3.57.98 \(\int \frac {e^{\frac {9}{-8-x^2+\log (x)}} (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+(-80+80 x-10 x^2+10 x^3) \log (x)+(5-5 x) \log ^2(x))}{e^x (64+16 x^2+x^4)+e^x (-16-2 x^2) \log (x)+e^x \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ 5 e^{-x-\frac {9}{8+x^2-\log (x)}} x \]

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Rubi [F]  time = 11.88, 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^{\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{e^x \left (64+16 x^2+x^4\right )+e^x \left (-16-2 x^2\right ) \log (x)+e^x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(9/(-8 - x^2 + Log[x]))*(275 - 320*x + 170*x^2 - 80*x^3 + 5*x^4 - 5*x^5 + (-80 + 80*x - 10*x^2 + 10*x^3
)*Log[x] + (5 - 5*x)*Log[x]^2))/(E^x*(64 + 16*x^2 + x^4) + E^x*(-16 - 2*x^2)*Log[x] + E^x*Log[x]^2),x]

[Out]

5*Defer[Int][E^(-x + 9/(-8 - x^2 + Log[x])), x] - 5*Defer[Int][E^(-x + 9/(-8 - x^2 + Log[x]))*x, x] - 45*Defer
[Int][E^(-x + 9/(-8 - x^2 + Log[x]))/(8 + x^2 - Log[x])^2, x] + 90*Defer[Int][(E^(-x + 9/(-8 - x^2 + Log[x]))*
x^2)/(8 + x^2 - Log[x])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x+\frac {9}{-8-x^2+\log (x)}} \left (275-320 x+170 x^2-80 x^3+5 x^4-5 x^5+\left (-80+80 x-10 x^2+10 x^3\right ) \log (x)+(5-5 x) \log ^2(x)\right )}{\left (8+x^2-\log (x)\right )^2} \, dx\\ &=\int \left (-5 e^{-x+\frac {9}{-8-x^2+\log (x)}} (-1+x)+\frac {45 e^{-x+\frac {9}{-8-x^2+\log (x)}} \left (-1+2 x^2\right )}{\left (8+x^2-\log (x)\right )^2}\right ) \, dx\\ &=-\left (5 \int e^{-x+\frac {9}{-8-x^2+\log (x)}} (-1+x) \, dx\right )+45 \int \frac {e^{-x+\frac {9}{-8-x^2+\log (x)}} \left (-1+2 x^2\right )}{\left (8+x^2-\log (x)\right )^2} \, dx\\ &=-\left (5 \int \left (-e^{-x+\frac {9}{-8-x^2+\log (x)}}+e^{-x+\frac {9}{-8-x^2+\log (x)}} x\right ) \, dx\right )+45 \int \left (-\frac {e^{-x+\frac {9}{-8-x^2+\log (x)}}}{\left (8+x^2-\log (x)\right )^2}+\frac {2 e^{-x+\frac {9}{-8-x^2+\log (x)}} x^2}{\left (8+x^2-\log (x)\right )^2}\right ) \, dx\\ &=5 \int e^{-x+\frac {9}{-8-x^2+\log (x)}} \, dx-5 \int e^{-x+\frac {9}{-8-x^2+\log (x)}} x \, dx-45 \int \frac {e^{-x+\frac {9}{-8-x^2+\log (x)}}}{\left (8+x^2-\log (x)\right )^2} \, dx+90 \int \frac {e^{-x+\frac {9}{-8-x^2+\log (x)}} x^2}{\left (8+x^2-\log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.27, size = 22, normalized size = 1.00 \begin {gather*} 5 e^{-x+\frac {9}{-8-x^2+\log (x)}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(9/(-8 - x^2 + Log[x]))*(275 - 320*x + 170*x^2 - 80*x^3 + 5*x^4 - 5*x^5 + (-80 + 80*x - 10*x^2 +
10*x^3)*Log[x] + (5 - 5*x)*Log[x]^2))/(E^x*(64 + 16*x^2 + x^4) + E^x*(-16 - 2*x^2)*Log[x] + E^x*Log[x]^2),x]

[Out]

5*E^(-x + 9/(-8 - x^2 + Log[x]))*x

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fricas [A]  time = 0.53, size = 21, normalized size = 0.95 \begin {gather*} 5 \, x e^{\left (-x - \frac {9}{x^{2} - \log \relax (x) + 8}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x+5)*log(x)^2+(10*x^3-10*x^2+80*x-80)*log(x)-5*x^5+5*x^4-80*x^3+170*x^2-320*x+275)/(exp(x)*log(
x)^2+(-2*x^2-16)*exp(x)*log(x)+(x^4+16*x^2+64)*exp(x))/exp(-9/(log(x)-x^2-8)),x, algorithm="fricas")

[Out]

5*x*e^(-x - 9/(x^2 - log(x) + 8))

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giac [A]  time = 0.32, size = 42, normalized size = 1.91 \begin {gather*} 5 \, x e^{\left (-\frac {8 \, x^{3} - 9 \, x^{2} - 8 \, x \log \relax (x) + 64 \, x + 9 \, \log \relax (x)}{8 \, {\left (x^{2} - \log \relax (x) + 8\right )}} - \frac {9}{8}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x+5)*log(x)^2+(10*x^3-10*x^2+80*x-80)*log(x)-5*x^5+5*x^4-80*x^3+170*x^2-320*x+275)/(exp(x)*log(
x)^2+(-2*x^2-16)*exp(x)*log(x)+(x^4+16*x^2+64)*exp(x))/exp(-9/(log(x)-x^2-8)),x, algorithm="giac")

[Out]

5*x*e^(-1/8*(8*x^3 - 9*x^2 - 8*x*log(x) + 64*x + 9*log(x))/(x^2 - log(x) + 8) - 9/8)

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maple [A]  time = 0.04, size = 32, normalized size = 1.45

method	result	size
risch	\(5 x \,{\mathrm e}^{-\frac {-x^{3}+x \ln \relax (x )-8 x -9}{\ln \relax (x )-x^{2}-8}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x+5)*ln(x)^2+(10*x^3-10*x^2+80*x-80)*ln(x)-5*x^5+5*x^4-80*x^3+170*x^2-320*x+275)/(exp(x)*ln(x)^2+(-2*
x^2-16)*exp(x)*ln(x)+(x^4+16*x^2+64)*exp(x))/exp(-9/(ln(x)-x^2-8)),x,method=_RETURNVERBOSE)

[Out]

5*x*exp(-(-x^3+x*ln(x)-8*x-9)/(ln(x)-x^2-8))

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maxima [A]  time = 0.48, size = 21, normalized size = 0.95 \begin {gather*} 5 \, x e^{\left (-x - \frac {9}{x^{2} - \log \relax (x) + 8}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x+5)*log(x)^2+(10*x^3-10*x^2+80*x-80)*log(x)-5*x^5+5*x^4-80*x^3+170*x^2-320*x+275)/(exp(x)*log(
x)^2+(-2*x^2-16)*exp(x)*log(x)+(x^4+16*x^2+64)*exp(x))/exp(-9/(log(x)-x^2-8)),x, algorithm="maxima")

[Out]

5*x*e^(-x - 9/(x^2 - log(x) + 8))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} -\int \frac {{\mathrm {e}}^{-\frac {9}{x^2-\ln \relax (x)+8}}\,\left (320\,x-170\,x^2+80\,x^3-5\,x^4+5\,x^5+{\ln \relax (x)}^2\,\left (5\,x-5\right )-\ln \relax (x)\,\left (10\,x^3-10\,x^2+80\,x-80\right )-275\right )}{{\mathrm {e}}^x\,{\ln \relax (x)}^2-{\mathrm {e}}^x\,\left (2\,x^2+16\right )\,\ln \relax (x)+{\mathrm {e}}^x\,\left (x^4+16\,x^2+64\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-9/(x^2 - log(x) + 8))*(320*x - 170*x^2 + 80*x^3 - 5*x^4 + 5*x^5 + log(x)^2*(5*x - 5) - log(x)*(80*x
- 10*x^2 + 10*x^3 - 80) - 275))/(exp(x)*log(x)^2 + exp(x)*(16*x^2 + x^4 + 64) - exp(x)*log(x)*(2*x^2 + 16)),x
)

[Out]

-int((exp(-9/(x^2 - log(x) + 8))*(320*x - 170*x^2 + 80*x^3 - 5*x^4 + 5*x^5 + log(x)^2*(5*x - 5) - log(x)*(80*x
- 10*x^2 + 10*x^3 - 80) - 275))/(exp(x)*log(x)^2 + exp(x)*(16*x^2 + x^4 + 64) - exp(x)*log(x)*(2*x^2 + 16)),
x)

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sympy [A]  time = 0.75, size = 17, normalized size = 0.77 \begin {gather*} 5 x e^{- x} e^{\frac {9}{- x^{2} + \log {\relax (x )} - 8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x+5)*ln(x)**2+(10*x**3-10*x**2+80*x-80)*ln(x)-5*x**5+5*x**4-80*x**3+170*x**2-320*x+275)/(exp(x)
*ln(x)**2+(-2*x**2-16)*exp(x)*ln(x)+(x**4+16*x**2+64)*exp(x))/exp(-9/(ln(x)-x**2-8)),x)

[Out]

5*x*exp(-x)*exp(9/(-x**2 + log(x) - 8))

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