\(\int \frac {9 e^x-9 x+(e^x (-9-18 x)+27 x) \log (2 x)}{2 e^{3 x} x^2-6 e^{2 x} x^3+6 e^x x^4-2 x^5} \, dx\) [3499]

Optimal result

Integrand size = 64, antiderivative size = 20 \[ \int \frac {9 e^x-9 x+\left (e^x (-9-18 x)+27 x\right ) \log (2 x)}{2 e^{3 x} x^2-6 e^{2 x} x^3+6 e^x x^4-2 x^5} \, dx=\frac {9 \log (2 x)}{2 x \left (-e^x+x\right )^2} \]

[Out]

9/2*ln(2*x)/x/(x-exp(x))^2

Rubi [F]

\[ \int \frac {9 e^x-9 x+\left (e^x (-9-18 x)+27 x\right ) \log (2 x)}{2 e^{3 x} x^2-6 e^{2 x} x^3+6 e^x x^4-2 x^5} \, dx=\int \frac {9 e^x-9 x+\left (e^x (-9-18 x)+27 x\right ) \log (2 x)}{2 e^{3 x} x^2-6 e^{2 x} x^3+6 e^x x^4-2 x^5} \, dx \]

[In]

Int[(9*E^x - 9*x + (E^x*(-9 - 18*x) + 27*x)*Log[2*x])/(2*E^(3*x)*x^2 - 6*E^(2*x)*x^3 + 6*E^x*x^4 - 2*x^5),x]

[Out]

-9*Log[2*x]*Defer[Int][(E^x - x)^(-3), x] + (9*Defer[Int][1/((E^x - x)^2*x^2), x])/2 - (9*Log[2*x]*Defer[Int][
1/((E^x - x)^2*x^2), x])/2 - 9*Log[2*x]*Defer[Int][1/((E^x - x)^2*x), x] - 9*Log[2*x]*Defer[Int][1/(x*(-E^x +
x)^3), x] + 9*Defer[Int][Defer[Int][(E^x - x)^(-3), x]/x, x] + (9*Defer[Int][Defer[Int][1/((E^x - x)^2*x^2), x
]/x, x])/2 + 9*Defer[Int][Defer[Int][1/((E^x - x)^2*x), x]/x, x] + 9*Defer[Int][Defer[Int][1/(x*(-E^x + x)^3),
x]/x, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {9 \left (e^x-x-\left (-3 x+e^x (1+2 x)\right ) \log (2 x)\right )}{2 \left (e^x-x\right )^3 x^2} \, dx \\ & = \frac {9}{2} \int \frac {e^x-x-\left (-3 x+e^x (1+2 x)\right ) \log (2 x)}{\left (e^x-x\right )^3 x^2} \, dx \\ & = \frac {9}{2} \int \left (-\frac {2 (-1+x) \log (2 x)}{\left (e^x-x\right )^3 x}-\frac {-1+\log (2 x)+2 x \log (2 x)}{\left (e^x-x\right )^2 x^2}\right ) \, dx \\ & = -\left (\frac {9}{2} \int \frac {-1+\log (2 x)+2 x \log (2 x)}{\left (e^x-x\right )^2 x^2} \, dx\right )-9 \int \frac {(-1+x) \log (2 x)}{\left (e^x-x\right )^3 x} \, dx \\ & = -\left (\frac {9}{2} \int \left (-\frac {1}{\left (e^x-x\right )^2 x^2}+\frac {\log (2 x)}{\left (e^x-x\right )^2 x^2}+\frac {2 \log (2 x)}{\left (e^x-x\right )^2 x}\right ) \, dx\right )+9 \int \frac {\int \frac {1}{\left (e^x-x\right )^3} \, dx+\int \frac {1}{x \left (-e^x+x\right )^3} \, dx}{x} \, dx-(9 \log (2 x)) \int \frac {1}{\left (e^x-x\right )^3} \, dx-(9 \log (2 x)) \int \frac {1}{x \left (-e^x+x\right )^3} \, dx \\ & = \frac {9}{2} \int \frac {1}{\left (e^x-x\right )^2 x^2} \, dx-\frac {9}{2} \int \frac {\log (2 x)}{\left (e^x-x\right )^2 x^2} \, dx-9 \int \frac {\log (2 x)}{\left (e^x-x\right )^2 x} \, dx+9 \int \left (\frac {\int \frac {1}{\left (e^x-x\right )^3} \, dx}{x}+\frac {\int \frac {1}{x \left (-e^x+x\right )^3} \, dx}{x}\right ) \, dx-(9 \log (2 x)) \int \frac {1}{\left (e^x-x\right )^3} \, dx-(9 \log (2 x)) \int \frac {1}{x \left (-e^x+x\right )^3} \, dx \\ & = \frac {9}{2} \int \frac {1}{\left (e^x-x\right )^2 x^2} \, dx+\frac {9}{2} \int \frac {\int \frac {1}{\left (e^x-x\right )^2 x^2} \, dx}{x} \, dx+9 \int \frac {\int \frac {1}{\left (e^x-x\right )^3} \, dx}{x} \, dx+9 \int \frac {\int \frac {1}{\left (e^x-x\right )^2 x} \, dx}{x} \, dx+9 \int \frac {\int \frac {1}{x \left (-e^x+x\right )^3} \, dx}{x} \, dx-\frac {1}{2} (9 \log (2 x)) \int \frac {1}{\left (e^x-x\right )^2 x^2} \, dx-(9 \log (2 x)) \int \frac {1}{\left (e^x-x\right )^3} \, dx-(9 \log (2 x)) \int \frac {1}{\left (e^x-x\right )^2 x} \, dx-(9 \log (2 x)) \int \frac {1}{x \left (-e^x+x\right )^3} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {9 e^x-9 x+\left (e^x (-9-18 x)+27 x\right ) \log (2 x)}{2 e^{3 x} x^2-6 e^{2 x} x^3+6 e^x x^4-2 x^5} \, dx=\frac {9 \log (2 x)}{2 \left (e^x-x\right )^2 x} \]

[In]

Integrate[(9*E^x - 9*x + (E^x*(-9 - 18*x) + 27*x)*Log[2*x])/(2*E^(3*x)*x^2 - 6*E^(2*x)*x^3 + 6*E^x*x^4 - 2*x^5
),x]

[Out]

(9*Log[2*x])/(2*(E^x - x)^2*x)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {9 \ln \left (2 x \right )}{2 x \left (x -{\mathrm e}^{x}\right )^{2}}\)	\(18\)
parallelrisch	\(\frac {9 \ln \left (2 x \right )}{2 x \left ({\mathrm e}^{2 x}-2 \,{\mathrm e}^{x} x +x^{2}\right )}\)	\(25\)

[In]

int((((-18*x-9)*exp(x)+27*x)*ln(2*x)+9*exp(x)-9*x)/(2*x^2*exp(x)^3-6*exp(x)^2*x^3+6*exp(x)*x^4-2*x^5),x,method
=_RETURNVERBOSE)

[Out]

9/2*ln(2*x)/x/(x-exp(x))^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {9 e^x-9 x+\left (e^x (-9-18 x)+27 x\right ) \log (2 x)}{2 e^{3 x} x^2-6 e^{2 x} x^3+6 e^x x^4-2 x^5} \, dx=\frac {9 \, \log \left (2 \, x\right )}{2 \, {\left (x^{3} - 2 \, x^{2} e^{x} + x e^{\left (2 \, x\right )}\right )}} \]

[In]

integrate((((-18*x-9)*exp(x)+27*x)*log(2*x)+9*exp(x)-9*x)/(2*x^2*exp(x)^3-6*exp(x)^2*x^3+6*exp(x)*x^4-2*x^5),x
, algorithm="fricas")

[Out]

9/2*log(2*x)/(x^3 - 2*x^2*e^x + x*e^(2*x))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {9 e^x-9 x+\left (e^x (-9-18 x)+27 x\right ) \log (2 x)}{2 e^{3 x} x^2-6 e^{2 x} x^3+6 e^x x^4-2 x^5} \, dx=\frac {9 \log {\left (2 x \right )}}{2 x^{3} - 4 x^{2} e^{x} + 2 x e^{2 x}} \]

[In]

integrate((((-18*x-9)*exp(x)+27*x)*ln(2*x)+9*exp(x)-9*x)/(2*x**2*exp(x)**3-6*exp(x)**2*x**3+6*exp(x)*x**4-2*x*
*5),x)

[Out]

9*log(2*x)/(2*x**3 - 4*x**2*exp(x) + 2*x*exp(2*x))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {9 e^x-9 x+\left (e^x (-9-18 x)+27 x\right ) \log (2 x)}{2 e^{3 x} x^2-6 e^{2 x} x^3+6 e^x x^4-2 x^5} \, dx=\frac {9 \, {\left (\log \left (2\right ) + \log \left (x\right )\right )}}{2 \, {\left (x^{3} - 2 \, x^{2} e^{x} + x e^{\left (2 \, x\right )}\right )}} \]

[In]

integrate((((-18*x-9)*exp(x)+27*x)*log(2*x)+9*exp(x)-9*x)/(2*x^2*exp(x)^3-6*exp(x)^2*x^3+6*exp(x)*x^4-2*x^5),x
, algorithm="maxima")

[Out]

9/2*(log(2) + log(x))/(x^3 - 2*x^2*e^x + x*e^(2*x))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {9 e^x-9 x+\left (e^x (-9-18 x)+27 x\right ) \log (2 x)}{2 e^{3 x} x^2-6 e^{2 x} x^3+6 e^x x^4-2 x^5} \, dx=\frac {9 \, \log \left (2 \, x\right )}{2 \, {\left (x^{3} - 2 \, x^{2} e^{x} + x e^{\left (2 \, x\right )}\right )}} \]

[In]

integrate((((-18*x-9)*exp(x)+27*x)*log(2*x)+9*exp(x)-9*x)/(2*x^2*exp(x)^3-6*exp(x)^2*x^3+6*exp(x)*x^4-2*x^5),x
, algorithm="giac")

[Out]

9/2*log(2*x)/(x^3 - 2*x^2*e^x + x*e^(2*x))

Mupad [B] (verification not implemented)

Time = 9.52 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {9 e^x-9 x+\left (e^x (-9-18 x)+27 x\right ) \log (2 x)}{2 e^{3 x} x^2-6 e^{2 x} x^3+6 e^x x^4-2 x^5} \, dx=\frac {9\,\ln \left (2\,x\right )}{2\,x\,{\mathrm {e}}^{2\,x}-4\,x^2\,{\mathrm {e}}^x+2\,x^3} \]

[In]

int((9*exp(x) - 9*x + log(2*x)*(27*x - exp(x)*(18*x + 9)))/(6*x^4*exp(x) + 2*x^2*exp(3*x) - 6*x^3*exp(2*x) - 2
*x^5),x)

[Out]

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