Integrand size = 48, antiderivative size = 18 \[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=\frac {10 e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x \log (x)} \] Output:
10*exp(1/x)/x/ln(x)*ln(1/x^2)
Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=\frac {10 e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x \log (x)} \] Input:
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]
Output:
(10*E^x^(-1)*Log[x^(-2)])/(x*Log[x])
Time = 0.53 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {7292, 27, 25, 2726}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (e^{\frac {1}{x}} (-10 x-10) \log \left (\frac {1}{x^2}\right )-20 e^{\frac {1}{x}} x\right ) \log (x)-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )}{x^3 \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {10 e^{\frac {1}{x}} \left (-x \log \left (\frac {1}{x^2}\right )-x \log (x) \log \left (\frac {1}{x^2}\right )-\log (x) \log \left (\frac {1}{x^2}\right )-2 x \log (x)\right )}{x^3 \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 10 \int -\frac {e^{\frac {1}{x}} \left (x \log \left (\frac {1}{x^2}\right )+x \log (x) \log \left (\frac {1}{x^2}\right )+\log (x) \log \left (\frac {1}{x^2}\right )+2 x \log (x)\right )}{x^3 \log ^2(x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -10 \int \frac {e^{\frac {1}{x}} \left (x \log \left (\frac {1}{x^2}\right )+x \log (x) \log \left (\frac {1}{x^2}\right )+\log (x) \log \left (\frac {1}{x^2}\right )+2 x \log (x)\right )}{x^3 \log ^2(x)}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {10 e^{\frac {1}{x}} \log \left (\frac {1}{x^2}\right )}{x \log (x)}\) |
Input:
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]
Output:
(10*E^x^(-1)*Log[x^(-2)])/(x*Log[x])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 0.52 (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 | \(\frac {\frac {\left (10 \ln \left (\frac {1}{x^{2}}\right )-20 \ln \left (\frac {1}{x}\right )\right ) {\mathrm e}^{\frac {1}{x}}}{x}+\frac {20 \,{\mathrm e}^{\frac {1}{x}} \ln \left (\frac {1}{x}\right )}{x}}{\ln \left (x \right )}\) | \(41\) |
default | \(\frac {\frac {\left (10 \ln \left (\frac {1}{x^{2}}\right )-20 \ln \left (\frac {1}{x}\right )\right ) {\mathrm e}^{\frac {1}{x}}}{x}+\frac {20 \,{\mathrm e}^{\frac {1}{x}} \ln \left (\frac {1}{x}\right )}{x}}{\ln \left (x \right )}\) | \(41\) |
risch | \(-\frac {20 \,{\mathrm e}^{\frac {1}{x}}}{x}+\frac {5 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\mathrm e}^{\frac {1}{x}} \left (\operatorname {csgn}\left (i x \right )^{2}-2 \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )+\operatorname {csgn}\left (i x^{2}\right )^{2}\right )}{x \ln \left (x \right )}\) | \(64\) |
Input:
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=_RETURNVERBOSE)
Output:
10*exp(1/x)/x/ln(x)*ln(1/x^2)
Time = 0.09 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.50 \[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=-\frac {20 \, e^{\frac {1}{x}}}{x} \] Input:
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, algorithm="fricas")
Output:
-20*e^(1/x)/x
Time = 0.07 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.44 \[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=- \frac {20 e^{\frac {1}{x}}}{x} \] Input:
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)
Output:
-20*exp(1/x)/x
\[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=\int { -\frac {10 \, {\left (x e^{\frac {1}{x}} \log \left (\frac {1}{x^{2}}\right ) + {\left ({\left (x + 1\right )} e^{\frac {1}{x}} \log \left (\frac {1}{x^{2}}\right ) + 2 \, x e^{\frac {1}{x}}\right )} \log \left (x\right )\right )}}{x^{3} \log \left (x\right )^{2}} \,d x } \] Input:
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, algorithm="maxima")
Output:
-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)
Time = 0.13 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.50 \[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=-\frac {20 \, e^{\frac {1}{x}}}{x} \] Input:
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, algorithm="giac")
Output:
-20*e^(1/x)/x
Time = 3.53 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=\frac {10\,\ln \left (\frac {1}{x^2}\right )\,{\mathrm {e}}^{1/x}}{x\,\ln \left (x\right )} \] Input:
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)
Output:
(10*log(1/x^2)*exp(1/x))/(x*log(x))
Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-10 e^{\frac {1}{x}} x \log \left (\frac {1}{x^2}\right )+\left (-20 e^{\frac {1}{x}} x+e^{\frac {1}{x}} (-10-10 x) \log \left (\frac {1}{x^2}\right )\right ) \log (x)}{x^3 \log ^2(x)} \, dx=-\frac {10 e^{\frac {1}{x}} \mathrm {log}\left (x^{2}\right )}{\mathrm {log}\left (x \right ) x} \] Input:
int((((-10*x-10)*exp(1/x)*log(1/x^2)-20*x*exp(1/x))*log(x)-10*x*exp(1/x)*l og(1/x^2))/x^3/log(x)^2,x)
Output:
( - 10*e**(1/x)*log(x**2))/(log(x)*x)