3.77.73 \(\int \frac {-x^2 \log (x)+e^{\frac {4+6 e^{25-5 e^x} x^2}{x}} (x+(-4 \log (x)+e^{25-5 e^x} (6 x^2-30 e^x x^3) \log (x)) \log (\log (x)))}{x^2 \log (x)} \, dx\)

Optimal. Leaf size=30 \[ -x+e^{\frac {4}{x}+6 e^{5 \left (5-e^x\right )} x} \log (\log (x)) \]

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Rubi [F]  time = 5.01, 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 {-x^2 \log (x)+e^{\frac {4+6 e^{25-5 e^x} x^2}{x}} \left (x+\left (-4 \log (x)+e^{25-5 e^x} \left (6 x^2-30 e^x x^3\right ) \log (x)\right ) \log (\log (x))\right )}{x^2 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-(x^2*Log[x]) + E^((4 + 6*E^(25 - 5*E^x)*x^2)/x)*(x + (-4*Log[x] + E^(25 - 5*E^x)*(6*x^2 - 30*E^x*x^3)*Lo
g[x])*Log[Log[x]]))/(x^2*Log[x]),x]

[Out]

-x + Defer[Int][E^(4/x + 6*E^(25 - 5*E^x)*x)/(x*Log[x]), x] + 6*Defer[Int][E^(25 - 5*E^x + 4/x + 6*E^(25 - 5*E
^x)*x)*Log[Log[x]], x] - 4*Defer[Int][(E^(4/x + 6*E^(25 - 5*E^x)*x)*Log[Log[x]])/x^2, x] - 30*Defer[Int][E^(25
- 5*E^x + 4/x + x + 6*E^(25 - 5*E^x)*x)*x*Log[Log[x]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1+\frac {e^{-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \left (e^{5 e^x} x-4 e^{5 e^x} \log (x) \log (\log (x))+6 e^{25} x^2 \log (x) \log (\log (x))-30 e^{25+x} x^3 \log (x) \log (\log (x))\right )}{x^2 \log (x)}\right ) \, dx\\ &=-x+\int \frac {e^{-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \left (e^{5 e^x} x-4 e^{5 e^x} \log (x) \log (\log (x))+6 e^{25} x^2 \log (x) \log (\log (x))-30 e^{25+x} x^3 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx\\ &=-x+\int \left (-30 e^{25-5 e^x+\frac {4}{x}+x+6 e^{25-5 e^x} x} x \log (\log (x))+\frac {e^{-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \left (e^{5 e^x} x-4 e^{5 e^x} \log (x) \log (\log (x))+6 e^{25} x^2 \log (x) \log (\log (x))\right )}{x^2 \log (x)}\right ) \, dx\\ &=-x-30 \int e^{25-5 e^x+\frac {4}{x}+x+6 e^{25-5 e^x} x} x \log (\log (x)) \, dx+\int \frac {e^{-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \left (e^{5 e^x} x-4 e^{5 e^x} \log (x) \log (\log (x))+6 e^{25} x^2 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx\\ &=-x-30 \int e^{25-5 e^x+\frac {4}{x}+x+6 e^{25-5 e^x} x} x \log (\log (x)) \, dx+\int \left (6 e^{25-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x))+\frac {e^{\frac {4}{x}+6 e^{25-5 e^x} x} (x-4 \log (x) \log (\log (x)))}{x^2 \log (x)}\right ) \, dx\\ &=-x+6 \int e^{25-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x)) \, dx-30 \int e^{25-5 e^x+\frac {4}{x}+x+6 e^{25-5 e^x} x} x \log (\log (x)) \, dx+\int \frac {e^{\frac {4}{x}+6 e^{25-5 e^x} x} (x-4 \log (x) \log (\log (x)))}{x^2 \log (x)} \, dx\\ &=-x+6 \int e^{25-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x)) \, dx-30 \int e^{25-5 e^x+\frac {4}{x}+x+6 e^{25-5 e^x} x} x \log (\log (x)) \, dx+\int \left (\frac {e^{\frac {4}{x}+6 e^{25-5 e^x} x}}{x \log (x)}-\frac {4 e^{\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x))}{x^2}\right ) \, dx\\ &=-x-4 \int \frac {e^{\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x))}{x^2} \, dx+6 \int e^{25-5 e^x+\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x)) \, dx-30 \int e^{25-5 e^x+\frac {4}{x}+x+6 e^{25-5 e^x} x} x \log (\log (x)) \, dx+\int \frac {e^{\frac {4}{x}+6 e^{25-5 e^x} x}}{x \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 28, normalized size = 0.93 \begin {gather*} -x+e^{\frac {4}{x}+6 e^{25-5 e^x} x} \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(x^2*Log[x]) + E^((4 + 6*E^(25 - 5*E^x)*x^2)/x)*(x + (-4*Log[x] + E^(25 - 5*E^x)*(6*x^2 - 30*E^x*x
^3)*Log[x])*Log[Log[x]]))/(x^2*Log[x]),x]

[Out]

-x + E^(4/x + 6*E^(25 - 5*E^x)*x)*Log[Log[x]]

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fricas [A]  time = 0.60, size = 28, normalized size = 0.93 \begin {gather*} e^{\left (\frac {2 \, {\left (3 \, x^{2} e^{\left (-5 \, e^{x} + 25\right )} + 2\right )}}{x}\right )} \log \left (\log \relax (x)\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-30*exp(x)*x^3+6*x^2)*log(x)*exp(-5*exp(x)+25)-4*log(x))*log(log(x))+x)*exp((6*x^2*exp(-5*exp(x)
+25)+4)/x)-x^2*log(x))/x^2/log(x),x, algorithm="fricas")

[Out]

e^(2*(3*x^2*e^(-5*e^x + 25) + 2)/x)*log(log(x)) - x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {x^{2} \log \relax (x) + {\left (2 \, {\left (3 \, {\left (5 \, x^{3} e^{x} - x^{2}\right )} e^{\left (-5 \, e^{x} + 25\right )} \log \relax (x) + 2 \, \log \relax (x)\right )} \log \left (\log \relax (x)\right ) - x\right )} e^{\left (\frac {2 \, {\left (3 \, x^{2} e^{\left (-5 \, e^{x} + 25\right )} + 2\right )}}{x}\right )}}{x^{2} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-30*exp(x)*x^3+6*x^2)*log(x)*exp(-5*exp(x)+25)-4*log(x))*log(log(x))+x)*exp((6*x^2*exp(-5*exp(x)
+25)+4)/x)-x^2*log(x))/x^2/log(x),x, algorithm="giac")

[Out]

integrate(-(x^2*log(x) + (2*(3*(5*x^3*e^x - x^2)*e^(-5*e^x + 25)*log(x) + 2*log(x))*log(log(x)) - x)*e^(2*(3*x
^2*e^(-5*e^x + 25) + 2)/x))/(x^2*log(x)), x)

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maple [A]  time = 0.07, size = 29, normalized size = 0.97

method	result	size
risch	\({\mathrm e}^{\frac {6 x^{2} {\mathrm e}^{-5 \,{\mathrm e}^{x}+25}+4}{x}} \ln \left (\ln \relax (x )\right )-x\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-30*exp(x)*x^3+6*x^2)*ln(x)*exp(-5*exp(x)+25)-4*ln(x))*ln(ln(x))+x)*exp((6*x^2*exp(-5*exp(x)+25)+4)/x)
-x^2*ln(x))/x^2/ln(x),x,method=_RETURNVERBOSE)

[Out]

exp(2*(3*x^2*exp(-5*exp(x)+25)+2)/x)*ln(ln(x))-x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -x - \int \frac {{\left (6 \, {\left (5 \, x^{3} e^{\left (x + 25\right )} \log \relax (x) - x^{2} e^{25} \log \relax (x)\right )} e^{\frac {4}{x}} \log \left (\log \relax (x)\right ) + {\left (4 \, \log \relax (x) \log \left (\log \relax (x)\right ) - x\right )} e^{\left (\frac {4}{x} + 5 \, e^{x}\right )}\right )} e^{\left (6 \, x e^{\left (-5 \, e^{x} + 25\right )} - 5 \, e^{x}\right )}}{x^{2} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-30*exp(x)*x^3+6*x^2)*log(x)*exp(-5*exp(x)+25)-4*log(x))*log(log(x))+x)*exp((6*x^2*exp(-5*exp(x)
+25)+4)/x)-x^2*log(x))/x^2/log(x),x, algorithm="maxima")

[Out]

-x - integrate((6*(5*x^3*e^(x + 25)*log(x) - x^2*e^25*log(x))*e^(4/x)*log(log(x)) + (4*log(x)*log(log(x)) - x)
*e^(4/x + 5*e^x))*e^(6*x*e^(-5*e^x + 25) - 5*e^x)/(x^2*log(x)), x)

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mupad [B]  time = 4.63, size = 25, normalized size = 0.83 \begin {gather*} \ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{6\,x\,{\mathrm {e}}^{25}\,{\mathrm {e}}^{-5\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{4/x}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2*log(x) - exp((6*x^2*exp(25 - 5*exp(x)) + 4)/x)*(x - log(log(x))*(4*log(x) + exp(25 - 5*exp(x))*log(x
)*(30*x^3*exp(x) - 6*x^2))))/(x^2*log(x)),x)

[Out]

log(log(x))*exp(6*x*exp(25)*exp(-5*exp(x)))*exp(4/x) - x

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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(((((-30*exp(x)*x**3+6*x**2)*ln(x)*exp(-5*exp(x)+25)-4*ln(x))*ln(ln(x))+x)*exp((6*x**2*exp(-5*exp(x)+
25)+4)/x)-x**2*ln(x))/x**2/ln(x),x)

[Out]

Timed out

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