3.86.94 \(\int \frac {e^{\frac {-16 x-4 x^2+8 x^3+2 x^4+e^x (32 x+8 x^2)+(-64-16 x+32 x^2+8 x^3) \log (4)+(-64-16 x+32 x^2+8 x^3) \log (x)}{x^2+4 x \log (4)+4 x \log (x)}} (16 x^2+8 x^4+4 x^5+(128 x+64 x^3+32 x^4) \log (4)+(256+128 x^2+64 x^3) \log ^2(4)+e^x (-128 x-64 x^2+32 x^3+8 x^4+(160 x^2+32 x^3) \log (4))+(128 x+64 x^3+32 x^4+e^x (160 x^2+32 x^3)+(512+256 x^2+128 x^3) \log (4)) \log (x)+(256+128 x^2+64 x^3) \log ^2(x))}{x^4+8 x^3 \log (4)+16 x^2 \log ^2(4)+(8 x^3+32 x^2 \log (4)) \log (x)+16 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=32 \[ -5+e^{2 (4+x) \left (-\frac {2}{x}+x+\frac {e^x}{\frac {x}{4}+\log (4)+\log (x)}\right )} \]

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Rubi [F]  time = 180.00, 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*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-16*x - 4*x^2 + 8*x^3 + 2*x^4 + E^x*(32*x + 8*x^2) + (-64 - 16*x + 32*x^2 + 8*x^3)*Log[4] + (-64 - 16
*x + 32*x^2 + 8*x^3)*Log[x])/(x^2 + 4*x*Log[4] + 4*x*Log[x]))*(16*x^2 + 8*x^4 + 4*x^5 + (128*x + 64*x^3 + 32*x
^4)*Log[4] + (256 + 128*x^2 + 64*x^3)*Log[4]^2 + E^x*(-128*x - 64*x^2 + 32*x^3 + 8*x^4 + (160*x^2 + 32*x^3)*Lo
g[4]) + (128*x + 64*x^3 + 32*x^4 + E^x*(160*x^2 + 32*x^3) + (512 + 256*x^2 + 128*x^3)*Log[4])*Log[x] + (256 +
128*x^2 + 64*x^3)*Log[x]^2))/(x^4 + 8*x^3*Log[4] + 16*x^2*Log[4]^2 + (8*x^3 + 32*x^2*Log[4])*Log[x] + 16*x^2*L
og[x]^2),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [F]  time = 37.48, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {-16 x-4 x^2+8 x^3+2 x^4+e^x \left (32 x+8 x^2\right )+\left (-64-16 x+32 x^2+8 x^3\right ) \log (4)+\left (-64-16 x+32 x^2+8 x^3\right ) \log (x)}{x^2+4 x \log (4)+4 x \log (x)}} \left (16 x^2+8 x^4+4 x^5+\left (128 x+64 x^3+32 x^4\right ) \log (4)+\left (256+128 x^2+64 x^3\right ) \log ^2(4)+e^x \left (-128 x-64 x^2+32 x^3+8 x^4+\left (160 x^2+32 x^3\right ) \log (4)\right )+\left (128 x+64 x^3+32 x^4+e^x \left (160 x^2+32 x^3\right )+\left (512+256 x^2+128 x^3\right ) \log (4)\right ) \log (x)+\left (256+128 x^2+64 x^3\right ) \log ^2(x)\right )}{x^4+8 x^3 \log (4)+16 x^2 \log ^2(4)+\left (8 x^3+32 x^2 \log (4)\right ) \log (x)+16 x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((-16*x - 4*x^2 + 8*x^3 + 2*x^4 + E^x*(32*x + 8*x^2) + (-64 - 16*x + 32*x^2 + 8*x^3)*Log[4] + (-6
4 - 16*x + 32*x^2 + 8*x^3)*Log[x])/(x^2 + 4*x*Log[4] + 4*x*Log[x]))*(16*x^2 + 8*x^4 + 4*x^5 + (128*x + 64*x^3
+ 32*x^4)*Log[4] + (256 + 128*x^2 + 64*x^3)*Log[4]^2 + E^x*(-128*x - 64*x^2 + 32*x^3 + 8*x^4 + (160*x^2 + 32*x
^3)*Log[4]) + (128*x + 64*x^3 + 32*x^4 + E^x*(160*x^2 + 32*x^3) + (512 + 256*x^2 + 128*x^3)*Log[4])*Log[x] + (
256 + 128*x^2 + 64*x^3)*Log[x]^2))/(x^4 + 8*x^3*Log[4] + 16*x^2*Log[4]^2 + (8*x^3 + 32*x^2*Log[4])*Log[x] + 16
*x^2*Log[x]^2),x]

[Out]

Integrate[(E^((-16*x - 4*x^2 + 8*x^3 + 2*x^4 + E^x*(32*x + 8*x^2) + (-64 - 16*x + 32*x^2 + 8*x^3)*Log[4] + (-6
4 - 16*x + 32*x^2 + 8*x^3)*Log[x])/(x^2 + 4*x*Log[4] + 4*x*Log[x]))*(16*x^2 + 8*x^4 + 4*x^5 + (128*x + 64*x^3
+ 32*x^4)*Log[4] + (256 + 128*x^2 + 64*x^3)*Log[4]^2 + E^x*(-128*x - 64*x^2 + 32*x^3 + 8*x^4 + (160*x^2 + 32*x
^3)*Log[4]) + (128*x + 64*x^3 + 32*x^4 + E^x*(160*x^2 + 32*x^3) + (512 + 256*x^2 + 128*x^3)*Log[4])*Log[x] + (
256 + 128*x^2 + 64*x^3)*Log[x]^2))/(x^4 + 8*x^3*Log[4] + 16*x^2*Log[4]^2 + (8*x^3 + 32*x^2*Log[4])*Log[x] + 16
*x^2*Log[x]^2), x]

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fricas [B]  time = 0.73, size = 81, normalized size = 2.53 \begin {gather*} e^{\left (\frac {2 \, {\left (x^{4} + 4 \, x^{3} - 2 \, x^{2} + 4 \, {\left (x^{2} + 4 \, x\right )} e^{x} + 8 \, {\left (x^{3} + 4 \, x^{2} - 2 \, x - 8\right )} \log \relax (2) + 4 \, {\left (x^{3} + 4 \, x^{2} - 2 \, x - 8\right )} \log \relax (x) - 8 \, x\right )}}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^3+128*x^2+256)*log(x)^2+((32*x^3+160*x^2)*exp(x)+2*(128*x^3+256*x^2+512)*log(2)+32*x^4+64*x^3
+128*x)*log(x)+(2*(32*x^3+160*x^2)*log(2)+8*x^4+32*x^3-64*x^2-128*x)*exp(x)+4*(64*x^3+128*x^2+256)*log(2)^2+2*
(32*x^4+64*x^3+128*x)*log(2)+4*x^5+8*x^4+16*x^2)*exp(((8*x^3+32*x^2-16*x-64)*log(x)+(8*x^2+32*x)*exp(x)+2*(8*x
^3+32*x^2-16*x-64)*log(2)+2*x^4+8*x^3-4*x^2-16*x)/(4*x*log(x)+8*x*log(2)+x^2))/(16*x^2*log(x)^2+(64*x^2*log(2)
+8*x^3)*log(x)+64*x^2*log(2)^2+16*x^3*log(2)+x^4),x, algorithm="fricas")

[Out]

e^(2*(x^4 + 4*x^3 - 2*x^2 + 4*(x^2 + 4*x)*e^x + 8*(x^3 + 4*x^2 - 2*x - 8)*log(2) + 4*(x^3 + 4*x^2 - 2*x - 8)*l
og(x) - 8*x)/(x^2 + 8*x*log(2) + 4*x*log(x)))

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giac [B]  time = 3.47, size = 302, normalized size = 9.44 \begin {gather*} e^{\left (\frac {2 \, x^{4}}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)} + \frac {16 \, x^{3} \log \relax (2)}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)} + \frac {8 \, x^{3} \log \relax (x)}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)} + \frac {8 \, x^{3}}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)} + \frac {8 \, x^{2} e^{x}}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)} + \frac {64 \, x^{2} \log \relax (2)}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)} + \frac {32 \, x^{2} \log \relax (x)}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)} - \frac {4 \, x^{2}}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)} + \frac {32 \, x e^{x}}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)} - \frac {32 \, x \log \relax (2)}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)} - \frac {16 \, x \log \relax (x)}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)} - \frac {16 \, x}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)} - \frac {128 \, \log \relax (2)}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)} - \frac {64 \, \log \relax (x)}{x^{2} + 8 \, x \log \relax (2) + 4 \, x \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^3+128*x^2+256)*log(x)^2+((32*x^3+160*x^2)*exp(x)+2*(128*x^3+256*x^2+512)*log(2)+32*x^4+64*x^3
+128*x)*log(x)+(2*(32*x^3+160*x^2)*log(2)+8*x^4+32*x^3-64*x^2-128*x)*exp(x)+4*(64*x^3+128*x^2+256)*log(2)^2+2*
(32*x^4+64*x^3+128*x)*log(2)+4*x^5+8*x^4+16*x^2)*exp(((8*x^3+32*x^2-16*x-64)*log(x)+(8*x^2+32*x)*exp(x)+2*(8*x
^3+32*x^2-16*x-64)*log(2)+2*x^4+8*x^3-4*x^2-16*x)/(4*x*log(x)+8*x*log(2)+x^2))/(16*x^2*log(x)^2+(64*x^2*log(2)
+8*x^3)*log(x)+64*x^2*log(2)^2+16*x^3*log(2)+x^4),x, algorithm="giac")

[Out]

e^(2*x^4/(x^2 + 8*x*log(2) + 4*x*log(x)) + 16*x^3*log(2)/(x^2 + 8*x*log(2) + 4*x*log(x)) + 8*x^3*log(x)/(x^2 +
8*x*log(2) + 4*x*log(x)) + 8*x^3/(x^2 + 8*x*log(2) + 4*x*log(x)) + 8*x^2*e^x/(x^2 + 8*x*log(2) + 4*x*log(x))
+ 64*x^2*log(2)/(x^2 + 8*x*log(2) + 4*x*log(x)) + 32*x^2*log(x)/(x^2 + 8*x*log(2) + 4*x*log(x)) - 4*x^2/(x^2 +
8*x*log(2) + 4*x*log(x)) + 32*x*e^x/(x^2 + 8*x*log(2) + 4*x*log(x)) - 32*x*log(2)/(x^2 + 8*x*log(2) + 4*x*log
(x)) - 16*x*log(x)/(x^2 + 8*x*log(2) + 4*x*log(x)) - 16*x/(x^2 + 8*x*log(2) + 4*x*log(x)) - 128*log(2)/(x^2 +
8*x*log(2) + 4*x*log(x)) - 64*log(x)/(x^2 + 8*x*log(2) + 4*x*log(x)))

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maple [A]  time = 0.16, size = 56, normalized size = 1.75

method	result	size
risch	\({\mathrm e}^{\frac {2 \left (4+x \right ) \left (4 x^{2} \ln \relax (x )+8 x^{2} \ln \relax (2)+x^{3}+4 \,{\mathrm e}^{x} x -8 \ln \relax (x )-16 \ln \relax (2)-2 x \right )}{x \left (4 \ln \relax (x )+8 \ln \relax (2)+x \right )}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((64*x^3+128*x^2+256)*ln(x)^2+((32*x^3+160*x^2)*exp(x)+2*(128*x^3+256*x^2+512)*ln(2)+32*x^4+64*x^3+128*x)*
ln(x)+(2*(32*x^3+160*x^2)*ln(2)+8*x^4+32*x^3-64*x^2-128*x)*exp(x)+4*(64*x^3+128*x^2+256)*ln(2)^2+2*(32*x^4+64*
x^3+128*x)*ln(2)+4*x^5+8*x^4+16*x^2)*exp(((8*x^3+32*x^2-16*x-64)*ln(x)+(8*x^2+32*x)*exp(x)+2*(8*x^3+32*x^2-16*
x-64)*ln(2)+2*x^4+8*x^3-4*x^2-16*x)/(4*x*ln(x)+8*x*ln(2)+x^2))/(16*x^2*ln(x)^2+(64*x^2*ln(2)+8*x^3)*ln(x)+64*x
^2*ln(2)^2+16*x^3*ln(2)+x^4),x,method=_RETURNVERBOSE)

[Out]

exp(2*(4+x)*(4*x^2*ln(x)+8*x^2*ln(2)+x^3+4*exp(x)*x-8*ln(x)-16*ln(2)-2*x)/x/(4*ln(x)+8*ln(2)+x))

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maxima [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(((64*x^3+128*x^2+256)*log(x)^2+((32*x^3+160*x^2)*exp(x)+2*(128*x^3+256*x^2+512)*log(2)+32*x^4+64*x^3
+128*x)*log(x)+(2*(32*x^3+160*x^2)*log(2)+8*x^4+32*x^3-64*x^2-128*x)*exp(x)+4*(64*x^3+128*x^2+256)*log(2)^2+2*
(32*x^4+64*x^3+128*x)*log(2)+4*x^5+8*x^4+16*x^2)*exp(((8*x^3+32*x^2-16*x-64)*log(x)+(8*x^2+32*x)*exp(x)+2*(8*x
^3+32*x^2-16*x-64)*log(2)+2*x^4+8*x^3-4*x^2-16*x)/(4*x*log(x)+8*x*log(2)+x^2))/(16*x^2*log(x)^2+(64*x^2*log(2)
+8*x^3)*log(x)+64*x^2*log(2)^2+16*x^3*log(2)+x^4),x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 6.72, size = 223, normalized size = 6.97 \begin {gather*} 2^{\frac {64\,\left (x^2-2\right )}{8\,x\,\ln \relax (2)+4\,x\,\ln \relax (x)+x^2}+\frac {16\,\left (x^2-2\right )}{x+8\,\ln \relax (2)+4\,\ln \relax (x)}}\,x^{\frac {8\,\left (x^2+4\,x\right )}{x+8\,\ln \relax (2)+4\,\ln \relax (x)}-\frac {16\,\left (x+4\right )}{8\,x\,\ln \relax (2)+4\,x\,\ln \relax (x)+x^2}}\,{\mathrm {e}}^{\frac {8\,x^2\,{\mathrm {e}}^x}{8\,x\,\ln \relax (2)+4\,x\,\ln \relax (x)+x^2}}\,{\mathrm {e}}^{-\frac {4\,x^2}{8\,x\,\ln \relax (2)+4\,x\,\ln \relax (x)+x^2}}\,{\mathrm {e}}^{\frac {2\,x^4}{8\,x\,\ln \relax (2)+4\,x\,\ln \relax (x)+x^2}}\,{\mathrm {e}}^{\frac {8\,x^3}{8\,x\,\ln \relax (2)+4\,x\,\ln \relax (x)+x^2}}\,{\mathrm {e}}^{\frac {32\,x\,{\mathrm {e}}^x}{8\,x\,\ln \relax (2)+4\,x\,\ln \relax (x)+x^2}}\,{\mathrm {e}}^{-\frac {16\,x}{8\,x\,\ln \relax (2)+4\,x\,\ln \relax (x)+x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(16*x + 2*log(2)*(16*x - 32*x^2 - 8*x^3 + 64) - exp(x)*(32*x + 8*x^2) + 4*x^2 - 8*x^3 - 2*x^4 + log(
x)*(16*x - 32*x^2 - 8*x^3 + 64))/(8*x*log(2) + 4*x*log(x) + x^2))*(2*log(2)*(128*x + 64*x^3 + 32*x^4) + log(x)
^2*(128*x^2 + 64*x^3 + 256) + log(x)*(128*x + exp(x)*(160*x^2 + 32*x^3) + 2*log(2)*(256*x^2 + 128*x^3 + 512) +
64*x^3 + 32*x^4) + exp(x)*(2*log(2)*(160*x^2 + 32*x^3) - 128*x - 64*x^2 + 32*x^3 + 8*x^4) + 16*x^2 + 8*x^4 +
4*x^5 + 4*log(2)^2*(128*x^2 + 64*x^3 + 256)))/(64*x^2*log(2)^2 + 16*x^2*log(x)^2 + log(x)*(64*x^2*log(2) + 8*x
^3) + 16*x^3*log(2) + x^4),x)

[Out]

2^((64*(x^2 - 2))/(8*x*log(2) + 4*x*log(x) + x^2) + (16*(x^2 - 2))/(x + 8*log(2) + 4*log(x)))*x^((8*(4*x + x^2
))/(x + 8*log(2) + 4*log(x)) - (16*(x + 4))/(8*x*log(2) + 4*x*log(x) + x^2))*exp((8*x^2*exp(x))/(8*x*log(2) +
4*x*log(x) + x^2))*exp(-(4*x^2)/(8*x*log(2) + 4*x*log(x) + x^2))*exp((2*x^4)/(8*x*log(2) + 4*x*log(x) + x^2))*
exp((8*x^3)/(8*x*log(2) + 4*x*log(x) + x^2))*exp((32*x*exp(x))/(8*x*log(2) + 4*x*log(x) + x^2))*exp(-(16*x)/(8
*x*log(2) + 4*x*log(x) + x^2))

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sympy [B]  time = 2.86, size = 85, normalized size = 2.66 \begin {gather*} e^{\frac {2 x^{4} + 8 x^{3} - 4 x^{2} - 16 x + \left (8 x^{2} + 32 x\right ) e^{x} + \left (8 x^{3} + 32 x^{2} - 16 x - 64\right ) \log {\relax (x )} + \left (16 x^{3} + 64 x^{2} - 32 x - 128\right ) \log {\relax (2 )}}{x^{2} + 4 x \log {\relax (x )} + 8 x \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x**3+128*x**2+256)*ln(x)**2+((32*x**3+160*x**2)*exp(x)+2*(128*x**3+256*x**2+512)*ln(2)+32*x**4+
64*x**3+128*x)*ln(x)+(2*(32*x**3+160*x**2)*ln(2)+8*x**4+32*x**3-64*x**2-128*x)*exp(x)+4*(64*x**3+128*x**2+256)
*ln(2)**2+2*(32*x**4+64*x**3+128*x)*ln(2)+4*x**5+8*x**4+16*x**2)*exp(((8*x**3+32*x**2-16*x-64)*ln(x)+(8*x**2+3
2*x)*exp(x)+2*(8*x**3+32*x**2-16*x-64)*ln(2)+2*x**4+8*x**3-4*x**2-16*x)/(4*x*ln(x)+8*x*ln(2)+x**2))/(16*x**2*l
n(x)**2+(64*x**2*ln(2)+8*x**3)*ln(x)+64*x**2*ln(2)**2+16*x**3*ln(2)+x**4),x)

[Out]

exp((2*x**4 + 8*x**3 - 4*x**2 - 16*x + (8*x**2 + 32*x)*exp(x) + (8*x**3 + 32*x**2 - 16*x - 64)*log(x) + (16*x*
*3 + 64*x**2 - 32*x - 128)*log(2))/(x**2 + 4*x*log(x) + 8*x*log(2)))

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