3.41.91 \(\int \frac {e^{-2 e^{-x+x^2}} (e^{-x+x^2} (8 x-16 x^2)-e^{2 e^{-x+x^2}} x \log ^2(16)+2 e^{2 e^{-x+x^2}+\log ^2(x)} \log ^2(16) \log (x))}{x \log ^2(16)} \, dx\)

Optimal. Leaf size=29 \[ e^{\log ^2(x)}-x+\frac {4 e^{-2 e^{-x+x^2}}}{\log ^2(16)} \]

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Rubi [F]  time = 1.26, 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^{-2 e^{-x+x^2}} \left (e^{-x+x^2} \left (8 x-16 x^2\right )-e^{2 e^{-x+x^2}} x \log ^2(16)+2 e^{2 e^{-x+x^2}+\log ^2(x)} \log ^2(16) \log (x)\right )}{x \log ^2(16)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-x + x^2)*(8*x - 16*x^2) - E^(2*E^(-x + x^2))*x*Log[16]^2 + 2*E^(2*E^(-x + x^2) + Log[x]^2)*Log[16]^2*
Log[x])/(E^(2*E^(-x + x^2))*x*Log[16]^2),x]

[Out]

E^Log[x]^2 - x + (8*Defer[Int][E^(-2*E^((-1 + x)*x) + (-1 + x)*x), x])/Log[16]^2 - (16*Defer[Int][E^(-2*E^((-1
+ x)*x) + (-1 + x)*x)*x, x])/Log[16]^2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{-2 e^{-x+x^2}} \left (e^{-x+x^2} \left (8 x-16 x^2\right )-e^{2 e^{-x+x^2}} x \log ^2(16)+2 e^{2 e^{-x+x^2}+\log ^2(x)} \log ^2(16) \log (x)\right )}{x} \, dx}{\log ^2(16)}\\ &=\frac {\int \left (e^{-2 e^{(-1+x) x}+(-1+x) x} (8-16 x)-\log ^2(16)+\frac {2 e^{\log ^2(x)} \log ^2(16) \log (x)}{x}\right ) \, dx}{\log ^2(16)}\\ &=-x+2 \int \frac {e^{\log ^2(x)} \log (x)}{x} \, dx+\frac {\int e^{-2 e^{(-1+x) x}+(-1+x) x} (8-16 x) \, dx}{\log ^2(16)}\\ &=-x+2 \operatorname {Subst}\left (\int e^{x^2} x \, dx,x,\log (x)\right )+\frac {\int \left (8 e^{-2 e^{(-1+x) x}+(-1+x) x}-16 e^{-2 e^{(-1+x) x}+(-1+x) x} x\right ) \, dx}{\log ^2(16)}\\ &=e^{\log ^2(x)}-x+\frac {8 \int e^{-2 e^{(-1+x) x}+(-1+x) x} \, dx}{\log ^2(16)}-\frac {16 \int e^{-2 e^{(-1+x) x}+(-1+x) x} x \, dx}{\log ^2(16)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 29, normalized size = 1.00 \begin {gather*} e^{\log ^2(x)}-x+\frac {4 e^{-2 e^{-x+x^2}}}{\log ^2(16)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-x + x^2)*(8*x - 16*x^2) - E^(2*E^(-x + x^2))*x*Log[16]^2 + 2*E^(2*E^(-x + x^2) + Log[x]^2)*Log[
16]^2*Log[x])/(E^(2*E^(-x + x^2))*x*Log[16]^2),x]

[Out]

E^Log[x]^2 - x + 4/(E^(2*E^(-x + x^2))*Log[16]^2)

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fricas [B]  time = 0.80, size = 59, normalized size = 2.03 \begin {gather*} -\frac {{\left (4 \, x e^{\left (2 \, e^{\left (x^{2} - x\right )}\right )} \log \relax (2)^{2} - 4 \, e^{\left (\log \relax (x)^{2} + 2 \, e^{\left (x^{2} - x\right )}\right )} \log \relax (2)^{2} - 1\right )} e^{\left (-2 \, e^{\left (x^{2} - x\right )}\right )}}{4 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(32*log(2)^2*log(x)*exp(exp(x^2-x))^2*exp(log(x)^2)-16*x*log(2)^2*exp(exp(x^2-x))^2+(-16*x^2+8*
x)*exp(x^2-x))/x/log(2)^2/exp(exp(x^2-x))^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.21, size = 68, normalized size = 2.34 \begin {gather*} -\frac {{\left (4 \, x e^{\left (x^{2} - x\right )} \log \relax (2)^{2} - 4 \, e^{\left (x^{2} + \log \relax (x)^{2} - x\right )} \log \relax (2)^{2} - e^{\left (x^{2} - x - 2 \, e^{\left (x^{2} - x\right )}\right )}\right )} e^{\left (-x^{2} + x\right )}}{4 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(32*log(2)^2*log(x)*exp(exp(x^2-x))^2*exp(log(x)^2)-16*x*log(2)^2*exp(exp(x^2-x))^2+(-16*x^2+8*
x)*exp(x^2-x))/x/log(2)^2/exp(exp(x^2-x))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 25, normalized size = 0.86

method	result	size
risch	\({\mathrm e}^{\ln \relax (x )^{2}}+\frac {{\mathrm e}^{-2 \,{\mathrm e}^{x \left (x -1\right )}}}{4 \ln \relax (2)^{2}}-x\)	\(25\)
default	\(\frac {4 \,{\mathrm e}^{-2 \,{\mathrm e}^{x^{2}-x}}-16 x \ln \relax (2)^{2}+16 \ln \relax (2)^{2} {\mathrm e}^{\ln \relax (x )^{2}}}{16 \ln \relax (2)^{2}}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(32*ln(2)^2*ln(x)*exp(exp(x^2-x))^2*exp(ln(x)^2)-16*x*ln(2)^2*exp(exp(x^2-x))^2+(-16*x^2+8*x)*exp(x^2
-x))/x/ln(2)^2/exp(exp(x^2-x))^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.60, size = 51, normalized size = 1.76 \begin {gather*} -\frac {4 \, x \log \relax (2)^{2} - {\left (4 \, e^{\left (\log \relax (x)^{2} + 2 \, e^{\left (x^{2} - x\right )}\right )} \log \relax (2)^{2} + 1\right )} e^{\left (-2 \, e^{\left (x^{2} - x\right )}\right )}}{4 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(32*log(2)^2*log(x)*exp(exp(x^2-x))^2*exp(log(x)^2)-16*x*log(2)^2*exp(exp(x^2-x))^2+(-16*x^2+8*
x)*exp(x^2-x))/x/log(2)^2/exp(exp(x^2-x))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.44, size = 26, normalized size = 0.90 \begin {gather*} {\mathrm {e}}^{{\ln \relax (x)}^2}-x+\frac {{\mathrm {e}}^{-2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^2}}}{4\,{\ln \relax (2)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2*exp(x^2 - x))*((exp(x^2 - x)*(8*x - 16*x^2))/16 - x*exp(2*exp(x^2 - x))*log(2)^2 + 2*exp(2*exp(x^2
- x))*exp(log(x)^2)*log(2)^2*log(x)))/(x*log(2)^2),x)

[Out]

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

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sympy [A]  time = 0.92, size = 26, normalized size = 0.90 \begin {gather*} - x + e^{\log {\relax (x )}^{2}} + \frac {e^{- 2 e^{x^{2} - x}}}{4 \log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(32*ln(2)**2*ln(x)*exp(exp(x**2-x))**2*exp(ln(x)**2)-16*x*ln(2)**2*exp(exp(x**2-x))**2+(-16*x**
2+8*x)*exp(x**2-x))/x/ln(2)**2/exp(exp(x**2-x))**2,x)

[Out]

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

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