3.4.68 \(\int \frac {-125-15 e^{2 x^2}-e^{3 x^2}+e^{x^2} (-75+e^8 (4 x^2-4 x^3))+(-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)) \log (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}) \log (\log (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}))}{(-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)) \log (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x})} \, dx\)

Optimal. Leaf size=29 \[ x \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right ) \]

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Rubi [F]  time = 3.95, 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 {-125-15 e^{2 x^2}-e^{3 x^2}+e^{x^2} \left (-75+e^8 \left (4 x^2-4 x^3\right )\right )+\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )\right )}{\left (-125+e^{3 x^2} (-1+x)+125 x+e^{2 x^2} (-15+15 x)+e^{x^2} (-75+75 x)\right ) \log \left (\frac {e^{\frac {e^8}{25+10 e^{x^2}+e^{2 x^2}}} x}{-4+4 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-125 - 15*E^(2*x^2) - E^(3*x^2) + E^x^2*(-75 + E^8*(4*x^2 - 4*x^3)) + (-125 + E^(3*x^2)*(-1 + x) + 125*x
+ E^(2*x^2)*(-15 + 15*x) + E^x^2*(-75 + 75*x))*Log[(E^(E^8/(25 + 10*E^x^2 + E^(2*x^2)))*x)/(-4 + 4*x)]*Log[Log
[(E^(E^8/(25 + 10*E^x^2 + E^(2*x^2)))*x)/(-4 + 4*x)]])/((-125 + E^(3*x^2)*(-1 + x) + 125*x + E^(2*x^2)*(-15 +
15*x) + E^x^2*(-75 + 75*x))*Log[(E^(E^8/(25 + 10*E^x^2 + E^(2*x^2)))*x)/(-4 + 4*x)]),x]

[Out]

-Defer[Int][1/((-1 + x)*Log[(E^(E^8/(5 + E^x^2)^2)*x)/(4*(-1 + x))]), x] + 20*E^8*Defer[Int][x^2/((5 + E^x^2)^
3*Log[(E^(E^8/(5 + E^x^2)^2)*x)/(4*(-1 + x))]), x] - 4*E^8*Defer[Int][x^2/((5 + E^x^2)^2*Log[(E^(E^8/(5 + E^x^
2)^2)*x)/(4*(-1 + x))]), x] + Defer[Int][Log[Log[(E^(E^8/(5 + E^x^2)^2)*x)/(4*(-1 + x))]], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {125+15 e^{2 x^2}+e^{3 x^2}-e^{x^2} \left (-75-4 e^8 (-1+x) x^2\right )-\left (5+e^{x^2}\right )^3 (-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )}{\left (5+e^{x^2}\right )^3 (1-x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx\\ &=\int \left (\frac {20 e^8 x^2}{\left (5+e^{x^2}\right )^3 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )}-\frac {4 e^8 x^2}{\left (5+e^{x^2}\right )^2 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )}+\frac {-1-\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )+x \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )}{(-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )}\right ) \, dx\\ &=-\left (\left (4 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^2 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx\right )+\left (20 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^3 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx+\int \frac {-1-\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )+x \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right ) \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )}{(-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx\\ &=-\left (\left (4 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^2 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx\right )+\left (20 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^3 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx+\int \left (-\frac {1}{(-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )}+\log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right )\right ) \, dx\\ &=-\left (\left (4 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^2 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx\right )+\left (20 e^8\right ) \int \frac {x^2}{\left (5+e^{x^2}\right )^3 \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx-\int \frac {1}{(-1+x) \log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )} \, dx+\int \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 29, normalized size = 1.00 \begin {gather*} x \log \left (\log \left (\frac {e^{\frac {e^8}{\left (5+e^{x^2}\right )^2}} x}{4 (-1+x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-125 - 15*E^(2*x^2) - E^(3*x^2) + E^x^2*(-75 + E^8*(4*x^2 - 4*x^3)) + (-125 + E^(3*x^2)*(-1 + x) +
125*x + E^(2*x^2)*(-15 + 15*x) + E^x^2*(-75 + 75*x))*Log[(E^(E^8/(25 + 10*E^x^2 + E^(2*x^2)))*x)/(-4 + 4*x)]*L
og[Log[(E^(E^8/(25 + 10*E^x^2 + E^(2*x^2)))*x)/(-4 + 4*x)]])/((-125 + E^(3*x^2)*(-1 + x) + 125*x + E^(2*x^2)*(
-15 + 15*x) + E^x^2*(-75 + 75*x))*Log[(E^(E^8/(25 + 10*E^x^2 + E^(2*x^2)))*x)/(-4 + 4*x)]),x]

[Out]

x*Log[Log[(E^(E^8/(5 + E^x^2)^2)*x)/(4*(-1 + x))]]

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fricas [A]  time = 0.63, size = 32, normalized size = 1.10 \begin {gather*} x \log \left (\log \left (\frac {x e^{\left (\frac {e^{8}}{e^{\left (2 \, x^{2}\right )} + 10 \, e^{\left (x^{2}\right )} + 25}\right )}}{4 \, {\left (x - 1\right )}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x-1)*exp(x^2)^3+(15*x-15)*exp(x^2)^2+(75*x-75)*exp(x^2)+125*x-125)*log(x*exp(exp(4)^2/(exp(x^2)^2
+10*exp(x^2)+25))/(4*x-4))*log(log(x*exp(exp(4)^2/(exp(x^2)^2+10*exp(x^2)+25))/(4*x-4)))-exp(x^2)^3-15*exp(x^2
)^2+((-4*x^3+4*x^2)*exp(4)^2-75)*exp(x^2)-125)/((x-1)*exp(x^2)^3+(15*x-15)*exp(x^2)^2+(75*x-75)*exp(x^2)+125*x
-125)/log(x*exp(exp(4)^2/(exp(x^2)^2+10*exp(x^2)+25))/(4*x-4)),x, algorithm="fricas")

[Out]

x*log(log(1/4*x*e^(e^8/(e^(2*x^2) + 10*e^(x^2) + 25))/(x - 1)))

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giac [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((((x-1)*exp(x^2)^3+(15*x-15)*exp(x^2)^2+(75*x-75)*exp(x^2)+125*x-125)*log(x*exp(exp(4)^2/(exp(x^2)^2
+10*exp(x^2)+25))/(4*x-4))*log(log(x*exp(exp(4)^2/(exp(x^2)^2+10*exp(x^2)+25))/(4*x-4)))-exp(x^2)^3-15*exp(x^2
)^2+((-4*x^3+4*x^2)*exp(4)^2-75)*exp(x^2)-125)/((x-1)*exp(x^2)^3+(15*x-15)*exp(x^2)^2+(75*x-75)*exp(x^2)+125*x
-125)/log(x*exp(exp(4)^2/(exp(x^2)^2+10*exp(x^2)+25))/(4*x-4)),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.66, size = 302, normalized size = 10.41

method	result	size
risch	\(\ln \left (-2 \ln \relax (2)+\ln \relax (x )+\ln \left ({\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}\right )-\ln \left (x -1\right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right )+\mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right )+\mathrm {csgn}\left (\frac {i}{x -1}\right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right ) \left (-\mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (\frac {i x \,{\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right )+\mathrm {csgn}\left (\frac {i {\mathrm e}^{\frac {{\mathrm e}^{8}}{{\mathrm e}^{2 x^{2}}+10 \,{\mathrm e}^{x^{2}}+25}}}{x -1}\right )\right )}{2}\right ) x\)	\(302\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x-1)*exp(x^2)^3+(15*x-15)*exp(x^2)^2+(75*x-75)*exp(x^2)+125*x-125)*ln(x*exp(exp(4)^2/(exp(x^2)^2+10*exp
(x^2)+25))/(4*x-4))*ln(ln(x*exp(exp(4)^2/(exp(x^2)^2+10*exp(x^2)+25))/(4*x-4)))-exp(x^2)^3-15*exp(x^2)^2+((-4*
x^3+4*x^2)*exp(4)^2-75)*exp(x^2)-125)/((x-1)*exp(x^2)^3+(15*x-15)*exp(x^2)^2+(75*x-75)*exp(x^2)+125*x-125)/ln(
x*exp(exp(4)^2/(exp(x^2)^2+10*exp(x^2)+25))/(4*x-4)),x,method=_RETURNVERBOSE)

[Out]

ln(-2*ln(2)+ln(x)+ln(exp(exp(8)/(exp(2*x^2)+10*exp(x^2)+25)))-ln(x-1)-1/2*I*Pi*csgn(I*exp(exp(8)/(exp(2*x^2)+1
0*exp(x^2)+25))/(x-1))*(-csgn(I*exp(exp(8)/(exp(2*x^2)+10*exp(x^2)+25))/(x-1))+csgn(I*exp(exp(8)/(exp(2*x^2)+1
0*exp(x^2)+25))))*(-csgn(I*exp(exp(8)/(exp(2*x^2)+10*exp(x^2)+25))/(x-1))+csgn(I/(x-1)))-1/2*I*Pi*csgn(I*x/(x-
1)*exp(exp(8)/(exp(2*x^2)+10*exp(x^2)+25)))*(-csgn(I*x/(x-1)*exp(exp(8)/(exp(2*x^2)+10*exp(x^2)+25)))+csgn(I*x
))*(-csgn(I*x/(x-1)*exp(exp(8)/(exp(2*x^2)+10*exp(x^2)+25)))+csgn(I*exp(exp(8)/(exp(2*x^2)+10*exp(x^2)+25))/(x
-1))))*x

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maxima [B]  time = 0.85, size = 76, normalized size = 2.62 \begin {gather*} x \log \left (-{\left (2 \, \log \relax (2) + \log \left (x - 1\right )\right )} e^{\left (2 \, x^{2}\right )} - 10 \, {\left (2 \, \log \relax (2) + \log \left (x - 1\right )\right )} e^{\left (x^{2}\right )} + {\left (e^{\left (2 \, x^{2}\right )} + 10 \, e^{\left (x^{2}\right )} + 25\right )} \log \relax (x) + e^{8} - 50 \, \log \relax (2) - 25 \, \log \left (x - 1\right )\right ) - 2 \, x \log \left (e^{\left (x^{2}\right )} + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x-1)*exp(x^2)^3+(15*x-15)*exp(x^2)^2+(75*x-75)*exp(x^2)+125*x-125)*log(x*exp(exp(4)^2/(exp(x^2)^2
+10*exp(x^2)+25))/(4*x-4))*log(log(x*exp(exp(4)^2/(exp(x^2)^2+10*exp(x^2)+25))/(4*x-4)))-exp(x^2)^3-15*exp(x^2
)^2+((-4*x^3+4*x^2)*exp(4)^2-75)*exp(x^2)-125)/((x-1)*exp(x^2)^3+(15*x-15)*exp(x^2)^2+(75*x-75)*exp(x^2)+125*x
-125)/log(x*exp(exp(4)^2/(exp(x^2)^2+10*exp(x^2)+25))/(4*x-4)),x, algorithm="maxima")

[Out]

x*log(-(2*log(2) + log(x - 1))*e^(2*x^2) - 10*(2*log(2) + log(x - 1))*e^(x^2) + (e^(2*x^2) + 10*e^(x^2) + 25)*
log(x) + e^8 - 50*log(2) - 25*log(x - 1)) - 2*x*log(e^(x^2) + 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {15\,{\mathrm {e}}^{2\,x^2}+{\mathrm {e}}^{3\,x^2}-{\mathrm {e}}^{x^2}\,\left ({\mathrm {e}}^8\,\left (4\,x^2-4\,x^3\right )-75\right )-\ln \left (\ln \left (\frac {x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^8}{10\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{2\,x^2}+25}}}{4\,x-4}\right )\right )\,\ln \left (\frac {x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^8}{10\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{2\,x^2}+25}}}{4\,x-4}\right )\,\left (125\,x+{\mathrm {e}}^{2\,x^2}\,\left (15\,x-15\right )+{\mathrm {e}}^{3\,x^2}\,\left (x-1\right )+{\mathrm {e}}^{x^2}\,\left (75\,x-75\right )-125\right )+125}{\ln \left (\frac {x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^8}{10\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{2\,x^2}+25}}}{4\,x-4}\right )\,\left (125\,x+{\mathrm {e}}^{2\,x^2}\,\left (15\,x-15\right )+{\mathrm {e}}^{3\,x^2}\,\left (x-1\right )+{\mathrm {e}}^{x^2}\,\left (75\,x-75\right )-125\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*exp(2*x^2) + exp(3*x^2) - exp(x^2)*(exp(8)*(4*x^2 - 4*x^3) - 75) - log(log((x*exp(exp(8)/(10*exp(x^2)
+ exp(2*x^2) + 25)))/(4*x - 4)))*log((x*exp(exp(8)/(10*exp(x^2) + exp(2*x^2) + 25)))/(4*x - 4))*(125*x + exp(
2*x^2)*(15*x - 15) + exp(3*x^2)*(x - 1) + exp(x^2)*(75*x - 75) - 125) + 125)/(log((x*exp(exp(8)/(10*exp(x^2) +
exp(2*x^2) + 25)))/(4*x - 4))*(125*x + exp(2*x^2)*(15*x - 15) + exp(3*x^2)*(x - 1) + exp(x^2)*(75*x - 75) - 1
25)),x)

[Out]

-int((15*exp(2*x^2) + exp(3*x^2) - exp(x^2)*(exp(8)*(4*x^2 - 4*x^3) - 75) - log(log((x*exp(exp(8)/(10*exp(x^2)
+ exp(2*x^2) + 25)))/(4*x - 4)))*log((x*exp(exp(8)/(10*exp(x^2) + exp(2*x^2) + 25)))/(4*x - 4))*(125*x + exp(
2*x^2)*(15*x - 15) + exp(3*x^2)*(x - 1) + exp(x^2)*(75*x - 75) - 125) + 125)/(log((x*exp(exp(8)/(10*exp(x^2) +
exp(2*x^2) + 25)))/(4*x - 4))*(125*x + exp(2*x^2)*(15*x - 15) + exp(3*x^2)*(x - 1) + exp(x^2)*(75*x - 75) - 1
25)), 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((((x-1)*exp(x**2)**3+(15*x-15)*exp(x**2)**2+(75*x-75)*exp(x**2)+125*x-125)*ln(x*exp(exp(4)**2/(exp(x
**2)**2+10*exp(x**2)+25))/(4*x-4))*ln(ln(x*exp(exp(4)**2/(exp(x**2)**2+10*exp(x**2)+25))/(4*x-4)))-exp(x**2)**
3-15*exp(x**2)**2+((-4*x**3+4*x**2)*exp(4)**2-75)*exp(x**2)-125)/((x-1)*exp(x**2)**3+(15*x-15)*exp(x**2)**2+(7
5*x-75)*exp(x**2)+125*x-125)/ln(x*exp(exp(4)**2/(exp(x**2)**2+10*exp(x**2)+25))/(4*x-4)),x)

[Out]

Timed out

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