\(\int \frac {e^x x^2+(4 x-2 e^x x-2 x \log (3)) \log (-2+e^x+\log (3))}{(-2+e^x+\log (3)) \log ^2(-2+e^x+\log (3))} \, dx\) [6302]

Optimal result

Integrand size = 52, antiderivative size = 17 \[ \int \frac {e^x x^2+\left (4 x-2 e^x x-2 x \log (3)\right ) \log \left (-2+e^x+\log (3)\right )}{\left (-2+e^x+\log (3)\right ) \log ^2\left (-2+e^x+\log (3)\right )} \, dx=7-\frac {x^2}{\log \left (-2+e^x+\log (3)\right )} \]

[Out]

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

Rubi [F]

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

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^x x^2+\left (4 x-2 e^x x-2 x \log (3)\right ) \log \left (-2+e^x+\log (3)\right )}{\left (-2+e^x+\log (3)\right ) \log ^2\left (-2+e^x+\log (3)\right )} \, dx=-\frac {x^2}{\log \left (-2+e^x+\log (3)\right )} \]

[In]

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

[Out]

-(x^2/Log[-2 + E^x + Log[3]])

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

method	result	size
norman	\(-\frac {x^{2}}{\ln \left ({\mathrm e}^{x}+\ln \left (3\right )-2\right )}\)	\(15\)
risch	\(-\frac {x^{2}}{\ln \left ({\mathrm e}^{x}+\ln \left (3\right )-2\right )}\)	\(15\)
parallelrisch	\(-\frac {x^{2}}{\ln \left ({\mathrm e}^{x}+\ln \left (3\right )-2\right )}\)	\(15\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^x x^2+\left (4 x-2 e^x x-2 x \log (3)\right ) \log \left (-2+e^x+\log (3)\right )}{\left (-2+e^x+\log (3)\right ) \log ^2\left (-2+e^x+\log (3)\right )} \, dx=-\frac {x^{2}}{\log \left (e^{x} + \log \left (3\right ) - 2\right )} \]

[In]

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

[Out]

-x^2/log(e^x + log(3) - 2)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^x x^2+\left (4 x-2 e^x x-2 x \log (3)\right ) \log \left (-2+e^x+\log (3)\right )}{\left (-2+e^x+\log (3)\right ) \log ^2\left (-2+e^x+\log (3)\right )} \, dx=- \frac {x^{2}}{\log {\left (e^{x} - 2 + \log {\left (3 \right )} \right )}} \]

[In]

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

[Out]

-x**2/log(exp(x) - 2 + log(3))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^x x^2+\left (4 x-2 e^x x-2 x \log (3)\right ) \log \left (-2+e^x+\log (3)\right )}{\left (-2+e^x+\log (3)\right ) \log ^2\left (-2+e^x+\log (3)\right )} \, dx=-\frac {x^{2}}{\log \left (e^{x} + \log \left (3\right ) - 2\right )} \]

[In]

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

[Out]

-x^2/log(e^x + log(3) - 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^x x^2+\left (4 x-2 e^x x-2 x \log (3)\right ) \log \left (-2+e^x+\log (3)\right )}{\left (-2+e^x+\log (3)\right ) \log ^2\left (-2+e^x+\log (3)\right )} \, dx=-\frac {x^{2}}{\log \left (e^{x} + \log \left (3\right ) - 2\right )} \]

[In]

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

[Out]

-x^2/log(e^x + log(3) - 2)

Mupad [B] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^x x^2+\left (4 x-2 e^x x-2 x \log (3)\right ) \log \left (-2+e^x+\log (3)\right )}{\left (-2+e^x+\log (3)\right ) \log ^2\left (-2+e^x+\log (3)\right )} \, dx=-\frac {x^2}{\ln \left (\ln \left (3\right )+{\mathrm {e}}^x-2\right )} \]

[In]

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

[Out]

-x^2/log(log(3) + exp(x) - 2)