3.88.78 \(\int \frac {-144 x-288 e^x x-144 x^2+(-144 x-72 x^3+e^x (-144-72 x^2)) \log (e^{2 x}+e^x x)}{(5 e^x x^5+5 x^6) \log (e^{2 x}+e^x x)+(40 e^x x^3+40 x^4) \log (e^{2 x}+e^x x) \log (x \log (e^{2 x}+e^x x))+(80 e^x x+80 x^2) \log (e^{2 x}+e^x x) \log ^2(x \log (e^{2 x}+e^x x))} \, dx\)

Optimal. Leaf size=27 \[ \frac {9}{5 \left (\frac {x^2}{4}+\log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )} \]

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Rubi [F]  time = 3.03, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-144 x-288 e^x x-144 x^2+\left (-144 x-72 x^3+e^x \left (-144-72 x^2\right )\right ) \log \left (e^{2 x}+e^x x\right )}{\left (5 e^x x^5+5 x^6\right ) \log \left (e^{2 x}+e^x x\right )+\left (40 e^x x^3+40 x^4\right ) \log \left (e^{2 x}+e^x x\right ) \log \left (x \log \left (e^{2 x}+e^x x\right )\right )+\left (80 e^x x+80 x^2\right ) \log \left (e^{2 x}+e^x x\right ) \log ^2\left (x \log \left (e^{2 x}+e^x x\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-144*x - 288*E^x*x - 144*x^2 + (-144*x - 72*x^3 + E^x*(-144 - 72*x^2))*Log[E^(2*x) + E^x*x])/((5*E^x*x^5
+ 5*x^6)*Log[E^(2*x) + E^x*x] + (40*E^x*x^3 + 40*x^4)*Log[E^(2*x) + E^x*x]*Log[x*Log[E^(2*x) + E^x*x]] + (80*E
^x*x + 80*x^2)*Log[E^(2*x) + E^x*x]*Log[x*Log[E^(2*x) + E^x*x]]^2),x]

[Out]

(-144*Defer[Int][1/(x*(x^2 + 4*Log[x*Log[E^x*(E^x + x)]])^2), x])/5 - (72*Defer[Int][x/(x^2 + 4*Log[x*Log[E^x*
(E^x + x)]])^2, x])/5 - (288*Defer[Int][1/(Log[E^x*(E^x + x)]*(x^2 + 4*Log[x*Log[E^x*(E^x + x)]])^2), x])/5 -
(144*Defer[Int][1/((E^x + x)*Log[E^x*(E^x + x)]*(x^2 + 4*Log[x*Log[E^x*(E^x + x)]])^2), x])/5 + (144*Defer[Int
][x/((E^x + x)*Log[E^x*(E^x + x)]*(x^2 + 4*Log[x*Log[E^x*(E^x + x)]])^2), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {72 \left (-2 x \left (1+2 e^x+x\right )-\left (e^x+x\right ) \left (2+x^2\right ) \log \left (e^x \left (e^x+x\right )\right )\right )}{5 x \left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx\\ &=\frac {72}{5} \int \frac {-2 x \left (1+2 e^x+x\right )-\left (e^x+x\right ) \left (2+x^2\right ) \log \left (e^x \left (e^x+x\right )\right )}{x \left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx\\ &=\frac {72}{5} \int \left (\frac {2 (-1+x)}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}+\frac {-4 x-2 \log \left (e^x \left (e^x+x\right )\right )-x^2 \log \left (e^x \left (e^x+x\right )\right )}{x \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}\right ) \, dx\\ &=\frac {72}{5} \int \frac {-4 x-2 \log \left (e^x \left (e^x+x\right )\right )-x^2 \log \left (e^x \left (e^x+x\right )\right )}{x \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx+\frac {144}{5} \int \frac {-1+x}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx\\ &=\frac {72}{5} \int \frac {-4 x-\left (2+x^2\right ) \log \left (e^x \left (e^x+x\right )\right )}{x \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx+\frac {144}{5} \int \left (-\frac {1}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}+\frac {x}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}\right ) \, dx\\ &=\frac {72}{5} \int \left (-\frac {2}{x \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}-\frac {x}{\left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}-\frac {4}{\log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2}\right ) \, dx-\frac {144}{5} \int \frac {1}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx+\frac {144}{5} \int \frac {x}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx\\ &=-\left (\frac {72}{5} \int \frac {x}{\left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx\right )-\frac {144}{5} \int \frac {1}{x \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx-\frac {144}{5} \int \frac {1}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx+\frac {144}{5} \int \frac {x}{\left (e^x+x\right ) \log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx-\frac {288}{5} \int \frac {1}{\log \left (e^x \left (e^x+x\right )\right ) \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.43, size = 25, normalized size = 0.93 \begin {gather*} \frac {36}{5 \left (x^2+4 \log \left (x \log \left (e^x \left (e^x+x\right )\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-144*x - 288*E^x*x - 144*x^2 + (-144*x - 72*x^3 + E^x*(-144 - 72*x^2))*Log[E^(2*x) + E^x*x])/((5*E^
x*x^5 + 5*x^6)*Log[E^(2*x) + E^x*x] + (40*E^x*x^3 + 40*x^4)*Log[E^(2*x) + E^x*x]*Log[x*Log[E^(2*x) + E^x*x]] +
(80*E^x*x + 80*x^2)*Log[E^(2*x) + E^x*x]*Log[x*Log[E^(2*x) + E^x*x]]^2),x]

[Out]

36/(5*(x^2 + 4*Log[x*Log[E^x*(E^x + x)]]))

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fricas [A]  time = 0.48, size = 23, normalized size = 0.85 \begin {gather*} \frac {36}{5 \, {\left (x^{2} + 4 \, \log \left (x \log \left (x e^{x} + e^{\left (2 \, x\right )}\right )\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-72*x^2-144)*exp(x)-72*x^3-144*x)*log(exp(x)^2+exp(x)*x)-288*exp(x)*x-144*x^2-144*x)/((80*exp(x)*
x+80*x^2)*log(exp(x)^2+exp(x)*x)*log(x*log(exp(x)^2+exp(x)*x))^2+(40*exp(x)*x^3+40*x^4)*log(exp(x)^2+exp(x)*x)
*log(x*log(exp(x)^2+exp(x)*x))+(5*x^5*exp(x)+5*x^6)*log(exp(x)^2+exp(x)*x)),x, algorithm="fricas")

[Out]

36/5/(x^2 + 4*log(x*log(x*e^x + e^(2*x))))

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giac [A]  time = 2.99, size = 23, normalized size = 0.85 \begin {gather*} \frac {36}{5 \, {\left (x^{2} + 4 \, \log \left (x \log \left (x e^{x} + e^{\left (2 \, x\right )}\right )\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-72*x^2-144)*exp(x)-72*x^3-144*x)*log(exp(x)^2+exp(x)*x)-288*exp(x)*x-144*x^2-144*x)/((80*exp(x)*
x+80*x^2)*log(exp(x)^2+exp(x)*x)*log(x*log(exp(x)^2+exp(x)*x))^2+(40*exp(x)*x^3+40*x^4)*log(exp(x)^2+exp(x)*x)
*log(x*log(exp(x)^2+exp(x)*x))+(5*x^5*exp(x)+5*x^6)*log(exp(x)^2+exp(x)*x)),x, algorithm="giac")

[Out]

36/5/(x^2 + 4*log(x*log(x*e^x + e^(2*x))))

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (-72 x^{2}-144\right ) {\mathrm e}^{x}-72 x^{3}-144 x \right ) \ln \left ({\mathrm e}^{x} x +{\mathrm e}^{2 x}\right )-288 \,{\mathrm e}^{x} x -144 x^{2}-144 x}{\left (80 \,{\mathrm e}^{x} x +80 x^{2}\right ) \ln \left ({\mathrm e}^{x} x +{\mathrm e}^{2 x}\right ) \ln \left (x \ln \left ({\mathrm e}^{x} x +{\mathrm e}^{2 x}\right )\right )^{2}+\left (40 \,{\mathrm e}^{x} x^{3}+40 x^{4}\right ) \ln \left ({\mathrm e}^{x} x +{\mathrm e}^{2 x}\right ) \ln \left (x \ln \left ({\mathrm e}^{x} x +{\mathrm e}^{2 x}\right )\right )+\left (5 x^{5} {\mathrm e}^{x}+5 x^{6}\right ) \ln \left ({\mathrm e}^{x} x +{\mathrm e}^{2 x}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-72*x^2-144)*exp(x)-72*x^3-144*x)*ln(exp(x)^2+exp(x)*x)-288*exp(x)*x-144*x^2-144*x)/((80*exp(x)*x+80*x^
2)*ln(exp(x)^2+exp(x)*x)*ln(x*ln(exp(x)^2+exp(x)*x))^2+(40*exp(x)*x^3+40*x^4)*ln(exp(x)^2+exp(x)*x)*ln(x*ln(ex
p(x)^2+exp(x)*x))+(5*x^5*exp(x)+5*x^6)*ln(exp(x)^2+exp(x)*x)),x)

[Out]

int((((-72*x^2-144)*exp(x)-72*x^3-144*x)*ln(exp(x)^2+exp(x)*x)-288*exp(x)*x-144*x^2-144*x)/((80*exp(x)*x+80*x^
2)*ln(exp(x)^2+exp(x)*x)*ln(x*ln(exp(x)^2+exp(x)*x))^2+(40*exp(x)*x^3+40*x^4)*ln(exp(x)^2+exp(x)*x)*ln(x*ln(ex
p(x)^2+exp(x)*x))+(5*x^5*exp(x)+5*x^6)*ln(exp(x)^2+exp(x)*x)),x)

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maxima [A]  time = 0.58, size = 22, normalized size = 0.81 \begin {gather*} \frac {36}{5 \, {\left (x^{2} + 4 \, \log \left (x + \log \left (x + e^{x}\right )\right ) + 4 \, \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-72*x^2-144)*exp(x)-72*x^3-144*x)*log(exp(x)^2+exp(x)*x)-288*exp(x)*x-144*x^2-144*x)/((80*exp(x)*
x+80*x^2)*log(exp(x)^2+exp(x)*x)*log(x*log(exp(x)^2+exp(x)*x))^2+(40*exp(x)*x^3+40*x^4)*log(exp(x)^2+exp(x)*x)
*log(x*log(exp(x)^2+exp(x)*x))+(5*x^5*exp(x)+5*x^6)*log(exp(x)^2+exp(x)*x)),x, algorithm="maxima")

[Out]

36/5/(x^2 + 4*log(x + log(x + e^x)) + 4*log(x))

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mupad [B]  time = 5.67, size = 25, normalized size = 0.93 \begin {gather*} \frac {36}{5\,\left (4\,\ln \left (x\,\ln \left ({\mathrm {e}}^{2\,x}+x\,{\mathrm {e}}^x\right )\right )+x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(144*x + log(exp(2*x) + x*exp(x))*(144*x + exp(x)*(72*x^2 + 144) + 72*x^3) + 288*x*exp(x) + 144*x^2)/(log
(exp(2*x) + x*exp(x))*(5*x^5*exp(x) + 5*x^6) + log(x*log(exp(2*x) + x*exp(x)))*log(exp(2*x) + x*exp(x))*(40*x^
3*exp(x) + 40*x^4) + log(x*log(exp(2*x) + x*exp(x)))^2*log(exp(2*x) + x*exp(x))*(80*x*exp(x) + 80*x^2)),x)

[Out]

36/(5*(4*log(x*log(exp(2*x) + x*exp(x))) + x^2))

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sympy [A]  time = 1.11, size = 22, normalized size = 0.81 \begin {gather*} \frac {36}{5 x^{2} + 20 \log {\left (x \log {\left (x e^{x} + e^{2 x} \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-72*x**2-144)*exp(x)-72*x**3-144*x)*ln(exp(x)**2+exp(x)*x)-288*exp(x)*x-144*x**2-144*x)/((80*exp(
x)*x+80*x**2)*ln(exp(x)**2+exp(x)*x)*ln(x*ln(exp(x)**2+exp(x)*x))**2+(40*exp(x)*x**3+40*x**4)*ln(exp(x)**2+exp
(x)*x)*ln(x*ln(exp(x)**2+exp(x)*x))+(5*x**5*exp(x)+5*x**6)*ln(exp(x)**2+exp(x)*x)),x)

[Out]

36/(5*x**2 + 20*log(x*log(x*exp(x) + exp(2*x))))

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