\(\int \frac {3 x+3 \log (3)+e^x (x+\log (3))+(-6 x-3 \log (3)+e^x (-2 x+x^2+(-1+x) \log (3))) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+(3 x^2+3 x \log (3)+e^x (x^2+x \log (3))) \log (x)} \, dx\) [7691]

Optimal result

Integrand size = 95, antiderivative size = 20 \[ \int \frac {3 x+3 \log (3)+e^x (x+\log (3))+\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+\left (3 x^2+3 x \log (3)+e^x \left (x^2+x \log (3)\right )\right ) \log (x)} \, dx=\log \left (-10+\frac {\left (3+e^x\right ) \log (x)}{x (x+\log (3))}\right ) \]

[Out]

ln((3+exp(x))*ln(x)/(ln(3)+x)/x-10)

Rubi [F]

\[ \int \frac {3 x+3 \log (3)+e^x (x+\log (3))+\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+\left (3 x^2+3 x \log (3)+e^x \left (x^2+x \log (3)\right )\right ) \log (x)} \, dx=\int \frac {3 x+3 \log (3)+e^x (x+\log (3))+\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+\left (3 x^2+3 x \log (3)+e^x \left (x^2+x \log (3)\right )\right ) \log (x)} \, dx \]

[In]

Int[(3*x + 3*Log[3] + E^x*(x + Log[3]) + (-6*x - 3*Log[3] + E^x*(-2*x + x^2 + (-1 + x)*Log[3]))*Log[x])/(-10*x
^4 - 20*x^3*Log[3] - 10*x^2*Log[3]^2 + (3*x^2 + 3*x*Log[3] + E^x*(x^2 + x*Log[3]))*Log[x]),x]

[Out]

x - Log[x] - Log[x + Log[3]] + Log[Log[x]] + 10*Log[3]*Defer[Int][(10*x^2 + 10*x*Log[3] - 3*Log[x] - E^x*Log[x
])^(-1), x] + 10*(2 - Log[3])*Defer[Int][x/(10*x^2 + 10*x*Log[3] - 3*Log[x] - E^x*Log[x]), x] - 10*Defer[Int][
x^2/(10*x^2 + 10*x*Log[3] - 3*Log[x] - E^x*Log[x]), x] - 10*Log[3]*Defer[Int][1/(Log[x]*(10*x^2 + 10*x*Log[3]
- 3*Log[x] - E^x*Log[x])), x] - 10*Defer[Int][x/(Log[x]*(10*x^2 + 10*x*Log[3] - 3*Log[x] - E^x*Log[x])), x] -
3*Defer[Int][Log[x]/(-10*x^2 - 10*x*Log[3] + 3*Log[x] + E^x*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-3 x-3 \log (3)-e^x (x+\log (3))-\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{x (x+\log (3)) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx \\ & = \int \left (\frac {x+\log (3)+x^2 \log (x)-2 x \left (1-\frac {\log (3)}{2}\right ) \log (x)-\log (3) \log (x)}{x (x+\log (3)) \log (x)}+\frac {-10 x-10 \log (3)-10 x^2 \log (x)+20 x \left (1-\frac {\log (3)}{2}\right ) \log (x)+10 \log (3) \log (x)+3 \log ^2(x)}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )}\right ) \, dx \\ & = \int \frac {x+\log (3)+x^2 \log (x)-2 x \left (1-\frac {\log (3)}{2}\right ) \log (x)-\log (3) \log (x)}{x (x+\log (3)) \log (x)} \, dx+\int \frac {-10 x-10 \log (3)-10 x^2 \log (x)+20 x \left (1-\frac {\log (3)}{2}\right ) \log (x)+10 \log (3) \log (x)+3 \log ^2(x)}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx \\ & = \int \frac {x+\log (3)+\left (x^2+x (-2+\log (3))-\log (3)\right ) \log (x)}{x (x+\log (3)) \log (x)} \, dx+\int \left (-\frac {10 x^2}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)}-\frac {10 x (-2+\log (3))}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)}+\frac {10 \log (3)}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)}-\frac {10 x}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )}-\frac {10 \log (3)}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )}-\frac {3 \log (x)}{-10 x^2-10 x \log (3)+3 \log (x)+e^x \log (x)}\right ) \, dx \\ & = -\left (3 \int \frac {\log (x)}{-10 x^2-10 x \log (3)+3 \log (x)+e^x \log (x)} \, dx\right )-10 \int \frac {x^2}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-10 \int \frac {x}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+(10 (2-\log (3))) \int \frac {x}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx+(10 \log (3)) \int \frac {1}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-(10 \log (3)) \int \frac {1}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+\int \left (\frac {x^2-x (2-\log (3))-\log (3)}{x (x+\log (3))}+\frac {1}{x \log (x)}\right ) \, dx \\ & = -\left (3 \int \frac {\log (x)}{-10 x^2-10 x \log (3)+3 \log (x)+e^x \log (x)} \, dx\right )-10 \int \frac {x^2}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-10 \int \frac {x}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+(10 (2-\log (3))) \int \frac {x}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx+(10 \log (3)) \int \frac {1}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-(10 \log (3)) \int \frac {1}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+\int \frac {x^2-x (2-\log (3))-\log (3)}{x (x+\log (3))} \, dx+\int \frac {1}{x \log (x)} \, dx \\ & = -\left (3 \int \frac {\log (x)}{-10 x^2-10 x \log (3)+3 \log (x)+e^x \log (x)} \, dx\right )-10 \int \frac {x^2}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-10 \int \frac {x}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+(10 (2-\log (3))) \int \frac {x}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx+(10 \log (3)) \int \frac {1}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-(10 \log (3)) \int \frac {1}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+\int \left (1-\frac {1}{x}+\frac {1}{-x-\log (3)}\right ) \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = x-\log (x)-\log (x+\log (3))+\log (\log (x))-3 \int \frac {\log (x)}{-10 x^2-10 x \log (3)+3 \log (x)+e^x \log (x)} \, dx-10 \int \frac {x^2}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-10 \int \frac {x}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+(10 (2-\log (3))) \int \frac {x}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx+(10 \log (3)) \int \frac {1}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-(10 \log (3)) \int \frac {1}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx \\ \end{align*}

Mathematica [F]

\[ \int \frac {3 x+3 \log (3)+e^x (x+\log (3))+\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+\left (3 x^2+3 x \log (3)+e^x \left (x^2+x \log (3)\right )\right ) \log (x)} \, dx=\int \frac {3 x+3 \log (3)+e^x (x+\log (3))+\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+\left (3 x^2+3 x \log (3)+e^x \left (x^2+x \log (3)\right )\right ) \log (x)} \, dx \]

[In]

Integrate[(3*x + 3*Log[3] + E^x*(x + Log[3]) + (-6*x - 3*Log[3] + E^x*(-2*x + x^2 + (-1 + x)*Log[3]))*Log[x])/
(-10*x^4 - 20*x^3*Log[3] - 10*x^2*Log[3]^2 + (3*x^2 + 3*x*Log[3] + E^x*(x^2 + x*Log[3]))*Log[x]),x]

[Out]

Integrate[(3*x + 3*Log[3] + E^x*(x + Log[3]) + (-6*x - 3*Log[3] + E^x*(-2*x + x^2 + (-1 + x)*Log[3]))*Log[x])/
(-10*x^4 - 20*x^3*Log[3] - 10*x^2*Log[3]^2 + (3*x^2 + 3*x*Log[3] + E^x*(x^2 + x*Log[3]))*Log[x]), x]

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60

method	result	size
parallelrisch	\(-\ln \left (\ln \left (3\right )+x \right )+\ln \left (x \ln \left (3\right )+x^{2}-\frac {{\mathrm e}^{x} \ln \left (x \right )}{10}-\frac {3 \ln \left (x \right )}{10}\right )-\ln \left (x \right )\)	\(32\)
norman	\(-\ln \left (x \right )-\ln \left (\ln \left (3\right )+x \right )+\ln \left (10 x \ln \left (3\right )-{\mathrm e}^{x} \ln \left (x \right )+10 x^{2}-3 \ln \left (x \right )\right )\)	\(35\)
risch	\(-\ln \left (x \ln \left (3\right )+x^{2}\right )+\ln \left (3+{\mathrm e}^{x}\right )+\ln \left (\ln \left (x \right )-\frac {10 x \left (\ln \left (3\right )+x \right )}{3+{\mathrm e}^{x}}\right )\)	\(35\)

[In]

int(((((-1+x)*ln(3)+x^2-2*x)*exp(x)-3*ln(3)-6*x)*ln(x)+(ln(3)+x)*exp(x)+3*ln(3)+3*x)/(((x*ln(3)+x^2)*exp(x)+3*
x*ln(3)+3*x^2)*ln(x)-10*x^2*ln(3)^2-20*x^3*ln(3)-10*x^4),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(3)+x)+ln(x*ln(3)+x^2-1/10*exp(x)*ln(x)-3/10*ln(x))-ln(x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (19) = 38\).

Time = 0.32 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.25 \[ \int \frac {3 x+3 \log (3)+e^x (x+\log (3))+\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+\left (3 x^2+3 x \log (3)+e^x \left (x^2+x \log (3)\right )\right ) \log (x)} \, dx=-\log \left (x^{2} + x \log \left (3\right )\right ) + \log \left (-\frac {10 \, x^{2} + 10 \, x \log \left (3\right ) - {\left (e^{x} + 3\right )} \log \left (x\right )}{e^{x} + 3}\right ) + \log \left (e^{x} + 3\right ) \]

[In]

integrate(((((-1+x)*log(3)+x^2-2*x)*exp(x)-3*log(3)-6*x)*log(x)+(log(3)+x)*exp(x)+3*log(3)+3*x)/(((x*log(3)+x^
2)*exp(x)+3*x*log(3)+3*x^2)*log(x)-10*x^2*log(3)^2-20*x^3*log(3)-10*x^4),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (17) = 34\).

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.95 \[ \int \frac {3 x+3 \log (3)+e^x (x+\log (3))+\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+\left (3 x^2+3 x \log (3)+e^x \left (x^2+x \log (3)\right )\right ) \log (x)} \, dx=- \log {\left (x^{2} + x \log {\left (3 \right )} \right )} + \log {\left (\frac {- 10 x^{2} - 10 x \log {\left (3 \right )} + 3 \log {\left (x \right )}}{\log {\left (x \right )}} + e^{x} \right )} + \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate(((((-1+x)*ln(3)+x**2-2*x)*exp(x)-3*ln(3)-6*x)*ln(x)+(ln(3)+x)*exp(x)+3*ln(3)+3*x)/(((x*ln(3)+x**2)*e
xp(x)+3*x*ln(3)+3*x**2)*ln(x)-10*x**2*ln(3)**2-20*x**3*ln(3)-10*x**4),x)

[Out]

-log(x**2 + x*log(3)) + log((-10*x**2 - 10*x*log(3) + 3*log(x))/log(x) + exp(x)) + log(log(x))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (19) = 38\).

Time = 0.36 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.15 \[ \int \frac {3 x+3 \log (3)+e^x (x+\log (3))+\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+\left (3 x^2+3 x \log (3)+e^x \left (x^2+x \log (3)\right )\right ) \log (x)} \, dx=-\log \left (x + \log \left (3\right )\right ) - \log \left (x\right ) + \log \left (-\frac {10 \, x^{2} + 10 \, x \log \left (3\right ) - e^{x} \log \left (x\right ) - 3 \, \log \left (x\right )}{\log \left (x\right )}\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((((-1+x)*log(3)+x^2-2*x)*exp(x)-3*log(3)-6*x)*log(x)+(log(3)+x)*exp(x)+3*log(3)+3*x)/(((x*log(3)+x^
2)*exp(x)+3*x*log(3)+3*x^2)*log(x)-10*x^2*log(3)^2-20*x^3*log(3)-10*x^4),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int \frac {3 x+3 \log (3)+e^x (x+\log (3))+\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+\left (3 x^2+3 x \log (3)+e^x \left (x^2+x \log (3)\right )\right ) \log (x)} \, dx=\log \left (-10 \, x^{2} - 10 \, x \log \left (3\right ) + e^{x} \log \left (x\right ) + 3 \, \log \left (x\right )\right ) - \log \left (x + \log \left (3\right )\right ) - \log \left (x\right ) \]

[In]

integrate(((((-1+x)*log(3)+x^2-2*x)*exp(x)-3*log(3)-6*x)*log(x)+(log(3)+x)*exp(x)+3*log(3)+3*x)/(((x*log(3)+x^
2)*exp(x)+3*x*log(3)+3*x^2)*log(x)-10*x^2*log(3)^2-20*x^3*log(3)-10*x^4),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {3 x+3 \log (3)+e^x (x+\log (3))+\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+\left (3 x^2+3 x \log (3)+e^x \left (x^2+x \log (3)\right )\right ) \log (x)} \, dx=-\int \frac {3\,x+3\,\ln \left (3\right )-\ln \left (x\right )\,\left (6\,x+3\,\ln \left (3\right )-{\mathrm {e}}^x\,\left (\ln \left (3\right )\,\left (x-1\right )-2\,x+x^2\right )\right )+{\mathrm {e}}^x\,\left (x+\ln \left (3\right )\right )}{10\,x^2\,{\ln \left (3\right )}^2+20\,x^3\,\ln \left (3\right )-\ln \left (x\right )\,\left (3\,x\,\ln \left (3\right )+{\mathrm {e}}^x\,\left (x^2+\ln \left (3\right )\,x\right )+3\,x^2\right )+10\,x^4} \,d x \]

[In]

int(-(3*x + 3*log(3) - log(x)*(6*x + 3*log(3) - exp(x)*(log(3)*(x - 1) - 2*x + x^2)) + exp(x)*(x + log(3)))/(1
0*x^2*log(3)^2 + 20*x^3*log(3) - log(x)*(3*x*log(3) + exp(x)*(x*log(3) + x^2) + 3*x^2) + 10*x^4),x)

[Out]

-int((3*x + 3*log(3) - log(x)*(6*x + 3*log(3) - exp(x)*(log(3)*(x - 1) - 2*x + x^2)) + exp(x)*(x + log(3)))/(1
0*x^2*log(3)^2 + 20*x^3*log(3) - log(x)*(3*x*log(3) + exp(x)*(x*log(3) + x^2) + 3*x^2) + 10*x^4), x)