3.78.16 \(\int \frac {-e^x+x+3 x^2+3 x \log (x)+e^{-2+\frac {e^x-x-3 x^2-3 x \log (x)}{e^2}} (4-e^x+6 x+3 \log (x)) (e^x-x-3 x^2-3 x \log (x))}{-e^x+x+3 x^2+3 x \log (x)} \, dx\)

Optimal. Leaf size=22 \[ e^{\frac {e^x-x-3 x (x+\log (x))}{e^2}}+x \]

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Rubi [F]  time = 1.74, 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+x+3 x^2+3 x \log (x)+e^{-2+\frac {e^x-x-3 x^2-3 x \log (x)}{e^2}} \left (4-e^x+6 x+3 \log (x)\right ) \left (e^x-x-3 x^2-3 x \log (x)\right )}{-e^x+x+3 x^2+3 x \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-E^x + x + 3*x^2 + 3*x*Log[x] + E^(-2 + (E^x - x - 3*x^2 - 3*x*Log[x])/E^2)*(4 - E^x + 6*x + 3*Log[x])*(E
^x - x - 3*x^2 - 3*x*Log[x]))/(-E^x + x + 3*x^2 + 3*x*Log[x]),x]

[Out]

x - 4*Defer[Int][E^(-2 + E^(-2 + x) - (x*(1 + 3*x))/E^2)/x^((3*x)/E^2), x] - 3*Log[x]*Defer[Int][E^(-2 + E^(-2
+ x) - (x*(1 + 3*x))/E^2)/x^((3*x)/E^2), x] + Defer[Int][E^(-2 + E^(-2 + x) + x - (x*(1 + 3*x))/E^2)/x^((3*x)
/E^2), x] - 6*Defer[Int][E^(-2 + E^(-2 + x) - (x*(1 + 3*x))/E^2)*x^(1 - (3*x)/E^2), x] + 3*Defer[Int][Defer[In
t][E^(-2 + E^(-2 + x) - (x*(1 + 3*x))/E^2)/x^((3*x)/E^2), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1+e^{-2+e^{-2+x}-\frac {x (1+3 x)}{e^2}} x^{-\frac {3 x}{e^2}} \left (-4+e^x-6 x-3 \log (x)\right )\right ) \, dx\\ &=x+\int e^{-2+e^{-2+x}-\frac {x (1+3 x)}{e^2}} x^{-\frac {3 x}{e^2}} \left (-4+e^x-6 x-3 \log (x)\right ) \, dx\\ &=x+\int \left (-4 e^{-2+e^{-2+x}-\frac {x (1+3 x)}{e^2}} x^{-\frac {3 x}{e^2}}+e^{-2+e^{-2+x}+x-\frac {x (1+3 x)}{e^2}} x^{-\frac {3 x}{e^2}}-6 e^{-2+e^{-2+x}-\frac {x (1+3 x)}{e^2}} x^{1-\frac {3 x}{e^2}}-3 e^{-2+e^{-2+x}-\frac {x (1+3 x)}{e^2}} x^{-\frac {3 x}{e^2}} \log (x)\right ) \, dx\\ &=x-3 \int e^{-2+e^{-2+x}-\frac {x (1+3 x)}{e^2}} x^{-\frac {3 x}{e^2}} \log (x) \, dx-4 \int e^{-2+e^{-2+x}-\frac {x (1+3 x)}{e^2}} x^{-\frac {3 x}{e^2}} \, dx-6 \int e^{-2+e^{-2+x}-\frac {x (1+3 x)}{e^2}} x^{1-\frac {3 x}{e^2}} \, dx+\int e^{-2+e^{-2+x}+x-\frac {x (1+3 x)}{e^2}} x^{-\frac {3 x}{e^2}} \, dx\\ &=x+3 \int \frac {\int e^{-2+e^{-2+x}-\frac {x (1+3 x)}{e^2}} x^{-\frac {3 x}{e^2}} \, dx}{x} \, dx-4 \int e^{-2+e^{-2+x}-\frac {x (1+3 x)}{e^2}} x^{-\frac {3 x}{e^2}} \, dx-6 \int e^{-2+e^{-2+x}-\frac {x (1+3 x)}{e^2}} x^{1-\frac {3 x}{e^2}} \, dx-(3 \log (x)) \int e^{-2+e^{-2+x}-\frac {x (1+3 x)}{e^2}} x^{-\frac {3 x}{e^2}} \, dx+\int e^{-2+e^{-2+x}+x-\frac {x (1+3 x)}{e^2}} x^{-\frac {3 x}{e^2}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 29, normalized size = 1.32 \begin {gather*} x+e^{\frac {e^x-x (1+3 x)}{e^2}} x^{-\frac {3 x}{e^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^x + x + 3*x^2 + 3*x*Log[x] + E^(-2 + (E^x - x - 3*x^2 - 3*x*Log[x])/E^2)*(4 - E^x + 6*x + 3*Log[
x])*(E^x - x - 3*x^2 - 3*x*Log[x]))/(-E^x + x + 3*x^2 + 3*x*Log[x]),x]

[Out]

x + E^((E^x - x*(1 + 3*x))/E^2)/x^((3*x)/E^2)

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fricas [B]  time = 0.73, size = 90, normalized size = 4.09 \begin {gather*} \frac {3 \, x^{3} + 3 \, x^{2} \log \relax (x) + x^{2} - x e^{x} - e^{\left (-{\left (3 \, x^{2} - e^{2} \log \left (-3 \, x^{2} - 3 \, x \log \relax (x) - x + e^{x}\right ) + 3 \, x \log \relax (x) + x + 2 \, e^{2} - e^{x}\right )} e^{\left (-2\right )} + 2\right )}}{3 \, x^{2} + 3 \, x \log \relax (x) + x - e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*log(x)-exp(x)+6*x+4)*exp(log(-3*x*log(x)+exp(x)-3*x^2-x)-2)*exp(exp(log(-3*x*log(x)+exp(x)-3*x^2
-x)-2))+3*x*log(x)-exp(x)+3*x^2+x)/(3*x*log(x)-exp(x)+3*x^2+x),x, algorithm="fricas")

[Out]

(3*x^3 + 3*x^2*log(x) + x^2 - x*e^x - e^(-(3*x^2 - e^2*log(-3*x^2 - 3*x*log(x) - x + e^x) + 3*x*log(x) + x + 2
*e^2 - e^x)*e^(-2) + 2))/(3*x^2 + 3*x*log(x) + x - e^x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 \, x^{2} + {\left (6 \, x - e^{x} + 3 \, \log \relax (x) + 4\right )} e^{\left (e^{\left (\log \left (-3 \, x^{2} - 3 \, x \log \relax (x) - x + e^{x}\right ) - 2\right )} + \log \left (-3 \, x^{2} - 3 \, x \log \relax (x) - x + e^{x}\right ) - 2\right )} + 3 \, x \log \relax (x) + x - e^{x}}{3 \, x^{2} + 3 \, x \log \relax (x) + x - e^{x}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*log(x)-exp(x)+6*x+4)*exp(log(-3*x*log(x)+exp(x)-3*x^2-x)-2)*exp(exp(log(-3*x*log(x)+exp(x)-3*x^2
-x)-2))+3*x*log(x)-exp(x)+3*x^2+x)/(3*x*log(x)-exp(x)+3*x^2+x),x, algorithm="giac")

[Out]

integrate((3*x^2 + (6*x - e^x + 3*log(x) + 4)*e^(e^(log(-3*x^2 - 3*x*log(x) - x + e^x) - 2) + log(-3*x^2 - 3*x
*log(x) - x + e^x) - 2) + 3*x*log(x) + x - e^x)/(3*x^2 + 3*x*log(x) + x - e^x), x)

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maple [A]  time = 0.05, size = 23, normalized size = 1.05

method	result	size
risch	\(x +{\mathrm e}^{\left (-3 x \ln \relax (x )+{\mathrm e}^{x}-3 x^{2}-x \right ) {\mathrm e}^{-2}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*ln(x)-exp(x)+6*x+4)*exp(ln(-3*x*ln(x)+exp(x)-3*x^2-x)-2)*exp(exp(ln(-3*x*ln(x)+exp(x)-3*x^2-x)-2))+3*x
*ln(x)-exp(x)+3*x^2+x)/(3*x*ln(x)-exp(x)+3*x^2+x),x,method=_RETURNVERBOSE)

[Out]

x+exp((-3*x*ln(x)+exp(x)-3*x^2-x)*exp(-2))

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maxima [B]  time = 0.40, size = 44, normalized size = 2.00 \begin {gather*} {\left (x e^{\left (3 \, x^{2} e^{\left (-2\right )} + x e^{\left (-2\right )}\right )} + e^{\left (-3 \, x e^{\left (-2\right )} \log \relax (x) + e^{\left (x - 2\right )}\right )}\right )} e^{\left (-3 \, x^{2} e^{\left (-2\right )} - x e^{\left (-2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*log(x)-exp(x)+6*x+4)*exp(log(-3*x*log(x)+exp(x)-3*x^2-x)-2)*exp(exp(log(-3*x*log(x)+exp(x)-3*x^2
-x)-2))+3*x*log(x)-exp(x)+3*x^2+x)/(3*x*log(x)-exp(x)+3*x^2+x),x, algorithm="maxima")

[Out]

(x*e^(3*x^2*e^(-2) + x*e^(-2)) + e^(-3*x*e^(-2)*log(x) + e^(x - 2)))*e^(-3*x^2*e^(-2) - x*e^(-2))

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mupad [B]  time = 6.19, size = 32, normalized size = 1.45 \begin {gather*} x+\frac {{\mathrm {e}}^{-3\,x^2\,{\mathrm {e}}^{-2}}\,{\mathrm {e}}^{{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{-2}}}{x^{3\,x\,{\mathrm {e}}^{-2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - exp(x) + 3*x*log(x) + 3*x^2 + exp(log(exp(x) - x - 3*x*log(x) - 3*x^2) - 2)*exp(exp(log(exp(x) - x -
3*x*log(x) - 3*x^2) - 2))*(6*x - exp(x) + 3*log(x) + 4))/(x - exp(x) + 3*x*log(x) + 3*x^2),x)

[Out]

x + (exp(-3*x^2*exp(-2))*exp(exp(-2)*exp(x))*exp(-x*exp(-2)))/x^(3*x*exp(-2))

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sympy [A]  time = 0.58, size = 22, normalized size = 1.00 \begin {gather*} x + e^{\frac {- 3 x^{2} - 3 x \log {\relax (x )} - x + e^{x}}{e^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*ln(x)-exp(x)+6*x+4)*exp(ln(-3*x*ln(x)+exp(x)-3*x**2-x)-2)*exp(exp(ln(-3*x*ln(x)+exp(x)-3*x**2-x)
-2))+3*x*ln(x)-exp(x)+3*x**2+x)/(3*x*ln(x)-exp(x)+3*x**2+x),x)

[Out]

x + exp((-3*x**2 - 3*x*log(x) - x + exp(x))*exp(-2))

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