3.94.72 \(\int \frac {50 x+e^{-2 e^x+2 x} (-2 e^{2+x} x+e^2 (1+2 x))}{-400+4 e^{2-2 e^x+2 x} x+100 x^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{4} \log \left (4-x \left (\frac {1}{25} e^{2-2 e^x+2 x}+x\right )\right ) \]

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Rubi [A]  time = 0.13, antiderivative size = 46, normalized size of antiderivative = 1.64, number of steps used = 1, number of rules used = 1, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6684} \begin {gather*} \frac {1}{4} \log \left (-e^{-2 e^x} \left (-25 e^{2 e^x} x^2-e^{2 x+2} x+100 e^{2 e^x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(50*x + E^(-2*E^x + 2*x)*(-2*E^(2 + x)*x + E^2*(1 + 2*x)))/(-400 + 4*E^(2 - 2*E^x + 2*x)*x + 100*x^2),x]

[Out]

Log[-((100*E^(2*E^x) - E^(2 + 2*x)*x - 25*E^(2*E^x)*x^2)/E^(2*E^x))]/4

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \log \left (-e^{-2 e^x} \left (100 e^{2 e^x}-e^{2+2 x} x-25 e^{2 e^x} x^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.10, size = 35, normalized size = 1.25 \begin {gather*} \frac {1}{4} \left (-2 e^x+\log \left (e^{2+2 x} x+25 e^{2 e^x} \left (-4+x^2\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(50*x + E^(-2*E^x + 2*x)*(-2*E^(2 + x)*x + E^2*(1 + 2*x)))/(-400 + 4*E^(2 - 2*E^x + 2*x)*x + 100*x^2
),x]

[Out]

(-2*E^x + Log[E^(2 + 2*x)*x + 25*E^(2*E^x)*(-4 + x^2)])/4

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fricas [A]  time = 0.49, size = 39, normalized size = 1.39 \begin {gather*} \frac {1}{4} \, \log \relax (x) + \frac {1}{4} \, \log \left (\frac {25 \, x^{2} + x e^{\left (2 \, {\left ({\left (x + 1\right )} e^{2} - e^{\left (x + 2\right )}\right )} e^{\left (-2\right )}\right )} - 100}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)^2*exp(x)+(2*x+1)*exp(1)^2)*exp(x-exp(x))^2+50*x)/(4*x*exp(1)^2*exp(x-exp(x))^2+100*x^2
-400),x, algorithm="fricas")

[Out]

1/4*log(x) + 1/4*log((25*x^2 + x*e^(2*((x + 1)*e^2 - e^(x + 2))*e^(-2)) - 100)/x)

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giac [A]  time = 0.24, size = 34, normalized size = 1.21 \begin {gather*} -\frac {1}{2} \, e^{x} + \frac {1}{4} \, \log \left (25 \, x^{2} e^{\left (2 \, e^{x}\right )} + x e^{\left (2 \, x + 2\right )} - 100 \, e^{\left (2 \, e^{x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)^2*exp(x)+(2*x+1)*exp(1)^2)*exp(x-exp(x))^2+50*x)/(4*x*exp(1)^2*exp(x-exp(x))^2+100*x^2
-400),x, algorithm="giac")

[Out]

-1/2*e^x + 1/4*log(25*x^2*e^(2*e^x) + x*e^(2*x + 2) - 100*e^(2*e^x))

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maple [A]  time = 0.15, size = 27, normalized size = 0.96

method	result	size
norman	\(\frac {\ln \left (4 x \,{\mathrm e}^{2} {\mathrm e}^{-2 \,{\mathrm e}^{x}+2 x}+100 x^{2}-400\right )}{4}\)	\(27\)
risch	\(\frac {\ln \relax (x )}{4}+\frac {\ln \left ({\mathrm e}^{-2 \,{\mathrm e}^{x}+2 x}+\frac {25 \left (x^{2}-4\right ) {\mathrm e}^{-2}}{x}\right )}{4}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*exp(1)^2*exp(x)+(2*x+1)*exp(1)^2)*exp(x-exp(x))^2+50*x)/(4*x*exp(1)^2*exp(x-exp(x))^2+100*x^2-400),
x,method=_RETURNVERBOSE)

[Out]

1/4*ln(4*x*exp(1)^2*exp(x-exp(x))^2+100*x^2-400)

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maxima [B]  time = 0.40, size = 50, normalized size = 1.79 \begin {gather*} -\frac {1}{2} \, e^{x} + \frac {1}{4} \, \log \left (x + 2\right ) + \frac {1}{4} \, \log \left (x - 2\right ) + \frac {1}{4} \, \log \left (\frac {x e^{\left (2 \, x + 2\right )} + 25 \, {\left (x^{2} - 4\right )} e^{\left (2 \, e^{x}\right )}}{25 \, {\left (x^{2} - 4\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)^2*exp(x)+(2*x+1)*exp(1)^2)*exp(x-exp(x))^2+50*x)/(4*x*exp(1)^2*exp(x-exp(x))^2+100*x^2
-400),x, algorithm="maxima")

[Out]

-1/2*e^x + 1/4*log(x + 2) + 1/4*log(x - 2) + 1/4*log(1/25*(x*e^(2*x + 2) + 25*(x^2 - 4)*e^(2*e^x))/(x^2 - 4))

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mupad [B]  time = 0.49, size = 22, normalized size = 0.79 \begin {gather*} \frac {\ln \left (x^2+\frac {x\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{-2\,{\mathrm {e}}^x}}{25}-4\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x + exp(2*x - 2*exp(x))*(exp(2)*(2*x + 1) - 2*x*exp(2)*exp(x)))/(100*x^2 + 4*x*exp(2)*exp(2*x - 2*exp(
x)) - 400),x)

[Out]

log(x^2 + (x*exp(2*x)*exp(2)*exp(-2*exp(x)))/25 - 4)/4

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sympy [A]  time = 0.33, size = 29, normalized size = 1.04 \begin {gather*} \frac {\log {\relax (x )}}{4} + \frac {\log {\left (e^{2 x - 2 e^{x}} + \frac {25 x^{2} - 100}{x e^{2}} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(1)**2*exp(x)+(2*x+1)*exp(1)**2)*exp(x-exp(x))**2+50*x)/(4*x*exp(1)**2*exp(x-exp(x))**2+10
0*x**2-400),x)

[Out]

log(x)/4 + log(exp(2*x - 2*exp(x)) + (25*x**2 - 100)*exp(-2)/x)/4

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