3.8.16 \(\int \frac {e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}} (-40 x^2+40 x \log (2)+e^x (-40 x^2+(20+20 x) \log (2)))}{e^{2 x} x^2+2 e^x x^3+x^4} \, dx\)

Optimal. Leaf size=32 \[ e^2+e^{1+4 \left (2+\frac {5 (2 x-\log (2))}{x \left (e^x+x\right )}\right )} \]

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Rubi [F]  time = 12.46, 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 {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}} \left (-40 x^2+40 x \log (2)+e^x \left (-40 x^2+(20+20 x) \log (2)\right )\right )}{e^{2 x} x^2+2 e^x x^3+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((40*x + 9*E^x*x + 9*x^2 - 20*Log[2])/(E^x*x + x^2))*(-40*x^2 + 40*x*Log[2] + E^x*(-40*x^2 + (20 + 20*x
)*Log[2])))/(E^(2*x)*x^2 + 2*E^x*x^3 + x^4),x]

[Out]

-20*(2 + Log[2])*Defer[Int][E^((40*x + 9*E^x*x + 9*x^2 - 20*Log[2])/(E^x*x + x^2))/(E^x + x)^2, x] + 20*Log[2]
*Defer[Int][E^((40*x + 9*E^x*x + 9*x^2 - 20*Log[2])/(E^x*x + x^2))/(x*(E^x + x)^2), x] + 40*Defer[Int][(E^((40
*x + 9*E^x*x + 9*x^2 - 20*Log[2])/(E^x*x + x^2))*x)/(E^x + x)^2, x] - 40*Defer[Int][E^((40*x + 9*E^x*x + 9*x^2
- 20*Log[2])/(E^x*x + x^2))/(E^x + x), x] + 20*Log[2]*Defer[Int][E^((40*x + 9*E^x*x + 9*x^2 - 20*Log[2])/(E^x
*x + x^2))/(x^2*(E^x + x)), x] + 20*Log[2]*Defer[Int][E^((40*x + 9*E^x*x + 9*x^2 - 20*Log[2])/(E^x*x + x^2))/(
x*(E^x + x)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}} \left (-40 x^2+40 x \log (2)+e^x \left (-40 x^2+(20+20 x) \log (2)\right )\right )}{x^2 \left (e^x+x\right )^2} \, dx\\ &=\int \left (\frac {20 e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}} (-1+x) (2 x-\log (2))}{x \left (e^x+x\right )^2}-\frac {20 e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}} \left (2 x^2-\log (2)-x \log (2)\right )}{x^2 \left (e^x+x\right )}\right ) \, dx\\ &=20 \int \frac {e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}} (-1+x) (2 x-\log (2))}{x \left (e^x+x\right )^2} \, dx-20 \int \frac {e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}} \left (2 x^2-\log (2)-x \log (2)\right )}{x^2 \left (e^x+x\right )} \, dx\\ &=20 \int \left (\frac {2 e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}} x}{\left (e^x+x\right )^2}-\frac {2 e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}} \left (1+\frac {\log (2)}{2}\right )}{\left (e^x+x\right )^2}+\frac {e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}} \log (2)}{x \left (e^x+x\right )^2}\right ) \, dx-20 \int \left (\frac {2 e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}}}{e^x+x}-\frac {e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}} \log (2)}{x^2 \left (e^x+x\right )}-\frac {e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}} \log (2)}{x \left (e^x+x\right )}\right ) \, dx\\ &=40 \int \frac {e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}} x}{\left (e^x+x\right )^2} \, dx-40 \int \frac {e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}}}{e^x+x} \, dx+(20 \log (2)) \int \frac {e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}}}{x \left (e^x+x\right )^2} \, dx+(20 \log (2)) \int \frac {e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}}}{x^2 \left (e^x+x\right )} \, dx+(20 \log (2)) \int \frac {e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}}}{x \left (e^x+x\right )} \, dx-(20 (2+\log (2))) \int \frac {e^{\frac {40 x+9 e^x x+9 x^2-20 \log (2)}{e^x x+x^2}}}{\left (e^x+x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.59, size = 36, normalized size = 1.12 \begin {gather*} 2^{-\frac {20}{e^x x+x^2}} e^{\frac {40+9 e^x+9 x}{e^x+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((40*x + 9*E^x*x + 9*x^2 - 20*Log[2])/(E^x*x + x^2))*(-40*x^2 + 40*x*Log[2] + E^x*(-40*x^2 + (20
+ 20*x)*Log[2])))/(E^(2*x)*x^2 + 2*E^x*x^3 + x^4),x]

[Out]

E^((40 + 9*E^x + 9*x)/(E^x + x))/2^(20/(E^x*x + x^2))

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fricas [A]  time = 0.70, size = 30, normalized size = 0.94 \begin {gather*} e^{\left (\frac {9 \, x^{2} + 9 \, x e^{x} + 40 \, x - 20 \, \log \relax (2)}{x^{2} + x e^{x}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x+20)*log(2)-40*x^2)*exp(x)+40*x*log(2)-40*x^2)*exp((9*exp(x)*x-20*log(2)+9*x^2+40*x)/(exp(x)*
x+x^2))/(exp(x)^2*x^2+2*exp(x)*x^3+x^4),x, algorithm="fricas")

[Out]

e^((9*x^2 + 9*x*e^x + 40*x - 20*log(2))/(x^2 + x*e^x))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x+20)*log(2)-40*x^2)*exp(x)+40*x*log(2)-40*x^2)*exp((9*exp(x)*x-20*log(2)+9*x^2+40*x)/(exp(x)*
x+x^2))/(exp(x)^2*x^2+2*exp(x)*x^3+x^4),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{640,[1,15,0]%%%}+%%%{-1600,[1,14,1]%%%}+%%%{-1920,[1,14,
0]%%%}+%%%{

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maple [A]  time = 0.27, size = 30, normalized size = 0.94

method	result	size
risch	\({\mathrm e}^{\frac {9 \,{\mathrm e}^{x} x -20 \ln \relax (2)+9 x^{2}+40 x}{\left ({\mathrm e}^{x}+x \right ) x}}\)	\(30\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {9 \,{\mathrm e}^{x} x -20 \ln \relax (2)+9 x^{2}+40 x}{{\mathrm e}^{x} x +x^{2}}}+{\mathrm e}^{x} x \,{\mathrm e}^{\frac {9 \,{\mathrm e}^{x} x -20 \ln \relax (2)+9 x^{2}+40 x}{{\mathrm e}^{x} x +x^{2}}}}{\left ({\mathrm e}^{x}+x \right ) x}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((20*x+20)*ln(2)-40*x^2)*exp(x)+40*x*ln(2)-40*x^2)*exp((9*exp(x)*x-20*ln(2)+9*x^2+40*x)/(exp(x)*x+x^2))/(
exp(x)^2*x^2+2*exp(x)*x^3+x^4),x,method=_RETURNVERBOSE)

[Out]

exp((9*exp(x)*x-20*ln(2)+9*x^2+40*x)/(exp(x)+x)/x)

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maxima [A]  time = 1.02, size = 37, normalized size = 1.16 \begin {gather*} e^{\left (-\frac {20 \, e^{\left (-x\right )} \log \relax (2)}{x} + \frac {20 \, \log \relax (2)}{x e^{x} + e^{\left (2 \, x\right )}} + \frac {40}{x + e^{x}} + 9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x+20)*log(2)-40*x^2)*exp(x)+40*x*log(2)-40*x^2)*exp((9*exp(x)*x-20*log(2)+9*x^2+40*x)/(exp(x)*
x+x^2))/(exp(x)^2*x^2+2*exp(x)*x^3+x^4),x, algorithm="maxima")

[Out]

e^(-20*e^(-x)*log(2)/x + 20*log(2)/(x*e^x + e^(2*x)) + 40/(x + e^x) + 9)

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mupad [B]  time = 1.02, size = 63, normalized size = 1.97 \begin {gather*} \frac {{\mathrm {e}}^{\frac {9\,x\,{\mathrm {e}}^x}{x\,{\mathrm {e}}^x+x^2}}\,{\mathrm {e}}^{\frac {40\,x}{x\,{\mathrm {e}}^x+x^2}}\,{\mathrm {e}}^{\frac {9\,x^2}{x\,{\mathrm {e}}^x+x^2}}}{2^{\frac {20}{x\,{\mathrm {e}}^x+x^2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((40*x - 20*log(2) + 9*x*exp(x) + 9*x^2)/(x*exp(x) + x^2))*(40*x*log(2) + exp(x)*(log(2)*(20*x + 20) -
40*x^2) - 40*x^2))/(2*x^3*exp(x) + x^2*exp(2*x) + x^4),x)

[Out]

(exp((9*x*exp(x))/(x*exp(x) + x^2))*exp((40*x)/(x*exp(x) + x^2))*exp((9*x^2)/(x*exp(x) + x^2)))/2^(20/(x*exp(x
) + x^2))

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sympy [A]  time = 0.39, size = 29, normalized size = 0.91 \begin {gather*} e^{\frac {9 x^{2} + 9 x e^{x} + 40 x - 20 \log {\relax (2 )}}{x^{2} + x e^{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x+20)*ln(2)-40*x**2)*exp(x)+40*x*ln(2)-40*x**2)*exp((9*exp(x)*x-20*ln(2)+9*x**2+40*x)/(exp(x)*
x+x**2))/(exp(x)**2*x**2+2*exp(x)*x**3+x**4),x)

[Out]

exp((9*x**2 + 9*x*exp(x) + 40*x - 20*log(2))/(x**2 + x*exp(x)))

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