\(\int \frac {-10 e^{\frac {1}{x}} x \log (\frac {1}{x^2})+(-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log (\frac {1}{x^2})) \log (x)}{x^3 \log ^2(x)} \, dx\) [8755]

Optimal result

Integrand size = 48, antiderivative size = 18 \[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=\frac {10 e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x \log (x)} \]

[Out]

10*exp(1/x)/x/ln(x)*ln(1/x^2)

Rubi [F]

\[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=\int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx \]

[In]

Int[(-10*E^x^(-1)*x*Log[x^(-2)] + (-20*E^x^(-1)*x + E^x^(-1)*(-10 - 10*x)*Log[x^(-2)])*Log[x])/(x^3*Log[x]^2),
x]

[Out]

-10*Defer[Int][(E^x^(-1)*Log[x^(-2)])/(x^2*Log[x]^2), x] - 20*Defer[Int][E^x^(-1)/(x^2*Log[x]), x] - 10*Defer[
Int][(E^x^(-1)*Log[x^(-2)])/(x^3*Log[x]), x] - 10*Defer[Int][(E^x^(-1)*Log[x^(-2)])/(x^2*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {10 e^{\frac {1}{x}} \left (-2 x \log (x)-\log \left (\frac {1}{x^2}\right ) (x+(1+x) \log (x))\right )}{x^3 \log ^2(x)} \, dx \\ & = 10 \int \frac {e^{\frac {1}{x}} \left (-2 x \log (x)-\log \left (\frac {1}{x^2}\right ) (x+(1+x) \log (x))\right )}{x^3 \log ^2(x)} \, dx \\ & = 10 \int \left (-\frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^2 \log ^2(x)}+\frac {e^{\frac {1}{x}} \left (-2 x-\log \left (\frac {1}{x^2}\right )-x \log \left (\frac {1}{x^2}\right )\right )}{x^3 \log (x)}\right ) \, dx \\ & = -\left (10 \int \frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^2 \log ^2(x)} \, dx\right )+10 \int \frac {e^{\frac {1}{x}} \left (-2 x-\log \left (\frac {1}{x^2}\right )-x \log \left (\frac {1}{x^2}\right )\right )}{x^3 \log (x)} \, dx \\ & = 10 \int \left (-\frac {2 e^{\frac {1}{x}}}{x^2 \log (x)}-\frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^3 \log (x)}-\frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^2 \log (x)}\right ) \, dx-10 \int \frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^2 \log ^2(x)} \, dx \\ & = -\left (10 \int \frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^2 \log ^2(x)} \, dx\right )-10 \int \frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^3 \log (x)} \, dx-10 \int \frac {e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x^2 \log (x)} \, dx-20 \int \frac {e^{\frac {1}{x}}}{x^2 \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=\frac {10 e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x \log (x)} \]

[In]

Integrate[(-10*E^x^(-1)*x*Log[x^(-2)] + (-20*E^x^(-1)*x + E^x^(-1)*(-10 - 10*x)*Log[x^(-2)])*Log[x])/(x^3*Log[
x]^2),x]

[Out]

(10*E^x^(-1)*Log[x^(-2)])/(x*Log[x])

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\frac {10 \,{\mathrm e}^{\frac {1}{x}} \ln \left (\frac {1}{x^{2}}\right )}{x \ln \left (x \right )}\)	\(18\)
derivativedivides	\(\frac {\frac {\left (-20 \ln \left (\frac {1}{x}\right )+10 \ln \left (\frac {1}{x^{2}}\right )\right ) {\mathrm e}^{\frac {1}{x}}}{x}+\frac {20 \,{\mathrm e}^{\frac {1}{x}} \ln \left (\frac {1}{x}\right )}{x}}{\ln \left (x \right )}\)	\(41\)
default	\(\frac {\frac {\left (-20 \ln \left (\frac {1}{x}\right )+10 \ln \left (\frac {1}{x^{2}}\right )\right ) {\mathrm e}^{\frac {1}{x}}}{x}+\frac {20 \,{\mathrm e}^{\frac {1}{x}} \ln \left (\frac {1}{x}\right )}{x}}{\ln \left (x \right )}\)	\(41\)
risch	\(-\frac {20 \,{\mathrm e}^{\frac {1}{x}}}{x}+\frac {5 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\mathrm e}^{\frac {1}{x}} \left (\operatorname {csgn}\left (i x \right )^{2}-2 \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )+\operatorname {csgn}\left (i x^{2}\right )^{2}\right )}{x \ln \left (x \right )}\)	\(64\)

[In]

int((((-10*x-10)*exp(1/x)*ln(1/x^2)-20*x*exp(1/x))*ln(x)-10*x*exp(1/x)*ln(1/x^2))/x^3/ln(x)^2,x,method=_RETURN
VERBOSE)

[Out]

10*exp(1/x)/x/ln(x)*ln(1/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.50 \[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=-\frac {20 \, e^{\frac {1}{x}}}{x} \]

[In]

integrate((((-10*x-10)*exp(1/x)*log(1/x^2)-20*x*exp(1/x))*log(x)-10*x*exp(1/x)*log(1/x^2))/x^3/log(x)^2,x, alg
orithm="fricas")

[Out]

-20*e^(1/x)/x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.44 \[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=- \frac {20 e^{\frac {1}{x}}}{x} \]

[In]

integrate((((-10*x-10)*exp(1/x)*ln(1/x**2)-20*x*exp(1/x))*ln(x)-10*x*exp(1/x)*ln(1/x**2))/x**3/ln(x)**2,x)

[Out]

-20*exp(1/x)/x

Maxima [F]

\[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=\int { -\frac {10 \, {\left (x e^{\frac {1}{x}} \log \left (\frac {1}{x^{2}}\right ) + {\left ({\left (x + 1\right )} e^{\frac {1}{x}} \log \left (\frac {1}{x^{2}}\right ) + 2 \, x e^{\frac {1}{x}}\right )} \log \left (x\right )\right )}}{x^{3} \log \left (x\right )^{2}} \,d x } \]

[In]

integrate((((-10*x-10)*exp(1/x)*log(1/x^2)-20*x*exp(1/x))*log(x)-10*x*exp(1/x)*log(1/x^2))/x^3/log(x)^2,x, alg
orithm="maxima")

[Out]

-10*integrate((x*e^(1/x)*log(x^(-2)) + ((x + 1)*e^(1/x)*log(x^(-2)) + 2*x*e^(1/x))*log(x))/(x^3*log(x)^2), x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.50 \[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=-\frac {20 \, e^{\frac {1}{x}}}{x} \]

[In]

integrate((((-10*x-10)*exp(1/x)*log(1/x^2)-20*x*exp(1/x))*log(x)-10*x*exp(1/x)*log(1/x^2))/x^3/log(x)^2,x, alg
orithm="giac")

[Out]

-20*e^(1/x)/x

Mupad [B] (verification not implemented)

Time = 14.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=\frac {10\,\ln \left (\frac {1}{x^2}\right )\,{\mathrm {e}}^{1/x}}{x\,\ln \left (x\right )} \]

[In]

int(-(log(x)*(20*x*exp(1/x) + log(1/x^2)*exp(1/x)*(10*x + 10)) + 10*x*log(1/x^2)*exp(1/x))/(x^3*log(x)^2),x)

[Out]

(10*log(1/x^2)*exp(1/x))/(x*log(x))