3.67.100 \(\int \frac {-e^{1+x} x^2+2 e^{10+e^2+e^{8-x^2}-x^2} x^3+(2 e^{2+e^2+e^{8-x^2}} x+2 e^{1+x} x) \log (e^{2+e^2+e^{8-x^2}}+e^{1+x})}{(e^{2+e^2+e^{8-x^2}}+e^{1+x}) \log ^2(e^{2+e^2+e^{8-x^2}}+e^{1+x})} \, dx\)

Optimal. Leaf size=29 \[ \frac {x^2}{\log \left (e^{2+e^2+e^{8-x^2}}+e^{1+x}\right )} \]

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Rubi [F]  time = 8.78, 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^{1+x} x^2+2 e^{10+e^2+e^{8-x^2}-x^2} x^3+\left (2 e^{2+e^2+e^{8-x^2}} x+2 e^{1+x} x\right ) \log \left (e^{2+e^2+e^{8-x^2}}+e^{1+x}\right )}{\left (e^{2+e^2+e^{8-x^2}}+e^{1+x}\right ) \log ^2\left (e^{2+e^2+e^{8-x^2}}+e^{1+x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-(E^(1 + x)*x^2) + 2*E^(10 + E^2 + E^(8 - x^2) - x^2)*x^3 + (2*E^(2 + E^2 + E^(8 - x^2))*x + 2*E^(1 + x)*
x)*Log[E^(2 + E^2 + E^(8 - x^2)) + E^(1 + x)])/((E^(2 + E^2 + E^(8 - x^2)) + E^(1 + x))*Log[E^(2 + E^2 + E^(8
- x^2)) + E^(1 + x)]^2),x]

[Out]

-Defer[Int][x^2/(1 + Log[E^(1 + E^2 + E^(8 - x^2)) + E^x])^2, x] + Defer[Int][(E^(1 + E^2 + E^(8 - x^2))*x^2)/
((E^(1 + E^2 + E^(8 - x^2)) + E^x)*(1 + Log[E^(1 + E^2 + E^(8 - x^2)) + E^x])^2), x] + (2*Defer[Int][(E^(E^(8
- x^2) + 10*(1 + E^2/10) - x^2)*x^3)/((E^(1 + E^2 + E^(8 - x^2)) + E^x)*(1 + Log[E^(1 + E^2 + E^(8 - x^2)) + E
^x])^2), x])/E + 2*Defer[Int][x/(1 + Log[E^(1 + E^2 + E^(8 - x^2)) + E^x]), x]

Rubi steps

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

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Mathematica [A]  time = 0.31, size = 29, normalized size = 1.00 \begin {gather*} \frac {x^2}{1+\log \left (e^{1+e^2+e^{8-x^2}}+e^x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^(1 + x)*x^2) + 2*E^(10 + E^2 + E^(8 - x^2) - x^2)*x^3 + (2*E^(2 + E^2 + E^(8 - x^2))*x + 2*E^(1
+ x)*x)*Log[E^(2 + E^2 + E^(8 - x^2)) + E^(1 + x)])/((E^(2 + E^2 + E^(8 - x^2)) + E^(1 + x))*Log[E^(2 + E^2 +
E^(8 - x^2)) + E^(1 + x)]^2),x]

[Out]

x^2/(1 + Log[E^(1 + E^2 + E^(8 - x^2)) + E^x])

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fricas [A]  time = 0.46, size = 25, normalized size = 0.86 \begin {gather*} \frac {x^{2}}{\log \left (e^{\left (x + 1\right )} + e^{\left (e^{2} + e^{\left (-x^{2} + 8\right )} + 2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(exp(-x^2+8)+exp(2)+2)+2*x*exp(x+1))*log(exp(exp(-x^2+8)+exp(2)+2)+exp(x+1))+2*x^3*exp(-x^2
+8)*exp(exp(-x^2+8)+exp(2)+2)-x^2*exp(x+1))/(exp(exp(-x^2+8)+exp(2)+2)+exp(x+1))/log(exp(exp(-x^2+8)+exp(2)+2)
+exp(x+1))^2,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{3} e^{\left (-x^{2} + e^{2} + e^{\left (-x^{2} + 8\right )} + 10\right )} - x^{2} e^{\left (x + 1\right )} + 2 \, {\left (x e^{\left (x + 1\right )} + x e^{\left (e^{2} + e^{\left (-x^{2} + 8\right )} + 2\right )}\right )} \log \left (e^{\left (x + 1\right )} + e^{\left (e^{2} + e^{\left (-x^{2} + 8\right )} + 2\right )}\right )}{{\left (e^{\left (x + 1\right )} + e^{\left (e^{2} + e^{\left (-x^{2} + 8\right )} + 2\right )}\right )} \log \left (e^{\left (x + 1\right )} + e^{\left (e^{2} + e^{\left (-x^{2} + 8\right )} + 2\right )}\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(exp(-x^2+8)+exp(2)+2)+2*x*exp(x+1))*log(exp(exp(-x^2+8)+exp(2)+2)+exp(x+1))+2*x^3*exp(-x^2
+8)*exp(exp(-x^2+8)+exp(2)+2)-x^2*exp(x+1))/(exp(exp(-x^2+8)+exp(2)+2)+exp(x+1))/log(exp(exp(-x^2+8)+exp(2)+2)
+exp(x+1))^2,x, algorithm="giac")

[Out]

integrate((2*x^3*e^(-x^2 + e^2 + e^(-x^2 + 8) + 10) - x^2*e^(x + 1) + 2*(x*e^(x + 1) + x*e^(e^2 + e^(-x^2 + 8)
+ 2))*log(e^(x + 1) + e^(e^2 + e^(-x^2 + 8) + 2)))/((e^(x + 1) + e^(e^2 + e^(-x^2 + 8) + 2))*log(e^(x + 1) +
e^(e^2 + e^(-x^2 + 8) + 2))^2), x)

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maple [A]  time = 0.04, size = 26, normalized size = 0.90

method	result	size
risch	\(\frac {x^{2}}{\ln \left ({\mathrm e}^{{\mathrm e}^{-x^{2}+8}+{\mathrm e}^{2}+2}+{\mathrm e}^{x +1}\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x*exp(exp(-x^2+8)+exp(2)+2)+2*x*exp(x+1))*ln(exp(exp(-x^2+8)+exp(2)+2)+exp(x+1))+2*x^3*exp(-x^2+8)*exp
(exp(-x^2+8)+exp(2)+2)-x^2*exp(x+1))/(exp(exp(-x^2+8)+exp(2)+2)+exp(x+1))/ln(exp(exp(-x^2+8)+exp(2)+2)+exp(x+1
))^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.45, size = 25, normalized size = 0.86 \begin {gather*} \frac {x^{2}}{\log \left (e^{x} + e^{\left (e^{2} + e^{\left (-x^{2} + 8\right )} + 1\right )}\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(exp(-x^2+8)+exp(2)+2)+2*x*exp(x+1))*log(exp(exp(-x^2+8)+exp(2)+2)+exp(x+1))+2*x^3*exp(-x^2
+8)*exp(exp(-x^2+8)+exp(2)+2)-x^2*exp(x+1))/(exp(exp(-x^2+8)+exp(2)+2)+exp(x+1))/log(exp(exp(-x^2+8)+exp(2)+2)
+exp(x+1))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.37, size = 286, normalized size = 9.86 \begin {gather*} \frac {x^2-\frac {2\,x\,\ln \left (\mathrm {e}\,{\mathrm {e}}^x+{\mathrm {e}}^2\,{\mathrm {e}}^{{\mathrm {e}}^8\,{\mathrm {e}}^{-x^2}}\,{\mathrm {e}}^{{\mathrm {e}}^2}\right )\,\left ({\mathrm {e}}^{x+1}+{\mathrm {e}}^{{\mathrm {e}}^2+{\mathrm {e}}^8\,{\mathrm {e}}^{-x^2}+2}\right )}{{\mathrm {e}}^{x+1}-2\,x\,{\mathrm {e}}^{{\mathrm {e}}^2+{\mathrm {e}}^8\,{\mathrm {e}}^{-x^2}-x^2+10}}}{\ln \left (\mathrm {e}\,{\mathrm {e}}^x+{\mathrm {e}}^2\,{\mathrm {e}}^{{\mathrm {e}}^8\,{\mathrm {e}}^{-x^2}}\,{\mathrm {e}}^{{\mathrm {e}}^2}\right )}-{\mathrm {e}}^{x^2-8}+\frac {4\,x^2\,{\mathrm {e}}^{-x^2+2\,x+10}-{\mathrm {e}}^{2\,x+2}+4\,x^3\,{\mathrm {e}}^{-x^2+2\,x+10}+4\,x^3\,{\mathrm {e}}^{-2\,x^2+2\,x+18}+x\,{\mathrm {e}}^{2\,x+2}-2\,x\,{\mathrm {e}}^{-x^2+2\,x+10}+2\,x^2\,{\mathrm {e}}^{2\,x+2}}{\left ({\mathrm {e}}^{x+1}-2\,x\,{\mathrm {e}}^{{\mathrm {e}}^2+{\mathrm {e}}^8\,{\mathrm {e}}^{-x^2}-x^2+10}\right )\,\left (x\,{\mathrm {e}}^{-x^2+x+9}-{\mathrm {e}}^{-x^2+x+9}+2\,x^2\,{\mathrm {e}}^{-x^2+x+9}+2\,x^2\,{\mathrm {e}}^{-2\,x^2+x+17}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(exp(2) + exp(8 - x^2) + 2) + exp(x + 1))*(2*x*exp(x + 1) + 2*x*exp(exp(2) + exp(8 - x^2) + 2)) -
x^2*exp(x + 1) + 2*x^3*exp(exp(2) + exp(8 - x^2) + 2)*exp(8 - x^2))/(log(exp(exp(2) + exp(8 - x^2) + 2) + exp(
x + 1))^2*(exp(exp(2) + exp(8 - x^2) + 2) + exp(x + 1))),x)

[Out]

(x^2 - (2*x*log(exp(1)*exp(x) + exp(2)*exp(exp(8)*exp(-x^2))*exp(exp(2)))*(exp(x + 1) + exp(exp(2) + exp(8)*ex
p(-x^2) + 2)))/(exp(x + 1) - 2*x*exp(exp(2) + exp(8)*exp(-x^2) - x^2 + 10)))/log(exp(1)*exp(x) + exp(2)*exp(ex
p(8)*exp(-x^2))*exp(exp(2))) - exp(x^2 - 8) + (4*x^2*exp(2*x - x^2 + 10) - exp(2*x + 2) + 4*x^3*exp(2*x - x^2
+ 10) + 4*x^3*exp(2*x - 2*x^2 + 18) + x*exp(2*x + 2) - 2*x*exp(2*x - x^2 + 10) + 2*x^2*exp(2*x + 2))/((exp(x +
1) - 2*x*exp(exp(2) + exp(8)*exp(-x^2) - x^2 + 10))*(x*exp(x - x^2 + 9) - exp(x - x^2 + 9) + 2*x^2*exp(x - x^
2 + 9) + 2*x^2*exp(x - 2*x^2 + 17)))

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sympy [A]  time = 2.89, size = 22, normalized size = 0.76 \begin {gather*} \frac {x^{2}}{\log {\left (e^{x + 1} + e^{e^{8 - x^{2}} + 2 + e^{2}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x*exp(exp(-x**2+8)+exp(2)+2)+2*x*exp(x+1))*ln(exp(exp(-x**2+8)+exp(2)+2)+exp(x+1))+2*x**3*exp(-x
**2+8)*exp(exp(-x**2+8)+exp(2)+2)-x**2*exp(x+1))/(exp(exp(-x**2+8)+exp(2)+2)+exp(x+1))/ln(exp(exp(-x**2+8)+exp
(2)+2)+exp(x+1))**2,x)

[Out]

x**2/log(exp(x + 1) + exp(exp(8 - x**2) + 2 + exp(2)))

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