3.83.66 \(\int \frac {-e^{10}+2 e^5 \log (20 x^2)-\log ^2(20 x^2)+e^{-\frac {x^2}{-e^5+\log (20 x^2)}} (2 x+2 e^5 x-2 x \log (20 x^2))}{-e^{10} x+2 e^5 x \log (20 x^2)-x \log ^2(20 x^2)+e^{-\frac {x^2}{-e^5+\log (20 x^2)}} (e^{10}-2 e^5 \log (20 x^2)+\log ^2(20 x^2))} \, dx\)

Optimal. Leaf size=26 \[ \log \left (e^{-\frac {x^2}{-e^5+\log \left (20 x^2\right )}}-x\right ) \]

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Rubi [A]  time = 0.82, antiderivative size = 25, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, integrand size = 146, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6741, 6684} \begin {gather*} \log \left (e^{\frac {x^2}{e^5-\log \left (20 x^2\right )}}-x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^10 + 2*E^5*Log[20*x^2] - Log[20*x^2]^2 + (2*x + 2*E^5*x - 2*x*Log[20*x^2])/E^(x^2/(-E^5 + Log[20*x^2])
))/(-(E^10*x) + 2*E^5*x*Log[20*x^2] - x*Log[20*x^2]^2 + (E^10 - 2*E^5*Log[20*x^2] + Log[20*x^2]^2)/E^(x^2/(-E^
5 + Log[20*x^2]))),x]

[Out]

Log[E^(x^2/(E^5 - Log[20*x^2])) - x]

Rule 6684

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

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^{10}+2 e^5 \log \left (20 x^2\right )-\log ^2\left (20 x^2\right )+e^{-\frac {x^2}{-e^5+\log \left (20 x^2\right )}} \left (2 x+2 e^5 x-2 x \log \left (20 x^2\right )\right )}{\left (e^{\frac {x^2}{e^5-\log \left (20 x^2\right )}}-x\right ) \left (e^5-\log \left (20 x^2\right )\right )^2} \, dx\\ &=\log \left (e^{\frac {x^2}{e^5-\log \left (20 x^2\right )}}-x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 46, normalized size = 1.77 \begin {gather*} -\frac {x^2}{-e^5+\log \left (20 x^2\right )}+\log \left (1-e^{\frac {x^2}{-e^5+\log \left (20 x^2\right )}} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^10 + 2*E^5*Log[20*x^2] - Log[20*x^2]^2 + (2*x + 2*E^5*x - 2*x*Log[20*x^2])/E^(x^2/(-E^5 + Log[20
*x^2])))/(-(E^10*x) + 2*E^5*x*Log[20*x^2] - x*Log[20*x^2]^2 + (E^10 - 2*E^5*Log[20*x^2] + Log[20*x^2]^2)/E^(x^
2/(-E^5 + Log[20*x^2]))),x]

[Out]

-(x^2/(-E^5 + Log[20*x^2])) + Log[1 - E^(x^2/(-E^5 + Log[20*x^2]))*x]

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fricas [A]  time = 0.53, size = 23, normalized size = 0.88 \begin {gather*} \log \left (-x + e^{\left (\frac {x^{2}}{e^{5} - \log \left (20 \, x^{2}\right )}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(20*x^2)+2*x*exp(5)+2*x)*exp(-x^2/(log(20*x^2)-exp(5)))-log(20*x^2)^2+2*exp(5)*log(20*x^2)
-exp(5)^2)/((log(20*x^2)^2-2*exp(5)*log(20*x^2)+exp(5)^2)*exp(-x^2/(log(20*x^2)-exp(5)))-x*log(20*x^2)^2+2*x*e
xp(5)*log(20*x^2)-x*exp(5)^2),x, algorithm="fricas")

[Out]

log(-x + e^(x^2/(e^5 - log(20*x^2))))

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giac [A]  time = 1.45, size = 23, normalized size = 0.88 \begin {gather*} \log \left (x - e^{\left (\frac {x^{2}}{e^{5} - \log \left (20 \, x^{2}\right )}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(20*x^2)+2*x*exp(5)+2*x)*exp(-x^2/(log(20*x^2)-exp(5)))-log(20*x^2)^2+2*exp(5)*log(20*x^2)
-exp(5)^2)/((log(20*x^2)^2-2*exp(5)*log(20*x^2)+exp(5)^2)*exp(-x^2/(log(20*x^2)-exp(5)))-x*log(20*x^2)^2+2*x*e
xp(5)*log(20*x^2)-x*exp(5)^2),x, algorithm="giac")

[Out]

log(x - e^(x^2/(e^5 - log(20*x^2))))

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maple [A]  time = 1.32, size = 25, normalized size = 0.96

method	result	size
default	\(\ln \left (x -{\mathrm e}^{-\frac {x^{2}}{\ln \left (20 x^{2}\right )-{\mathrm e}^{5}}}\right )\)	\(25\)
norman	\(\ln \left (x -{\mathrm e}^{-\frac {x^{2}}{\ln \left (20 x^{2}\right )-{\mathrm e}^{5}}}\right )\)	\(25\)
risch	\(\frac {x^{2}}{-\ln \left (20 x^{2}\right )+{\mathrm e}^{5}}+\frac {x^{2}}{\ln \left (20 x^{2}\right )-{\mathrm e}^{5}}+\ln \left ({\mathrm e}^{\frac {x^{2}}{-\ln \left (20 x^{2}\right )+{\mathrm e}^{5}}}-x \right )\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*ln(20*x^2)+2*x*exp(5)+2*x)*exp(-x^2/(ln(20*x^2)-exp(5)))-ln(20*x^2)^2+2*exp(5)*ln(20*x^2)-exp(5)^2)
/((ln(20*x^2)^2-2*exp(5)*ln(20*x^2)+exp(5)^2)*exp(-x^2/(ln(20*x^2)-exp(5)))-x*ln(20*x^2)^2+2*x*exp(5)*ln(20*x^
2)-x*exp(5)^2),x,method=_RETURNVERBOSE)

[Out]

ln(x-exp(-x^2/(ln(20*x^2)-exp(5))))

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maxima [B]  time = 0.73, size = 56, normalized size = 2.15 \begin {gather*} \frac {x^{2}}{e^{5} - \log \relax (5) - 2 \, \log \relax (2) - 2 \, \log \relax (x)} + \log \relax (x) + \log \left (\frac {x e^{\left (-\frac {x^{2}}{e^{5} - \log \relax (5) - 2 \, \log \relax (2) - 2 \, \log \relax (x)}\right )} - 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(20*x^2)+2*x*exp(5)+2*x)*exp(-x^2/(log(20*x^2)-exp(5)))-log(20*x^2)^2+2*exp(5)*log(20*x^2)
-exp(5)^2)/((log(20*x^2)^2-2*exp(5)*log(20*x^2)+exp(5)^2)*exp(-x^2/(log(20*x^2)-exp(5)))-x*log(20*x^2)^2+2*x*e
xp(5)*log(20*x^2)-x*exp(5)^2),x, algorithm="maxima")

[Out]

x^2/(e^5 - log(5) - 2*log(2) - 2*log(x)) + log(x) + log((x*e^(-x^2/(e^5 - log(5) - 2*log(2) - 2*log(x))) - 1)/
x)

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mupad [B]  time = 7.55, size = 23, normalized size = 0.88 \begin {gather*} \ln \left ({\mathrm {e}}^{\frac {x^2}{{\mathrm {e}}^5-\ln \left (20\,x^2\right )}}-x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(10) + log(20*x^2)^2 - exp(x^2/(exp(5) - log(20*x^2)))*(2*x + 2*x*exp(5) - 2*x*log(20*x^2)) - 2*exp(5)
*log(20*x^2))/(x*exp(10) - exp(x^2/(exp(5) - log(20*x^2)))*(exp(10) + log(20*x^2)^2 - 2*exp(5)*log(20*x^2)) +
x*log(20*x^2)^2 - 2*x*exp(5)*log(20*x^2)),x)

[Out]

log(exp(x^2/(exp(5) - log(20*x^2))) - x)

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sympy [A]  time = 0.76, size = 19, normalized size = 0.73 \begin {gather*} \log {\left (- x + e^{- \frac {x^{2}}{\log {\left (20 x^{2} \right )} - e^{5}}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*ln(20*x**2)+2*x*exp(5)+2*x)*exp(-x**2/(ln(20*x**2)-exp(5)))-ln(20*x**2)**2+2*exp(5)*ln(20*x**
2)-exp(5)**2)/((ln(20*x**2)**2-2*exp(5)*ln(20*x**2)+exp(5)**2)*exp(-x**2/(ln(20*x**2)-exp(5)))-x*ln(20*x**2)**
2+2*x*exp(5)*ln(20*x**2)-x*exp(5)**2),x)

[Out]

log(-x + exp(-x**2/(log(20*x**2) - exp(5))))

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