3.61.42 \(\int \frac {-320 e^{4 x}+1280 e^{4 x} x \log (x)-40 x \log ^2(x)+20 e^{e^x+x} x \log ^2(x)+(-64 e^{4 x}+256 e^{4 x} x \log (x)-8 x \log ^2(x)+4 e^{e^x+x} x \log ^2(x)) \log (\frac {256 e^{8 x}-64 e^{4 x} x \log (x)+e^{2 e^x} \log ^2(x)+4 x^2 \log ^2(x)+e^{e^x} (32 e^{4 x} \log (x)-4 x \log ^2(x))}{\log ^2(x)})}{16 e^{4 x} x \log (x)+e^{e^x} x \log ^2(x)-2 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ \left (5+\log \left (\left (-e^{e^x}+2 x-\frac {16 e^{4 x}}{\log (x)}\right )^2\right )\right )^2 \]

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Rubi [B]  time = 2.75, antiderivative size = 61, normalized size of antiderivative = 2.10, number of steps used = 4, number of rules used = 3, integrand size = 181, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6688, 12, 6708} \begin {gather*} \log ^2\left (\frac {\left (16 e^{4 x}+\left (e^{e^x}-2 x\right ) \log (x)\right )^2}{\log ^2(x)}\right )+10 \log \left (\frac {\left (16 e^{4 x}+\left (e^{e^x}-2 x\right ) \log (x)\right )^2}{\log ^2(x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-320*E^(4*x) + 1280*E^(4*x)*x*Log[x] - 40*x*Log[x]^2 + 20*E^(E^x + x)*x*Log[x]^2 + (-64*E^(4*x) + 256*E^(
4*x)*x*Log[x] - 8*x*Log[x]^2 + 4*E^(E^x + x)*x*Log[x]^2)*Log[(256*E^(8*x) - 64*E^(4*x)*x*Log[x] + E^(2*E^x)*Lo
g[x]^2 + 4*x^2*Log[x]^2 + E^E^x*(32*E^(4*x)*Log[x] - 4*x*Log[x]^2))/Log[x]^2])/(16*E^(4*x)*x*Log[x] + E^E^x*x*
Log[x]^2 - 2*x^2*Log[x]^2),x]

[Out]

10*Log[(16*E^(4*x) + (E^E^x - 2*x)*Log[x])^2/Log[x]^2] + Log[(16*E^(4*x) + (E^E^x - 2*x)*Log[x])^2/Log[x]^2]^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 6688

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

Rule 6708

Int[(u_)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(p_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(w*D[v, x] + v*D[w, x])
]}, Dist[c, Subst[Int[(a + b*x^p)^m, x], x, v*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p}, x] && IntegerQ[p]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \left (16 e^{4 x}-64 e^{4 x} x \log (x)-\left (-2+e^{e^x+x}\right ) x \log ^2(x)\right ) \left (-5-\log \left (\frac {\left (16 e^{4 x}+\left (e^{e^x}-2 x\right ) \log (x)\right )^2}{\log ^2(x)}\right )\right )}{x \log (x) \left (16 e^{4 x}+\left (e^{e^x}-2 x\right ) \log (x)\right )} \, dx\\ &=4 \int \frac {\left (16 e^{4 x}-64 e^{4 x} x \log (x)-\left (-2+e^{e^x+x}\right ) x \log ^2(x)\right ) \left (-5-\log \left (\frac {\left (16 e^{4 x}+\left (e^{e^x}-2 x\right ) \log (x)\right )^2}{\log ^2(x)}\right )\right )}{x \log (x) \left (16 e^{4 x}+\left (e^{e^x}-2 x\right ) \log (x)\right )} \, dx\\ &=2 \operatorname {Subst}\left (\int (-5+x) \, dx,x,-\log \left (\frac {\left (16 e^{4 x}+\left (e^{e^x}-2 x\right ) \log (x)\right )^2}{\log ^2(x)}\right )\right )\\ &=10 \log \left (\frac {\left (16 e^{4 x}+\left (e^{e^x}-2 x\right ) \log (x)\right )^2}{\log ^2(x)}\right )+\log ^2\left (\frac {\left (16 e^{4 x}+\left (e^{e^x}-2 x\right ) \log (x)\right )^2}{\log ^2(x)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.19, size = 59, normalized size = 2.03 \begin {gather*} \log \left (\frac {\left (16 e^{4 x}+\left (e^{e^x}-2 x\right ) \log (x)\right )^2}{\log ^2(x)}\right ) \left (10+\log \left (\frac {\left (16 e^{4 x}+\left (e^{e^x}-2 x\right ) \log (x)\right )^2}{\log ^2(x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-320*E^(4*x) + 1280*E^(4*x)*x*Log[x] - 40*x*Log[x]^2 + 20*E^(E^x + x)*x*Log[x]^2 + (-64*E^(4*x) + 2
56*E^(4*x)*x*Log[x] - 8*x*Log[x]^2 + 4*E^(E^x + x)*x*Log[x]^2)*Log[(256*E^(8*x) - 64*E^(4*x)*x*Log[x] + E^(2*E
^x)*Log[x]^2 + 4*x^2*Log[x]^2 + E^E^x*(32*E^(4*x)*Log[x] - 4*x*Log[x]^2))/Log[x]^2])/(16*E^(4*x)*x*Log[x] + E^
E^x*x*Log[x]^2 - 2*x^2*Log[x]^2),x]

[Out]

Log[(16*E^(4*x) + (E^E^x - 2*x)*Log[x])^2/Log[x]^2]*(10 + Log[(16*E^(4*x) + (E^E^x - 2*x)*Log[x])^2/Log[x]^2])

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fricas [B]  time = 0.65, size = 159, normalized size = 5.48 \begin {gather*} \log \left (\frac {{\left (4 \, x^{2} e^{\left (2 \, x\right )} \log \relax (x)^{2} - 64 \, x e^{\left (6 \, x\right )} \log \relax (x) + e^{\left (2 \, x + 2 \, e^{x}\right )} \log \relax (x)^{2} - 4 \, {\left (x e^{x} \log \relax (x)^{2} - 8 \, e^{\left (5 \, x\right )} \log \relax (x)\right )} e^{\left (x + e^{x}\right )} + 256 \, e^{\left (10 \, x\right )}\right )} e^{\left (-2 \, x\right )}}{\log \relax (x)^{2}}\right )^{2} + 10 \, \log \left (\frac {{\left (4 \, x^{2} e^{\left (2 \, x\right )} \log \relax (x)^{2} - 64 \, x e^{\left (6 \, x\right )} \log \relax (x) + e^{\left (2 \, x + 2 \, e^{x}\right )} \log \relax (x)^{2} - 4 \, {\left (x e^{x} \log \relax (x)^{2} - 8 \, e^{\left (5 \, x\right )} \log \relax (x)\right )} e^{\left (x + e^{x}\right )} + 256 \, e^{\left (10 \, x\right )}\right )} e^{\left (-2 \, x\right )}}{\log \relax (x)^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*exp(x)*log(x)^2*exp(exp(x))-8*x*log(x)^2+256*x*exp(x)^4*log(x)-64*exp(x)^4)*log((log(x)^2*exp(
exp(x))^2+(-4*x*log(x)^2+32*exp(x)^4*log(x))*exp(exp(x))+4*x^2*log(x)^2-64*x*exp(x)^4*log(x)+256*exp(x)^8)/log
(x)^2)+20*x*exp(x)*log(x)^2*exp(exp(x))-40*x*log(x)^2+1280*x*exp(x)^4*log(x)-320*exp(x)^4)/(x*log(x)^2*exp(exp
(x))-2*x^2*log(x)^2+16*x*exp(x)^4*log(x)),x, algorithm="fricas")

[Out]

log((4*x^2*e^(2*x)*log(x)^2 - 64*x*e^(6*x)*log(x) + e^(2*x + 2*e^x)*log(x)^2 - 4*(x*e^x*log(x)^2 - 8*e^(5*x)*l
og(x))*e^(x + e^x) + 256*e^(10*x))*e^(-2*x)/log(x)^2)^2 + 10*log((4*x^2*e^(2*x)*log(x)^2 - 64*x*e^(6*x)*log(x)
+ e^(2*x + 2*e^x)*log(x)^2 - 4*(x*e^x*log(x)^2 - 8*e^(5*x)*log(x))*e^(x + e^x) + 256*e^(10*x))*e^(-2*x)/log(x
)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {4 \, {\left (5 \, x e^{\left (x + e^{x}\right )} \log \relax (x)^{2} + 320 \, x e^{\left (4 \, x\right )} \log \relax (x) - 10 \, x \log \relax (x)^{2} + {\left (x e^{\left (x + e^{x}\right )} \log \relax (x)^{2} + 64 \, x e^{\left (4 \, x\right )} \log \relax (x) - 2 \, x \log \relax (x)^{2} - 16 \, e^{\left (4 \, x\right )}\right )} \log \left (\frac {4 \, x^{2} \log \relax (x)^{2} - 64 \, x e^{\left (4 \, x\right )} \log \relax (x) + e^{\left (2 \, e^{x}\right )} \log \relax (x)^{2} - 4 \, {\left (x \log \relax (x)^{2} - 8 \, e^{\left (4 \, x\right )} \log \relax (x)\right )} e^{\left (e^{x}\right )} + 256 \, e^{\left (8 \, x\right )}}{\log \relax (x)^{2}}\right ) - 80 \, e^{\left (4 \, x\right )}\right )}}{2 \, x^{2} \log \relax (x)^{2} - x e^{\left (e^{x}\right )} \log \relax (x)^{2} - 16 \, x e^{\left (4 \, x\right )} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*exp(x)*log(x)^2*exp(exp(x))-8*x*log(x)^2+256*x*exp(x)^4*log(x)-64*exp(x)^4)*log((log(x)^2*exp(
exp(x))^2+(-4*x*log(x)^2+32*exp(x)^4*log(x))*exp(exp(x))+4*x^2*log(x)^2-64*x*exp(x)^4*log(x)+256*exp(x)^8)/log
(x)^2)+20*x*exp(x)*log(x)^2*exp(exp(x))-40*x*log(x)^2+1280*x*exp(x)^4*log(x)-320*exp(x)^4)/(x*log(x)^2*exp(exp
(x))-2*x^2*log(x)^2+16*x*exp(x)^4*log(x)),x, algorithm="giac")

[Out]

integrate(-4*(5*x*e^(x + e^x)*log(x)^2 + 320*x*e^(4*x)*log(x) - 10*x*log(x)^2 + (x*e^(x + e^x)*log(x)^2 + 64*x
*e^(4*x)*log(x) - 2*x*log(x)^2 - 16*e^(4*x))*log((4*x^2*log(x)^2 - 64*x*e^(4*x)*log(x) + e^(2*e^x)*log(x)^2 -
4*(x*log(x)^2 - 8*e^(4*x)*log(x))*e^(e^x) + 256*e^(8*x))/log(x)^2) - 80*e^(4*x))/(2*x^2*log(x)^2 - x*e^(e^x)*l
og(x)^2 - 16*x*e^(4*x)*log(x)), x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (4 x \,{\mathrm e}^{x} \ln \relax (x )^{2} {\mathrm e}^{{\mathrm e}^{x}}-8 x \ln \relax (x )^{2}+256 x \,{\mathrm e}^{4 x} \ln \relax (x )-64 \,{\mathrm e}^{4 x}\right ) \ln \left (\frac {\ln \relax (x )^{2} {\mathrm e}^{2 \,{\mathrm e}^{x}}+\left (-4 x \ln \relax (x )^{2}+32 \,{\mathrm e}^{4 x} \ln \relax (x )\right ) {\mathrm e}^{{\mathrm e}^{x}}+4 x^{2} \ln \relax (x )^{2}-64 x \,{\mathrm e}^{4 x} \ln \relax (x )+256 \,{\mathrm e}^{8 x}}{\ln \relax (x )^{2}}\right )+20 x \,{\mathrm e}^{x} \ln \relax (x )^{2} {\mathrm e}^{{\mathrm e}^{x}}-40 x \ln \relax (x )^{2}+1280 x \,{\mathrm e}^{4 x} \ln \relax (x )-320 \,{\mathrm e}^{4 x}}{x \ln \relax (x )^{2} {\mathrm e}^{{\mathrm e}^{x}}-2 x^{2} \ln \relax (x )^{2}+16 x \,{\mathrm e}^{4 x} \ln \relax (x )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x*exp(x)*ln(x)^2*exp(exp(x))-8*x*ln(x)^2+256*x*exp(x)^4*ln(x)-64*exp(x)^4)*ln((ln(x)^2*exp(exp(x))^2+(
-4*x*ln(x)^2+32*exp(x)^4*ln(x))*exp(exp(x))+4*x^2*ln(x)^2-64*x*exp(x)^4*ln(x)+256*exp(x)^8)/ln(x)^2)+20*x*exp(
x)*ln(x)^2*exp(exp(x))-40*x*ln(x)^2+1280*x*exp(x)^4*ln(x)-320*exp(x)^4)/(x*ln(x)^2*exp(exp(x))-2*x^2*ln(x)^2+1
6*x*exp(x)^4*ln(x)),x)

[Out]

int(((4*x*exp(x)*ln(x)^2*exp(exp(x))-8*x*ln(x)^2+256*x*exp(x)^4*ln(x)-64*exp(x)^4)*ln((ln(x)^2*exp(exp(x))^2+(
-4*x*ln(x)^2+32*exp(x)^4*ln(x))*exp(exp(x))+4*x^2*ln(x)^2-64*x*exp(x)^4*ln(x)+256*exp(x)^8)/ln(x)^2)+20*x*exp(
x)*ln(x)^2*exp(exp(x))-40*x*ln(x)^2+1280*x*exp(x)^4*ln(x)-320*exp(x)^4)/(x*ln(x)^2*exp(exp(x))-2*x^2*ln(x)^2+1
6*x*exp(x)^4*ln(x)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -4 \, \int \frac {5 \, x e^{\left (x + e^{x}\right )} \log \relax (x)^{2} + 320 \, x e^{\left (4 \, x\right )} \log \relax (x) - 10 \, x \log \relax (x)^{2} + {\left (x e^{\left (x + e^{x}\right )} \log \relax (x)^{2} + 64 \, x e^{\left (4 \, x\right )} \log \relax (x) - 2 \, x \log \relax (x)^{2} - 16 \, e^{\left (4 \, x\right )}\right )} \log \left (\frac {4 \, x^{2} \log \relax (x)^{2} - 64 \, x e^{\left (4 \, x\right )} \log \relax (x) + e^{\left (2 \, e^{x}\right )} \log \relax (x)^{2} - 4 \, {\left (x \log \relax (x)^{2} - 8 \, e^{\left (4 \, x\right )} \log \relax (x)\right )} e^{\left (e^{x}\right )} + 256 \, e^{\left (8 \, x\right )}}{\log \relax (x)^{2}}\right ) - 80 \, e^{\left (4 \, x\right )}}{2 \, x^{2} \log \relax (x)^{2} - x e^{\left (e^{x}\right )} \log \relax (x)^{2} - 16 \, x e^{\left (4 \, x\right )} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x*exp(x)*log(x)^2*exp(exp(x))-8*x*log(x)^2+256*x*exp(x)^4*log(x)-64*exp(x)^4)*log((log(x)^2*exp(
exp(x))^2+(-4*x*log(x)^2+32*exp(x)^4*log(x))*exp(exp(x))+4*x^2*log(x)^2-64*x*exp(x)^4*log(x)+256*exp(x)^8)/log
(x)^2)+20*x*exp(x)*log(x)^2*exp(exp(x))-40*x*log(x)^2+1280*x*exp(x)^4*log(x)-320*exp(x)^4)/(x*log(x)^2*exp(exp
(x))-2*x^2*log(x)^2+16*x*exp(x)^4*log(x)),x, algorithm="maxima")

[Out]

-4*integrate((5*x*e^(x + e^x)*log(x)^2 + 320*x*e^(4*x)*log(x) - 10*x*log(x)^2 + (x*e^(x + e^x)*log(x)^2 + 64*x
*e^(4*x)*log(x) - 2*x*log(x)^2 - 16*e^(4*x))*log((4*x^2*log(x)^2 - 64*x*e^(4*x)*log(x) + e^(2*e^x)*log(x)^2 -
4*(x*log(x)^2 - 8*e^(4*x)*log(x))*e^(e^x) + 256*e^(8*x))/log(x)^2) - 80*e^(4*x))/(2*x^2*log(x)^2 - x*e^(e^x)*l
og(x)^2 - 16*x*e^(4*x)*log(x)), x)

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mupad [B]  time = 5.33, size = 91, normalized size = 3.14 \begin {gather*} {\ln \left (\frac {256\,{\mathrm {e}}^{8\,x}+{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,{\ln \relax (x)}^2+4\,x^2\,{\ln \relax (x)}^2-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (4\,x\,{\ln \relax (x)}^2-32\,{\mathrm {e}}^{4\,x}\,\ln \relax (x)\right )-64\,x\,{\mathrm {e}}^{4\,x}\,\ln \relax (x)}{{\ln \relax (x)}^2}\right )}^2+20\,\ln \left (\frac {16\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{{\mathrm {e}}^x}\,\ln \relax (x)-2\,x\,\ln \relax (x)}{\ln \relax (x)}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(320*exp(4*x) + log((256*exp(8*x) + exp(2*exp(x))*log(x)^2 + 4*x^2*log(x)^2 - exp(exp(x))*(4*x*log(x)^2 -
32*exp(4*x)*log(x)) - 64*x*exp(4*x)*log(x))/log(x)^2)*(64*exp(4*x) + 8*x*log(x)^2 - 256*x*exp(4*x)*log(x) - 4
*x*exp(exp(x))*exp(x)*log(x)^2) + 40*x*log(x)^2 - 1280*x*exp(4*x)*log(x) - 20*x*exp(exp(x))*exp(x)*log(x)^2)/(
x*exp(exp(x))*log(x)^2 - 2*x^2*log(x)^2 + 16*x*exp(4*x)*log(x)),x)

[Out]

20*log((16*exp(4*x) + exp(exp(x))*log(x) - 2*x*log(x))/log(x)) + log((256*exp(8*x) + exp(2*exp(x))*log(x)^2 +
4*x^2*log(x)^2 - exp(exp(x))*(4*x*log(x)^2 - 32*exp(4*x)*log(x)) - 64*x*exp(4*x)*log(x))/log(x)^2)^2

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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(((4*x*exp(x)*ln(x)**2*exp(exp(x))-8*x*ln(x)**2+256*x*exp(x)**4*ln(x)-64*exp(x)**4)*ln((ln(x)**2*exp(
exp(x))**2+(-4*x*ln(x)**2+32*exp(x)**4*ln(x))*exp(exp(x))+4*x**2*ln(x)**2-64*x*exp(x)**4*ln(x)+256*exp(x)**8)/
ln(x)**2)+20*x*exp(x)*ln(x)**2*exp(exp(x))-40*x*ln(x)**2+1280*x*exp(x)**4*ln(x)-320*exp(x)**4)/(x*ln(x)**2*exp
(exp(x))-2*x**2*ln(x)**2+16*x*exp(x)**4*ln(x)),x)

[Out]

Timed out

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