3.71.97 \(\int \frac {e^{x^8} (-5 x^2-40 x^{10}+40 x^{11}+e^2 (5 x^2+40 x^{10})) \log ^2(x)+e^{\frac {x}{5 \log (x)}} (e^{x^8} x^2-e^{x^8} x^2 \log (x)+e^{x^8} (10 x+40 x^9) \log ^2(x))}{5 e^{\frac {2 x}{5 \log (x)}} \log ^2(x)+e^{\frac {x}{5 \log (x)}} (-10 x+10 e^2 x+10 x^2) \log ^2(x)+(5 x^2+5 e^4 x^2-10 x^3+5 x^4+e^2 (-10 x^2+10 x^3)) \log ^2(x)} \, dx\)

Optimal. Leaf size=30 \[ \frac {e^{x^8} x}{-1+e^2+\frac {e^{\frac {x}{5 \log (x)}}}{x}+x} \]

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Rubi [F]  time = 15.50, 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^8} \left (-5 x^2-40 x^{10}+40 x^{11}+e^2 \left (5 x^2+40 x^{10}\right )\right ) \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (e^{x^8} x^2-e^{x^8} x^2 \log (x)+e^{x^8} \left (10 x+40 x^9\right ) \log ^2(x)\right )}{5 e^{\frac {2 x}{5 \log (x)}} \log ^2(x)+e^{\frac {x}{5 \log (x)}} \left (-10 x+10 e^2 x+10 x^2\right ) \log ^2(x)+\left (5 x^2+5 e^4 x^2-10 x^3+5 x^4+e^2 \left (-10 x^2+10 x^3\right )\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x^8*(-5*x^2 - 40*x^10 + 40*x^11 + E^2*(5*x^2 + 40*x^10))*Log[x]^2 + E^(x/(5*Log[x]))*(E^x^8*x^2 - E^x^8
*x^2*Log[x] + E^x^8*(10*x + 40*x^9)*Log[x]^2))/(5*E^((2*x)/(5*Log[x]))*Log[x]^2 + E^(x/(5*Log[x]))*(-10*x + 10
*E^2*x + 10*x^2)*Log[x]^2 + (5*x^2 + 5*E^4*x^2 - 10*x^3 + 5*x^4 + E^2*(-10*x^2 + 10*x^3))*Log[x]^2),x]

[Out]

(1 - E^2)*Defer[Int][(E^x^8*x^2)/(E^(x/(5*Log[x])) - (1 - E^2)*x + x^2)^2, x] - 2*Defer[Int][(E^x^8*x^3)/(E^(x
/(5*Log[x])) - (1 - E^2)*x + x^2)^2, x] + 2*Defer[Int][(E^x^8*x)/(E^(x/(5*Log[x])) - (1 - E^2)*x + x^2), x] +
8*Defer[Int][(E^x^8*x^9)/(E^(x/(5*Log[x])) - (1 - E^2)*x + x^2), x] + ((1 - E^2)*Defer[Int][(E^x^8*x^3)/((E^(x
/(5*Log[x])) - (1 - E^2)*x + x^2)^2*Log[x]^2), x])/5 - Defer[Int][(E^x^8*x^4)/((E^(x/(5*Log[x])) - (1 - E^2)*x
+ x^2)^2*Log[x]^2), x]/5 + Defer[Int][(E^x^8*x^2)/((E^(x/(5*Log[x])) - (1 - E^2)*x + x^2)*Log[x]^2), x]/5 + D
efer[Int][(E^x^8*x^2)/((-E^(x/(5*Log[x])) + (1 - E^2)*x - x^2)*Log[x]), x]/5 - ((1 - E^2)*Defer[Int][(E^x^8*x^
3)/((E^(x/(5*Log[x])) - (1 - E^2)*x + x^2)^2*Log[x]), x])/5 + Defer[Int][(E^x^8*x^4)/((E^(x/(5*Log[x])) - (1 -
E^2)*x + x^2)^2*Log[x]), x]/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{x^8} x \left (e^{\frac {x}{5 \log (x)}} x-e^{\frac {x}{5 \log (x)}} x \log (x)+5 \left (e^{\frac {x}{5 \log (x)}} \left (2+8 x^8\right )+e^2 \left (x+8 x^9\right )+x \left (-1-8 x^8+8 x^9\right )\right ) \log ^2(x)\right )}{5 \left (e^{\frac {x}{5 \log (x)}}+e^2 x+(-1+x) x\right )^2 \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \frac {e^{x^8} x \left (e^{\frac {x}{5 \log (x)}} x-e^{\frac {x}{5 \log (x)}} x \log (x)+5 \left (e^{\frac {x}{5 \log (x)}} \left (2+8 x^8\right )+e^2 \left (x+8 x^9\right )+x \left (-1-8 x^8+8 x^9\right )\right ) \log ^2(x)\right )}{\left (e^{\frac {x}{5 \log (x)}}+e^2 x+(-1+x) x\right )^2 \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {e^{x^8} x^2 \left (\left (1-e^2\right ) x-x^2-\left (1-e^2\right ) x \log (x)+x^2 \log (x)+5 \left (1-e^2\right ) \log ^2(x)-10 x \log ^2(x)\right )}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log ^2(x)}+\frac {e^{x^8} x \left (x-x \log (x)+10 \log ^2(x)+40 x^8 \log ^2(x)\right )}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right ) \log ^2(x)}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{x^8} x^2 \left (\left (1-e^2\right ) x-x^2-\left (1-e^2\right ) x \log (x)+x^2 \log (x)+5 \left (1-e^2\right ) \log ^2(x)-10 x \log ^2(x)\right )}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log ^2(x)} \, dx+\frac {1}{5} \int \frac {e^{x^8} x \left (x-x \log (x)+10 \log ^2(x)+40 x^8 \log ^2(x)\right )}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right ) \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {10 e^{x^8} x}{e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2}+\frac {40 e^{x^8} x^9}{e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2}+\frac {e^{x^8} x^2}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right ) \log ^2(x)}+\frac {e^{x^8} x^2}{\left (-e^{\frac {x}{5 \log (x)}}+\left (1-e^2\right ) x-x^2\right ) \log (x)}\right ) \, dx+\frac {1}{5} \int \frac {e^{x^8} x^2 \left (-x \left (-1+e^2+x\right )+x \left (-1+e^2+x\right ) \log (x)-5 \left (-1+e^2+2 x\right ) \log ^2(x)\right )}{\left (e^{\frac {x}{5 \log (x)}}+e^2 x+(-1+x) x\right )^2 \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {5 e^{x^8} \left (1-e^2\right ) x^2}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2}-\frac {10 e^{x^8} x^3}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2}+\frac {e^{x^8} \left (1-e^2\right ) x^3}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log ^2(x)}-\frac {e^{x^8} x^4}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log ^2(x)}-\frac {e^{x^8} \left (1-e^2\right ) x^3}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log (x)}+\frac {e^{x^8} x^4}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log (x)}\right ) \, dx+\frac {1}{5} \int \frac {e^{x^8} x^2}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right ) \log ^2(x)} \, dx+\frac {1}{5} \int \frac {e^{x^8} x^2}{\left (-e^{\frac {x}{5 \log (x)}}+\left (1-e^2\right ) x-x^2\right ) \log (x)} \, dx+2 \int \frac {e^{x^8} x}{e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2} \, dx+8 \int \frac {e^{x^8} x^9}{e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2} \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{x^8} x^4}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log ^2(x)} \, dx\right )+\frac {1}{5} \int \frac {e^{x^8} x^2}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right ) \log ^2(x)} \, dx+\frac {1}{5} \int \frac {e^{x^8} x^2}{\left (-e^{\frac {x}{5 \log (x)}}+\left (1-e^2\right ) x-x^2\right ) \log (x)} \, dx+\frac {1}{5} \int \frac {e^{x^8} x^4}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log (x)} \, dx-2 \int \frac {e^{x^8} x^3}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2} \, dx+2 \int \frac {e^{x^8} x}{e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2} \, dx+8 \int \frac {e^{x^8} x^9}{e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2} \, dx+\frac {1}{5} \left (1-e^2\right ) \int \frac {e^{x^8} x^3}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log ^2(x)} \, dx+\left (1-e^2\right ) \int \frac {e^{x^8} x^2}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2} \, dx+\frac {1}{5} \left (-1+e^2\right ) \int \frac {e^{x^8} x^3}{\left (e^{\frac {x}{5 \log (x)}}-\left (1-e^2\right ) x+x^2\right )^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 34, normalized size = 1.13 \begin {gather*} \frac {e^{x^8} x^2}{e^{\frac {x}{5 \log (x)}}-x+e^2 x+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x^8*(-5*x^2 - 40*x^10 + 40*x^11 + E^2*(5*x^2 + 40*x^10))*Log[x]^2 + E^(x/(5*Log[x]))*(E^x^8*x^2 -
E^x^8*x^2*Log[x] + E^x^8*(10*x + 40*x^9)*Log[x]^2))/(5*E^((2*x)/(5*Log[x]))*Log[x]^2 + E^(x/(5*Log[x]))*(-10*
x + 10*E^2*x + 10*x^2)*Log[x]^2 + (5*x^2 + 5*E^4*x^2 - 10*x^3 + 5*x^4 + E^2*(-10*x^2 + 10*x^3))*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.83, size = 29, normalized size = 0.97 \begin {gather*} \frac {x^{2} e^{\left (x^{8}\right )}}{x^{2} + x e^{2} - x + e^{\left (\frac {x}{5 \, \log \relax (x)}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((40*x^9+10*x)*exp(x^8)*log(x)^2-x^2*exp(x^8)*log(x)+x^2*exp(x^8))*exp(1/5*x/log(x))+((40*x^10+5*x^
2)*exp(2)+40*x^11-40*x^10-5*x^2)*exp(x^8)*log(x)^2)/(5*log(x)^2*exp(1/5*x/log(x))^2+(10*exp(2)*x+10*x^2-10*x)*
log(x)^2*exp(1/5*x/log(x))+(5*x^2*exp(2)^2+(10*x^3-10*x^2)*exp(2)+5*x^4-10*x^3+5*x^2)*log(x)^2),x, algorithm="
fricas")

[Out]

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

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giac [A]  time = 0.46, size = 29, normalized size = 0.97 \begin {gather*} \frac {x^{2} e^{\left (x^{8}\right )}}{x^{2} + x e^{2} - x + e^{\left (\frac {x}{5 \, \log \relax (x)}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((40*x^9+10*x)*exp(x^8)*log(x)^2-x^2*exp(x^8)*log(x)+x^2*exp(x^8))*exp(1/5*x/log(x))+((40*x^10+5*x^
2)*exp(2)+40*x^11-40*x^10-5*x^2)*exp(x^8)*log(x)^2)/(5*log(x)^2*exp(1/5*x/log(x))^2+(10*exp(2)*x+10*x^2-10*x)*
log(x)^2*exp(1/5*x/log(x))+(5*x^2*exp(2)^2+(10*x^3-10*x^2)*exp(2)+5*x^4-10*x^3+5*x^2)*log(x)^2),x, algorithm="
giac")

[Out]

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

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maple [A]  time = 0.05, size = 30, normalized size = 1.00

method	result	size
risch	\(\frac {x^{2} {\mathrm e}^{x^{8}}}{{\mathrm e}^{2} x +x^{2}+{\mathrm e}^{\frac {x}{5 \ln \relax (x )}}-x}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((40*x^9+10*x)*exp(x^8)*ln(x)^2-x^2*exp(x^8)*ln(x)+x^2*exp(x^8))*exp(1/5*x/ln(x))+((40*x^10+5*x^2)*exp(2)
+40*x^11-40*x^10-5*x^2)*exp(x^8)*ln(x)^2)/(5*ln(x)^2*exp(1/5*x/ln(x))^2+(10*exp(2)*x+10*x^2-10*x)*ln(x)^2*exp(
1/5*x/ln(x))+(5*x^2*exp(2)^2+(10*x^3-10*x^2)*exp(2)+5*x^4-10*x^3+5*x^2)*ln(x)^2),x,method=_RETURNVERBOSE)

[Out]

x^2*exp(x^8)/(exp(2)*x+x^2+exp(1/5*x/ln(x))-x)

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maxima [A]  time = 0.48, size = 28, normalized size = 0.93 \begin {gather*} \frac {x^{2} e^{\left (x^{8}\right )}}{x^{2} + x {\left (e^{2} - 1\right )} + e^{\left (\frac {x}{5 \, \log \relax (x)}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((40*x^9+10*x)*exp(x^8)*log(x)^2-x^2*exp(x^8)*log(x)+x^2*exp(x^8))*exp(1/5*x/log(x))+((40*x^10+5*x^
2)*exp(2)+40*x^11-40*x^10-5*x^2)*exp(x^8)*log(x)^2)/(5*log(x)^2*exp(1/5*x/log(x))^2+(10*exp(2)*x+10*x^2-10*x)*
log(x)^2*exp(1/5*x/log(x))+(5*x^2*exp(2)^2+(10*x^3-10*x^2)*exp(2)+5*x^4-10*x^3+5*x^2)*log(x)^2),x, algorithm="
maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {x}{5\,\ln \relax (x)}}\,\left (x^2\,{\mathrm {e}}^{x^8}+{\mathrm {e}}^{x^8}\,{\ln \relax (x)}^2\,\left (40\,x^9+10\,x\right )-x^2\,{\mathrm {e}}^{x^8}\,\ln \relax (x)\right )+{\mathrm {e}}^{x^8}\,{\ln \relax (x)}^2\,\left ({\mathrm {e}}^2\,\left (40\,x^{10}+5\,x^2\right )-5\,x^2-40\,x^{10}+40\,x^{11}\right )}{5\,{\mathrm {e}}^{\frac {2\,x}{5\,\ln \relax (x)}}\,{\ln \relax (x)}^2+{\ln \relax (x)}^2\,\left (5\,x^2\,{\mathrm {e}}^4-{\mathrm {e}}^2\,\left (10\,x^2-10\,x^3\right )+5\,x^2-10\,x^3+5\,x^4\right )+{\mathrm {e}}^{\frac {x}{5\,\ln \relax (x)}}\,{\ln \relax (x)}^2\,\left (10\,x\,{\mathrm {e}}^2-10\,x+10\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x/(5*log(x)))*(x^2*exp(x^8) + exp(x^8)*log(x)^2*(10*x + 40*x^9) - x^2*exp(x^8)*log(x)) + exp(x^8)*log
(x)^2*(exp(2)*(5*x^2 + 40*x^10) - 5*x^2 - 40*x^10 + 40*x^11))/(5*exp((2*x)/(5*log(x)))*log(x)^2 + log(x)^2*(5*
x^2*exp(4) - exp(2)*(10*x^2 - 10*x^3) + 5*x^2 - 10*x^3 + 5*x^4) + exp(x/(5*log(x)))*log(x)^2*(10*x*exp(2) - 10
*x + 10*x^2)),x)

[Out]

int((exp(x/(5*log(x)))*(x^2*exp(x^8) + exp(x^8)*log(x)^2*(10*x + 40*x^9) - x^2*exp(x^8)*log(x)) + exp(x^8)*log
(x)^2*(exp(2)*(5*x^2 + 40*x^10) - 5*x^2 - 40*x^10 + 40*x^11))/(5*exp((2*x)/(5*log(x)))*log(x)^2 + log(x)^2*(5*
x^2*exp(4) - exp(2)*(10*x^2 - 10*x^3) + 5*x^2 - 10*x^3 + 5*x^4) + exp(x/(5*log(x)))*log(x)^2*(10*x*exp(2) - 10
*x + 10*x^2)), x)

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sympy [A]  time = 0.55, size = 26, normalized size = 0.87 \begin {gather*} \frac {x^{2} e^{x^{8}}}{x^{2} - x + x e^{2} + e^{\frac {x}{5 \log {\relax (x )}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((40*x**9+10*x)*exp(x**8)*ln(x)**2-x**2*exp(x**8)*ln(x)+x**2*exp(x**8))*exp(1/5*x/ln(x))+((40*x**10
+5*x**2)*exp(2)+40*x**11-40*x**10-5*x**2)*exp(x**8)*ln(x)**2)/(5*ln(x)**2*exp(1/5*x/ln(x))**2+(10*exp(2)*x+10*
x**2-10*x)*ln(x)**2*exp(1/5*x/ln(x))+(5*x**2*exp(2)**2+(10*x**3-10*x**2)*exp(2)+5*x**4-10*x**3+5*x**2)*ln(x)**
2),x)

[Out]

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

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