3.55.84 \(\int \frac {e^{2 x} (-6 x^2+3 x^3)+e^x (2 x-4 x^2+2 x^3+(-24 x^2+12 x^3) \log (9))+(e^x (-2+4 x-2 x^2)+(-8+16 x-8 x^2) \log (9)) \log (e^x+4 \log (9))}{e^x (x^2-2 x^3+x^4)+(4 x^2-8 x^3+4 x^4) \log (9)} \, dx\)

Optimal. Leaf size=25 \[ \frac {3 e^x}{-1+x}+\frac {2 \log \left (e^x+4 \log (9)\right )}{x} \]

________________________________________________________________________________________

Rubi [F]  time = 0.51, 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 {e^{2 x} \left (-6 x^2+3 x^3\right )+e^x \left (2 x-4 x^2+2 x^3+\left (-24 x^2+12 x^3\right ) \log (9)\right )+\left (e^x \left (-2+4 x-2 x^2\right )+\left (-8+16 x-8 x^2\right ) \log (9)\right ) \log \left (e^x+4 \log (9)\right )}{e^x \left (x^2-2 x^3+x^4\right )+\left (4 x^2-8 x^3+4 x^4\right ) \log (9)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*x)*(-6*x^2 + 3*x^3) + E^x*(2*x - 4*x^2 + 2*x^3 + (-24*x^2 + 12*x^3)*Log[9]) + (E^x*(-2 + 4*x - 2*x^2
) + (-8 + 16*x - 8*x^2)*Log[9])*Log[E^x + 4*Log[9]])/(E^x*(x^2 - 2*x^3 + x^4) + (4*x^2 - 8*x^3 + 4*x^4)*Log[9]
),x]

[Out]

(-3*E^x)/(1 - x) + 2*Log[x] + (2*Log[E^x + 4*Log[9]])/x - 8*Log[9]*Defer[Int][1/(x*(E^x + 4*Log[9])), x] - 2*D
efer[Int][E^x/(x*(E^x + 4*Log[9])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {e^x x \left (2-2 x \left (2+3 e^x+12 \log (9)\right )+x^2 \left (2+3 e^x+12 \log (9)\right )\right )}{(-1+x)^2 \left (e^x+4 \log (9)\right )}-2 \log \left (e^x+4 \log (9)\right )}{x^2} \, dx\\ &=\int \left (\frac {3 e^x (-2+x)}{(-1+x)^2}-\frac {8 \log (9)}{x \left (e^x+4 \log (9)\right )}+\frac {2 \left (x-\log \left (e^x+4 \log (9)\right )\right )}{x^2}\right ) \, dx\\ &=2 \int \frac {x-\log \left (e^x+4 \log (9)\right )}{x^2} \, dx+3 \int \frac {e^x (-2+x)}{(-1+x)^2} \, dx-(8 \log (9)) \int \frac {1}{x \left (e^x+4 \log (9)\right )} \, dx\\ &=-\frac {3 e^x}{1-x}+2 \int \left (\frac {1}{x}-\frac {\log \left (e^x+4 \log (9)\right )}{x^2}\right ) \, dx-(8 \log (9)) \int \frac {1}{x \left (e^x+4 \log (9)\right )} \, dx\\ &=-\frac {3 e^x}{1-x}+2 \log (x)-2 \int \frac {\log \left (e^x+4 \log (9)\right )}{x^2} \, dx-(8 \log (9)) \int \frac {1}{x \left (e^x+4 \log (9)\right )} \, dx\\ &=-\frac {3 e^x}{1-x}+2 \log (x)+\frac {2 \log \left (e^x+4 \log (9)\right )}{x}-2 \int \frac {e^x}{x \left (e^x+4 \log (9)\right )} \, dx-(8 \log (9)) \int \frac {1}{x \left (e^x+4 \log (9)\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 25, normalized size = 1.00 \begin {gather*} \frac {3 e^x}{-1+x}+\frac {2 \log \left (e^x+4 \log (9)\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(-6*x^2 + 3*x^3) + E^x*(2*x - 4*x^2 + 2*x^3 + (-24*x^2 + 12*x^3)*Log[9]) + (E^x*(-2 + 4*x -
2*x^2) + (-8 + 16*x - 8*x^2)*Log[9])*Log[E^x + 4*Log[9]])/(E^x*(x^2 - 2*x^3 + x^4) + (4*x^2 - 8*x^3 + 4*x^4)*
Log[9]),x]

[Out]

(3*E^x)/(-1 + x) + (2*Log[E^x + 4*Log[9]])/x

________________________________________________________________________________________

fricas [A]  time = 2.03, size = 29, normalized size = 1.16 \begin {gather*} \frac {3 \, x e^{x} + 2 \, {\left (x - 1\right )} \log \left (e^{x} + 8 \, \log \relax (3)\right )}{x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+4*x-2)*exp(x)+2*(-8*x^2+16*x-8)*log(3))*log(exp(x)+8*log(3))+(3*x^3-6*x^2)*exp(x)^2+(2*(12
*x^3-24*x^2)*log(3)+2*x^3-4*x^2+2*x)*exp(x))/((x^4-2*x^3+x^2)*exp(x)+2*(4*x^4-8*x^3+4*x^2)*log(3)),x, algorith
m="fricas")

[Out]

(3*x*e^x + 2*(x - 1)*log(e^x + 8*log(3)))/(x^2 - x)

________________________________________________________________________________________

giac [A]  time = 1.76, size = 37, normalized size = 1.48 \begin {gather*} \frac {3 \, x e^{x} + 2 \, x \log \left (e^{x} + 8 \, \log \relax (3)\right ) - 2 \, \log \left (e^{x} + 8 \, \log \relax (3)\right )}{x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+4*x-2)*exp(x)+2*(-8*x^2+16*x-8)*log(3))*log(exp(x)+8*log(3))+(3*x^3-6*x^2)*exp(x)^2+(2*(12
*x^3-24*x^2)*log(3)+2*x^3-4*x^2+2*x)*exp(x))/((x^4-2*x^3+x^2)*exp(x)+2*(4*x^4-8*x^3+4*x^2)*log(3)),x, algorith
m="giac")

[Out]

(3*x*e^x + 2*x*log(e^x + 8*log(3)) - 2*log(e^x + 8*log(3)))/(x^2 - x)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 24, normalized size = 0.96

method	result	size
risch	\(\frac {2 \ln \left ({\mathrm e}^{x}+8 \ln \relax (3)\right )}{x}+\frac {3 \,{\mathrm e}^{x}}{x -1}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2+4*x-2)*exp(x)+2*(-8*x^2+16*x-8)*ln(3))*ln(exp(x)+8*ln(3))+(3*x^3-6*x^2)*exp(x)^2+(2*(12*x^3-24*x
^2)*ln(3)+2*x^3-4*x^2+2*x)*exp(x))/((x^4-2*x^3+x^2)*exp(x)+2*(4*x^4-8*x^3+4*x^2)*ln(3)),x,method=_RETURNVERBOS
E)

[Out]

2*ln(exp(x)+8*ln(3))/x+3/(x-1)*exp(x)

________________________________________________________________________________________

maxima [A]  time = 0.53, size = 29, normalized size = 1.16 \begin {gather*} \frac {3 \, x e^{x} + 2 \, {\left (x - 1\right )} \log \left (e^{x} + 8 \, \log \relax (3)\right )}{x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2+4*x-2)*exp(x)+2*(-8*x^2+16*x-8)*log(3))*log(exp(x)+8*log(3))+(3*x^3-6*x^2)*exp(x)^2+(2*(12
*x^3-24*x^2)*log(3)+2*x^3-4*x^2+2*x)*exp(x))/((x^4-2*x^3+x^2)*exp(x)+2*(4*x^4-8*x^3+4*x^2)*log(3)),x, algorith
m="maxima")

[Out]

(3*x*e^x + 2*(x - 1)*log(e^x + 8*log(3)))/(x^2 - x)

________________________________________________________________________________________

mupad [B]  time = 0.36, size = 23, normalized size = 0.92 \begin {gather*} \frac {3\,{\mathrm {e}}^x}{x-1}+\frac {2\,\ln \left (8\,\ln \relax (3)+{\mathrm {e}}^x\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(8*log(3) + exp(x))*(2*log(3)*(8*x^2 - 16*x + 8) + exp(x)*(2*x^2 - 4*x + 2)) + exp(2*x)*(6*x^2 - 3*x^
3) - exp(x)*(2*x - 2*log(3)*(24*x^2 - 12*x^3) - 4*x^2 + 2*x^3))/(exp(x)*(x^2 - 2*x^3 + x^4) + 2*log(3)*(4*x^2
- 8*x^3 + 4*x^4)),x)

[Out]

(3*exp(x))/(x - 1) + (2*log(8*log(3) + exp(x)))/x

________________________________________________________________________________________

sympy [A]  time = 0.27, size = 20, normalized size = 0.80 \begin {gather*} \frac {3 e^{x}}{x - 1} + \frac {2 \log {\left (e^{x} + 8 \log {\relax (3 )} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**2+4*x-2)*exp(x)+2*(-8*x**2+16*x-8)*ln(3))*ln(exp(x)+8*ln(3))+(3*x**3-6*x**2)*exp(x)**2+(2*(
12*x**3-24*x**2)*ln(3)+2*x**3-4*x**2+2*x)*exp(x))/((x**4-2*x**3+x**2)*exp(x)+2*(4*x**4-8*x**3+4*x**2)*ln(3)),x
)

[Out]

3*exp(x)/(x - 1) + 2*log(exp(x) + 8*log(3))/x

________________________________________________________________________________________