Integrand size = 61, antiderivative size = 21 \[ \int \frac {-25 e^6-10 e^3 \log (-5 x)-\log ^2(-5 x)+\left (10 e^3+2 \log (-5 x)\right ) \log (x)+2 e^6 x^2 \log ^2(x)}{e^6 x \log ^2(x)} \, dx=x^2+\frac {\left (5+\frac {\log (-5 x)}{e^3}\right )^2}{\log (x)} \]
Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {-25 e^6-10 e^3 \log (-5 x)-\log ^2(-5 x)+\left (10 e^3+2 \log (-5 x)\right ) \log (x)+2 e^6 x^2 \log ^2(x)}{e^6 x \log ^2(x)} \, dx=\frac {e^6 x^2+\frac {\left (5 e^3+\log (-5 x)-\log (x)\right )^2}{\log (x)}+\log (x)}{e^6} \]
Integrate[(-25*E^6 - 10*E^3*Log[-5*x] - Log[-5*x]^2 + (10*E^3 + 2*Log[-5*x ])*Log[x] + 2*E^6*x^2*Log[x]^2)/(E^6*x*Log[x]^2),x]
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.
\(\displaystyle \int \frac {2 e^6 x^2 \log ^2(x)-\log ^2(-5 x)-10 e^3 \log (-5 x)+\left (2 \log (-5 x)+10 e^3\right ) \log (x)-25 e^6}{e^6 x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {\log ^2(-5 x)+10 e^3 \log (-5 x)-2 e^6 x^2 \log ^2(x)-2 \left (\log (-5 x)+5 e^3\right ) \log (x)+25 e^6}{x \log ^2(x)}dx}{e^6}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\log ^2(-5 x)+10 e^3 \log (-5 x)-2 e^6 x^2 \log ^2(x)-2 \left (\log (-5 x)+5 e^3\right ) \log (x)+25 e^6}{x \log ^2(x)}dx}{e^6}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (\frac {\left (\log (-5 x)+5 e^3\right )^2}{x \log ^2(x)}-\frac {2 \left (\log (-5 x)+5 e^3\right )}{x \log (x)}-2 e^6 x\right )dx}{e^6}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\int \frac {\left (\log (-5 x)+5 e^3\right )^2}{x \log ^2(x)}dx-e^6 x^2-2 \log (x)-2 \left (\log (-5 x)+5 e^3\right ) \log (\log (x))+2 \log (x) \log (\log (x))}{e^6}\) |
Int[(-25*E^6 - 10*E^3*Log[-5*x] - Log[-5*x]^2 + (10*E^3 + 2*Log[-5*x])*Log [x] + 2*E^6*x^2*Log[x]^2)/(E^6*x*Log[x]^2),x]
3.20.67.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.82 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{-6} \left ({\mathrm e}^{6} \ln \left (x \right ) x^{2}+25 \,{\mathrm e}^{6}+10 \,{\mathrm e}^{3} \ln \left (-5 x \right )+\ln \left (-5 x \right )^{2}\right )}{\ln \left (x \right )}\) | \(41\) |
default | \({\mathrm e}^{-6} \left (\ln \left (5\right ) \left (\frac {2 \ln \left (-x \right )}{\ln \left (x \right )}+\frac {10 \,{\mathrm e}^{3}}{\ln \left (x \right )}\right )+10 \,{\mathrm e}^{3} \ln \left (\ln \left (x \right )\right )+\frac {\ln \left (-x \right )^{2}}{\ln \left (x \right )}+\frac {25 \,{\mathrm e}^{6}}{\ln \left (x \right )}-10 \,{\mathrm e}^{3} \left (-\frac {\ln \left (-x \right )}{\ln \left (x \right )}+\ln \left (\ln \left (x \right )\right )\right )+x^{2} {\mathrm e}^{6}+\frac {\ln \left (5\right )^{2}}{\ln \left (x \right )}\right )\) | \(92\) |
parts | \(2 \,{\mathrm e}^{-6} \left (\ln \left (-x \right ) \ln \left (\ln \left (x \right )\right )-\ln \left (x \right ) \ln \left (\ln \left (x \right )\right )+\ln \left (x \right )+5 \,{\mathrm e}^{3} \ln \left (\ln \left (x \right )\right )\right )+2 \,{\mathrm e}^{-6} \ln \left (5\right ) \ln \left (\ln \left (x \right )\right )-{\mathrm e}^{-6} \left (-\frac {\ln \left (-x \right )^{2}}{\ln \left (x \right )}+2 \ln \left (-x \right ) \ln \left (\ln \left (x \right )\right )-2 \ln \left (x \right ) \ln \left (\ln \left (x \right )\right )+2 \ln \left (x \right )-\frac {25 \,{\mathrm e}^{6}}{\ln \left (x \right )}+10 \,{\mathrm e}^{3} \left (-\frac {\ln \left (-x \right )}{\ln \left (x \right )}+\ln \left (\ln \left (x \right )\right )\right )\right )-2 \ln \left (5\right ) {\mathrm e}^{-6} \left (-\frac {\ln \left (-x \right )}{\ln \left (x \right )}+\ln \left (\ln \left (x \right )\right )-\frac {5 \,{\mathrm e}^{3}}{\ln \left (x \right )}\right )+\frac {{\mathrm e}^{-6} \ln \left (5\right )^{2}}{\ln \left (x \right )}+x^{2}\) | \(157\) |
risch | \(x^{2}+{\mathrm e}^{-6} \ln \left (x \right )+\frac {{\mathrm e}^{-6} \left (-4 \pi ^{2}+40 \,{\mathrm e}^{3} \ln \left (5\right )-8 \pi ^{2} \operatorname {csgn}\left (i x \right )^{3}-4 \pi ^{2} \operatorname {csgn}\left (i x \right )^{6}+40 i {\mathrm e}^{3} \pi +8 i \ln \left (5\right ) \pi +40 i {\mathrm e}^{3} \pi \operatorname {csgn}\left (i x \right )^{3}-40 i {\mathrm e}^{3} \pi \operatorname {csgn}\left (i x \right )^{2}+8 i \ln \left (5\right ) \pi \operatorname {csgn}\left (i x \right )^{3}-8 i \ln \left (5\right ) \pi \operatorname {csgn}\left (i x \right )^{2}+100 \,{\mathrm e}^{6}+4 \ln \left (5\right )^{2}-4 \pi ^{2} \operatorname {csgn}\left (i x \right )^{4}+8 \pi ^{2} \operatorname {csgn}\left (i x \right )^{5}+8 \pi ^{2} \operatorname {csgn}\left (i x \right )^{2}\right )}{4 \ln \left (x \right )}\) | \(164\) |
int((2*x^2*exp(3)^2*ln(x)^2+(2*ln(-5*x)+10*exp(3))*ln(x)-ln(-5*x)^2-10*exp (3)*ln(-5*x)-25*exp(3)^2)/x/exp(3)^2/ln(x)^2,x,method=_RETURNVERBOSE)
Result contains complex when optimal does not.
Time = 0.24 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.76 \[ \int \frac {-25 e^6-10 e^3 \log (-5 x)-\log ^2(-5 x)+\left (10 e^3+2 \log (-5 x)\right ) \log (x)+2 e^6 x^2 \log ^2(x)}{e^6 x \log ^2(x)} \, dx=-\frac {{\left (i \, \pi + \log \left (5\right )\right )}^{2} - {\left (i \, \pi + \log \left (5\right )\right )} {\left (x^{2} e^{6} - 10 \, e^{3}\right )} + {\left (-i \, \pi + x^{2} e^{6} - \log \left (5\right )\right )} \log \left (-5 \, x\right ) + \log \left (-5 \, x\right )^{2} + 25 \, e^{6}}{{\left (i \, \pi + \log \left (5\right )\right )} e^{6} - e^{6} \log \left (-5 \, x\right )} \]
integrate((2*x^2*exp(3)^2*log(x)^2+(2*log(-5*x)+10*exp(3))*log(x)-log(-5*x )^2-10*exp(3)*log(-5*x)-25*exp(3)^2)/x/exp(3)^2/log(x)^2,x, algorithm=\
-((I*pi + log(5))^2 - (I*pi + log(5))*(x^2*e^6 - 10*e^3) + (-I*pi + x^2*e^ 6 - log(5))*log(-5*x) + log(-5*x)^2 + 25*e^6)/((I*pi + log(5))*e^6 - e^6*l og(-5*x))
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.76 \[ \int \frac {-25 e^6-10 e^3 \log (-5 x)-\log ^2(-5 x)+\left (10 e^3+2 \log (-5 x)\right ) \log (x)+2 e^6 x^2 \log ^2(x)}{e^6 x \log ^2(x)} \, dx=\frac {x^{2} e^{6} + \log {\left (x \right )}}{e^{6}} + \frac {- \pi ^{2} + \log {\left (5 \right )}^{2} + 10 e^{3} \log {\left (5 \right )} + 25 e^{6} + 2 i \pi \log {\left (5 \right )} + 10 i \pi e^{3}}{e^{6} \log {\left (x \right )}} \]
integrate((2*x**2*exp(3)**2*ln(x)**2+(2*ln(-5*x)+10*exp(3))*ln(x)-ln(-5*x) **2-10*exp(3)*ln(-5*x)-25*exp(3)**2)/x/exp(3)**2/ln(x)**2,x)
(x**2*exp(6) + log(x))*exp(-6) + (-pi**2 + log(5)**2 + 10*exp(3)*log(5) + 25*exp(6) + 2*I*pi*log(5) + 10*I*pi*exp(3))*exp(-6)/log(x)
Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {-25 e^6-10 e^3 \log (-5 x)-\log ^2(-5 x)+\left (10 e^3+2 \log (-5 x)\right ) \log (x)+2 e^6 x^2 \log ^2(x)}{e^6 x \log ^2(x)} \, dx={\left (x^{2} e^{6} + \frac {10 \, e^{3} \log \left (-5 \, x\right )}{\log \left (x\right )} + \frac {\log \left (-5 \, x\right )^{2}}{\log \left (x\right )} + \frac {25 \, e^{6}}{\log \left (x\right )}\right )} e^{\left (-6\right )} \]
integrate((2*x^2*exp(3)^2*log(x)^2+(2*log(-5*x)+10*exp(3))*log(x)-log(-5*x )^2-10*exp(3)*log(-5*x)-25*exp(3)^2)/x/exp(3)^2/log(x)^2,x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (20) = 40\).
Time = 0.26 (sec) , antiderivative size = 139, normalized size of antiderivative = 6.62 \[ \int \frac {-25 e^6-10 e^3 \log (-5 x)-\log ^2(-5 x)+\left (10 e^3+2 \log (-5 x)\right ) \log (x)+2 e^6 x^2 \log ^2(x)}{e^6 x \log ^2(x)} \, dx=\frac {{\left (\pi ^{2} x^{2} e^{6} \mathrm {sgn}\left (x\right ) - \pi ^{2} x^{2} e^{6} - 2 \, x^{2} e^{6} \log \left ({\left | x \right |}\right )^{2} + 20 \, \pi ^{2} e^{3} \mathrm {sgn}\left (x\right ) + 4 \, \pi ^{2} \log \left (5\right ) \mathrm {sgn}\left (x\right ) + \pi ^{2} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (x\right ) - 20 \, \pi ^{2} e^{3} - 4 \, \pi ^{2} \log \left (5\right ) + 7 \, \pi ^{2} \log \left ({\left | x \right |}\right ) - 20 \, e^{3} \log \left (5\right ) \log \left ({\left | x \right |}\right ) - 2 \, \log \left (5\right )^{2} \log \left ({\left | x \right |}\right ) - 2 \, \log \left ({\left | x \right |}\right )^{3} - 50 \, e^{6} \log \left ({\left | x \right |}\right )\right )} e^{\left (-6\right )}}{\pi ^{2} \mathrm {sgn}\left (x\right ) - \pi ^{2} - 2 \, \log \left ({\left | x \right |}\right )^{2}} \]
integrate((2*x^2*exp(3)^2*log(x)^2+(2*log(-5*x)+10*exp(3))*log(x)-log(-5*x )^2-10*exp(3)*log(-5*x)-25*exp(3)^2)/x/exp(3)^2/log(x)^2,x, algorithm=\
(pi^2*x^2*e^6*sgn(x) - pi^2*x^2*e^6 - 2*x^2*e^6*log(abs(x))^2 + 20*pi^2*e^ 3*sgn(x) + 4*pi^2*log(5)*sgn(x) + pi^2*log(abs(x))*sgn(x) - 20*pi^2*e^3 - 4*pi^2*log(5) + 7*pi^2*log(abs(x)) - 20*e^3*log(5)*log(abs(x)) - 2*log(5)^ 2*log(abs(x)) - 2*log(abs(x))^3 - 50*e^6*log(abs(x)))*e^(-6)/(pi^2*sgn(x) - pi^2 - 2*log(abs(x))^2)
Time = 8.78 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.33 \[ \int \frac {-25 e^6-10 e^3 \log (-5 x)-\log ^2(-5 x)+\left (10 e^3+2 \log (-5 x)\right ) \log (x)+2 e^6 x^2 \log ^2(x)}{e^6 x \log ^2(x)} \, dx=2\,{\mathrm {e}}^{-6}\,\ln \left (x\right )-{\mathrm {e}}^{-6}\,\left (2\,\ln \left (x\right )-x^2\,{\mathrm {e}}^6\right )+\frac {{\mathrm {e}}^{-6}\,\left ({\ln \left (-5\,x\right )}^2+10\,{\mathrm {e}}^3\,\ln \left (-5\,x\right )+25\,{\mathrm {e}}^6\right )}{\ln \left (x\right )} \]
int(-(exp(-6)*(25*exp(6) + 10*log(-5*x)*exp(3) - log(x)*(2*log(-5*x) + 10* exp(3)) + log(-5*x)^2 - 2*x^2*exp(6)*log(x)^2))/(x*log(x)^2),x)