3.32.72 \(\int \frac {(-2-2 e^{4 x+2 x^2}+4 x+4 x^2) \log ^2(\frac {25 e^{-x}}{4})+(-2 x-2 e^{4 x+2 x^2} x) \log (\frac {25 e^{-x}}{4}) \log (e^{-4 x-2 x^2} (x^2+e^{4 x+2 x^2} x^2))}{(x+e^{4 x+2 x^2} x) \log ^2(e^{-4 x-2 x^2} (x^2+e^{4 x+2 x^2} x^2))} \, dx\)

Optimal. Leaf size=32 \[ \frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{\log \left (x \left (x+e^{-x (4+2 x)} x\right )\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 7.39, 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 {\left (-2-2 e^{4 x+2 x^2}+4 x+4 x^2\right ) \log ^2\left (\frac {25 e^{-x}}{4}\right )+\left (-2 x-2 e^{4 x+2 x^2} x\right ) \log \left (\frac {25 e^{-x}}{4}\right ) \log \left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )}{\left (x+e^{4 x+2 x^2} x\right ) \log ^2\left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-2 - 2*E^(4*x + 2*x^2) + 4*x + 4*x^2)*Log[25/(4*E^x)]^2 + (-2*x - 2*E^(4*x + 2*x^2)*x)*Log[25/(4*E^x)]*L
og[E^(-4*x - 2*x^2)*(x^2 + E^(4*x + 2*x^2)*x^2)])/((x + E^(4*x + 2*x^2)*x)*Log[E^(-4*x - 2*x^2)*(x^2 + E^(4*x
+ 2*x^2)*x^2)]^2),x]

[Out]

4*Defer[Int][Log[25/(4*E^x)]^2/((1 + E^(2*x*(2 + x)))*Log[x^2 + x^2/E^(2*x*(2 + x))]^2), x] - 2*Defer[Int][Log
[25/(4*E^x)]^2/(x*Log[x^2 + x^2/E^(2*x*(2 + x))]^2), x] + 4*Defer[Int][(x*Log[25/(4*E^x)]^2)/((1 + E^(2*x*(2 +
x)))*Log[x^2 + x^2/E^(2*x*(2 + x))]^2), x] - 2*Defer[Int][Log[25/(4*E^x)]/Log[x^2 + x^2/E^(2*x*(2 + x))], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \log \left (\frac {25 e^{-x}}{4}\right ) \left (-\frac {\left (1+e^{2 x (2+x)}-2 x-2 x^2\right ) \log \left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) x}-\log \left (\left (1+e^{-2 x (2+x)}\right ) x^2\right )\right )}{\log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx\\ &=2 \int \frac {\log \left (\frac {25 e^{-x}}{4}\right ) \left (-\frac {\left (1+e^{2 x (2+x)}-2 x-2 x^2\right ) \log \left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) x}-\log \left (\left (1+e^{-2 x (2+x)}\right ) x^2\right )\right )}{\log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx\\ &=2 \int \left (\frac {2 (1+x) \log ^2\left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )}+\frac {\log \left (\frac {25 e^{-x}}{4}\right ) \left (-\log \left (\frac {25 e^{-x}}{4}\right )-x \log \left (\left (1+e^{-2 x (2+x)}\right ) x^2\right )\right )}{x \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )}\right ) \, dx\\ &=2 \int \frac {\log \left (\frac {25 e^{-x}}{4}\right ) \left (-\log \left (\frac {25 e^{-x}}{4}\right )-x \log \left (\left (1+e^{-2 x (2+x)}\right ) x^2\right )\right )}{x \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx+4 \int \frac {(1+x) \log ^2\left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx\\ &=2 \int \left (-\frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{x \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )}-\frac {\log \left (\frac {25 e^{-x}}{4}\right )}{\log \left (x^2+e^{-2 x (2+x)} x^2\right )}\right ) \, dx+4 \int \left (\frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )}+\frac {x \log ^2\left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )}\right ) \, dx\\ &=-\left (2 \int \frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{x \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx\right )-2 \int \frac {\log \left (\frac {25 e^{-x}}{4}\right )}{\log \left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx+4 \int \frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx+4 \int \frac {x \log ^2\left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 33, normalized size = 1.03 \begin {gather*} \frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{\log \left (\left (1+e^{-4 x-2 x^2}\right ) x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-2 - 2*E^(4*x + 2*x^2) + 4*x + 4*x^2)*Log[25/(4*E^x)]^2 + (-2*x - 2*E^(4*x + 2*x^2)*x)*Log[25/(4*E
^x)]*Log[E^(-4*x - 2*x^2)*(x^2 + E^(4*x + 2*x^2)*x^2)])/((x + E^(4*x + 2*x^2)*x)*Log[E^(-4*x - 2*x^2)*(x^2 + E
^(4*x + 2*x^2)*x^2)]^2),x]

[Out]

Log[25/(4*E^x)]^2/Log[(1 + E^(-4*x - 2*x^2))*x^2]

________________________________________________________________________________________

fricas [A]  time = 0.55, size = 46, normalized size = 1.44 \begin {gather*} \frac {x^{2} - 2 \, x \log \left (\frac {25}{4}\right ) + \log \left (\frac {25}{4}\right )^{2}}{\log \left ({\left (x^{2} e^{\left (2 \, x^{2} + 4 \, x\right )} + x^{2}\right )} e^{\left (-2 \, x^{2} - 4 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(2*x^2+4*x)-2*x)*log(25/4/exp(x))*log((x^2*exp(2*x^2+4*x)+x^2)/exp(2*x^2+4*x))+(-2*exp(2*x
^2+4*x)+4*x^2+4*x-2)*log(25/4/exp(x))^2)/(x*exp(2*x^2+4*x)+x)/log((x^2*exp(2*x^2+4*x)+x^2)/exp(2*x^2+4*x))^2,x
, algorithm="fricas")

[Out]

(x^2 - 2*x*log(25/4) + log(25/4)^2)/log((x^2*e^(2*x^2 + 4*x) + x^2)*e^(-2*x^2 - 4*x))

________________________________________________________________________________________

giac [B]  time = 0.87, size = 65, normalized size = 2.03 \begin {gather*} \frac {x^{2} - 4 \, x \log \relax (5) + 4 \, \log \relax (5)^{2} + 4 \, x \log \relax (2) - 8 \, \log \relax (5) \log \relax (2) + 4 \, \log \relax (2)^{2}}{\log \left ({\left (x^{2} e^{\left (2 \, x^{2} + 4 \, x\right )} + x^{2}\right )} e^{\left (-2 \, x^{2} - 4 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(2*x^2+4*x)-2*x)*log(25/4/exp(x))*log((x^2*exp(2*x^2+4*x)+x^2)/exp(2*x^2+4*x))+(-2*exp(2*x
^2+4*x)+4*x^2+4*x-2)*log(25/4/exp(x))^2)/(x*exp(2*x^2+4*x)+x)/log((x^2*exp(2*x^2+4*x)+x^2)/exp(2*x^2+4*x))^2,x
, algorithm="giac")

[Out]

(x^2 - 4*x*log(5) + 4*log(5)^2 + 4*x*log(2) - 8*log(5)*log(2) + 4*log(2)^2)/log((x^2*e^(2*x^2 + 4*x) + x^2)*e^
(-2*x^2 - 4*x))

________________________________________________________________________________________

maple [C]  time = 0.69, size = 426, normalized size = 13.31

method	result	size
risch	\(\frac {i \left (16 \ln \relax (5)^{2}+16 \ln \relax (2)^{2}-32 \ln \relax (2) \ln \relax (5)-16 \ln \relax (5) \ln \left ({\mathrm e}^{x}\right )+16 \ln \relax (2) \ln \left ({\mathrm e}^{x}\right )+4 \ln \left ({\mathrm e}^{x}\right )^{2}\right )}{2 \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-4 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+2 \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right ) \mathrm {csgn}\left (i x^{2} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right ) {\mathrm e}^{-2 x \left (2+x \right )}\right )-2 \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right ) {\mathrm e}^{-2 x \left (2+x \right )}\right )^{2}+2 \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right )-2 \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right )^{2}-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right )^{2}+2 \pi \mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right )^{3}-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right ) \mathrm {csgn}\left (i x^{2} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right ) {\mathrm e}^{-2 x \left (2+x \right )}\right )^{2}+2 \pi \mathrm {csgn}\left (i x^{2} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right ) {\mathrm e}^{-2 x \left (2+x \right )}\right )^{3}+8 i \ln \relax (x )+4 i \ln \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )-4 i \ln \left ({\mathrm e}^{2 x \left (2+x \right )}\right )}\)	\(426\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*exp(2*x^2+4*x)-2*x)*ln(25/4/exp(x))*ln((x^2*exp(2*x^2+4*x)+x^2)/exp(2*x^2+4*x))+(-2*exp(2*x^2+4*x)+
4*x^2+4*x-2)*ln(25/4/exp(x))^2)/(x*exp(2*x^2+4*x)+x)/ln((x^2*exp(2*x^2+4*x)+x^2)/exp(2*x^2+4*x))^2,x,method=_R
ETURNVERBOSE)

[Out]

1/2*I*(16*ln(5)^2+16*ln(2)^2-32*ln(2)*ln(5)-16*ln(5)*ln(exp(x))+16*ln(2)*ln(exp(x))+4*ln(exp(x))^2)/(Pi*csgn(I
*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3+Pi*csgn(I*x^2)*csgn(I*exp(-2*x*(2+x))*(exp(2*x
*(2+x))+1))*csgn(I*x^2*(exp(2*x*(2+x))+1)*exp(-2*x*(2+x)))-Pi*csgn(I*x^2)*csgn(I*x^2*(exp(2*x*(2+x))+1)*exp(-2
*x*(2+x)))^2+Pi*csgn(I*(exp(2*x*(2+x))+1))*csgn(I*exp(-2*x*(2+x)))*csgn(I*exp(-2*x*(2+x))*(exp(2*x*(2+x))+1))-
Pi*csgn(I*(exp(2*x*(2+x))+1))*csgn(I*exp(-2*x*(2+x))*(exp(2*x*(2+x))+1))^2-Pi*csgn(I*exp(-2*x*(2+x)))*csgn(I*e
xp(-2*x*(2+x))*(exp(2*x*(2+x))+1))^2+Pi*csgn(I*exp(-2*x*(2+x))*(exp(2*x*(2+x))+1))^3-Pi*csgn(I*exp(-2*x*(2+x))
*(exp(2*x*(2+x))+1))*csgn(I*x^2*(exp(2*x*(2+x))+1)*exp(-2*x*(2+x)))^2+Pi*csgn(I*x^2*(exp(2*x*(2+x))+1)*exp(-2*
x*(2+x)))^3+4*I*ln(x)+2*I*ln(exp(2*x*(2+x))+1)-2*I*ln(exp(2*x*(2+x))))

________________________________________________________________________________________

maxima [B]  time = 0.59, size = 64, normalized size = 2.00 \begin {gather*} -\frac {x^{2} - 4 \, x {\left (\log \relax (5) - \log \relax (2)\right )} + 4 \, \log \relax (5)^{2} - 8 \, \log \relax (5) \log \relax (2) + 4 \, \log \relax (2)^{2}}{2 \, x^{2} + 4 \, x - 2 \, \log \relax (x) - \log \left (e^{\left (2 \, x^{2} + 4 \, x\right )} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(2*x^2+4*x)-2*x)*log(25/4/exp(x))*log((x^2*exp(2*x^2+4*x)+x^2)/exp(2*x^2+4*x))+(-2*exp(2*x
^2+4*x)+4*x^2+4*x-2)*log(25/4/exp(x))^2)/(x*exp(2*x^2+4*x)+x)/log((x^2*exp(2*x^2+4*x)+x^2)/exp(2*x^2+4*x))^2,x
, algorithm="maxima")

[Out]

-(x^2 - 4*x*(log(5) - log(2)) + 4*log(5)^2 - 8*log(5)*log(2) + 4*log(2)^2)/(2*x^2 + 4*x - 2*log(x) - log(e^(2*
x^2 + 4*x) + 1))

________________________________________________________________________________________

mupad [B]  time = 2.34, size = 226, normalized size = 7.06 \begin {gather*} \frac {{\left (x-\ln \left (\frac {25}{4}\right )\right )}^2+\frac {x\,\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{-2\,x^2}\,\left (x^2+x^2\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{2\,x^2}\right )\right )\,\left ({\mathrm {e}}^{2\,x^2+4\,x}+1\right )\,\left (x-\ln \left (\frac {25}{4}\right )\right )}{2\,x-{\mathrm {e}}^{2\,x^2+4\,x}+2\,x^2-1}}{\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{-2\,x^2}\,\left (x^2+x^2\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{2\,x^2}\right )\right )}-x\,\ln \left (\frac {25}{4}\right )+x^2-\frac {2\,\left (3\,x^2\,\ln \left (\frac {25}{4}\right )+3\,x^3\,\ln \left (\frac {25}{4}\right )-8\,x^4\,\ln \left (\frac {25}{4}\right )-12\,x^5\,\ln \left (\frac {25}{4}\right )-4\,x^6\,\ln \left (\frac {25}{4}\right )-3\,x^3-3\,x^4+8\,x^5+12\,x^6+4\,x^7\right )}{\left (4\,x^3+8\,x^2-3\right )\,\left (2\,x-{\mathrm {e}}^{2\,x^2+4\,x}+2\,x^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((25*exp(-x))/4)^2*(4*x - 2*exp(4*x + 2*x^2) + 4*x^2 - 2) - log((25*exp(-x))/4)*log(exp(- 4*x - 2*x^2)
*(x^2*exp(4*x + 2*x^2) + x^2))*(2*x + 2*x*exp(4*x + 2*x^2)))/(log(exp(- 4*x - 2*x^2)*(x^2*exp(4*x + 2*x^2) + x
^2))^2*(x + x*exp(4*x + 2*x^2))),x)

[Out]

((x - log(25/4))^2 + (x*log(exp(-4*x)*exp(-2*x^2)*(x^2 + x^2*exp(4*x)*exp(2*x^2)))*(exp(4*x + 2*x^2) + 1)*(x -
log(25/4)))/(2*x - exp(4*x + 2*x^2) + 2*x^2 - 1))/log(exp(-4*x)*exp(-2*x^2)*(x^2 + x^2*exp(4*x)*exp(2*x^2)))
- x*log(25/4) + x^2 - (2*(3*x^2*log(25/4) + 3*x^3*log(25/4) - 8*x^4*log(25/4) - 12*x^5*log(25/4) - 4*x^6*log(2
5/4) - 3*x^3 - 3*x^4 + 8*x^5 + 12*x^6 + 4*x^7))/((8*x^2 + 4*x^3 - 3)*(2*x - exp(4*x + 2*x^2) + 2*x^2 - 1))

________________________________________________________________________________________

sympy [B]  time = 0.67, size = 68, normalized size = 2.12 \begin {gather*} \frac {x^{2} - 4 x \log {\relax (5 )} + 4 x \log {\relax (2 )} - 8 \log {\relax (2 )} \log {\relax (5 )} + 4 \log {\relax (2 )}^{2} + 4 \log {\relax (5 )}^{2}}{\log {\left (\left (x^{2} e^{2 x^{2} + 4 x} + x^{2}\right ) e^{- 2 x^{2} - 4 x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(2*x**2+4*x)-2*x)*ln(25/4/exp(x))*ln((x**2*exp(2*x**2+4*x)+x**2)/exp(2*x**2+4*x))+(-2*exp(
2*x**2+4*x)+4*x**2+4*x-2)*ln(25/4/exp(x))**2)/(x*exp(2*x**2+4*x)+x)/ln((x**2*exp(2*x**2+4*x)+x**2)/exp(2*x**2+
4*x))**2,x)

[Out]

(x**2 - 4*x*log(5) + 4*x*log(2) - 8*log(2)*log(5) + 4*log(2)**2 + 4*log(5)**2)/log((x**2*exp(2*x**2 + 4*x) + x
**2)*exp(-2*x**2 - 4*x))

________________________________________________________________________________________