3.24.87 \(\int \frac {-4 \log ^2(2 x)+e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x} ((1+2 x) \log ^2(2 x)+e^{\frac {6 x^2}{\log (2 x)}} (6 x^2-12 x^2 \log (2 x)))}{\log ^2(2 x)} \, dx\)

Optimal. Leaf size=25 \[ \left (-4+e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x}\right ) x \]

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Rubi [F]  time = 4.37, 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 {-4 \log ^2(2 x)+e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x} \left ((1+2 x) \log ^2(2 x)+e^{\frac {6 x^2}{\log (2 x)}} \left (6 x^2-12 x^2 \log (2 x)\right )\right )}{\log ^2(2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*Log[2*x]^2 + E^(-E^((6*x^2)/Log[2*x]) + 2*x)*((1 + 2*x)*Log[2*x]^2 + E^((6*x^2)/Log[2*x])*(6*x^2 - 12*
x^2*Log[2*x])))/Log[2*x]^2,x]

[Out]

-4*x + Defer[Int][E^(-E^((6*x^2)/Log[2*x]) + 2*x), x] + 2*Defer[Int][E^(-E^((6*x^2)/Log[2*x]) + 2*x)*x, x] + 6
*Defer[Int][(E^(-E^((6*x^2)/Log[2*x]) + 2*x + (6*x^2)/Log[2*x])*x^2)/Log[2*x]^2, x] - 12*Defer[Int][(E^(-E^((6
*x^2)/Log[2*x]) + 2*x + (6*x^2)/Log[2*x])*x^2)/Log[2*x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-e^{-e^{\frac {6 x^2}{\log (2 x)}}} \left (4 e^{e^{\frac {6 x^2}{\log (2 x)}}}-e^{2 x}-2 e^{2 x} x\right )-\frac {6 e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x+\frac {6 x^2}{\log (2 x)}} x^2 (-1+2 \log (2 x))}{\log ^2(2 x)}\right ) \, dx\\ &=-\left (6 \int \frac {e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x+\frac {6 x^2}{\log (2 x)}} x^2 (-1+2 \log (2 x))}{\log ^2(2 x)} \, dx\right )-\int e^{-e^{\frac {6 x^2}{\log (2 x)}}} \left (4 e^{e^{\frac {6 x^2}{\log (2 x)}}}-e^{2 x}-2 e^{2 x} x\right ) \, dx\\ &=-\left (6 \int \left (-\frac {e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x+\frac {6 x^2}{\log (2 x)}} x^2}{\log ^2(2 x)}+\frac {2 e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x+\frac {6 x^2}{\log (2 x)}} x^2}{\log (2 x)}\right ) \, dx\right )-\int \left (4-e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x} (1+2 x)\right ) \, dx\\ &=-4 x+6 \int \frac {e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x+\frac {6 x^2}{\log (2 x)}} x^2}{\log ^2(2 x)} \, dx-12 \int \frac {e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x+\frac {6 x^2}{\log (2 x)}} x^2}{\log (2 x)} \, dx+\int e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x} (1+2 x) \, dx\\ &=-4 x+6 \int \frac {e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x+\frac {6 x^2}{\log (2 x)}} x^2}{\log ^2(2 x)} \, dx-12 \int \frac {e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x+\frac {6 x^2}{\log (2 x)}} x^2}{\log (2 x)} \, dx+\int \left (e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x}+2 e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x} x\right ) \, dx\\ &=-4 x+2 \int e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x} x \, dx+6 \int \frac {e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x+\frac {6 x^2}{\log (2 x)}} x^2}{\log ^2(2 x)} \, dx-12 \int \frac {e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x+\frac {6 x^2}{\log (2 x)}} x^2}{\log (2 x)} \, dx+\int e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.44, size = 27, normalized size = 1.08 \begin {gather*} -4 x+e^{-e^{\frac {6 x^2}{\log (2 x)}}+2 x} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*Log[2*x]^2 + E^(-E^((6*x^2)/Log[2*x]) + 2*x)*((1 + 2*x)*Log[2*x]^2 + E^((6*x^2)/Log[2*x])*(6*x^2
- 12*x^2*Log[2*x])))/Log[2*x]^2,x]

[Out]

-4*x + E^(-E^((6*x^2)/Log[2*x]) + 2*x)*x

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fricas [A]  time = 0.54, size = 25, normalized size = 1.00 \begin {gather*} x e^{\left (2 \, x - e^{\left (\frac {6 \, x^{2}}{\log \left (2 \, x\right )}\right )}\right )} - 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x^2*log(2*x)+6*x^2)*exp(6*x^2/log(2*x))+(2*x+1)*log(2*x)^2)*exp(-exp(6*x^2/log(2*x))+2*x)-4*l
og(2*x)^2)/log(2*x)^2,x, algorithm="fricas")

[Out]

x*e^(2*x - e^(6*x^2/log(2*x))) - 4*x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x^2*log(2*x)+6*x^2)*exp(6*x^2/log(2*x))+(2*x+1)*log(2*x)^2)*exp(-exp(6*x^2/log(2*x))+2*x)-4*l
og(2*x)^2)/log(2*x)^2,x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.07, size = 26, normalized size = 1.04

method	result	size
risch	\(x \,{\mathrm e}^{-{\mathrm e}^{\frac {6 x^{2}}{\ln \left (2 x \right )}}+2 x}-4 x\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-12*x^2*ln(2*x)+6*x^2)*exp(6*x^2/ln(2*x))+(2*x+1)*ln(2*x)^2)*exp(-exp(6*x^2/ln(2*x))+2*x)-4*ln(2*x)^2)/
ln(2*x)^2,x,method=_RETURNVERBOSE)

[Out]

x*exp(-exp(6*x^2/ln(2*x))+2*x)-4*x

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maxima [A]  time = 0.65, size = 26, normalized size = 1.04 \begin {gather*} x e^{\left (2 \, x - e^{\left (\frac {6 \, x^{2}}{\log \relax (2) + \log \relax (x)}\right )}\right )} - 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x^2*log(2*x)+6*x^2)*exp(6*x^2/log(2*x))+(2*x+1)*log(2*x)^2)*exp(-exp(6*x^2/log(2*x))+2*x)-4*l
og(2*x)^2)/log(2*x)^2,x, algorithm="maxima")

[Out]

x*e^(2*x - e^(6*x^2/(log(2) + log(x)))) - 4*x

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mupad [B]  time = 1.51, size = 25, normalized size = 1.00 \begin {gather*} x\,{\mathrm {e}}^{2\,x-{\mathrm {e}}^{\frac {6\,x^2}{\ln \left (2\,x\right )}}}-4\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*log(2*x)^2 - exp(2*x - exp((6*x^2)/log(2*x)))*(log(2*x)^2*(2*x + 1) - exp((6*x^2)/log(2*x))*(12*x^2*lo
g(2*x) - 6*x^2)))/log(2*x)^2,x)

[Out]

x*exp(2*x - exp((6*x^2)/log(2*x))) - 4*x

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x**2*ln(2*x)+6*x**2)*exp(6*x**2/ln(2*x))+(2*x+1)*ln(2*x)**2)*exp(-exp(6*x**2/ln(2*x))+2*x)-4*
ln(2*x)**2)/ln(2*x)**2,x)

[Out]

Timed out

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