3.94.16 \(\int \frac {e^{\frac {144 x}{80-40 \log (x)+5 \log ^2(x)}} (-864+144 \log (x))+e^{e^x} (-320 e^x+240 e^x \log (x)-60 e^x \log ^2(x)+5 e^x \log ^3(x))}{-320+240 \log (x)-60 \log ^2(x)+5 \log ^3(x)} \, dx\)

Optimal. Leaf size=20 \[ 5+e^{e^x}+e^{\frac {144 x}{5 (-4+\log (x))^2}} \]

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Rubi [A]  time = 1.27, antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {6741, 12, 6688, 6742, 2282, 2194, 6706} \begin {gather*} e^{e^x}+e^{\frac {144 x}{5 (4-\log (x))^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((144*x)/(80 - 40*Log[x] + 5*Log[x]^2))*(-864 + 144*Log[x]) + E^E^x*(-320*E^x + 240*E^x*Log[x] - 60*E^x
*Log[x]^2 + 5*E^x*Log[x]^3))/(-320 + 240*Log[x] - 60*Log[x]^2 + 5*Log[x]^3),x]

[Out]

E^E^x + E^((144*x)/(5*(4 - Log[x])^2))

Rule 12

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^{\frac {144 x}{80-40 \log (x)+5 \log ^2(x)}} (-864+144 \log (x))-e^{e^x} \left (-320 e^x+240 e^x \log (x)-60 e^x \log ^2(x)+5 e^x \log ^3(x)\right )}{5 (4-\log (x))^3} \, dx\\ &=\frac {1}{5} \int \frac {-e^{\frac {144 x}{80-40 \log (x)+5 \log ^2(x)}} (-864+144 \log (x))-e^{e^x} \left (-320 e^x+240 e^x \log (x)-60 e^x \log ^2(x)+5 e^x \log ^3(x)\right )}{(4-\log (x))^3} \, dx\\ &=\frac {1}{5} \int \frac {-144 e^{\frac {144 x}{5 (-4+\log (x))^2}} (-6+\log (x))-5 e^{e^x+x} (-4+\log (x))^3}{(4-\log (x))^3} \, dx\\ &=\frac {1}{5} \int \left (5 e^{e^x+x}+\frac {144 e^{\frac {144 x}{5 (-4+\log (x))^2}} (-6+\log (x))}{(-4+\log (x))^3}\right ) \, dx\\ &=\frac {144}{5} \int \frac {e^{\frac {144 x}{5 (-4+\log (x))^2}} (-6+\log (x))}{(-4+\log (x))^3} \, dx+\int e^{e^x+x} \, dx\\ &=e^{\frac {144 x}{5 (4-\log (x))^2}}+\operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )\\ &=e^{e^x}+e^{\frac {144 x}{5 (4-\log (x))^2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 27, normalized size = 1.35 \begin {gather*} \frac {1}{5} \left (5 e^{e^x}+5 e^{\frac {144 x}{5 (-4+\log (x))^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((144*x)/(80 - 40*Log[x] + 5*Log[x]^2))*(-864 + 144*Log[x]) + E^E^x*(-320*E^x + 240*E^x*Log[x] -
60*E^x*Log[x]^2 + 5*E^x*Log[x]^3))/(-320 + 240*Log[x] - 60*Log[x]^2 + 5*Log[x]^3),x]

[Out]

(5*E^E^x + 5*E^((144*x)/(5*(-4 + Log[x])^2)))/5

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fricas [A]  time = 0.68, size = 20, normalized size = 1.00 \begin {gather*} e^{\left (\frac {144 \, x}{5 \, {\left (\log \relax (x)^{2} - 8 \, \log \relax (x) + 16\right )}}\right )} + e^{\left (e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)*log(x)^3-60*exp(x)*log(x)^2+240*exp(x)*log(x)-320*exp(x))*exp(exp(x))+(144*log(x)-864)*ex
p(144*x/(5*log(x)^2-40*log(x)+80)))/(5*log(x)^3-60*log(x)^2+240*log(x)-320),x, algorithm="fricas")

[Out]

e^(144/5*x/(log(x)^2 - 8*log(x) + 16)) + e^(e^x)

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giac [A]  time = 0.21, size = 20, normalized size = 1.00 \begin {gather*} e^{\left (\frac {144 \, x}{5 \, {\left (\log \relax (x)^{2} - 8 \, \log \relax (x) + 16\right )}}\right )} + e^{\left (e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)*log(x)^3-60*exp(x)*log(x)^2+240*exp(x)*log(x)-320*exp(x))*exp(exp(x))+(144*log(x)-864)*ex
p(144*x/(5*log(x)^2-40*log(x)+80)))/(5*log(x)^3-60*log(x)^2+240*log(x)-320),x, algorithm="giac")

[Out]

e^(144/5*x/(log(x)^2 - 8*log(x) + 16)) + e^(e^x)

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maple [A]  time = 0.08, size = 15, normalized size = 0.75

method	result	size
risch	\({\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}^{\frac {144 x}{5 \left (\ln \relax (x )-4\right )^{2}}}\)	\(15\)
default	\(\frac {-40 \ln \relax (x ) {\mathrm e}^{\frac {144 x}{5 \ln \relax (x )^{2}-40 \ln \relax (x )+80}}+5 \ln \relax (x )^{2} {\mathrm e}^{\frac {144 x}{5 \ln \relax (x )^{2}-40 \ln \relax (x )+80}}+80 \,{\mathrm e}^{\frac {144 x}{5 \ln \relax (x )^{2}-40 \ln \relax (x )+80}}}{5 \left (\ln \relax (x )-4\right )^{2}}+{\mathrm e}^{{\mathrm e}^{x}}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*exp(x)*ln(x)^3-60*exp(x)*ln(x)^2+240*exp(x)*ln(x)-320*exp(x))*exp(exp(x))+(144*ln(x)-864)*exp(144*x/(5
*ln(x)^2-40*ln(x)+80)))/(5*ln(x)^3-60*ln(x)^2+240*ln(x)-320),x,method=_RETURNVERBOSE)

[Out]

exp(exp(x))+exp(144/5*x/(ln(x)-4)^2)

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maxima [A]  time = 0.49, size = 20, normalized size = 1.00 \begin {gather*} e^{\left (\frac {144 \, x}{5 \, {\left (\log \relax (x)^{2} - 8 \, \log \relax (x) + 16\right )}}\right )} + e^{\left (e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)*log(x)^3-60*exp(x)*log(x)^2+240*exp(x)*log(x)-320*exp(x))*exp(exp(x))+(144*log(x)-864)*ex
p(144*x/(5*log(x)^2-40*log(x)+80)))/(5*log(x)^3-60*log(x)^2+240*log(x)-320),x, algorithm="maxima")

[Out]

e^(144/5*x/(log(x)^2 - 8*log(x) + 16)) + e^(e^x)

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mupad [B]  time = 7.23, size = 22, normalized size = 1.10 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^x}+{\mathrm {e}}^{\frac {144\,x}{5\,{\ln \relax (x)}^2-40\,\ln \relax (x)+80}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(x))*(320*exp(x) - 240*exp(x)*log(x) + 60*exp(x)*log(x)^2 - 5*exp(x)*log(x)^3) - exp((144*x)/(5*l
og(x)^2 - 40*log(x) + 80))*(144*log(x) - 864))/(240*log(x) - 60*log(x)^2 + 5*log(x)^3 - 320),x)

[Out]

exp(exp(x)) + exp((144*x)/(5*log(x)^2 - 40*log(x) + 80))

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sympy [A]  time = 1.21, size = 22, normalized size = 1.10 \begin {gather*} e^{\frac {144 x}{5 \log {\relax (x )}^{2} - 40 \log {\relax (x )} + 80}} + e^{e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(x)*ln(x)**3-60*exp(x)*ln(x)**2+240*exp(x)*ln(x)-320*exp(x))*exp(exp(x))+(144*ln(x)-864)*exp(
144*x/(5*ln(x)**2-40*ln(x)+80)))/(5*ln(x)**3-60*ln(x)**2+240*ln(x)-320),x)

[Out]

exp(144*x/(5*log(x)**2 - 40*log(x) + 80)) + exp(exp(x))

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