3.92.41 \(\int (2 x+6 x^2+4 x^3+e^{\frac {8 (5 e^x-5 \log (x))}{x}} (-40 x^2+4 x^3+e^x (-40 x^2+40 x^3)+40 x^2 \log (x))+e^{\frac {4 (5 e^x-5 \log (x))}{x}} (40 x+34 x^2-8 x^3+e^x (40 x-40 x^3)+(-40 x-40 x^2) \log (x))) \, dx\)

Optimal. Leaf size=27 \[ \left (x+x^2-e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x^2\right )^2 \]

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Rubi [F]  time = 7.99, 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 \left (2 x+6 x^2+4 x^3+e^{\frac {8 \left (5 e^x-5 \log (x)\right )}{x}} \left (-40 x^2+4 x^3+e^x \left (-40 x^2+40 x^3\right )+40 x^2 \log (x)\right )+e^{\frac {4 \left (5 e^x-5 \log (x)\right )}{x}} \left (40 x+34 x^2-8 x^3+e^x \left (40 x-40 x^3\right )+\left (-40 x-40 x^2\right ) \log (x)\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[2*x + 6*x^2 + 4*x^3 + E^((8*(5*E^x - 5*Log[x]))/x)*(-40*x^2 + 4*x^3 + E^x*(-40*x^2 + 40*x^3) + 40*x^2*Log[
x]) + E^((4*(5*E^x - 5*Log[x]))/x)*(40*x + 34*x^2 - 8*x^3 + E^x*(40*x - 40*x^3) + (-40*x - 40*x^2)*Log[x]),x]

[Out]

x^2 + 2*x^3 + x^4 - (E^((40*E^x)/x)*(x^2 + E^x*(x^2 - x^3) - x^2*Log[x]))/(x^(40/x)*((E^x - x^(-1))/x - (E^x -
Log[x])/x^2)) + 40*Defer[Int][E^(x + (20*(E^x - Log[x]))/x)*x, x] + 40*Defer[Int][E^((20*(E^x - Log[x]))/x)*x
, x] + 34*Defer[Int][E^((20*(E^x - Log[x]))/x)*x^2, x] - 40*Defer[Int][E^(x + (20*(E^x - Log[x]))/x)*x^3, x] -
8*Defer[Int][E^((20*(E^x - Log[x]))/x)*x^3, x] - 40*Defer[Int][E^((20*(E^x - Log[x]))/x)*x*Log[x], x] - 40*De
fer[Int][E^((20*(E^x - Log[x]))/x)*x^2*Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x^2+2 x^3+x^4+\int e^{\frac {8 \left (5 e^x-5 \log (x)\right )}{x}} \left (-40 x^2+4 x^3+e^x \left (-40 x^2+40 x^3\right )+40 x^2 \log (x)\right ) \, dx+\int e^{\frac {4 \left (5 e^x-5 \log (x)\right )}{x}} \left (40 x+34 x^2-8 x^3+e^x \left (40 x-40 x^3\right )+\left (-40 x-40 x^2\right ) \log (x)\right ) \, dx\\ &=x^2+2 x^3+x^4-\frac {e^{\frac {40 e^x}{x}} x^{-40/x} \left (x^2+e^x \left (x^2-x^3\right )-x^2 \log (x)\right )}{\frac {e^x-\frac {1}{x}}{x}-\frac {e^x-\log (x)}{x^2}}+\int 2 e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x \left (20+20 e^x+17 x-4 x^2-20 e^x x^2-20 \log (x)-20 x \log (x)\right ) \, dx\\ &=x^2+2 x^3+x^4-\frac {e^{\frac {40 e^x}{x}} x^{-40/x} \left (x^2+e^x \left (x^2-x^3\right )-x^2 \log (x)\right )}{\frac {e^x-\frac {1}{x}}{x}-\frac {e^x-\log (x)}{x^2}}+2 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x \left (20+20 e^x+17 x-4 x^2-20 e^x x^2-20 \log (x)-20 x \log (x)\right ) \, dx\\ &=x^2+2 x^3+x^4-\frac {e^{\frac {40 e^x}{x}} x^{-40/x} \left (x^2+e^x \left (x^2-x^3\right )-x^2 \log (x)\right )}{\frac {e^x-\frac {1}{x}}{x}-\frac {e^x-\log (x)}{x^2}}+2 \int \left (-20 e^{x+\frac {20 \left (e^x-\log (x)\right )}{x}} x \left (-1+x^2\right )-e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x \left (-20-17 x+4 x^2+20 \log (x)+20 x \log (x)\right )\right ) \, dx\\ &=x^2+2 x^3+x^4-\frac {e^{\frac {40 e^x}{x}} x^{-40/x} \left (x^2+e^x \left (x^2-x^3\right )-x^2 \log (x)\right )}{\frac {e^x-\frac {1}{x}}{x}-\frac {e^x-\log (x)}{x^2}}-2 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x \left (-20-17 x+4 x^2+20 \log (x)+20 x \log (x)\right ) \, dx-40 \int e^{x+\frac {20 \left (e^x-\log (x)\right )}{x}} x \left (-1+x^2\right ) \, dx\\ &=x^2+2 x^3+x^4-\frac {e^{\frac {40 e^x}{x}} x^{-40/x} \left (x^2+e^x \left (x^2-x^3\right )-x^2 \log (x)\right )}{\frac {e^x-\frac {1}{x}}{x}-\frac {e^x-\log (x)}{x^2}}-2 \int \left (e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x \left (-20-17 x+4 x^2\right )+20 e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x (1+x) \log (x)\right ) \, dx-40 \int \left (-e^{x+\frac {20 \left (e^x-\log (x)\right )}{x}} x+e^{x+\frac {20 \left (e^x-\log (x)\right )}{x}} x^3\right ) \, dx\\ &=x^2+2 x^3+x^4-\frac {e^{\frac {40 e^x}{x}} x^{-40/x} \left (x^2+e^x \left (x^2-x^3\right )-x^2 \log (x)\right )}{\frac {e^x-\frac {1}{x}}{x}-\frac {e^x-\log (x)}{x^2}}-2 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x \left (-20-17 x+4 x^2\right ) \, dx+40 \int e^{x+\frac {20 \left (e^x-\log (x)\right )}{x}} x \, dx-40 \int e^{x+\frac {20 \left (e^x-\log (x)\right )}{x}} x^3 \, dx-40 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x (1+x) \log (x) \, dx\\ &=x^2+2 x^3+x^4-\frac {e^{\frac {40 e^x}{x}} x^{-40/x} \left (x^2+e^x \left (x^2-x^3\right )-x^2 \log (x)\right )}{\frac {e^x-\frac {1}{x}}{x}-\frac {e^x-\log (x)}{x^2}}-2 \int \left (-20 e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x-17 e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x^2+4 e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x^3\right ) \, dx+40 \int e^{x+\frac {20 \left (e^x-\log (x)\right )}{x}} x \, dx-40 \int e^{x+\frac {20 \left (e^x-\log (x)\right )}{x}} x^3 \, dx-40 \int \left (e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x \log (x)+e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x^2 \log (x)\right ) \, dx\\ &=x^2+2 x^3+x^4-\frac {e^{\frac {40 e^x}{x}} x^{-40/x} \left (x^2+e^x \left (x^2-x^3\right )-x^2 \log (x)\right )}{\frac {e^x-\frac {1}{x}}{x}-\frac {e^x-\log (x)}{x^2}}-8 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x^3 \, dx+34 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x^2 \, dx+40 \int e^{x+\frac {20 \left (e^x-\log (x)\right )}{x}} x \, dx+40 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x \, dx-40 \int e^{x+\frac {20 \left (e^x-\log (x)\right )}{x}} x^3 \, dx-40 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x \log (x) \, dx-40 \int e^{\frac {20 \left (e^x-\log (x)\right )}{x}} x^2 \log (x) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.76, size = 57, normalized size = 2.11 \begin {gather*} -2 e^{\frac {20 e^x}{x}} x^{3-\frac {20}{x}} (1+x)+x^2 \left (1+2 x+x^2+e^{\frac {40 e^x}{x}} x^{2-\frac {40}{x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2*x + 6*x^2 + 4*x^3 + E^((8*(5*E^x - 5*Log[x]))/x)*(-40*x^2 + 4*x^3 + E^x*(-40*x^2 + 40*x^3) + 40*x^
2*Log[x]) + E^((4*(5*E^x - 5*Log[x]))/x)*(40*x + 34*x^2 - 8*x^3 + E^x*(40*x - 40*x^3) + (-40*x - 40*x^2)*Log[x
]),x]

[Out]

-2*E^((20*E^x)/x)*x^(3 - 20/x)*(1 + x) + x^2*(1 + 2*x + x^2 + E^((40*E^x)/x)*x^(2 - 40/x))

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fricas [A]  time = 1.13, size = 51, normalized size = 1.89 \begin {gather*} x^{4} e^{\left (\frac {40 \, {\left (e^{x} - \log \relax (x)\right )}}{x}\right )} + x^{4} + 2 \, x^{3} + x^{2} - 2 \, {\left (x^{4} + x^{3}\right )} e^{\left (\frac {20 \, {\left (e^{x} - \log \relax (x)\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((40*x^2*log(x)+(40*x^3-40*x^2)*exp(x)+4*x^3-40*x^2)*exp((-5*log(x)+5*exp(x))/x)^8+((-40*x^2-40*x)*lo
g(x)+(-40*x^3+40*x)*exp(x)-8*x^3+34*x^2+40*x)*exp((-5*log(x)+5*exp(x))/x)^4+4*x^3+6*x^2+2*x,x, algorithm="fric
as")

[Out]

x^4*e^(40*(e^x - log(x))/x) + x^4 + 2*x^3 + x^2 - 2*(x^4 + x^3)*e^(20*(e^x - log(x))/x)

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giac [B]  time = 0.22, size = 65, normalized size = 2.41 \begin {gather*} x^{4} e^{\left (\frac {40 \, {\left (e^{x} - \log \relax (x)\right )}}{x}\right )} - 2 \, x^{4} e^{\left (\frac {20 \, {\left (e^{x} - \log \relax (x)\right )}}{x}\right )} + x^{4} - 2 \, x^{3} e^{\left (\frac {20 \, {\left (e^{x} - \log \relax (x)\right )}}{x}\right )} + 2 \, x^{3} + x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((40*x^2*log(x)+(40*x^3-40*x^2)*exp(x)+4*x^3-40*x^2)*exp((-5*log(x)+5*exp(x))/x)^8+((-40*x^2-40*x)*lo
g(x)+(-40*x^3+40*x)*exp(x)-8*x^3+34*x^2+40*x)*exp((-5*log(x)+5*exp(x))/x)^4+4*x^3+6*x^2+2*x,x, algorithm="giac
")

[Out]

x^4*e^(40*(e^x - log(x))/x) - 2*x^4*e^(20*(e^x - log(x))/x) + x^4 - 2*x^3*e^(20*(e^x - log(x))/x) + 2*x^3 + x^
2

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maple [A]  time = 0.15, size = 51, normalized size = 1.89

method	result	size
risch	\(x^{4} {\mathrm e}^{\frac {40 \,{\mathrm e}^{x}-40 \ln \relax (x )}{x}}-2 x^{3} \left (x +1\right ) {\mathrm e}^{\frac {-20 \ln \relax (x )+20 \,{\mathrm e}^{x}}{x}}+x^{4}+2 x^{3}+x^{2}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((40*x^2*ln(x)+(40*x^3-40*x^2)*exp(x)+4*x^3-40*x^2)*exp((-5*ln(x)+5*exp(x))/x)^8+((-40*x^2-40*x)*ln(x)+(-40
*x^3+40*x)*exp(x)-8*x^3+34*x^2+40*x)*exp((-5*ln(x)+5*exp(x))/x)^4+4*x^3+6*x^2+2*x,x,method=_RETURNVERBOSE)

[Out]

x^4*exp(40*(exp(x)-ln(x))/x)-2*x^3*(x+1)*exp(20*(exp(x)-ln(x))/x)+x^4+2*x^3+x^2

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maxima [B]  time = 0.37, size = 57, normalized size = 2.11 \begin {gather*} x^{4} e^{\left (\frac {40 \, e^{x}}{x} - \frac {40 \, \log \relax (x)}{x}\right )} + x^{4} + 2 \, x^{3} + x^{2} - 2 \, {\left (x^{4} + x^{3}\right )} e^{\left (\frac {20 \, e^{x}}{x} - \frac {20 \, \log \relax (x)}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((40*x^2*log(x)+(40*x^3-40*x^2)*exp(x)+4*x^3-40*x^2)*exp((-5*log(x)+5*exp(x))/x)^8+((-40*x^2-40*x)*lo
g(x)+(-40*x^3+40*x)*exp(x)-8*x^3+34*x^2+40*x)*exp((-5*log(x)+5*exp(x))/x)^4+4*x^3+6*x^2+2*x,x, algorithm="maxi
ma")

[Out]

x^4*e^(40*e^x/x - 40*log(x)/x) + x^4 + 2*x^3 + x^2 - 2*(x^4 + x^3)*e^(20*e^x/x - 20*log(x)/x)

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mupad [B]  time = 8.20, size = 63, normalized size = 2.33 \begin {gather*} x^2+2\,x^3+x^4-\frac {{\mathrm {e}}^{\frac {20\,{\mathrm {e}}^x}{x}}\,\left (2\,x^4+2\,x^3\right )}{x^{20/x}}+\frac {x^4\,{\mathrm {e}}^{\frac {40\,{\mathrm {e}}^x}{x}}}{x^{40/x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x - exp((8*(5*exp(x) - 5*log(x)))/x)*(exp(x)*(40*x^2 - 40*x^3) - 40*x^2*log(x) + 40*x^2 - 4*x^3) + 6*x^2
+ 4*x^3 + exp((4*(5*exp(x) - 5*log(x)))/x)*(40*x + exp(x)*(40*x - 40*x^3) - log(x)*(40*x + 40*x^2) + 34*x^2 -
8*x^3),x)

[Out]

x^2 + 2*x^3 + x^4 - (exp((20*exp(x))/x)*(2*x^3 + 2*x^4))/x^(20/x) + (x^4*exp((40*exp(x))/x))/x^(40/x)

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sympy [B]  time = 0.59, size = 53, normalized size = 1.96 \begin {gather*} x^{4} e^{\frac {8 \left (5 e^{x} - 5 \log {\relax (x )}\right )}{x}} + x^{4} + 2 x^{3} + x^{2} + \left (- 2 x^{4} - 2 x^{3}\right ) e^{\frac {4 \left (5 e^{x} - 5 \log {\relax (x )}\right )}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((40*x**2*ln(x)+(40*x**3-40*x**2)*exp(x)+4*x**3-40*x**2)*exp((-5*ln(x)+5*exp(x))/x)**8+((-40*x**2-40*
x)*ln(x)+(-40*x**3+40*x)*exp(x)-8*x**3+34*x**2+40*x)*exp((-5*ln(x)+5*exp(x))/x)**4+4*x**3+6*x**2+2*x,x)

[Out]

x**4*exp(8*(5*exp(x) - 5*log(x))/x) + x**4 + 2*x**3 + x**2 + (-2*x**4 - 2*x**3)*exp(4*(5*exp(x) - 5*log(x))/x)

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