3.96.81 \(\int \frac {e^x (40-20 x)+e^x (10-5 x) \log (-2+x)+\frac {e^{4+\log ^2(4+\log (-2+x))} (e^x (-12+4 x)+e^x (-2+x) \log (-2+x)+2 e^x \log (4+\log (-2+x)))}{(4+\log (-2+x))^4}}{-8+4 x+(-2+x) \log (-2+x)} \, dx\)

Optimal. Leaf size=19 \[ e^x \left (-5+e^{(-2+\log (4+\log (-2+x)))^2}\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 8.46, 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^x (40-20 x)+e^x (10-5 x) \log (-2+x)+\frac {e^{4+\log ^2(4+\log (-2+x))} \left (e^x (-12+4 x)+e^x (-2+x) \log (-2+x)+2 e^x \log (4+\log (-2+x))\right )}{(4+\log (-2+x))^4}}{-8+4 x+(-2+x) \log (-2+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(40 - 20*x) + E^x*(10 - 5*x)*Log[-2 + x] + (E^(4 + Log[4 + Log[-2 + x]]^2)*(E^x*(-12 + 4*x) + E^x*(-2
+ x)*Log[-2 + x] + 2*E^x*Log[4 + Log[-2 + x]]))/(4 + Log[-2 + x])^4)/(-8 + 4*x + (-2 + x)*Log[-2 + x]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^x (40-20 x)-e^x (10-5 x) \log (-2+x)-\frac {e^{4+\log ^2(4+\log (-2+x))} \left (e^x (-12+4 x)+e^x (-2+x) \log (-2+x)+2 e^x \log (4+\log (-2+x))\right )}{(4+\log (-2+x))^4}}{(2-x) (4+\log (-2+x))} \, dx\\ &=\int \left (-5 e^x+\frac {e^{4+x+\log ^2(4+\log (-2+x))} (-12+4 x-2 \log (-2+x)+x \log (-2+x)+2 \log (4+\log (-2+x)))}{(-2+x) (4+\log (-2+x))^5}\right ) \, dx\\ &=-\left (5 \int e^x \, dx\right )+\int \frac {e^{4+x+\log ^2(4+\log (-2+x))} (-12+4 x-2 \log (-2+x)+x \log (-2+x)+2 \log (4+\log (-2+x)))}{(-2+x) (4+\log (-2+x))^5} \, dx\\ &=-5 e^x+\int \left (-\frac {12 e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^5}+\frac {4 e^{4+x+\log ^2(4+\log (-2+x))} x}{(-2+x) (4+\log (-2+x))^5}-\frac {2 e^{4+x+\log ^2(4+\log (-2+x))} \log (-2+x)}{(-2+x) (4+\log (-2+x))^5}+\frac {e^{4+x+\log ^2(4+\log (-2+x))} x \log (-2+x)}{(-2+x) (4+\log (-2+x))^5}+\frac {2 e^{4+x+\log ^2(4+\log (-2+x))} \log (4+\log (-2+x))}{(-2+x) (4+\log (-2+x))^5}\right ) \, dx\\ &=-5 e^x-2 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))} \log (-2+x)}{(-2+x) (4+\log (-2+x))^5} \, dx+2 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))} \log (4+\log (-2+x))}{(-2+x) (4+\log (-2+x))^5} \, dx+4 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))} x}{(-2+x) (4+\log (-2+x))^5} \, dx-12 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^5} \, dx+\int \frac {e^{4+x+\log ^2(4+\log (-2+x))} x \log (-2+x)}{(-2+x) (4+\log (-2+x))^5} \, dx\\ &=-5 e^x-2 \int \left (-\frac {4 e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^5}+\frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^4}\right ) \, dx+2 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))} \log (4+\log (-2+x))}{(-2+x) (4+\log (-2+x))^5} \, dx+4 \int \left (\frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(4+\log (-2+x))^5}+\frac {2 e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^5}\right ) \, dx-12 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^5} \, dx+\int \left (-\frac {4 e^{4+x+\log ^2(4+\log (-2+x))} x}{(-2+x) (4+\log (-2+x))^5}+\frac {e^{4+x+\log ^2(4+\log (-2+x))} x}{(-2+x) (4+\log (-2+x))^4}\right ) \, dx\\ &=-5 e^x-2 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^4} \, dx+2 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))} \log (4+\log (-2+x))}{(-2+x) (4+\log (-2+x))^5} \, dx+4 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(4+\log (-2+x))^5} \, dx-4 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))} x}{(-2+x) (4+\log (-2+x))^5} \, dx+2 \left (8 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^5} \, dx\right )-12 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^5} \, dx+\int \frac {e^{4+x+\log ^2(4+\log (-2+x))} x}{(-2+x) (4+\log (-2+x))^4} \, dx\\ &=-5 e^x-2 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^4} \, dx+2 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))} \log (4+\log (-2+x))}{(-2+x) (4+\log (-2+x))^5} \, dx+4 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(4+\log (-2+x))^5} \, dx-4 \int \left (\frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(4+\log (-2+x))^5}+\frac {2 e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^5}\right ) \, dx+2 \left (8 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^5} \, dx\right )-12 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^5} \, dx+\int \left (\frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(4+\log (-2+x))^4}+\frac {2 e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^4}\right ) \, dx\\ &=-5 e^x+2 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))} \log (4+\log (-2+x))}{(-2+x) (4+\log (-2+x))^5} \, dx+8 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^5} \, dx-12 \int \frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(-2+x) (4+\log (-2+x))^5} \, dx+\int \frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(4+\log (-2+x))^4} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 2.48, size = 29, normalized size = 1.53 \begin {gather*} -5 e^x+\frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(4+\log (-2+x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(40 - 20*x) + E^x*(10 - 5*x)*Log[-2 + x] + (E^(4 + Log[4 + Log[-2 + x]]^2)*(E^x*(-12 + 4*x) + E
^x*(-2 + x)*Log[-2 + x] + 2*E^x*Log[4 + Log[-2 + x]]))/(4 + Log[-2 + x])^4)/(-8 + 4*x + (-2 + x)*Log[-2 + x]),
x]

[Out]

-5*E^x + E^(4 + x + Log[4 + Log[-2 + x]]^2)/(4 + Log[-2 + x])^4

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 27, normalized size = 1.42 \begin {gather*} e^{\left (\log \left (\log \left (x - 2\right ) + 4\right )^{2} + x - 4 \, \log \left (\log \left (x - 2\right ) + 4\right ) + 4\right )} - 5 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*log(log(x-2)+4)+(x-2)*exp(x)*log(x-2)+(4*x-12)*exp(x))*exp(log(log(x-2)+4)^2-4*log(log(x-
2)+4)+4)+(-5*x+10)*exp(x)*log(x-2)+(-20*x+40)*exp(x))/((x-2)*log(x-2)+4*x-8),x, algorithm="fricas")

[Out]

e^(log(log(x - 2) + 4)^2 + x - 4*log(log(x - 2) + 4) + 4) - 5*e^x

________________________________________________________________________________________

giac [B]  time = 0.92, size = 92, normalized size = 4.84 \begin {gather*} -\frac {5 \, e^{x} \log \left (x - 2\right )^{4} + 80 \, e^{x} \log \left (x - 2\right )^{3} + 480 \, e^{x} \log \left (x - 2\right )^{2} + 1280 \, e^{x} \log \left (x - 2\right ) - e^{\left (\log \left (\log \left (x - 2\right ) + 4\right )^{2} + x + 4\right )} + 1280 \, e^{x}}{\log \left (x - 2\right )^{4} + 16 \, \log \left (x - 2\right )^{3} + 96 \, \log \left (x - 2\right )^{2} + 256 \, \log \left (x - 2\right ) + 256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*log(log(x-2)+4)+(x-2)*exp(x)*log(x-2)+(4*x-12)*exp(x))*exp(log(log(x-2)+4)^2-4*log(log(x-
2)+4)+4)+(-5*x+10)*exp(x)*log(x-2)+(-20*x+40)*exp(x))/((x-2)*log(x-2)+4*x-8),x, algorithm="giac")

[Out]

-(5*e^x*log(x - 2)^4 + 80*e^x*log(x - 2)^3 + 480*e^x*log(x - 2)^2 + 1280*e^x*log(x - 2) - e^(log(log(x - 2) +
4)^2 + x + 4) + 1280*e^x)/(log(x - 2)^4 + 16*log(x - 2)^3 + 96*log(x - 2)^2 + 256*log(x - 2) + 256)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 28, normalized size = 1.47

method	result	size
risch	\(\frac {{\mathrm e}^{x +\ln \left (\ln \left (x -2\right )+4\right )^{2}+4}}{\left (\ln \left (x -2\right )+4\right )^{4}}-5 \,{\mathrm e}^{x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)*ln(ln(x-2)+4)+(x-2)*exp(x)*ln(x-2)+(4*x-12)*exp(x))*exp(ln(ln(x-2)+4)^2-4*ln(ln(x-2)+4)+4)+(-5*
x+10)*exp(x)*ln(x-2)+(-20*x+40)*exp(x))/((x-2)*ln(x-2)+4*x-8),x,method=_RETURNVERBOSE)

[Out]

1/(ln(x-2)+4)^4*exp(x+ln(ln(x-2)+4)^2+4)-5*exp(x)

________________________________________________________________________________________

maxima [B]  time = 0.40, size = 92, normalized size = 4.84 \begin {gather*} -\frac {5 \, e^{x} \log \left (x - 2\right )^{4} + 80 \, e^{x} \log \left (x - 2\right )^{3} + 480 \, e^{x} \log \left (x - 2\right )^{2} + 1280 \, e^{x} \log \left (x - 2\right ) - e^{\left (\log \left (\log \left (x - 2\right ) + 4\right )^{2} + x + 4\right )} + 1280 \, e^{x}}{\log \left (x - 2\right )^{4} + 16 \, \log \left (x - 2\right )^{3} + 96 \, \log \left (x - 2\right )^{2} + 256 \, \log \left (x - 2\right ) + 256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*log(log(x-2)+4)+(x-2)*exp(x)*log(x-2)+(4*x-12)*exp(x))*exp(log(log(x-2)+4)^2-4*log(log(x-
2)+4)+4)+(-5*x+10)*exp(x)*log(x-2)+(-20*x+40)*exp(x))/((x-2)*log(x-2)+4*x-8),x, algorithm="maxima")

[Out]

-(5*e^x*log(x - 2)^4 + 80*e^x*log(x - 2)^3 + 480*e^x*log(x - 2)^2 + 1280*e^x*log(x - 2) - e^(log(log(x - 2) +
4)^2 + x + 4) + 1280*e^x)/(log(x - 2)^4 + 16*log(x - 2)^3 + 96*log(x - 2)^2 + 256*log(x - 2) + 256)

________________________________________________________________________________________

mupad [B]  time = 8.21, size = 58, normalized size = 3.05 \begin {gather*} -\frac {{\mathrm {e}}^x\,\left (1280\,\ln \left (x-2\right )-{\mathrm {e}}^4\,{\mathrm {e}}^{{\ln \left (\ln \left (x-2\right )+4\right )}^2}+480\,{\ln \left (x-2\right )}^2+80\,{\ln \left (x-2\right )}^3+5\,{\ln \left (x-2\right )}^4+1280\right )}{{\left (\ln \left (x-2\right )+4\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(20*x - 40) - exp(log(log(x - 2) + 4)^2 - 4*log(log(x - 2) + 4) + 4)*(2*log(log(x - 2) + 4)*exp(x
) + exp(x)*(4*x - 12) + log(x - 2)*exp(x)*(x - 2)) + log(x - 2)*exp(x)*(5*x - 10))/(4*x + log(x - 2)*(x - 2) -
8),x)

[Out]

-(exp(x)*(1280*log(x - 2) - exp(4)*exp(log(log(x - 2) + 4)^2) + 480*log(x - 2)^2 + 80*log(x - 2)^3 + 5*log(x -
2)^4 + 1280))/(log(x - 2) + 4)^4

________________________________________________________________________________________

sympy [B]  time = 0.98, size = 53, normalized size = 2.79 \begin {gather*} - 5 e^{x} + \frac {e^{x} e^{\log {\left (\log {\left (x - 2 \right )} + 4 \right )}^{2} + 4}}{\log {\left (x - 2 \right )}^{4} + 16 \log {\left (x - 2 \right )}^{3} + 96 \log {\left (x - 2 \right )}^{2} + 256 \log {\left (x - 2 \right )} + 256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*ln(ln(x-2)+4)+(x-2)*exp(x)*ln(x-2)+(4*x-12)*exp(x))*exp(ln(ln(x-2)+4)**2-4*ln(ln(x-2)+4)+
4)+(-5*x+10)*exp(x)*ln(x-2)+(-20*x+40)*exp(x))/((x-2)*ln(x-2)+4*x-8),x)

[Out]

-5*exp(x) + exp(x)*exp(log(log(x - 2) + 4)**2 + 4)/(log(x - 2)**4 + 16*log(x - 2)**3 + 96*log(x - 2)**2 + 256*
log(x - 2) + 256)

________________________________________________________________________________________