Integrand size = 195, antiderivative size = 27 \[ \int \frac {2-15 x-15 e^4 x+30 x^2+\left (e^4-2 x\right ) \log \left (x^2\right )+\left (-15 x+\log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )}{\left (15 e^4 x^2-15 x^3\right ) \log ^2(5)+\left (-e^4 x+x^2\right ) \log ^2(5) \log \left (x^2\right )+\left (15 x^2 \log ^2(5)-x \log ^2(5) \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )+\left (-15 e^4 x^2+15 x^3+\left (e^4 x-x^2\right ) \log \left (x^2\right )+\left (-15 x^2+x \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )\right ) \log \left (e^4 x-x^2+x \log \left (-15 x+\log \left (x^2\right )\right )\right )} \, dx=\log \left (\log ^2(5)-\log \left (x \left (e^4-x+\log \left (-15 x+\log \left (x^2\right )\right )\right )\right )\right ) \]
Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {2-15 x-15 e^4 x+30 x^2+\left (e^4-2 x\right ) \log \left (x^2\right )+\left (-15 x+\log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )}{\left (15 e^4 x^2-15 x^3\right ) \log ^2(5)+\left (-e^4 x+x^2\right ) \log ^2(5) \log \left (x^2\right )+\left (15 x^2 \log ^2(5)-x \log ^2(5) \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )+\left (-15 e^4 x^2+15 x^3+\left (e^4 x-x^2\right ) \log \left (x^2\right )+\left (-15 x^2+x \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )\right ) \log \left (e^4 x-x^2+x \log \left (-15 x+\log \left (x^2\right )\right )\right )} \, dx=\log \left (\log ^2(5)-\log \left (x \left (e^4-x+\log \left (-15 x+\log \left (x^2\right )\right )\right )\right )\right ) \]
Integrate[(2 - 15*x - 15*E^4*x + 30*x^2 + (E^4 - 2*x)*Log[x^2] + (-15*x + Log[x^2])*Log[-15*x + Log[x^2]])/((15*E^4*x^2 - 15*x^3)*Log[5]^2 + (-(E^4* x) + x^2)*Log[5]^2*Log[x^2] + (15*x^2*Log[5]^2 - x*Log[5]^2*Log[x^2])*Log[ -15*x + Log[x^2]] + (-15*E^4*x^2 + 15*x^3 + (E^4*x - x^2)*Log[x^2] + (-15* x^2 + x*Log[x^2])*Log[-15*x + Log[x^2]])*Log[E^4*x - x^2 + x*Log[-15*x + L og[x^2]]]),x]
Time = 0.71 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {6, 7239, 7235}
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.
| \(\displaystyle \int \frac {30 x^2+\left (e^4-2 x\right ) \log \left (x^2\right )+\left (\log \left (x^2\right )-15 x\right ) \log \left (\log \left (x^2\right )-15 x\right )-15 e^4 x-15 x+2}{\left (x^2-e^4 x\right ) \log ^2(5) \log \left (x^2\right )+\left (15 x^2 \log ^2(5)-x \log ^2(5) \log \left (x^2\right )\right ) \log \left (\log \left (x^2\right )-15 x\right )+\left (15 e^4 x^2-15 x^3\right ) \log ^2(5)+\left (15 x^3-15 e^4 x^2+\left (e^4 x-x^2\right ) \log \left (x^2\right )+\left (x \log \left (x^2\right )-15 x^2\right ) \log \left (\log \left (x^2\right )-15 x\right )\right ) \log \left (-x^2+x \log \left (\log \left (x^2\right )-15 x\right )+e^4 x\right )} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {30 x^2+\left (e^4-2 x\right ) \log \left (x^2\right )+\left (\log \left (x^2\right )-15 x\right ) \log \left (\log \left (x^2\right )-15 x\right )+\left (-15-15 e^4\right ) x+2}{\left (x^2-e^4 x\right ) \log ^2(5) \log \left (x^2\right )+\left (15 x^2 \log ^2(5)-x \log ^2(5) \log \left (x^2\right )\right ) \log \left (\log \left (x^2\right )-15 x\right )+\left (15 e^4 x^2-15 x^3\right ) \log ^2(5)+\left (15 x^3-15 e^4 x^2+\left (e^4 x-x^2\right ) \log \left (x^2\right )+\left (x \log \left (x^2\right )-15 x^2\right ) \log \left (\log \left (x^2\right )-15 x\right )\right ) \log \left (-x^2+x \log \left (\log \left (x^2\right )-15 x\right )+e^4 x\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {30 x^2-15 x \log \left (\log \left (x^2\right )-15 x\right )+\log \left (x^2\right ) \left (\log \left (\log \left (x^2\right )-15 x\right )-2 x+e^4\right )-15 \left (1+e^4\right ) x+2}{x \left (15 x-\log \left (x^2\right )\right ) \left (\log \left (\log \left (x^2\right )-15 x\right )-x+e^4\right ) \left (\log ^2(5)-\log \left (x \left (\log \left (\log \left (x^2\right )-15 x\right )-x+e^4\right )\right )\right )}dx\) |
| \(\Big \downarrow \) 7235 |
| \(\displaystyle \log \left (\log ^2(5)-\log \left (x \left (\log \left (\log \left (x^2\right )-15 x\right )-x+e^4\right )\right )\right )\) |
Int[(2 - 15*x - 15*E^4*x + 30*x^2 + (E^4 - 2*x)*Log[x^2] + (-15*x + Log[x^ 2])*Log[-15*x + Log[x^2]])/((15*E^4*x^2 - 15*x^3)*Log[5]^2 + (-(E^4*x) + x ^2)*Log[5]^2*Log[x^2] + (15*x^2*Log[5]^2 - x*Log[5]^2*Log[x^2])*Log[-15*x + Log[x^2]] + (-15*E^4*x^2 + 15*x^3 + (E^4*x - x^2)*Log[x^2] + (-15*x^2 + x*Log[x^2])*Log[-15*x + Log[x^2]])*Log[E^4*x - x^2 + x*Log[-15*x + Log[x^2 ]]]),x]
3.10.94.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 25.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
| method | result | size |
| parallelrisch | \(\ln \left (-\ln \left (5\right )^{2}+\ln \left (x \left ({\mathrm e}^{4}+\ln \left (\ln \left (x^{2}\right )-15 x \right )-x \right )\right )\right )\) | \(27\) |
int(((ln(x^2)-15*x)*ln(ln(x^2)-15*x)+(exp(4)-2*x)*ln(x^2)-15*x*exp(4)+30*x ^2-15*x+2)/(((x*ln(x^2)-15*x^2)*ln(ln(x^2)-15*x)+(x*exp(4)-x^2)*ln(x^2)-15 *x^2*exp(4)+15*x^3)*ln(x*ln(ln(x^2)-15*x)+x*exp(4)-x^2)+(-x*ln(5)^2*ln(x^2 )+15*x^2*ln(5)^2)*ln(ln(x^2)-15*x)+(-x*exp(4)+x^2)*ln(5)^2*ln(x^2)+(15*x^2 *exp(4)-15*x^3)*ln(5)^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {2-15 x-15 e^4 x+30 x^2+\left (e^4-2 x\right ) \log \left (x^2\right )+\left (-15 x+\log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )}{\left (15 e^4 x^2-15 x^3\right ) \log ^2(5)+\left (-e^4 x+x^2\right ) \log ^2(5) \log \left (x^2\right )+\left (15 x^2 \log ^2(5)-x \log ^2(5) \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )+\left (-15 e^4 x^2+15 x^3+\left (e^4 x-x^2\right ) \log \left (x^2\right )+\left (-15 x^2+x \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )\right ) \log \left (e^4 x-x^2+x \log \left (-15 x+\log \left (x^2\right )\right )\right )} \, dx=\log \left (-\log \left (5\right )^{2} + \log \left (-x^{2} + x e^{4} + x \log \left (-15 \, x + \log \left (x^{2}\right )\right )\right )\right ) \]
integrate(((log(x^2)-15*x)*log(log(x^2)-15*x)+(exp(4)-2*x)*log(x^2)-15*x*e xp(4)+30*x^2-15*x+2)/(((x*log(x^2)-15*x^2)*log(log(x^2)-15*x)+(x*exp(4)-x^ 2)*log(x^2)-15*x^2*exp(4)+15*x^3)*log(x*log(log(x^2)-15*x)+x*exp(4)-x^2)+( -x*log(5)^2*log(x^2)+15*x^2*log(5)^2)*log(log(x^2)-15*x)+(-x*exp(4)+x^2)*l og(5)^2*log(x^2)+(15*x^2*exp(4)-15*x^3)*log(5)^2),x, algorithm=\
Time = 18.55 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {2-15 x-15 e^4 x+30 x^2+\left (e^4-2 x\right ) \log \left (x^2\right )+\left (-15 x+\log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )}{\left (15 e^4 x^2-15 x^3\right ) \log ^2(5)+\left (-e^4 x+x^2\right ) \log ^2(5) \log \left (x^2\right )+\left (15 x^2 \log ^2(5)-x \log ^2(5) \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )+\left (-15 e^4 x^2+15 x^3+\left (e^4 x-x^2\right ) \log \left (x^2\right )+\left (-15 x^2+x \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )\right ) \log \left (e^4 x-x^2+x \log \left (-15 x+\log \left (x^2\right )\right )\right )} \, dx=\log {\left (\log {\left (- x^{2} + x \log {\left (- 15 x + \log {\left (x^{2} \right )} \right )} + x e^{4} \right )} - \log {\left (5 \right )}^{2} \right )} \]
integrate(((ln(x**2)-15*x)*ln(ln(x**2)-15*x)+(exp(4)-2*x)*ln(x**2)-15*x*ex p(4)+30*x**2-15*x+2)/(((x*ln(x**2)-15*x**2)*ln(ln(x**2)-15*x)+(x*exp(4)-x* *2)*ln(x**2)-15*x**2*exp(4)+15*x**3)*ln(x*ln(ln(x**2)-15*x)+x*exp(4)-x**2) +(-x*ln(5)**2*ln(x**2)+15*x**2*ln(5)**2)*ln(ln(x**2)-15*x)+(-x*exp(4)+x**2 )*ln(5)**2*ln(x**2)+(15*x**2*exp(4)-15*x**3)*ln(5)**2),x)
Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {2-15 x-15 e^4 x+30 x^2+\left (e^4-2 x\right ) \log \left (x^2\right )+\left (-15 x+\log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )}{\left (15 e^4 x^2-15 x^3\right ) \log ^2(5)+\left (-e^4 x+x^2\right ) \log ^2(5) \log \left (x^2\right )+\left (15 x^2 \log ^2(5)-x \log ^2(5) \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )+\left (-15 e^4 x^2+15 x^3+\left (e^4 x-x^2\right ) \log \left (x^2\right )+\left (-15 x^2+x \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )\right ) \log \left (e^4 x-x^2+x \log \left (-15 x+\log \left (x^2\right )\right )\right )} \, dx=\log \left (-\log \left (5\right )^{2} + \log \left (x\right ) + \log \left (-x + e^{4} + \log \left (-15 \, x + 2 \, \log \left (x\right )\right )\right )\right ) \]
integrate(((log(x^2)-15*x)*log(log(x^2)-15*x)+(exp(4)-2*x)*log(x^2)-15*x*e xp(4)+30*x^2-15*x+2)/(((x*log(x^2)-15*x^2)*log(log(x^2)-15*x)+(x*exp(4)-x^ 2)*log(x^2)-15*x^2*exp(4)+15*x^3)*log(x*log(log(x^2)-15*x)+x*exp(4)-x^2)+( -x*log(5)^2*log(x^2)+15*x^2*log(5)^2)*log(log(x^2)-15*x)+(-x*exp(4)+x^2)*l og(5)^2*log(x^2)+(15*x^2*exp(4)-15*x^3)*log(5)^2),x, algorithm=\
\[ \int \frac {2-15 x-15 e^4 x+30 x^2+\left (e^4-2 x\right ) \log \left (x^2\right )+\left (-15 x+\log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )}{\left (15 e^4 x^2-15 x^3\right ) \log ^2(5)+\left (-e^4 x+x^2\right ) \log ^2(5) \log \left (x^2\right )+\left (15 x^2 \log ^2(5)-x \log ^2(5) \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )+\left (-15 e^4 x^2+15 x^3+\left (e^4 x-x^2\right ) \log \left (x^2\right )+\left (-15 x^2+x \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )\right ) \log \left (e^4 x-x^2+x \log \left (-15 x+\log \left (x^2\right )\right )\right )} \, dx=\int { \frac {30 \, x^{2} - 15 \, x e^{4} - {\left (2 \, x - e^{4}\right )} \log \left (x^{2}\right ) - {\left (15 \, x - \log \left (x^{2}\right )\right )} \log \left (-15 \, x + \log \left (x^{2}\right )\right ) - 15 \, x + 2}{{\left (x^{2} - x e^{4}\right )} \log \left (5\right )^{2} \log \left (x^{2}\right ) - 15 \, {\left (x^{3} - x^{2} e^{4}\right )} \log \left (5\right )^{2} + {\left (15 \, x^{3} - 15 \, x^{2} e^{4} - {\left (x^{2} - x e^{4}\right )} \log \left (x^{2}\right ) - {\left (15 \, x^{2} - x \log \left (x^{2}\right )\right )} \log \left (-15 \, x + \log \left (x^{2}\right )\right )\right )} \log \left (-x^{2} + x e^{4} + x \log \left (-15 \, x + \log \left (x^{2}\right )\right )\right ) + {\left (15 \, x^{2} \log \left (5\right )^{2} - x \log \left (5\right )^{2} \log \left (x^{2}\right )\right )} \log \left (-15 \, x + \log \left (x^{2}\right )\right )} \,d x } \]
integrate(((log(x^2)-15*x)*log(log(x^2)-15*x)+(exp(4)-2*x)*log(x^2)-15*x*e xp(4)+30*x^2-15*x+2)/(((x*log(x^2)-15*x^2)*log(log(x^2)-15*x)+(x*exp(4)-x^ 2)*log(x^2)-15*x^2*exp(4)+15*x^3)*log(x*log(log(x^2)-15*x)+x*exp(4)-x^2)+( -x*log(5)^2*log(x^2)+15*x^2*log(5)^2)*log(log(x^2)-15*x)+(-x*exp(4)+x^2)*l og(5)^2*log(x^2)+(15*x^2*exp(4)-15*x^3)*log(5)^2),x, algorithm=\
integrate((30*x^2 - 15*x*e^4 - (2*x - e^4)*log(x^2) - (15*x - log(x^2))*lo g(-15*x + log(x^2)) - 15*x + 2)/((x^2 - x*e^4)*log(5)^2*log(x^2) - 15*(x^3 - x^2*e^4)*log(5)^2 + (15*x^3 - 15*x^2*e^4 - (x^2 - x*e^4)*log(x^2) - (15 *x^2 - x*log(x^2))*log(-15*x + log(x^2)))*log(-x^2 + x*e^4 + x*log(-15*x + log(x^2))) + (15*x^2*log(5)^2 - x*log(5)^2*log(x^2))*log(-15*x + log(x^2) )), x)
Time = 17.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {2-15 x-15 e^4 x+30 x^2+\left (e^4-2 x\right ) \log \left (x^2\right )+\left (-15 x+\log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )}{\left (15 e^4 x^2-15 x^3\right ) \log ^2(5)+\left (-e^4 x+x^2\right ) \log ^2(5) \log \left (x^2\right )+\left (15 x^2 \log ^2(5)-x \log ^2(5) \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )+\left (-15 e^4 x^2+15 x^3+\left (e^4 x-x^2\right ) \log \left (x^2\right )+\left (-15 x^2+x \log \left (x^2\right )\right ) \log \left (-15 x+\log \left (x^2\right )\right )\right ) \log \left (e^4 x-x^2+x \log \left (-15 x+\log \left (x^2\right )\right )\right )} \, dx=\ln \left (\ln \left (x\,\left ({\mathrm {e}}^4-x+\ln \left (\ln \left (x^2\right )-15\,x\right )\right )\right )-{\ln \left (5\right )}^2\right ) \]
int(-(15*x + log(log(x^2) - 15*x)*(15*x - log(x^2)) + 15*x*exp(4) + log(x^ 2)*(2*x - exp(4)) - 30*x^2 - 2)/(log(5)^2*(15*x^2*exp(4) - 15*x^3) + log(l og(x^2) - 15*x)*(15*x^2*log(5)^2 - x*log(x^2)*log(5)^2) + log(x*exp(4) + x *log(log(x^2) - 15*x) - x^2)*(log(x^2)*(x*exp(4) - x^2) - 15*x^2*exp(4) + 15*x^3 + log(log(x^2) - 15*x)*(x*log(x^2) - 15*x^2)) - log(x^2)*log(5)^2*( x*exp(4) - x^2)),x)