\(\int \frac {3-3 x+3 e^x x-6 x^2-9 x^3+(9-3 e^x+3 x+3 x^2+3 x^3-3 \log (x)) \log (3-e^x+x+x^2+x^3-\log (x))}{-3 x^2+e^x x^2-x^3-x^4-x^5+x^2 \log (x)} \, dx\) [2577]

Optimal result

Integrand size = 102, antiderivative size = 25 \[ \int \frac {3-3 x+3 e^x x-6 x^2-9 x^3+\left (9-3 e^x+3 x+3 x^2+3 x^3-3 \log (x)\right ) \log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{-3 x^2+e^x x^2-x^3-x^4-x^5+x^2 \log (x)} \, dx=\frac {3 \log \left (3-e^x+x+x^2 (1+x)-\log (x)\right )}{x} \]

[Out]

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

Rubi [F]

\[ \int \frac {3-3 x+3 e^x x-6 x^2-9 x^3+\left (9-3 e^x+3 x+3 x^2+3 x^3-3 \log (x)\right ) \log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{-3 x^2+e^x x^2-x^3-x^4-x^5+x^2 \log (x)} \, dx=\int \frac {3-3 x+3 e^x x-6 x^2-9 x^3+\left (9-3 e^x+3 x+3 x^2+3 x^3-3 \log (x)\right ) \log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{-3 x^2+e^x x^2-x^3-x^4-x^5+x^2 \log (x)} \, dx \]

[In]

Int[(3 - 3*x + 3*E^x*x - 6*x^2 - 9*x^3 + (9 - 3*E^x + 3*x + 3*x^2 + 3*x^3 - 3*Log[x])*Log[3 - E^x + x + x^2 +
x^3 - Log[x]])/(-3*x^2 + E^x*x^2 - x^3 - x^4 - x^5 + x^2*Log[x]),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {3 \left (1+2 x-x^2-2 x^3+x^4-x \log (x)\right )}{x^2 \left (3-e^x+x+x^2+x^3-\log (x)\right )}+\frac {3 \left (x-\log \left (3-e^x+x+x^2+x^3-\log (x)\right )\right )}{x^2}\right ) \, dx \\ & = -\left (3 \int \frac {1+2 x-x^2-2 x^3+x^4-x \log (x)}{x^2 \left (3-e^x+x+x^2+x^3-\log (x)\right )} \, dx\right )+3 \int \frac {x-\log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{x^2} \, dx \\ & = -\left (3 \int \frac {\left (1+x-x^2\right )^2-x \log (x)}{x^2 \left (3-e^x+x+x^2+x^3-\log (x)\right )} \, dx\right )+3 \int \left (\frac {1}{x}-\frac {\log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{x^2}\right ) \, dx \\ & = 3 \log (x)-3 \int \left (\frac {1}{x^2 \left (3-e^x+x+x^2+x^3-\log (x)\right )}+\frac {2}{x \left (3-e^x+x+x^2+x^3-\log (x)\right )}-\frac {2 x}{3-e^x+x+x^2+x^3-\log (x)}+\frac {x^2}{3-e^x+x+x^2+x^3-\log (x)}-\frac {\log (x)}{x \left (3-e^x+x+x^2+x^3-\log (x)\right )}+\frac {1}{-3+e^x-x-x^2-x^3+\log (x)}\right ) \, dx-3 \int \frac {\log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{x^2} \, dx \\ & = 3 \log (x)-3 \int \frac {1}{x^2 \left (3-e^x+x+x^2+x^3-\log (x)\right )} \, dx-3 \int \frac {x^2}{3-e^x+x+x^2+x^3-\log (x)} \, dx+3 \int \frac {\log (x)}{x \left (3-e^x+x+x^2+x^3-\log (x)\right )} \, dx-3 \int \frac {1}{-3+e^x-x-x^2-x^3+\log (x)} \, dx-3 \int \frac {\log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{x^2} \, dx-6 \int \frac {1}{x \left (3-e^x+x+x^2+x^3-\log (x)\right )} \, dx+6 \int \frac {x}{3-e^x+x+x^2+x^3-\log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {3-3 x+3 e^x x-6 x^2-9 x^3+\left (9-3 e^x+3 x+3 x^2+3 x^3-3 \log (x)\right ) \log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{-3 x^2+e^x x^2-x^3-x^4-x^5+x^2 \log (x)} \, dx=\frac {3 \log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{x} \]

[In]

Integrate[(3 - 3*x + 3*E^x*x - 6*x^2 - 9*x^3 + (9 - 3*E^x + 3*x + 3*x^2 + 3*x^3 - 3*Log[x])*Log[3 - E^x + x +
x^2 + x^3 - Log[x]])/(-3*x^2 + E^x*x^2 - x^3 - x^4 - x^5 + x^2*Log[x]),x]

[Out]

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

Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {3 \ln \left (-\ln \left (x \right )-{\mathrm e}^{x}+x^{3}+x^{2}+x +3\right )}{x}\)	\(24\)
parallelrisch	\(\frac {3 \ln \left (-\ln \left (x \right )-{\mathrm e}^{x}+x^{3}+x^{2}+x +3\right )}{x}\)	\(24\)

[In]

int(((-3*ln(x)-3*exp(x)+3*x^3+3*x^2+3*x+9)*ln(-ln(x)-exp(x)+x^3+x^2+x+3)+3*exp(x)*x-9*x^3-6*x^2-3*x+3)/(x^2*ln
(x)+exp(x)*x^2-x^5-x^4-x^3-3*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {3-3 x+3 e^x x-6 x^2-9 x^3+\left (9-3 e^x+3 x+3 x^2+3 x^3-3 \log (x)\right ) \log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{-3 x^2+e^x x^2-x^3-x^4-x^5+x^2 \log (x)} \, dx=\frac {3 \, \log \left (x^{3} + x^{2} + x - e^{x} - \log \left (x\right ) + 3\right )}{x} \]

[In]

integrate(((-3*log(x)-3*exp(x)+3*x^3+3*x^2+3*x+9)*log(-log(x)-exp(x)+x^3+x^2+x+3)+3*exp(x)*x-9*x^3-6*x^2-3*x+3
)/(x^2*log(x)+exp(x)*x^2-x^5-x^4-x^3-3*x^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {3-3 x+3 e^x x-6 x^2-9 x^3+\left (9-3 e^x+3 x+3 x^2+3 x^3-3 \log (x)\right ) \log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{-3 x^2+e^x x^2-x^3-x^4-x^5+x^2 \log (x)} \, dx=\frac {3 \log {\left (x^{3} + x^{2} + x - e^{x} - \log {\left (x \right )} + 3 \right )}}{x} \]

[In]

integrate(((-3*ln(x)-3*exp(x)+3*x**3+3*x**2+3*x+9)*ln(-ln(x)-exp(x)+x**3+x**2+x+3)+3*exp(x)*x-9*x**3-6*x**2-3*
x+3)/(x**2*ln(x)+exp(x)*x**2-x**5-x**4-x**3-3*x**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {3-3 x+3 e^x x-6 x^2-9 x^3+\left (9-3 e^x+3 x+3 x^2+3 x^3-3 \log (x)\right ) \log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{-3 x^2+e^x x^2-x^3-x^4-x^5+x^2 \log (x)} \, dx=\frac {3 \, \log \left (x^{3} + x^{2} + x - e^{x} - \log \left (x\right ) + 3\right )}{x} \]

[In]

integrate(((-3*log(x)-3*exp(x)+3*x^3+3*x^2+3*x+9)*log(-log(x)-exp(x)+x^3+x^2+x+3)+3*exp(x)*x-9*x^3-6*x^2-3*x+3
)/(x^2*log(x)+exp(x)*x^2-x^5-x^4-x^3-3*x^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {3-3 x+3 e^x x-6 x^2-9 x^3+\left (9-3 e^x+3 x+3 x^2+3 x^3-3 \log (x)\right ) \log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{-3 x^2+e^x x^2-x^3-x^4-x^5+x^2 \log (x)} \, dx=\frac {3 \, \log \left (x^{3} + x^{2} + x - e^{x} - \log \left (x\right ) + 3\right )}{x} \]

[In]

integrate(((-3*log(x)-3*exp(x)+3*x^3+3*x^2+3*x+9)*log(-log(x)-exp(x)+x^3+x^2+x+3)+3*exp(x)*x-9*x^3-6*x^2-3*x+3
)/(x^2*log(x)+exp(x)*x^2-x^5-x^4-x^3-3*x^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {3-3 x+3 e^x x-6 x^2-9 x^3+\left (9-3 e^x+3 x+3 x^2+3 x^3-3 \log (x)\right ) \log \left (3-e^x+x+x^2+x^3-\log (x)\right )}{-3 x^2+e^x x^2-x^3-x^4-x^5+x^2 \log (x)} \, dx=\frac {3\,\ln \left (x-{\mathrm {e}}^x-\ln \left (x\right )+x^2+x^3+3\right )}{x} \]

[In]

int((3*x - 3*x*exp(x) + 6*x^2 + 9*x^3 - log(x - exp(x) - log(x) + x^2 + x^3 + 3)*(3*x - 3*exp(x) - 3*log(x) +
3*x^2 + 3*x^3 + 9) - 3)/(3*x^2 - x^2*log(x) - x^2*exp(x) + x^3 + x^4 + x^5),x)

[Out]

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