\(\int \frac {-25 x+(100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x) \log (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x) \log (\log (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x))}{(4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x) \log (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x)} \, dx\) [7616]

Optimal result

Integrand size = 134, antiderivative size = 25 \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=25 x \log \left (\log \left (4 e^{2 e^{4+\frac {\log (5)}{e^5}}}-x\right )\right ) \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(75\) vs. \(2(25)=50\).

Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 3.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6820, 12, 2458, 2395, 2335, 2339, 29, 2600} \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=100 e^{2\ 5^{\frac {1}{e^5}} e^4} \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )-25 \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right ) \]

[In]

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^
5)) - x]),x]

[Out]

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^(
-5)*E^4) - x]]

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2339

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

Rule 2395

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
]))

Rule 2458

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[
r, 0]) && IntegerQ[2*r]

Rule 2600

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[
d*x^n], x], x] /; FreeQ[{c, d, n, p}, x]

Rule 6820

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

Rubi steps \begin{align*} \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 \text {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 \text {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 \text {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 \text {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 \operatorname {LogIntegral}\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )-25 \text {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 ) \text {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 ) \text {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{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=25 x \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right ) \]

[In]

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[
5])/E^5)) - x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84

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	\(25 \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{\left (\ln \left (5\right )+4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}}-x \right )\right ) x\)	\(28\)
norman	\(25 \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{\left (\ln \left (5\right )+4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}}-x \right )\right ) x\)	\(28\)
parallelrisch	\(25 \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{\left (\ln \left (5\right )+4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}}-x \right )\right ) x\)	\(28\)
parts	\(25 \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{\left (\ln \left (5\right )+4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}}-x \right )\right ) x\)	\(28\)

[In]

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)
)/exp(5)))^2-x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=25 \, x \log \left (\log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right )\right ) \]

[In]

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((
log(5)+4*exp(5))/exp(5)))^2-x),x, algorithm="fricas")

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=\text {Exception raised: CoercionFailed} \]

[In]

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
)+4*exp(5))/exp(5)))**2-x),x)

[Out]

Exception raised: CoercionFailed >> Cannot convert x - 4*exp(2*5**exp(-5)*exp(4)) of type <class 'sympy.core.a
dd.Add'> to QQ[x,exp(5**exp(-5)*exp(4))]

Maxima [F]

\[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=\int { \frac {25 \, {\left ({\left (x - 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right )} \log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right ) \log \left (\log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right )\right ) + x\right )}}{{\left (x - 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right )} \log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right )} \,d x } \]

[In]

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((
log(5)+4*exp(5))/exp(5)))^2-x),x, algorithm="maxima")

[Out]

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
og(-x + 4*e^(2*5^e^(-5)*e^4))), x)

Giac [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=25 \, x \log \left (\log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right )\right ) \]

[In]

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((
log(5)+4*exp(5))/exp(5)))^2-x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.85 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=25\,x\,\ln \left (\ln \left (4\,{\mathrm {e}}^{2\,5^{{\mathrm {e}}^{-5}}\,{\mathrm {e}}^4}-x\right )\right ) \]

[In]

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))))
- x)*(x - 4*exp(2*exp(exp(-5)*(4*exp(5) + log(5)))))),x)

[Out]

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