3.78.85 \(\int \frac {e^{-4 x^2} (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} (31+5 x+(-31-5 x+248 x^2+40 x^3) \log (x)-5 e^{4 x^2} \log ^2(x)))}{\log ^2(x)} \, dx\) [7785]

Optimal. Leaf size=25 \[ x+e^{\frac {e^{-4 x^2} x}{\log (x)}} (4-5 (7+x)) \]

[Out]

x+exp(x/exp(4*x^2)/ln(x))*(-31-5*x)

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Rubi [F]
time = 3.32, 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^{-4 x^2} \left (e^{4 x^2} \log ^2(x)+e^{\frac {e^{-4 x^2} x}{\log (x)}} \left (31+5 x+\left (-31-5 x+248 x^2+40 x^3\right ) \log (x)-5 e^{4 x^2} \log ^2(x)\right )\right )}{\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(4*x^2)*Log[x]^2 + E^(x/(E^(4*x^2)*Log[x]))*(31 + 5*x + (-31 - 5*x + 248*x^2 + 40*x^3)*Log[x] - 5*E^(4*
x^2)*Log[x]^2))/(E^(4*x^2)*Log[x]^2),x]

[Out]

x - 5*Defer[Int][E^(x/(E^(4*x^2)*Log[x])), x] + 31*Defer[Int][E^(-4*x^2 + x/(E^(4*x^2)*Log[x]))/Log[x]^2, x] +
5*Defer[Int][(E^(-4*x^2 + x/(E^(4*x^2)*Log[x]))*x)/Log[x]^2, x] - 31*Defer[Int][E^(-4*x^2 + x/(E^(4*x^2)*Log[
x]))/Log[x], x] - 5*Defer[Int][(E^(-4*x^2 + x/(E^(4*x^2)*Log[x]))*x)/Log[x], x] + 248*Defer[Int][(E^(-4*x^2 +
x/(E^(4*x^2)*Log[x]))*x^2)/Log[x], x] + 40*Defer[Int][(E^(-4*x^2 + x/(E^(4*x^2)*Log[x]))*x^3)/Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1-5 e^{\frac {e^{-4 x^2} x}{\log (x)}}+\frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} (31+5 x) \left (1-\log (x)+8 x^2 \log (x)\right )}{\log ^2(x)}\right ) \, dx\\ &=x-5 \int e^{\frac {e^{-4 x^2} x}{\log (x)}} \, dx+\int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} (31+5 x) \left (1-\log (x)+8 x^2 \log (x)\right )}{\log ^2(x)} \, dx\\ &=x-5 \int e^{\frac {e^{-4 x^2} x}{\log (x)}} \, dx+\int \left (\frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} (31+5 x)}{\log ^2(x)}+\frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} \left (-31-5 x+248 x^2+40 x^3\right )}{\log (x)}\right ) \, dx\\ &=x-5 \int e^{\frac {e^{-4 x^2} x}{\log (x)}} \, dx+\int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} (31+5 x)}{\log ^2(x)} \, dx+\int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} \left (-31-5 x+248 x^2+40 x^3\right )}{\log (x)} \, dx\\ &=x-5 \int e^{\frac {e^{-4 x^2} x}{\log (x)}} \, dx+\int \left (\frac {31 e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}}}{\log ^2(x)}+\frac {5 e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x}{\log ^2(x)}\right ) \, dx+\int \left (-\frac {31 e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}}}{\log (x)}-\frac {5 e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x}{\log (x)}+\frac {248 e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x^2}{\log (x)}+\frac {40 e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x^3}{\log (x)}\right ) \, dx\\ &=x-5 \int e^{\frac {e^{-4 x^2} x}{\log (x)}} \, dx+5 \int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x}{\log ^2(x)} \, dx-5 \int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x}{\log (x)} \, dx+31 \int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}}}{\log ^2(x)} \, dx-31 \int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}}}{\log (x)} \, dx+40 \int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x^3}{\log (x)} \, dx+248 \int \frac {e^{-4 x^2+\frac {e^{-4 x^2} x}{\log (x)}} x^2}{\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.69, size = 23, normalized size = 0.92 \begin {gather*} e^{\frac {e^{-4 x^2} x}{\log (x)}} (-31-5 x)+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4*x^2)*Log[x]^2 + E^(x/(E^(4*x^2)*Log[x]))*(31 + 5*x + (-31 - 5*x + 248*x^2 + 40*x^3)*Log[x] - 5
*E^(4*x^2)*Log[x]^2))/(E^(4*x^2)*Log[x]^2),x]

[Out]

E^(x/(E^(4*x^2)*Log[x]))*(-31 - 5*x) + x

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Maple [A]
time = 0.32, size = 22, normalized size = 0.88

method	result	size
risch	\(x +{\mathrm e}^{\frac {x \,{\mathrm e}^{-4 x^{2}}}{\ln \left (x \right )}} \left (-31-5 x \right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*exp(4*x^2)*ln(x)^2+(40*x^3+248*x^2-5*x-31)*ln(x)+5*x+31)*exp(x/exp(4*x^2)/ln(x))+exp(4*x^2)*ln(x)^2)/
exp(4*x^2)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

x+exp(x*exp(-4*x^2)/ln(x))*(-31-5*x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(4*x^2)*log(x)^2+(40*x^3+248*x^2-5*x-31)*log(x)+5*x+31)*exp(x/exp(4*x^2)/log(x))+exp(4*x^2)*
log(x)^2)/exp(4*x^2)/log(x)^2,x, algorithm="maxima")

[Out]

x - integrate((5*e^(4*x^2)*log(x)^2 - (40*x^3 + 248*x^2 - 5*x - 31)*log(x) - 5*x - 31)*e^(-4*x^2 + x*e^(-4*x^2
)/log(x))/log(x)^2, x)

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Fricas [A]
time = 0.39, size = 22, normalized size = 0.88 \begin {gather*} -{\left (5 \, x + 31\right )} e^{\left (\frac {x e^{\left (-4 \, x^{2}\right )}}{\log \left (x\right )}\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(4*x^2)*log(x)^2+(40*x^3+248*x^2-5*x-31)*log(x)+5*x+31)*exp(x/exp(4*x^2)/log(x))+exp(4*x^2)*
log(x)^2)/exp(4*x^2)/log(x)^2,x, algorithm="fricas")

[Out]

-(5*x + 31)*e^(x*e^(-4*x^2)/log(x)) + x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(4*x**2)*ln(x)**2+(40*x**3+248*x**2-5*x-31)*ln(x)+5*x+31)*exp(x/exp(4*x**2)/ln(x))+exp(4*x**
2)*ln(x)**2)/exp(4*x**2)/ln(x)**2,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6438 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*exp(4*x^2)*log(x)^2+(40*x^3+248*x^2-5*x-31)*log(x)+5*x+31)*exp(x/exp(4*x^2)/log(x))+exp(4*x^2)*
log(x)^2)/exp(4*x^2)/log(x)^2,x, algorithm="giac")

[Out]

integrate((e^(4*x^2)*log(x)^2 - (5*e^(4*x^2)*log(x)^2 - (40*x^3 + 248*x^2 - 5*x - 31)*log(x) - 5*x - 31)*e^(x*
e^(-4*x^2)/log(x)))*e^(-4*x^2)/log(x)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{-4\,x^2}\,\left ({\mathrm {e}}^{4\,x^2}\,{\ln \left (x\right )}^2+{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-4\,x^2}}{\ln \left (x\right )}}\,\left (-5\,{\mathrm {e}}^{4\,x^2}\,{\ln \left (x\right )}^2+\left (40\,x^3+248\,x^2-5\,x-31\right )\,\ln \left (x\right )+5\,x+31\right )\right )}{{\ln \left (x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-4*x^2)*(exp(4*x^2)*log(x)^2 + exp((x*exp(-4*x^2))/log(x))*(5*x - 5*exp(4*x^2)*log(x)^2 - log(x)*(5*x
- 248*x^2 - 40*x^3 + 31) + 31)))/log(x)^2,x)

[Out]

int((exp(-4*x^2)*(exp(4*x^2)*log(x)^2 + exp((x*exp(-4*x^2))/log(x))*(5*x - 5*exp(4*x^2)*log(x)^2 - log(x)*(5*x
- 248*x^2 - 40*x^3 + 31) + 31)))/log(x)^2, x)

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