\(\int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} (22+e^{4 x^2} (-160 x^2+8 x^3)+(-2+16 e^{4 x^2} x^2) \log (x))}{2 x^2} \, dx\) [6073]

Optimal result

Integrand size = 73, antiderivative size = 30 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\frac {\left (-10+\frac {x}{2}+\log (x)\right ) \left (e^{e^{4 x^2}}-x+3 \log (x)\right )}{x} \]

[Out]

(1/2*x-10+ln(x))*(exp(exp(4*x^2))-x+3*ln(x))/x

Rubi [F]

\[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx \]

[In]

Int[(-60 + x - x^2 + 72*Log[x] - 6*Log[x]^2 + E^E^(4*x^2)*(22 + E^(4*x^2)*(-160*x^2 + 8*x^3) + (-2 + 16*E^(4*x
^2)*x^2)*Log[x]))/(2*x^2),x]

[Out]

E^E^(4*x^2)/2 - x/2 + Log[x]/2 - (30*Log[x])/x + (3*Log[x]^2)/x - 80*Defer[Int][E^(E^(4*x^2) + 4*x^2), x] + 8*
Log[x]*Defer[Int][E^(E^(4*x^2) + 4*x^2), x] + 11*Defer[Int][E^E^(4*x^2)/x^2, x] - Log[x]*Defer[Int][E^E^(4*x^2
)/x^2, x] - 8*Defer[Int][Defer[Int][E^(E^(4*x^2) + 4*x^2), x]/x, x] + Defer[Int][Defer[Int][E^E^(4*x^2)/x^2, x
]/x, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{x^2} \, dx \\ & = \frac {1}{2} \int \left (8 e^{e^{4 x^2}+4 x^2} (-20+x+2 \log (x))-\frac {60-22 e^{e^{4 x^2}}-x+x^2-72 \log (x)+2 e^{e^{4 x^2}} \log (x)+6 \log ^2(x)}{x^2}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {60-22 e^{e^{4 x^2}}-x+x^2-72 \log (x)+2 e^{e^{4 x^2}} \log (x)+6 \log ^2(x)}{x^2} \, dx\right )+4 \int e^{e^{4 x^2}+4 x^2} (-20+x+2 \log (x)) \, dx \\ & = -\left (\frac {1}{2} \int \left (\frac {2 e^{e^{4 x^2}} (-11+\log (x))}{x^2}+\frac {60-x+x^2-72 \log (x)+6 \log ^2(x)}{x^2}\right ) \, dx\right )+4 \int \left (-20 e^{e^{4 x^2}+4 x^2}+e^{e^{4 x^2}+4 x^2} x+2 e^{e^{4 x^2}+4 x^2} \log (x)\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {60-x+x^2-72 \log (x)+6 \log ^2(x)}{x^2} \, dx\right )+4 \int e^{e^{4 x^2}+4 x^2} x \, dx+8 \int e^{e^{4 x^2}+4 x^2} \log (x) \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx-\int \frac {e^{e^{4 x^2}} (-11+\log (x))}{x^2} \, dx \\ & = -\left (\frac {1}{2} \int \left (\frac {60-x+x^2}{x^2}-\frac {72 \log (x)}{x^2}+\frac {6 \log ^2(x)}{x^2}\right ) \, dx\right )+2 \text {Subst}\left (\int e^{e^{4 x}+4 x} \, dx,x,x^2\right )-8 \int \frac {\int e^{e^{4 x^2}+4 x^2} \, dx}{x} \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx+(8 \log (x)) \int e^{e^{4 x^2}+4 x^2} \, dx-\int \left (-\frac {11 e^{e^{4 x^2}}}{x^2}+\frac {e^{e^{4 x^2}} \log (x)}{x^2}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {60-x+x^2}{x^2} \, dx\right )+\frac {1}{2} \text {Subst}\left (\int e^x \, dx,x,e^{4 x^2}\right )-3 \int \frac {\log ^2(x)}{x^2} \, dx-8 \int \frac {\int e^{e^{4 x^2}+4 x^2} \, dx}{x} \, dx+11 \int \frac {e^{e^{4 x^2}}}{x^2} \, dx+36 \int \frac {\log (x)}{x^2} \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx+(8 \log (x)) \int e^{e^{4 x^2}+4 x^2} \, dx-\int \frac {e^{e^{4 x^2}} \log (x)}{x^2} \, dx \\ & = \frac {1}{2} e^{e^{4 x^2}}-\frac {36}{x}-\frac {36 \log (x)}{x}+\frac {3 \log ^2(x)}{x}-\frac {1}{2} \int \left (1+\frac {60}{x^2}-\frac {1}{x}\right ) \, dx-6 \int \frac {\log (x)}{x^2} \, dx-8 \int \frac {\int e^{e^{4 x^2}+4 x^2} \, dx}{x} \, dx+11 \int \frac {e^{e^{4 x^2}}}{x^2} \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx-\log (x) \int \frac {e^{e^{4 x^2}}}{x^2} \, dx+(8 \log (x)) \int e^{e^{4 x^2}+4 x^2} \, dx+\int \frac {\int \frac {e^{e^{4 x^2}}}{x^2} \, dx}{x} \, dx \\ & = \frac {1}{2} e^{e^{4 x^2}}-\frac {x}{2}+\frac {\log (x)}{2}-\frac {30 \log (x)}{x}+\frac {3 \log ^2(x)}{x}-8 \int \frac {\int e^{e^{4 x^2}+4 x^2} \, dx}{x} \, dx+11 \int \frac {e^{e^{4 x^2}}}{x^2} \, dx-80 \int e^{e^{4 x^2}+4 x^2} \, dx-\log (x) \int \frac {e^{e^{4 x^2}}}{x^2} \, dx+(8 \log (x)) \int e^{e^{4 x^2}+4 x^2} \, dx+\int \frac {\int \frac {e^{e^{4 x^2}}}{x^2} \, dx}{x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\frac {e^{e^{4 x^2}} (-20+x)-x^2+\left (-60+2 e^{e^{4 x^2}}+x\right ) \log (x)+6 \log ^2(x)}{2 x} \]

[In]

Integrate[(-60 + x - x^2 + 72*Log[x] - 6*Log[x]^2 + E^E^(4*x^2)*(22 + E^(4*x^2)*(-160*x^2 + 8*x^3) + (-2 + 16*
E^(4*x^2)*x^2)*Log[x]))/(2*x^2),x]

[Out]

(E^E^(4*x^2)*(-20 + x) - x^2 + (-60 + 2*E^E^(4*x^2) + x)*Log[x] + 6*Log[x]^2)/(2*x)

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.47

method	result	size
risch	\(\frac {3 \ln \left (x \right )^{2}}{x}-\frac {30 \ln \left (x \right )}{x}+\frac {\ln \left (x \right )}{2}-\frac {x}{2}+\frac {\left (x -20+2 \ln \left (x \right )\right ) {\mathrm e}^{{\mathrm e}^{4 x^{2}}}}{2 x}\)	\(44\)
parallelrisch	\(\frac {6 \ln \left (x \right )^{2}+2 \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{4 x^{2}}}+x \ln \left (x \right )+x \,{\mathrm e}^{{\mathrm e}^{4 x^{2}}}-x^{2}-60 \ln \left (x \right )-20 \,{\mathrm e}^{{\mathrm e}^{4 x^{2}}}}{2 x}\)	\(55\)

[In]

int(1/2*(((16*x^2*exp(4*x^2)-2)*ln(x)+(8*x^3-160*x^2)*exp(4*x^2)+22)*exp(exp(4*x^2))-6*ln(x)^2+72*ln(x)-x^2+x-
60)/x^2,x,method=_RETURNVERBOSE)

[Out]

3*ln(x)^2/x-30*ln(x)/x+1/2*ln(x)-1/2*x+1/2/x*(x-20+2*ln(x))*exp(exp(4*x^2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=-\frac {x^{2} - {\left (x + 2 \, \log \left (x\right ) - 20\right )} e^{\left (e^{\left (4 \, x^{2}\right )}\right )} - {\left (x - 60\right )} \log \left (x\right ) - 6 \, \log \left (x\right )^{2}}{2 \, x} \]

[In]

integrate(1/2*(((16*x^2*exp(4*x^2)-2)*log(x)+(8*x^3-160*x^2)*exp(4*x^2)+22)*exp(exp(4*x^2))-6*log(x)^2+72*log(
x)-x^2+x-60)/x^2,x, algorithm="fricas")

[Out]

-1/2*(x^2 - (x + 2*log(x) - 20)*e^(e^(4*x^2)) - (x - 60)*log(x) - 6*log(x)^2)/x

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=- \frac {x}{2} + \frac {\log {\left (x \right )}}{2} + \frac {\left (x + 2 \log {\left (x \right )} - 20\right ) e^{e^{4 x^{2}}}}{2 x} + \frac {3 \log {\left (x \right )}^{2}}{x} - \frac {30 \log {\left (x \right )}}{x} \]

[In]

integrate(1/2*(((16*x**2*exp(4*x**2)-2)*ln(x)+(8*x**3-160*x**2)*exp(4*x**2)+22)*exp(exp(4*x**2))-6*ln(x)**2+72
*ln(x)-x**2+x-60)/x**2,x)

[Out]

-x/2 + log(x)/2 + (x + 2*log(x) - 20)*exp(exp(4*x**2))/(2*x) + 3*log(x)**2/x - 30*log(x)/x

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=-\frac {1}{2} \, x + \frac {{\left (x + 2 \, \log \left (x\right ) - 20\right )} e^{\left (e^{\left (4 \, x^{2}\right )}\right )} + 6 \, \log \left (x\right )^{2} - 60 \, \log \left (x\right ) - 60}{2 \, x} + \frac {30}{x} + \frac {1}{2} \, \log \left (x\right ) \]

[In]

integrate(1/2*(((16*x^2*exp(4*x^2)-2)*log(x)+(8*x^3-160*x^2)*exp(4*x^2)+22)*exp(exp(4*x^2))-6*log(x)^2+72*log(
x)-x^2+x-60)/x^2,x, algorithm="maxima")

[Out]

-1/2*x + 1/2*((x + 2*log(x) - 20)*e^(e^(4*x^2)) + 6*log(x)^2 - 60*log(x) - 60)/x + 30/x + 1/2*log(x)

Giac [F]

\[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\int { -\frac {x^{2} - 2 \, {\left (4 \, {\left (x^{3} - 20 \, x^{2}\right )} e^{\left (4 \, x^{2}\right )} + {\left (8 \, x^{2} e^{\left (4 \, x^{2}\right )} - 1\right )} \log \left (x\right ) + 11\right )} e^{\left (e^{\left (4 \, x^{2}\right )}\right )} + 6 \, \log \left (x\right )^{2} - x - 72 \, \log \left (x\right ) + 60}{2 \, x^{2}} \,d x } \]

[In]

integrate(1/2*(((16*x^2*exp(4*x^2)-2)*log(x)+(8*x^3-160*x^2)*exp(4*x^2)+22)*exp(exp(4*x^2))-6*log(x)^2+72*log(
x)-x^2+x-60)/x^2,x, algorithm="giac")

[Out]

integrate(-1/2*(x^2 - 2*(4*(x^3 - 20*x^2)*e^(4*x^2) + (8*x^2*e^(4*x^2) - 1)*log(x) + 11)*e^(e^(4*x^2)) + 6*log
(x)^2 - x - 72*log(x) + 60)/x^2, x)

Mupad [B] (verification not implemented)

Time = 12.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {-60+x-x^2+72 \log (x)-6 \log ^2(x)+e^{e^{4 x^2}} \left (22+e^{4 x^2} \left (-160 x^2+8 x^3\right )+\left (-2+16 e^{4 x^2} x^2\right ) \log (x)\right )}{2 x^2} \, dx=\frac {\ln \left (x\right )}{2}-\frac {x}{2}-\frac {30\,\ln \left (x\right )}{x}+\frac {3\,{\ln \left (x\right )}^2}{x}+\frac {{\mathrm {e}}^{{\mathrm {e}}^{4\,x^2}}\,\left (\frac {x}{2}+\ln \left (x\right )-10\right )}{x} \]

[In]

int((x/2 + 36*log(x) + (exp(exp(4*x^2))*(log(x)*(16*x^2*exp(4*x^2) - 2) - exp(4*x^2)*(160*x^2 - 8*x^3) + 22))/
2 - 3*log(x)^2 - x^2/2 - 30)/x^2,x)

[Out]

log(x)/2 - x/2 - (30*log(x))/x + (3*log(x)^2)/x + (exp(exp(4*x^2))*(x/2 + log(x) - 10))/x