3.3.32 \(\int \frac {e^{5+e^x+x} (2-\log (\frac {\log ^2(-2+x)}{x^2})) (2 x+(8-2 x^2+e^x (4 x-2 x^2)) \log (-2+x)+(-2-x+x^2+e^x (-2 x+x^2)) \log (-2+x) \log (\frac {\log ^2(-2+x)}{x^2}))}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log (\frac {\log ^2(-2+x)}{x^2})} \, dx\)

Optimal. Leaf size=25 \[ e^{5+e^x+x} x \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 2.12, 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^{5+e^x+x} \left (2-\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \left (2 x+\left (8-2 x^2+e^x \left (4 x-2 x^2\right )\right ) \log (-2+x)+\left (-2-x+x^2+e^x \left (-2 x+x^2\right )\right ) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right )}{(4-2 x) \log (-2+x)+(-2+x) \log (-2+x) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

2*E^(5 + E^x) - (E^(5 + E^x + x)*(x + E^x*x)*Log[Log[-2 + x]^2/x^2])/(1 + E^x) + 2*Defer[Int][E^(5 + E^x + x)*
x, x] + 2*Defer[Int][E^(5 + E^x + 2*x)*x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int e^{5+e^x+x} \left (2 \left (2+x+e^x x\right )-\frac {2 x}{(-2+x) \log (-2+x)}-\left (1+x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \, dx\\ &=\int \left (2 e^{5+e^x+x} \left (2+x+e^x x\right )-\frac {2 e^{5+e^x+x} x}{(-2+x) \log (-2+x)}-e^{5+e^x+x} \left (1+x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \, dx\\ &=2 \int e^{5+e^x+x} \left (2+x+e^x x\right ) \, dx-2 \int \frac {e^{5+e^x+x} x}{(-2+x) \log (-2+x)} \, dx-\int e^{5+e^x+x} \left (1+x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right ) \, dx\\ &=-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int \left (2 e^{5+e^x+x}+e^{5+e^x+x} x+e^{5+e^x+2 x} x\right ) \, dx-2 \int \left (\frac {e^{5+e^x+x}}{\log (-2+x)}+\frac {2 e^{5+e^x+x}}{(-2+x) \log (-2+x)}\right ) \, dx+\int 2 e^{5+e^x+x} \left (-1+\frac {x}{(-2+x) \log (-2+x)}\right ) \, dx\\ &=-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx+2 \int e^{5+e^x+x} \left (-1+\frac {x}{(-2+x) \log (-2+x)}\right ) \, dx-2 \int \frac {e^{5+e^x+x}}{\log (-2+x)} \, dx+4 \int e^{5+e^x+x} \, dx-4 \int \frac {e^{5+e^x+x}}{(-2+x) \log (-2+x)} \, dx\\ &=-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx+2 \int \left (-e^{5+e^x+x}+\frac {e^{5+e^x+x} x}{(-2+x) \log (-2+x)}\right ) \, dx-2 \int \frac {e^{5+e^x+x}}{\log (-2+x)} \, dx-4 \int \frac {e^{5+e^x+x}}{(-2+x) \log (-2+x)} \, dx+4 \operatorname {Subst}\left (\int e^{5+x} \, dx,x,e^x\right )\\ &=4 e^{5+e^x}-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}-2 \int e^{5+e^x+x} \, dx+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx-2 \int \frac {e^{5+e^x+x}}{\log (-2+x)} \, dx+2 \int \frac {e^{5+e^x+x} x}{(-2+x) \log (-2+x)} \, dx-4 \int \frac {e^{5+e^x+x}}{(-2+x) \log (-2+x)} \, dx\\ &=4 e^{5+e^x}-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx+2 \int \left (\frac {e^{5+e^x+x}}{\log (-2+x)}+\frac {2 e^{5+e^x+x}}{(-2+x) \log (-2+x)}\right ) \, dx-2 \int \frac {e^{5+e^x+x}}{\log (-2+x)} \, dx-2 \operatorname {Subst}\left (\int e^{5+x} \, dx,x,e^x\right )-4 \int \frac {e^{5+e^x+x}}{(-2+x) \log (-2+x)} \, dx\\ &=2 e^{5+e^x}-\frac {e^{5+e^x+x} \left (x+e^x x\right ) \log \left (\frac {\log ^2(-2+x)}{x^2}\right )}{1+e^x}+2 \int e^{5+e^x+x} x \, dx+2 \int e^{5+e^x+2 x} x \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 1.67, size = 24, normalized size = 0.96 \begin {gather*} -e^{5+e^x+x} x \left (-2+\log \left (\frac {\log ^2(-2+x)}{x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.84, size = 24, normalized size = 0.96 \begin {gather*} x e^{\left (x + e^{x} + \log \left (-\log \left (\frac {\log \left (x - 2\right )^{2}}{x^{2}}\right ) + 2\right ) + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-2*x)*exp(x)+x^2-x-2)*log(x-2)*log(log(x-2)^2/x^2)+((-2*x^2+4*x)*exp(x)-2*x^2+8)*log(x-2)+2*x)
*exp(log(-log(log(x-2)^2/x^2)+2)+exp(x)+5+x)/((x-2)*log(x-2)*log(log(x-2)^2/x^2)+(4-2*x)*log(x-2)),x, algorith
m="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-2*x)*exp(x)+x^2-x-2)*log(x-2)*log(log(x-2)^2/x^2)+((-2*x^2+4*x)*exp(x)-2*x^2+8)*log(x-2)+2*x)
*exp(log(-log(log(x-2)^2/x^2)+2)+exp(x)+5+x)/((x-2)*log(x-2)*log(log(x-2)^2/x^2)+(4-2*x)*log(x-2)),x, algorith
m="giac")

[Out]

undef

________________________________________________________________________________________

maple [C]  time = 0.38, size = 627, normalized size = 25.08

method	result	size
risch	\(\frac {2 i \left (-\frac {i \pi x \,\mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )^{2}}{2}+\frac {i \pi x \mathrm {csgn}\left (i \ln \left (x -2\right )\right )^{2} \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )}{2}-i \pi x \,\mathrm {csgn}\left (i \ln \left (x -2\right )\right ) \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )^{2}-\frac {i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}+i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )^{2}}{2}+\frac {i \pi x \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )^{3}}{2}+\frac {i \pi x \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )^{3}}{2}-\frac {i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+2 x +\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )}{2}+2 x \ln \relax (x )-2 x \ln \left (\ln \left (x -2\right )\right )\right ) \left (2 \ln \relax (x )-2 \ln \left (\ln \left (x -2\right )\right )+\frac {i \pi \,\mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right ) \left (-\mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )+\mathrm {csgn}\left (i \ln \left (x -2\right )\right )\right )^{2}}{2}-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}+\frac {i \pi \,\mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right ) \left (-\mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )+\mathrm {csgn}\left (\frac {i}{x^{2}}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )+\mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )\right )}{2}+2\right ) {\mathrm e}^{{\mathrm e}^{x}+5+x}}{4 i \ln \relax (x )+4 i-4 i \ln \left (\ln \left (x -2\right )\right )-\pi \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )^{3}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-\pi \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )-\pi \mathrm {csgn}\left (i \ln \left (x -2\right )\right )^{2} \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )+2 \pi \,\mathrm {csgn}\left (i \ln \left (x -2\right )\right ) \mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )^{2}+\pi \,\mathrm {csgn}\left (i \ln \left (x -2\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \left (x -2\right )^{2}}{x^{2}}\right )^{2}}\)	\(627\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2-2*x)*exp(x)+x^2-x-2)*ln(x-2)*ln(ln(x-2)^2/x^2)+((-2*x^2+4*x)*exp(x)-2*x^2+8)*ln(x-2)+2*x)*exp(ln(-l
n(ln(x-2)^2/x^2)+2)+exp(x)+5+x)/((x-2)*ln(x-2)*ln(ln(x-2)^2/x^2)+(4-2*x)*ln(x-2)),x,method=_RETURNVERBOSE)

[Out]

2*I*(-1/2*I*Pi*x*csgn(I*ln(x-2)^2)*csgn(I/x^2*ln(x-2)^2)^2+1/2*I*Pi*x*csgn(I*ln(x-2))^2*csgn(I*ln(x-2)^2)-I*Pi
*x*csgn(I*ln(x-2))*csgn(I*ln(x-2)^2)^2-1/2*I*Pi*x*csgn(I*x)^2*csgn(I*x^2)+I*Pi*x*csgn(I*x)*csgn(I*x^2)^2-1/2*I
*Pi*x*csgn(I/x^2)*csgn(I/x^2*ln(x-2)^2)^2+1/2*I*Pi*x*csgn(I/x^2*ln(x-2)^2)^3+1/2*I*Pi*x*csgn(I*ln(x-2)^2)^3-1/
2*I*Pi*x*csgn(I*x^2)^3+2*x+1/2*I*Pi*x*csgn(I/x^2)*csgn(I*ln(x-2)^2)*csgn(I/x^2*ln(x-2)^2)+2*x*ln(x)-2*x*ln(ln(
x-2)))/(4*I*ln(x)+4*I-4*I*ln(ln(x-2))-Pi*csgn(I/x^2*ln(x-2)^2)^3+Pi*csgn(I*x^2)^3-Pi*csgn(I*ln(x-2)^2)^3-Pi*cs
gn(I/x^2)*csgn(I*ln(x-2)^2)*csgn(I/x^2*ln(x-2)^2)-Pi*csgn(I*ln(x-2))^2*csgn(I*ln(x-2)^2)+2*Pi*csgn(I*ln(x-2))*
csgn(I*ln(x-2)^2)^2+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I/x^2)*csgn(I/x^2*ln(x-2)^
2)^2+Pi*csgn(I*ln(x-2)^2)*csgn(I/x^2*ln(x-2)^2)^2)*(2*ln(x)-2*ln(ln(x-2))+1/2*I*Pi*csgn(I*ln(x-2)^2)*(-csgn(I*
ln(x-2)^2)+csgn(I*ln(x-2)))^2-1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2+1/2*I*Pi*csgn(I/x^2*ln(x-2)^2)*(
-csgn(I/x^2*ln(x-2)^2)+csgn(I/x^2))*(-csgn(I/x^2*ln(x-2)^2)+csgn(I*ln(x-2)^2))+2)*exp(exp(x)+5+x)

________________________________________________________________________________________

maxima [A]  time = 0.74, size = 33, normalized size = 1.32 \begin {gather*} -2 \, x e^{\left (x + e^{x} + 5\right )} \log \left (\log \left (x - 2\right )\right ) + 2 \, {\left (x e^{5} \log \relax (x) + x e^{5}\right )} e^{\left (x + e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-2*x)*exp(x)+x^2-x-2)*log(x-2)*log(log(x-2)^2/x^2)+((-2*x^2+4*x)*exp(x)-2*x^2+8)*log(x-2)+2*x)
*exp(log(-log(log(x-2)^2/x^2)+2)+exp(x)+5+x)/((x-2)*log(x-2)*log(log(x-2)^2/x^2)+(4-2*x)*log(x-2)),x, algorith
m="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {{\mathrm {e}}^{x+\ln \left (2-\ln \left (\frac {{\ln \left (x-2\right )}^2}{x^2}\right )\right )+{\mathrm {e}}^x+5}\,\left (2\,x+\ln \left (x-2\right )\,\left ({\mathrm {e}}^x\,\left (4\,x-2\,x^2\right )-2\,x^2+8\right )-\ln \left (x-2\right )\,\ln \left (\frac {{\ln \left (x-2\right )}^2}{x^2}\right )\,\left (x+{\mathrm {e}}^x\,\left (2\,x-x^2\right )-x^2+2\right )\right )}{\ln \left (x-2\right )\,\left (2\,x-4\right )-\ln \left (x-2\right )\,\ln \left (\frac {{\ln \left (x-2\right )}^2}{x^2}\right )\,\left (x-2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x + log(2 - log(log(x - 2)^2/x^2)) + exp(x) + 5)*(2*x + log(x - 2)*(exp(x)*(4*x - 2*x^2) - 2*x^2 + 8
) - log(x - 2)*log(log(x - 2)^2/x^2)*(x + exp(x)*(2*x - x^2) - x^2 + 2)))/(log(x - 2)*(2*x - 4) - log(x - 2)*l
og(log(x - 2)^2/x^2)*(x - 2)),x)

[Out]

-int((exp(x + log(2 - log(log(x - 2)^2/x^2)) + exp(x) + 5)*(2*x + log(x - 2)*(exp(x)*(4*x - 2*x^2) - 2*x^2 + 8
) - log(x - 2)*log(log(x - 2)^2/x^2)*(x + exp(x)*(2*x - x^2) - x^2 + 2)))/(log(x - 2)*(2*x - 4) - log(x - 2)*l
og(log(x - 2)^2/x^2)*(x - 2)), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2-2*x)*exp(x)+x**2-x-2)*ln(x-2)*ln(ln(x-2)**2/x**2)+((-2*x**2+4*x)*exp(x)-2*x**2+8)*ln(x-2)+2*
x)*exp(ln(-ln(ln(x-2)**2/x**2)+2)+exp(x)+5+x)/((x-2)*ln(x-2)*ln(ln(x-2)**2/x**2)+(4-2*x)*ln(x-2)),x)

[Out]

Timed out

________________________________________________________________________________________