∫ −25x+(100e2e4e5+log(5) ______________ e5 −25x) log(4e2e4e5+log(5) ______________ e5 −x) log(log(4e2e4e5+log(5) ______________ e5 −x)) ___________________________________________________________________________________________________________ (4e2e4e5+log(5) ______________ e5 −x) log(4e2e4e5+log(5) _____________

Optimal. Leaf size=25 \[ 25 x \log \left (\log \left (4 e^{2 e^{4+\frac {\log (5)}{e^5}}}-x\right )\right ) \]

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Rubi [B] time = 0.53, antiderivative size = 75, normalized size of antiderivative = 3.00, number of steps used = 12, number of rules used = 8, integrand size = 134, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6688, 12, 2411, 2353, 2298, 2302, 29, 2520} \begin {gather*} 100 e^{2\ 5^{\frac {1}{e^5}} e^4} \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )-25 \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right ) \end {gather*}

Int[(-25*x + (100*E^(2*E^((4*E^5 + Log[5])/E^5)) - 25*x)*Log[4*E^(2*E^((4*E^5 + Log[5])/E^5)) - x]*Log[Log[4*E

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

100*E^(2*5^E^(-5)*E^4)*Log[Log[4*E^(2*5^E^(-5)*E^4) - x]] - 25*(4*E^(2*5^E^(-5)*E^4) - x)*Log[Log[4*E^(2*5^E^(

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

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]

:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[

{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))

^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,

d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[

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

\begin {gather*} \begin {aligned} \text {integral} &=\int 25 \left (-\frac {x}{\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )}+\log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )\right ) \, dx\\ &=25 \int \left (-\frac {x}{\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )}+\log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )\right ) \, dx\\ &=-\left (25 \int \frac {x}{\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )} \, dx\right )+25 \int \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right ) \, dx\\ &=25 \operatorname {Subst}\left (\int \frac {4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x}{x \log (x)} \, dx,x,4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )-25 \operatorname {Subst}\left (\int \log (\log (x)) \, dx,x,4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\\ &=-25 \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )+25 \operatorname {Subst}\left (\int \left (-\frac {1}{\log (x)}+\frac {4 e^{2\ 5^{\frac {1}{e^5}} e^4}}{x \log (x)}\right ) \, dx,x,4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )+25 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\\ &=-25 \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )+25 \text {li}\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )-25 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )+\left (100 e^{2\ 5^{\frac {1}{e^5}} e^4}\right ) \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\\ &=-25 \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )+\left (100 e^{2\ 5^{\frac {1}{e^5}} e^4}\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )\\ &=100 e^{2\ 5^{\frac {1}{e^5}} e^4} \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )-25 \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [C] time = 0.14, size = 66, normalized size = 2.64 \begin {gather*} 25 \left (\text {Ei}\left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )+x \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )-\text {li}\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right ) \end {gather*}

Integrate[(-25*x + (100*E^(2*E^((4*E^5 + Log[5])/E^5)) - 25*x)*Log[4*E^(2*E^((4*E^5 + Log[5])/E^5)) - x]*Log[L

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

25*(ExpIntegralEi[Log[4*E^(2*5^E^(-5)*E^4) - x]] + x*Log[Log[4*E^(2*5^E^(-5)*E^4) - x]] - LogIntegral[4*E^(2*5

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fricas [A] time = 1.98, size = 25, normalized size = 1.00 \begin {gather*} 25 \, x \log \left (\log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \relax (5)\right )} e^{\left (-5\right )}\right )}\right )}\right )\right ) \end {gather*}

integrate(((100*exp(exp((log(5)+4*exp(5))/exp(5)))^2-25*x)*log(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x)*log(l

og(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x))-25*x)/(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x)/log(4*exp(exp((

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giac [A] time = 0.41, size = 25, normalized size = 1.00 \begin {gather*} 25 \, x \log \left (\log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \relax (5)\right )} e^{\left (-5\right )}\right )}\right )}\right )\right ) \end {gather*}

integrate(((100*exp(exp((log(5)+4*exp(5))/exp(5)))^2-25*x)*log(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x)*log(l

og(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x))-25*x)/(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x)/log(4*exp(exp((

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method	result	size

risch	\(25 x \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,5^{{\mathrm e}^{-5}} {\mathrm e}^{4}}-x \right )\right )\)	\(21\)
default	\(25 \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{\left (\ln \relax (5)+4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}}-x \right )\right ) x\)	\(28\)
norman	\(25 \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{\left (\ln \relax (5)+4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}}-x \right )\right ) x\)	\(28\)

int(((100*exp(exp((ln(5)+4*exp(5))/exp(5)))^2-25*x)*ln(4*exp(exp((ln(5)+4*exp(5))/exp(5)))^2-x)*ln(ln(4*exp(ex

p((ln(5)+4*exp(5))/exp(5)))^2-x))-25*x)/(4*exp(exp((ln(5)+4*exp(5))/exp(5)))^2-x)/ln(4*exp(exp((ln(5)+4*exp(5)

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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 100 \, {\left (\log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right ) \log \left (\log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right )\right ) - \log \left (-x + 4 \, e^{\left (2 \, e^{\left (e^{\left (-5\right )} \log \relax (5) + 4\right )}\right )}\right ) \log \left (\log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right )\right ) - \log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right )\right )} e^{\left (2 \, e^{\left (e^{\left (-5\right )} \log \relax (5) + 4\right )}\right )} \log \left (\log \left (-x + 4 \, e^{\left (2 \, e^{\left (e^{\left (-5\right )} \log \relax (5) + 4\right )}\right )}\right )\right ) + 100 \, e^{\left (2 \, e^{\left (e^{\left (-5\right )} \log \relax (5) + 4\right )}\right )} \log \left (x - 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right ) + 25 \, \int \frac {x \log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right ) \log \left (\log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right )\right ) + x}{{\left (x - 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right )} \log \left (-x + 4 \, e^{\left (2 \cdot 5^{e^{\left (-5\right )}} e^{4}\right )}\right )}\,{d x} \end {gather*}

integrate(((100*exp(exp((log(5)+4*exp(5))/exp(5)))^2-25*x)*log(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x)*log(l

og(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x))-25*x)/(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x)/log(4*exp(exp((

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

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

-x + 4*e^(2*e^(e^(-5)*log(5) + 4)))) + 100*e^(2*e^(e^(-5)*log(5) + 4))*log(x - 4*e^(2*5^e^(-5)*e^4)) + 25*inte

grate((x*log(-x + 4*e^(2*5^e^(-5)*e^4))*log(log(-x + 4*e^(2*5^e^(-5)*e^4))) + x)/((x - 4*e^(2*5^e^(-5)*e^4))*l

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mupad [B] time = 6.94, size = 20, normalized size = 0.80 \begin {gather*} 25\,x\,\ln \left (\ln \left (4\,{\mathrm {e}}^{2\,5^{{\mathrm {e}}^{-5}}\,{\mathrm {e}}^4}-x\right )\right ) \end {gather*}

int((25*x + log(4*exp(2*exp(exp(-5)*(4*exp(5) + log(5)))) - x)*log(log(4*exp(2*exp(exp(-5)*(4*exp(5) + log(5))

)) - x))*(25*x - 100*exp(2*exp(exp(-5)*(4*exp(5) + log(5))))))/(log(4*exp(2*exp(exp(-5)*(4*exp(5) + log(5))))

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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: CoercionFailed} \end {gather*}

integrate(((100*exp(exp((ln(5)+4*exp(5))/exp(5)))**2-25*x)*ln(4*exp(exp((ln(5)+4*exp(5))/exp(5)))**2-x)*ln(ln(

4*exp(exp((ln(5)+4*exp(5))/exp(5)))**2-x))-25*x)/(4*exp(exp((ln(5)+4*exp(5))/exp(5)))**2-x)/ln(4*exp(exp((ln(5

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3.77.16 \(\int \frac {-25 x+(100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x) \log (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x) \log (\log (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x))}{(4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x) \log (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x)} \, dx\)