3.59.11 \(\int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+(15 x-10 x^2) \log (x)+(-125+50 e^x-5 e^{2 x}+(-15 x+5 x^2) \log (x)) \log (25-10 e^x+e^{2 x}+(3 x-x^2) \log (x)) \log (\log (25-10 e^x+e^{2 x}+(3 x-x^2) \log (x)))}{(-25 x+10 e^x x-e^{2 x} x+(-3 x^2+x^3) \log (x)) \log (25-10 e^x+e^{2 x}+(3 x-x^2) \log (x)) \log (\log (25-10 e^x+e^{2 x}+(3 x-x^2) \log (x))) \log ^2(\frac {\log (\log (25-10 e^x+e^{2 x}+(3 x-x^2) \log (x)))}{x})} \, dx\)

Optimal. Leaf size=30 \[ \frac {5}{\log \left (\frac {\log \left (\log \left (\left (5-e^x\right )^2+(3-x) x \log (x)\right )\right )}{x}\right )} \]

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Rubi [F]  time = 24.51, 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 {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(15*x - 50*E^x*x + 10*E^(2*x)*x - 5*x^2 + (15*x - 10*x^2)*Log[x] + (-125 + 50*E^x - 5*E^(2*x) + (-15*x + 5
*x^2)*Log[x])*Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]*Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[
x]]])/((-25*x + 10*E^x*x - E^(2*x)*x + (-3*x^2 + x^3)*Log[x])*Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]*
Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]]*Log[Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]]/
x]^2),x]

[Out]

5*Defer[Int][1/(x*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]/x]^2), x] - 10*Defer[Int][1/(Log[(-5 + E^x)^2
- (-3 + x)*x*Log[x]]*Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]
]/x]^2), x] + 235*Defer[Int][1/(((-5 + E^x)^2 - (-3 + x)*x*Log[x])*Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]*Log[L
og[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]/x]^2), x] - 50*Defer[Int]
[E^x/(((-5 + E^x)^2 - (-3 + x)*x*Log[x])*Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]*Log[Log[(-5 + E^x)^2 - (-3 + x)
*x*Log[x]]]*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]/x]^2), x] - 5*Defer[Int][x/((-(-5 + E^x)^2 + (-3 +
x)*x*Log[x])*Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]*Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]*Log[Log[Log[(-5
+ E^x)^2 - (-3 + x)*x*Log[x]]]/x]^2), x] + 15*Defer[Int][Log[x]/((-(-5 + E^x)^2 + (-3 + x)*x*Log[x])*Log[(-5 +
E^x)^2 - (-3 + x)*x*Log[x]]*Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*
Log[x]]]/x]^2), x] - 40*Defer[Int][(x*Log[x])/((-(-5 + E^x)^2 + (-3 + x)*x*Log[x])*Log[(-5 + E^x)^2 - (-3 + x)
*x*Log[x]]*Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]/x]^2), x]
+ 10*Defer[Int][(x^2*Log[x])/((-(-5 + E^x)^2 + (-3 + x)*x*Log[x])*Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]*Log[L
og[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]*Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]/x]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {5 \left (-47+10 e^x-x+3 \log (x)-8 x \log (x)+2 x^2 \log (x)\right )}{\left (25-10 e^x+e^{2 x}+3 x \log (x)-x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )}-\frac {5 \left (2 x-\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )\right )}{x \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )}\right ) \, dx\\ &=-\left (5 \int \frac {-47+10 e^x-x+3 \log (x)-8 x \log (x)+2 x^2 \log (x)}{\left (25-10 e^x+e^{2 x}+3 x \log (x)-x^2 \log (x)\right ) \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx\right )-5 \int \frac {2 x-\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x \log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 27, normalized size = 0.90 \begin {gather*} \frac {5}{\log \left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15*x - 50*E^x*x + 10*E^(2*x)*x - 5*x^2 + (15*x - 10*x^2)*Log[x] + (-125 + 50*E^x - 5*E^(2*x) + (-15
*x + 5*x^2)*Log[x])*Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]*Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2
)*Log[x]]])/((-25*x + 10*E^x*x - E^(2*x)*x + (-3*x^2 + x^3)*Log[x])*Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Lo
g[x]]*Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]]*Log[Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log
[x]]]/x]^2),x]

[Out]

5/Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]/x]

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fricas [A]  time = 0.53, size = 32, normalized size = 1.07 \begin {gather*} \frac {5}{\log \left (\frac {\log \left (\log \left (-{\left (x^{2} - 3 \, x\right )} \log \relax (x) + e^{\left (2 \, x\right )} - 10 \, e^{x} + 25\right )\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^2-15*x)*log(x)-5*exp(x)^2+50*exp(x)-125)*log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25)*log(log
((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))+(-10*x^2+15*x)*log(x)+10*x*exp(x)^2-50*exp(x)*x-5*x^2+15*x)/((x^3-3
*x^2)*log(x)-x*exp(x)^2+10*exp(x)*x-25*x)/log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25)/log(log((-x^2+3*x)*log(
x)+exp(x)^2-10*exp(x)+25))/log(log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))/x)^2,x, algorithm="fricas")

[Out]

5/log(log(log(-(x^2 - 3*x)*log(x) + e^(2*x) - 10*e^x + 25))/x)

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giac [A]  time = 1.28, size = 34, normalized size = 1.13 \begin {gather*} -\frac {5}{\log \relax (x) - \log \left (\log \left (\log \left (-x^{2} \log \relax (x) + 3 \, x \log \relax (x) + e^{\left (2 \, x\right )} - 10 \, e^{x} + 25\right )\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^2-15*x)*log(x)-5*exp(x)^2+50*exp(x)-125)*log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25)*log(log
((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))+(-10*x^2+15*x)*log(x)+10*x*exp(x)^2-50*exp(x)*x-5*x^2+15*x)/((x^3-3
*x^2)*log(x)-x*exp(x)^2+10*exp(x)*x-25*x)/log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25)/log(log((-x^2+3*x)*log(
x)+exp(x)^2-10*exp(x)+25))/log(log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))/x)^2,x, algorithm="giac")

[Out]

-5/(log(x) - log(log(log(-x^2*log(x) + 3*x*log(x) + e^(2*x) - 10*e^x + 25))))

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maple [C]  time = 0.31, size = 250, normalized size = 8.33

method	result	size
risch	\(-\frac {10 i}{\pi \,\mathrm {csgn}\left (i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \relax (x )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )\right ) \mathrm {csgn}\left (\frac {i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \relax (x )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \relax (x )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )\right ) \mathrm {csgn}\left (\frac {i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \relax (x )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right )-\pi \mathrm {csgn}\left (\frac {i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \relax (x )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )}{x}\right )^{3}+\pi \mathrm {csgn}\left (\frac {i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \relax (x )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )+2 i \ln \relax (x )-2 i \ln \left (\ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \relax (x )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )\right )}\)	\(250\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5*x^2-15*x)*ln(x)-5*exp(x)^2+50*exp(x)-125)*ln((-x^2+3*x)*ln(x)+exp(x)^2-10*exp(x)+25)*ln(ln((-x^2+3*x)
*ln(x)+exp(x)^2-10*exp(x)+25))+(-10*x^2+15*x)*ln(x)+10*x*exp(x)^2-50*exp(x)*x-5*x^2+15*x)/((x^3-3*x^2)*ln(x)-x
*exp(x)^2+10*exp(x)*x-25*x)/ln((-x^2+3*x)*ln(x)+exp(x)^2-10*exp(x)+25)/ln(ln((-x^2+3*x)*ln(x)+exp(x)^2-10*exp(
x)+25))/ln(ln(ln((-x^2+3*x)*ln(x)+exp(x)^2-10*exp(x)+25))/x)^2,x,method=_RETURNVERBOSE)

[Out]

-10*I/(Pi*csgn(I*ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25)))*csgn(I/x*ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*e
xp(x)+25)))^2-Pi*csgn(I*ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25)))*csgn(I/x*ln(ln((-x^2+3*x)*ln(x)+exp(2*
x)-10*exp(x)+25)))*csgn(I/x)-Pi*csgn(I/x*ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25)))^3+Pi*csgn(I/x*ln(ln((
-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25)))^2*csgn(I/x)+2*I*ln(x)-2*I*ln(ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x
)+25))))

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maxima [A]  time = 0.89, size = 33, normalized size = 1.10 \begin {gather*} -\frac {5}{\log \relax (x) - \log \left (\log \left (\log \left (-{\left (x^{2} - 3 \, x\right )} \log \relax (x) + e^{\left (2 \, x\right )} - 10 \, e^{x} + 25\right )\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x^2-15*x)*log(x)-5*exp(x)^2+50*exp(x)-125)*log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25)*log(log
((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))+(-10*x^2+15*x)*log(x)+10*x*exp(x)^2-50*exp(x)*x-5*x^2+15*x)/((x^3-3
*x^2)*log(x)-x*exp(x)^2+10*exp(x)*x-25*x)/log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25)/log(log((-x^2+3*x)*log(
x)+exp(x)^2-10*exp(x)+25))/log(log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))/x)^2,x, algorithm="maxima")

[Out]

-5/(log(x) - log(log(log(-(x^2 - 3*x)*log(x) + e^(2*x) - 10*e^x + 25))))

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mupad [B]  time = 6.93, size = 33, normalized size = 1.10 \begin {gather*} \frac {5}{\ln \left (\frac {\ln \left (\ln \left ({\mathrm {e}}^{2\,x}-10\,{\mathrm {e}}^x+\ln \relax (x)\,\left (3\,x-x^2\right )+25\right )\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*x + 10*x*exp(2*x) + log(x)*(15*x - 10*x^2) - 50*x*exp(x) - 5*x^2 - log(log(exp(2*x) - 10*exp(x) + log
(x)*(3*x - x^2) + 25))*log(exp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25)*(5*exp(2*x) - 50*exp(x) + log(x)*(1
5*x - 5*x^2) + 125))/(log(log(log(exp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25))/x)^2*log(log(exp(2*x) - 10*
exp(x) + log(x)*(3*x - x^2) + 25))*log(exp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25)*(25*x + log(x)*(3*x^2 -
x^3) + x*exp(2*x) - 10*x*exp(x))),x)

[Out]

5/log(log(log(exp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25))/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((((5*x**2-15*x)*ln(x)-5*exp(x)**2+50*exp(x)-125)*ln((-x**2+3*x)*ln(x)+exp(x)**2-10*exp(x)+25)*ln(ln(
(-x**2+3*x)*ln(x)+exp(x)**2-10*exp(x)+25))+(-10*x**2+15*x)*ln(x)+10*x*exp(x)**2-50*exp(x)*x-5*x**2+15*x)/((x**
3-3*x**2)*ln(x)-x*exp(x)**2+10*exp(x)*x-25*x)/ln((-x**2+3*x)*ln(x)+exp(x)**2-10*exp(x)+25)/ln(ln((-x**2+3*x)*l
n(x)+exp(x)**2-10*exp(x)+25))/ln(ln(ln((-x**2+3*x)*ln(x)+exp(x)**2-10*exp(x)+25))/x)**2,x)

[Out]

Timed out

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