3.48.11 \(\int \frac {e^{\frac {3 x^2}{4 e^x x-x \log (4)}} (-64 e^{2 x} x^2-12 x^3 \log (4)-4 x^2 \log ^2(4)+e^x x (48 x^2-48 x^3+32 x \log (4)))}{16 e^{2 x} x^4-8 e^x x^4 \log (4)+x^4 \log ^2(4)} \, dx\)

Optimal. Leaf size=25 \[ \frac {4 e^{-\frac {3 x^2}{-4 e^x x+x \log (4)}}}{x} \]

________________________________________________________________________________________

Rubi [F]  time = 3.91, 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^{\frac {3 x^2}{4 e^x x-x \log (4)}} \left (-64 e^{2 x} x^2-12 x^3 \log (4)-4 x^2 \log ^2(4)+e^x x \left (48 x^2-48 x^3+32 x \log (4)\right )\right )}{16 e^{2 x} x^4-8 e^x x^4 \log (4)+x^4 \log ^2(4)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((3*x^2)/(4*E^x*x - x*Log[4]))*(-64*E^(2*x)*x^2 - 12*x^3*Log[4] - 4*x^2*Log[4]^2 + E^x*x*(48*x^2 - 48*x
^3 + 32*x*Log[4])))/(16*E^(2*x)*x^4 - 8*E^x*x^4*Log[4] + x^4*Log[4]^2),x]

[Out]

-4*Defer[Int][E^((3*x)/(4*E^x - Log[4]))/x^2, x] - 12*Log[4]*Defer[Int][E^((3*x)/(4*E^x - Log[4]))/(4*E^x - Lo
g[4])^2, x] - 12*Defer[Int][E^((3*x)/(4*E^x - Log[4]))/(4*E^x - Log[4]), x] + 12*Defer[Int][E^((3*x)/(4*E^x -
Log[4]))/(x*(4*E^x - Log[4])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 e^{\frac {3 x}{4 e^x-\log (4)}} \left (-16 e^{2 x}-4 e^x \left (-3 x+3 x^2-2 \log (4)\right )-\log (4) (3 x+\log (4))\right )}{x^2 \left (4 e^x-\log (4)\right )^2} \, dx\\ &=4 \int \frac {e^{\frac {3 x}{4 e^x-\log (4)}} \left (-16 e^{2 x}-4 e^x \left (-3 x+3 x^2-2 \log (4)\right )-\log (4) (3 x+\log (4))\right )}{x^2 \left (4 e^x-\log (4)\right )^2} \, dx\\ &=4 \int \left (-\frac {e^{\frac {3 x}{4 e^x-\log (4)}}}{x^2}-\frac {3 e^{\frac {3 x}{4 e^x-\log (4)}} (-1+x)}{x \left (4 e^x-\log (4)\right )}-\frac {3 e^{\frac {3 x}{4 e^x-\log (4)}} \log (4)}{\left (4 e^x-\log (4)\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {e^{\frac {3 x}{4 e^x-\log (4)}}}{x^2} \, dx\right )-12 \int \frac {e^{\frac {3 x}{4 e^x-\log (4)}} (-1+x)}{x \left (4 e^x-\log (4)\right )} \, dx-(12 \log (4)) \int \frac {e^{\frac {3 x}{4 e^x-\log (4)}}}{\left (4 e^x-\log (4)\right )^2} \, dx\\ &=-\left (4 \int \frac {e^{\frac {3 x}{4 e^x-\log (4)}}}{x^2} \, dx\right )-12 \int \left (\frac {e^{\frac {3 x}{4 e^x-\log (4)}}}{4 e^x-\log (4)}-\frac {e^{\frac {3 x}{4 e^x-\log (4)}}}{x \left (4 e^x-\log (4)\right )}\right ) \, dx-(12 \log (4)) \int \frac {e^{\frac {3 x}{4 e^x-\log (4)}}}{\left (4 e^x-\log (4)\right )^2} \, dx\\ &=-\left (4 \int \frac {e^{\frac {3 x}{4 e^x-\log (4)}}}{x^2} \, dx\right )-12 \int \frac {e^{\frac {3 x}{4 e^x-\log (4)}}}{4 e^x-\log (4)} \, dx+12 \int \frac {e^{\frac {3 x}{4 e^x-\log (4)}}}{x \left (4 e^x-\log (4)\right )} \, dx-(12 \log (4)) \int \frac {e^{\frac {3 x}{4 e^x-\log (4)}}}{\left (4 e^x-\log (4)\right )^2} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.73, size = 22, normalized size = 0.88 \begin {gather*} \frac {4 e^{\frac {3 x}{4 e^x-\log (4)}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((3*x^2)/(4*E^x*x - x*Log[4]))*(-64*E^(2*x)*x^2 - 12*x^3*Log[4] - 4*x^2*Log[4]^2 + E^x*x*(48*x^2
- 48*x^3 + 32*x*Log[4])))/(16*E^(2*x)*x^4 - 8*E^x*x^4*Log[4] + x^4*Log[4]^2),x]

[Out]

(4*E^((3*x)/(4*E^x - Log[4])))/x

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 25, normalized size = 1.00 \begin {gather*} \frac {4 \, e^{\left (-\frac {3 \, x^{2}}{2 \, {\left (x \log \relax (2) - 2 \, e^{\left (x + \log \relax (x)\right )}\right )}}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-64*exp(x+log(x))^2+(64*x*log(2)-48*x^3+48*x^2)*exp(x+log(x))-16*x^2*log(2)^2-24*x^3*log(2))/(16*x^
2*exp(x+log(x))^2-16*x^3*log(2)*exp(x+log(x))+4*x^4*log(2)^2)/exp(-3*x^2/(4*exp(x+log(x))-2*x*log(2))),x, algo
rithm="fricas")

[Out]

4*e^(-3/2*x^2/(x*log(2) - 2*e^(x + log(x))))/x

________________________________________________________________________________________

giac [A]  time = 0.62, size = 35, normalized size = 1.40 \begin {gather*} \frac {4 \, e^{\left (x - \frac {4 \, x e^{x} - 2 \, x \log \relax (2) - 3 \, x}{2 \, {\left (2 \, e^{x} - \log \relax (2)\right )}}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-64*exp(x+log(x))^2+(64*x*log(2)-48*x^3+48*x^2)*exp(x+log(x))-16*x^2*log(2)^2-24*x^3*log(2))/(16*x^
2*exp(x+log(x))^2-16*x^3*log(2)*exp(x+log(x))+4*x^4*log(2)^2)/exp(-3*x^2/(4*exp(x+log(x))-2*x*log(2))),x, algo
rithm="giac")

[Out]

4*e^(x - 1/2*(4*x*e^x - 2*x*log(2) - 3*x)/(2*e^x - log(2)))/x

________________________________________________________________________________________

maple [A]  time = 0.26, size = 21, normalized size = 0.84

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{\frac {3 x}{2 \left (2 \,{\mathrm e}^{x}-\ln \relax (2)\right )}}}{x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-64*exp(x+ln(x))^2+(64*x*ln(2)-48*x^3+48*x^2)*exp(x+ln(x))-16*x^2*ln(2)^2-24*x^3*ln(2))/(16*x^2*exp(x+ln(
x))^2-16*x^3*ln(2)*exp(x+ln(x))+4*x^4*ln(2)^2)/exp(-3*x^2/(4*exp(x+ln(x))-2*x*ln(2))),x,method=_RETURNVERBOSE)

[Out]

4/x*exp(3/2*x/(2*exp(x)-ln(2)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -2 \, \int -\frac {{\left (3 \, x^{3} \log \relax (2) + 2 \, x^{2} \log \relax (2)^{2} + 8 \, x^{2} e^{\left (2 \, x\right )} + 2 \, {\left (3 \, x^{3} - 3 \, x^{2} - 4 \, x \log \relax (2)\right )} x e^{x}\right )} e^{\left (\frac {3 \, x^{2}}{2 \, {\left (2 \, x e^{x} - x \log \relax (2)\right )}}\right )}}{4 \, x^{4} e^{x} \log \relax (2) - x^{4} \log \relax (2)^{2} - 4 \, x^{4} e^{\left (2 \, x\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-64*exp(x+log(x))^2+(64*x*log(2)-48*x^3+48*x^2)*exp(x+log(x))-16*x^2*log(2)^2-24*x^3*log(2))/(16*x^
2*exp(x+log(x))^2-16*x^3*log(2)*exp(x+log(x))+4*x^4*log(2)^2)/exp(-3*x^2/(4*exp(x+log(x))-2*x*log(2))),x, algo
rithm="maxima")

[Out]

-2*integrate(-(3*x^3*log(2) + 2*x^2*log(2)^2 + 8*x^2*e^(2*x) + 2*(3*x^3 - 3*x^2 - 4*x*log(2))*x*e^x)*e^(3/2*x^
2/(2*x*e^x - x*log(2)))/(4*x^4*e^x*log(2) - x^4*log(2)^2 - 4*x^4*e^(2*x)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^{\frac {3\,x^2}{4\,{\mathrm {e}}^{x+\ln \relax (x)}-2\,x\,\ln \relax (2)}}\,\left (64\,{\mathrm {e}}^{2\,x+2\,\ln \relax (x)}+16\,x^2\,{\ln \relax (2)}^2-{\mathrm {e}}^{x+\ln \relax (x)}\,\left (-48\,x^3+48\,x^2+64\,\ln \relax (2)\,x\right )+24\,x^3\,\ln \relax (2)\right )}{4\,x^4\,{\ln \relax (2)}^2+16\,x^2\,{\mathrm {e}}^{2\,x+2\,\ln \relax (x)}-16\,x^3\,{\mathrm {e}}^{x+\ln \relax (x)}\,\ln \relax (2)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((3*x^2)/(4*exp(x + log(x)) - 2*x*log(2)))*(64*exp(2*x + 2*log(x)) + 16*x^2*log(2)^2 - exp(x + log(x)
)*(64*x*log(2) + 48*x^2 - 48*x^3) + 24*x^3*log(2)))/(4*x^4*log(2)^2 + 16*x^2*exp(2*x + 2*log(x)) - 16*x^3*exp(
x + log(x))*log(2)),x)

[Out]

int(-(exp((3*x^2)/(4*exp(x + log(x)) - 2*x*log(2)))*(64*exp(2*x + 2*log(x)) + 16*x^2*log(2)^2 - exp(x + log(x)
)*(64*x*log(2) + 48*x^2 - 48*x^3) + 24*x^3*log(2)))/(4*x^4*log(2)^2 + 16*x^2*exp(2*x + 2*log(x)) - 16*x^3*exp(
x + log(x))*log(2)), x)

________________________________________________________________________________________

sympy [A]  time = 0.27, size = 22, normalized size = 0.88 \begin {gather*} \frac {4 e^{\frac {3 x^{2}}{4 x e^{x} - 2 x \log {\relax (2 )}}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-64*exp(x+ln(x))**2+(64*x*ln(2)-48*x**3+48*x**2)*exp(x+ln(x))-16*x**2*ln(2)**2-24*x**3*ln(2))/(16*x
**2*exp(x+ln(x))**2-16*x**3*ln(2)*exp(x+ln(x))+4*x**4*ln(2)**2)/exp(-3*x**2/(4*exp(x+ln(x))-2*x*ln(2))),x)

[Out]

4*exp(3*x**2/(4*x*exp(x) - 2*x*log(2)))/x

________________________________________________________________________________________