\right )}\right )} e^{\left (x + 5 \, e^{x}\right )} - x e^{\left (x - 3\right )}\right )} e^{\left (2 \, e^{\left (5 \, e^{x}\right )} - 4\right )} \log \left (-x + e^{\left (x - 3\right )}\right ) \log \left (\log \left (-x + e^{\left (x - 3\right )}\right )\right ) - {\left (x^{2} e^{\left (x - 3\right )} - x^{2}\right )} e^{\left (e^{\left (5 \, e^{x}\right )} - 2\right )} + {\left ({\left (2 \, x^{2} + 5 \, {\left (x^{3} - x^{2} e^{\left (x - 3\right )}\right )} e^{\left (x + 5 \, e^{x}\right )} - 2 \, x e^{\left (x - 3\right )}\right )} e^{\left (e^{\left (5 \, e^{x}\right )} - 2\right )} \log \left (-x + e^{\left (x - 3\right )}\right ) \log \left (\log \left (-x + e^{\left (x - 3\right )}\right )\right ) - x^{2} e^{\left (x - 3\right )} + x^{2}\right )} \log \left (\log \left (\log \left (-x + e^{\left (x - 3\right )}\right )\right )\right )\right )}}{{\left (x - e^{\left (x - 3\right )}\right )} \log \left (-x + e^{\left (x - 3\right )}\right ) \log \left (\log \left (-x + e^{\left (x - 3\right )}\right )\right )}\,{d x} \end {gather*}

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	\(25 x^{2} {\mathrm e}^{2 \,{\mathrm e}^{5 \,{\mathrm e}^{x}}-4}+50 x^{2} {\mathrm e}^{{\mathrm e}^{5 \,{\mathrm e}^{x}}-2} \ln \left (\ln \left (\ln \left ({\mathrm e}^{x -3}-x \right )\right )\right )+25 \ln \left (\ln \left (\ln \left ({\mathrm e}^{x -3}-x \right )\right )\right )^{2} x^{2}\)	\(59\)

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 {gather*} 25 \, {\left (x^{2} e^{4} \log \left (\log \left (\log \left (-x e^{3} + e^{x}\right ) - 3\right )\right )^{2} + 2 \, x^{2} e^{\left (e^{\left (5 \, e^{x}\right )} + 2\right )} \log \left (\log \left (\log \left (-x e^{3} + e^{x}\right ) - 3\right )\right ) + x^{2} e^{\left (2 \, e^{\left (5 \, e^{x}\right )}\right )}\right )} e^{\left (-4\right )} \end {gather*}

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 {gather*} -\int \frac {\ln \left (\ln \left ({\mathrm {e}}^{x-3}-x\right )\right )\,\ln \left ({\mathrm {e}}^{x-3}-x\right )\,\left (50\,x\,{\mathrm {e}}^{x-3}-50\,x^2\right )\,{\ln \left (\ln \left (\ln \left ({\mathrm {e}}^{x-3}-x\right )\right )\right )}^2+\left (50\,x^2\,{\mathrm {e}}^{x-3}-50\,x^2+\ln \left (\ln \left ({\mathrm {e}}^{x-3}-x\right )\right )\,{\mathrm {e}}^{{\mathrm {e}}^{5\,{\mathrm {e}}^x}-2}\,\ln \left ({\mathrm {e}}^{x-3}-x\right )\,\left (100\,x\,{\mathrm {e}}^{x-3}-100\,x^2+{\mathrm {e}}^{x+5\,{\mathrm {e}}^x}\,\left (250\,x^2\,{\mathrm {e}}^{x-3}-250\,x^3\right )\right )\right )\,\ln \left (\ln \left (\ln \left ({\mathrm {e}}^{x-3}-x\right )\right )\right )+{\mathrm {e}}^{{\mathrm {e}}^{5\,{\mathrm {e}}^x}-2}\,\left (50\,x^2\,{\mathrm {e}}^{x-3}-50\,x^2\right )+\ln \left (\ln \left ({\mathrm {e}}^{x-3}-x\right )\right )\,{\mathrm {e}}^{2\,{\mathrm {e}}^{5\,{\mathrm {e}}^x}-4}\,\ln \left ({\mathrm {e}}^{x-3}-x\right )\,\left (50\,x\,{\mathrm {e}}^{x-3}-50\,x^2+{\mathrm {e}}^{x+5\,{\mathrm {e}}^x}\,\left (250\,x^2\,{\mathrm {e}}^{x-3}-250\,x^3\right )\right )}{\ln \left (\ln \left ({\mathrm {e}}^{x-3}-x\right )\right )\,\ln \left ({\mathrm {e}}^{x-3}-x\right )\,\left (x-{\mathrm {e}}^{x-3}\right )} \,d x \end {gather*}

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 {gather*} \text {Timed out} \end {gather*}

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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