3.93.32 \(\int \frac {(16+8 x+x^2+e^x (-32-16 x-2 x^2)+e^{2 x} (16+8 x+x^2)) \log ^2(x)+e^{\frac {-4 x-8 \log (x)}{(-4-x+e^x (4+x)) \log (x)}} (-16-4 x+e^x (16+4 x)+(16+e^x (-16+16 x+4 x^2)) \log (x)+(-8+e^x (40+8 x)) \log ^2(x))}{(16+8 x+x^2+e^x (-32-16 x-2 x^2)+e^{2 x} (16+8 x+x^2)) \log ^2(x)} \, dx\)

Optimal. Leaf size=33 \[ e^{\frac {2 x \left (\frac {4}{x}+\frac {2}{\log (x)}\right )}{\left (1-e^x\right ) (4+x)}}+x \]

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Rubi [F]  time = 41.19, 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 {\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)+\exp \left (\frac {-4 x-8 \log (x)}{\left (-4-x+e^x (4+x)\right ) \log (x)}\right ) \left (-16-4 x+e^x (16+4 x)+\left (16+e^x \left (-16+16 x+4 x^2\right )\right ) \log (x)+\left (-8+e^x (40+8 x)\right ) \log ^2(x)\right )}{\left (16+8 x+x^2+e^x \left (-32-16 x-2 x^2\right )+e^{2 x} \left (16+8 x+x^2\right )\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((16 + 8*x + x^2 + E^x*(-32 - 16*x - 2*x^2) + E^(2*x)*(16 + 8*x + x^2))*Log[x]^2 + E^((-4*x - 8*Log[x])/((
-4 - x + E^x*(4 + x))*Log[x]))*(-16 - 4*x + E^x*(16 + 4*x) + (16 + E^x*(-16 + 16*x + 4*x^2))*Log[x] + (-8 + E^
x*(40 + 8*x))*Log[x]^2))/((16 + 8*x + x^2 + E^x*(-32 - 16*x - 2*x^2) + E^(2*x)*(16 + 8*x + x^2))*Log[x]^2),x]

[Out]

x + 8*Defer[Int][1/(E^((4*(x + 2*Log[x]))/((-1 + E^x)*(4 + x)*Log[x]))*(-1 + E^x)*(4 + x)^2), x] + 8*Defer[Int
][1/(E^((4*(x + 2*Log[x]))/((-1 + E^x)*(4 + x)*Log[x]))*(-1 + E^x)^2*(4 + x)), x] + 8*Defer[Int][1/(E^((4*(x +
2*Log[x]))/((-1 + E^x)*(4 + x)*Log[x]))*(-1 + E^x)*(4 + x)), x] + 4*Defer[Int][1/(E^((4*(x + 2*Log[x]))/((-1
+ E^x)*(4 + x)*Log[x]))*(-1 + E^x)*(4 + x)*Log[x]^2), x] + 4*Defer[Int][1/(E^((4*(x + 2*Log[x]))/((-1 + E^x)*(
4 + x)*Log[x]))*(-1 + E^x)^2*Log[x]), x] + 4*Defer[Int][1/(E^((4*(x + 2*Log[x]))/((-1 + E^x)*(4 + x)*Log[x]))*
(-1 + E^x)*Log[x]), x] - 16*Defer[Int][1/(E^((4*(x + 2*Log[x]))/((-1 + E^x)*(4 + x)*Log[x]))*(-1 + E^x)*(4 + x
)^2*Log[x]), x] - 16*Defer[Int][1/(E^((4*(x + 2*Log[x]))/((-1 + E^x)*(4 + x)*Log[x]))*(-1 + E^x)^2*(4 + x)*Log
[x]), x] - 16*Defer[Int][1/(E^((4*(x + 2*Log[x]))/((-1 + E^x)*(4 + x)*Log[x]))*(-1 + E^x)*(4 + x)*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1+\frac {4 \exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) \left (\left (-1+e^x\right ) (4+x)+\left (4+e^x \left (-4+4 x+x^2\right )\right ) \log (x)+2 \left (-1+e^x (5+x)\right ) \log ^2(x)\right )}{\left (-1+e^x\right )^2 (4+x)^2 \log ^2(x)}\right ) \, dx\\ &=x+4 \int \frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) \left (\left (-1+e^x\right ) (4+x)+\left (4+e^x \left (-4+4 x+x^2\right )\right ) \log (x)+2 \left (-1+e^x (5+x)\right ) \log ^2(x)\right )}{\left (-1+e^x\right )^2 (4+x)^2 \log ^2(x)} \, dx\\ &=x+4 \int \left (\frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) (x+2 \log (x))}{\left (-1+e^x\right )^2 (4+x) \log (x)}+\frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) \left (4+x-4 \log (x)+4 x \log (x)+x^2 \log (x)+10 \log ^2(x)+2 x \log ^2(x)\right )}{\left (-1+e^x\right ) (4+x)^2 \log ^2(x)}\right ) \, dx\\ &=x+4 \int \frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) (x+2 \log (x))}{\left (-1+e^x\right )^2 (4+x) \log (x)} \, dx+4 \int \frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) \left (4+x-4 \log (x)+4 x \log (x)+x^2 \log (x)+10 \log ^2(x)+2 x \log ^2(x)\right )}{\left (-1+e^x\right ) (4+x)^2 \log ^2(x)} \, dx\\ &=x+4 \int \left (\frac {10 \exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right )}{\left (-1+e^x\right ) (4+x)^2}+\frac {2 \exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) x}{\left (-1+e^x\right ) (4+x)^2}+\frac {4 \exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right )}{\left (-1+e^x\right ) (4+x)^2 \log ^2(x)}+\frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) x}{\left (-1+e^x\right ) (4+x)^2 \log ^2(x)}-\frac {4 \exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right )}{\left (-1+e^x\right ) (4+x)^2 \log (x)}+\frac {4 \exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) x}{\left (-1+e^x\right ) (4+x)^2 \log (x)}+\frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) x^2}{\left (-1+e^x\right ) (4+x)^2 \log (x)}\right ) \, dx+4 \int \left (\frac {2 \exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right )}{\left (-1+e^x\right )^2 (4+x)}+\frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) x}{\left (-1+e^x\right )^2 (4+x) \log (x)}\right ) \, dx\\ &=x+4 \int \frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) x}{\left (-1+e^x\right ) (4+x)^2 \log ^2(x)} \, dx+4 \int \frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) x^2}{\left (-1+e^x\right ) (4+x)^2 \log (x)} \, dx+4 \int \frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) x}{\left (-1+e^x\right )^2 (4+x) \log (x)} \, dx+8 \int \frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) x}{\left (-1+e^x\right ) (4+x)^2} \, dx+8 \int \frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right )}{\left (-1+e^x\right )^2 (4+x)} \, dx+16 \int \frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right )}{\left (-1+e^x\right ) (4+x)^2 \log ^2(x)} \, dx-16 \int \frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right )}{\left (-1+e^x\right ) (4+x)^2 \log (x)} \, dx+16 \int \frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right ) x}{\left (-1+e^x\right ) (4+x)^2 \log (x)} \, dx+40 \int \frac {\exp \left (-\frac {4 (x+2 \log (x))}{\left (-1+e^x\right ) (4+x) \log (x)}\right )}{\left (-1+e^x\right ) (4+x)^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 38, normalized size = 1.15 \begin {gather*} e^{-\frac {8}{\left (-1+e^x\right ) (4+x)}-\frac {4 x}{\left (-1+e^x\right ) (4+x) \log (x)}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((16 + 8*x + x^2 + E^x*(-32 - 16*x - 2*x^2) + E^(2*x)*(16 + 8*x + x^2))*Log[x]^2 + E^((-4*x - 8*Log[
x])/((-4 - x + E^x*(4 + x))*Log[x]))*(-16 - 4*x + E^x*(16 + 4*x) + (16 + E^x*(-16 + 16*x + 4*x^2))*Log[x] + (-
8 + E^x*(40 + 8*x))*Log[x]^2))/((16 + 8*x + x^2 + E^x*(-32 - 16*x - 2*x^2) + E^(2*x)*(16 + 8*x + x^2))*Log[x]^
2),x]

[Out]

E^(-8/((-1 + E^x)*(4 + x)) - (4*x)/((-1 + E^x)*(4 + x)*Log[x])) + x

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fricas [A]  time = 0.86, size = 28, normalized size = 0.85 \begin {gather*} x + e^{\left (-\frac {4 \, {\left (x + 2 \, \log \relax (x)\right )}}{{\left ({\left (x + 4\right )} e^{x} - x - 4\right )} \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((8*x+40)*exp(x)-8)*log(x)^2+((4*x^2+16*x-16)*exp(x)+16)*log(x)+(4*x+16)*exp(x)-16-4*x)*exp((-8*lo
g(x)-4*x)/((4+x)*exp(x)-x-4)/log(x))+((x^2+8*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)*log(x)^2)/((x^
2+8*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)/log(x)^2,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((8*x+40)*exp(x)-8)*log(x)^2+((4*x^2+16*x-16)*exp(x)+16)*log(x)+(4*x+16)*exp(x)-16-4*x)*exp((-8*lo
g(x)-4*x)/((4+x)*exp(x)-x-4)/log(x))+((x^2+8*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)*log(x)^2)/((x^
2+8*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)/log(x)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:exp(sageVARx)^2=exp(2*sageVARx)exp(sageVARx)^2=exp(2*sageVARx)exp(sageVARx)^2=exp(2*sageVARx)exp(sageVARx)^
2=exp(2*sag

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maple [A]  time = 0.06, size = 27, normalized size = 0.82

method	result	size
risch	\(x +{\mathrm e}^{-\frac {4 \left (2 \ln \relax (x )+x \right )}{\left (4+x \right ) \left ({\mathrm e}^{x}-1\right ) \ln \relax (x )}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((8*x+40)*exp(x)-8)*ln(x)^2+((4*x^2+16*x-16)*exp(x)+16)*ln(x)+(4*x+16)*exp(x)-16-4*x)*exp((-8*ln(x)-4*x)
/((4+x)*exp(x)-x-4)/ln(x))+((x^2+8*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)*ln(x)^2)/((x^2+8*x+16)*e
xp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.54, size = 82, normalized size = 2.48 \begin {gather*} {\left (x e^{\left (\frac {8}{{\left (x + 4\right )} e^{x} - x - 4} + \frac {4}{{\left (e^{x} - 1\right )} \log \relax (x)}\right )} + e^{\left (\frac {16}{{\left ({\left (x + 4\right )} e^{x} - x - 4\right )} \log \relax (x)}\right )}\right )} e^{\left (-\frac {8}{{\left (x + 4\right )} e^{x} - x - 4} - \frac {4}{{\left (e^{x} - 1\right )} \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((8*x+40)*exp(x)-8)*log(x)^2+((4*x^2+16*x-16)*exp(x)+16)*log(x)+(4*x+16)*exp(x)-16-4*x)*exp((-8*lo
g(x)-4*x)/((4+x)*exp(x)-x-4)/log(x))+((x^2+8*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)*log(x)^2)/((x^
2+8*x+16)*exp(x)^2+(-2*x^2-16*x-32)*exp(x)+x^2+8*x+16)/log(x)^2,x, algorithm="maxima")

[Out]

(x*e^(8/((x + 4)*e^x - x - 4) + 4/((e^x - 1)*log(x))) + e^(16/(((x + 4)*e^x - x - 4)*log(x))))*e^(-8/((x + 4)*
e^x - x - 4) - 4/((e^x - 1)*log(x)))

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mupad [B]  time = 7.69, size = 59, normalized size = 1.79 \begin {gather*} x+x^{\frac {8}{4\,\ln \relax (x)-4\,{\mathrm {e}}^x\,\ln \relax (x)+x\,\ln \relax (x)-x\,{\mathrm {e}}^x\,\ln \relax (x)}}\,{\mathrm {e}}^{\frac {4\,x}{4\,\ln \relax (x)-4\,{\mathrm {e}}^x\,\ln \relax (x)+x\,\ln \relax (x)-x\,{\mathrm {e}}^x\,\ln \relax (x)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)^2*(8*x + exp(2*x)*(8*x + x^2 + 16) - exp(x)*(16*x + 2*x^2 + 32) + x^2 + 16) + exp((4*x + 8*log(x))
/(log(x)*(x - exp(x)*(x + 4) + 4)))*(log(x)*(exp(x)*(16*x + 4*x^2 - 16) + 16) - 4*x + exp(x)*(4*x + 16) + log(
x)^2*(exp(x)*(8*x + 40) - 8) - 16))/(log(x)^2*(8*x + exp(2*x)*(8*x + x^2 + 16) - exp(x)*(16*x + 2*x^2 + 32) +
x^2 + 16)),x)

[Out]

x + x^(8/(4*log(x) - 4*exp(x)*log(x) + x*log(x) - x*exp(x)*log(x)))*exp((4*x)/(4*log(x) - 4*exp(x)*log(x) + x*
log(x) - x*exp(x)*log(x)))

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sympy [A]  time = 4.64, size = 26, normalized size = 0.79 \begin {gather*} x + e^{\frac {- 4 x - 8 \log {\relax (x )}}{\left (- x + \left (x + 4\right ) e^{x} - 4\right ) \log {\relax (x )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((8*x+40)*exp(x)-8)*ln(x)**2+((4*x**2+16*x-16)*exp(x)+16)*ln(x)+(4*x+16)*exp(x)-16-4*x)*exp((-8*ln
(x)-4*x)/((4+x)*exp(x)-x-4)/ln(x))+((x**2+8*x+16)*exp(x)**2+(-2*x**2-16*x-32)*exp(x)+x**2+8*x+16)*ln(x)**2)/((
x**2+8*x+16)*exp(x)**2+(-2*x**2-16*x-32)*exp(x)+x**2+8*x+16)/ln(x)**2,x)

[Out]

x + exp((-4*x - 8*log(x))/((-x + (x + 4)*exp(x) - 4)*log(x)))

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