Integrand size = 102, antiderivative size = 30 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {x^2}{\left (-\frac {5}{e}+\frac {x^3 (4+x)}{5 \log (4)}\right ) \log (x)} \]
Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {5 e x^2 \log (4)}{\left (e x^3 (4+x)-25 \log (4)\right ) \log (x)} \]
Integrate[(E^2*(-20*x^4 - 5*x^5)*Log[4] + 125*E*x*Log[4]^2 + (E^2*(-20*x^4 - 10*x^5)*Log[4] - 250*E*x*Log[4]^2)*Log[x])/((E^2*(16*x^6 + 8*x^7 + x^8) + E*(-200*x^3 - 50*x^4)*Log[4] + 625*Log[4]^2)*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.
\(\displaystyle \int \frac {\left (e^2 \left (-10 x^5-20 x^4\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)+e^2 \left (-5 x^5-20 x^4\right ) \log (4)+125 e x \log ^2(4)}{\left (e \left (-50 x^4-200 x^3\right ) \log (4)+e^2 \left (x^8+8 x^7+16 x^6\right )+625 \log ^2(4)\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \frac {\left (e^2 \left (-10 x^5-20 x^4\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)+e^2 \left (-5 x^5-20 x^4\right ) \log (4)+125 e x \log ^2(4)}{\left (e x^4+4 e x^3-25 \log (4)\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 e x \log (4) \left (-e x^4-2 e x^4 \log (x)-4 e x^3-4 e x^3 \log (x)-50 \log (4) \log (x)+25 \log (4)\right )}{\left (e x^4+4 e x^3-25 \log (4)\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 e \log (4) \int -\frac {x \left (2 e \log (x) x^4+e x^4+4 e \log (x) x^3+4 e x^3+50 \log (4) \log (x)-25 \log (4)\right )}{\left (e x^4+4 e x^3-25 \log (4)\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -5 e \log (4) \int \frac {x \left (2 e \log (x) x^4+e x^4+4 e \log (x) x^3+4 e x^3+50 \log (4) \log (x)-25 \log (4)\right )}{\left (e x^4+4 e x^3-25 \log (4)\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -5 e \log (4) \int \left (\frac {2 \left (e x^4+2 e x^3+25 \log (4)\right ) x}{\left (e x^4+4 e x^3-25 \log (4)\right )^2 \log (x)}+\frac {x}{\left (e x^4+4 e x^3-25 \log (4)\right ) \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 e \log (4) \left (\int \frac {x}{\left (e x^4+4 e x^3-25 \log (4)\right ) \log ^2(x)}dx+2 \int \frac {x \left (e x^4+2 e x^3+25 \log (4)\right )}{\left (e x^4+4 e x^3-25 \log (4)\right )^2 \log (x)}dx\right )\) |
Int[(E^2*(-20*x^4 - 5*x^5)*Log[4] + 125*E*x*Log[4]^2 + (E^2*(-20*x^4 - 10* x^5)*Log[4] - 250*E*x*Log[4]^2)*Log[x])/((E^2*(16*x^6 + 8*x^7 + x^8) + E*( -200*x^3 - 50*x^4)*Log[4] + 625*Log[4]^2)*Log[x]^2),x]
3.28.40.3.1 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]]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 1.85 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17
method | result | size |
default | \(-\frac {10 \,{\mathrm e} \ln \left (2\right ) x^{2}}{\left (-x^{4} {\mathrm e}-4 x^{3} {\mathrm e}+50 \ln \left (2\right )\right ) \ln \left (x \right )}\) | \(35\) |
risch | \(-\frac {10 \,{\mathrm e} \ln \left (2\right ) x^{2}}{\left (-x^{4} {\mathrm e}-4 x^{3} {\mathrm e}+50 \ln \left (2\right )\right ) \ln \left (x \right )}\) | \(35\) |
parallelrisch | \(-\frac {10 \,{\mathrm e} \ln \left (2\right ) x^{2}}{\left (-x^{4} {\mathrm e}-4 x^{3} {\mathrm e}+50 \ln \left (2\right )\right ) \ln \left (x \right )}\) | \(35\) |
int(((-1000*ln(2)^2*exp(1)*x+2*(-10*x^5-20*x^4)*exp(1)^2*ln(2))*ln(x)+500* ln(2)^2*exp(1)*x+2*(-5*x^5-20*x^4)*exp(1)^2*ln(2))/(2500*ln(2)^2+2*(-50*x^ 4-200*x^3)*exp(1)*ln(2)+(x^8+8*x^7+16*x^6)*exp(1)^2)/ln(x)^2,x,method=_RET URNVERBOSE)
Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {10 \, x^{2} e \log \left (2\right )}{{\left ({\left (x^{4} + 4 \, x^{3}\right )} e - 50 \, \log \left (2\right )\right )} \log \left (x\right )} \]
integrate(((-1000*log(2)^2*exp(1)*x+2*(-10*x^5-20*x^4)*exp(1)^2*log(2))*lo g(x)+500*log(2)^2*exp(1)*x+2*(-5*x^5-20*x^4)*exp(1)^2*log(2))/(2500*log(2) ^2+2*(-50*x^4-200*x^3)*exp(1)*log(2)+(x^8+8*x^7+16*x^6)*exp(1)^2)/log(x)^2 ,x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {10 e x^{2} \log {\left (2 \right )}}{\left (e x^{4} + 4 e x^{3} - 50 \log {\left (2 \right )}\right ) \log {\left (x \right )}} \]
integrate(((-1000*ln(2)**2*exp(1)*x+2*(-10*x**5-20*x**4)*exp(1)**2*ln(2))* ln(x)+500*ln(2)**2*exp(1)*x+2*(-5*x**5-20*x**4)*exp(1)**2*ln(2))/(2500*ln( 2)**2+2*(-50*x**4-200*x**3)*exp(1)*ln(2)+(x**8+8*x**7+16*x**6)*exp(1)**2)/ ln(x)**2,x)
Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {10 \, x^{2} e \log \left (2\right )}{{\left (x^{4} e + 4 \, x^{3} e - 50 \, \log \left (2\right )\right )} \log \left (x\right )} \]
integrate(((-1000*log(2)^2*exp(1)*x+2*(-10*x^5-20*x^4)*exp(1)^2*log(2))*lo g(x)+500*log(2)^2*exp(1)*x+2*(-5*x^5-20*x^4)*exp(1)^2*log(2))/(2500*log(2) ^2+2*(-50*x^4-200*x^3)*exp(1)*log(2)+(x^8+8*x^7+16*x^6)*exp(1)^2)/log(x)^2 ,x, algorithm=\
Timed out. \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\text {Timed out} \]
integrate(((-1000*log(2)^2*exp(1)*x+2*(-10*x^5-20*x^4)*exp(1)^2*log(2))*lo g(x)+500*log(2)^2*exp(1)*x+2*(-5*x^5-20*x^4)*exp(1)^2*log(2))/(2500*log(2) ^2+2*(-50*x^4-200*x^3)*exp(1)*log(2)+(x^8+8*x^7+16*x^6)*exp(1)^2)/log(x)^2 ,x, algorithm=\
Time = 14.95 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^2 \left (-20 x^4-5 x^5\right ) \log (4)+125 e x \log ^2(4)+\left (e^2 \left (-20 x^4-10 x^5\right ) \log (4)-250 e x \log ^2(4)\right ) \log (x)}{\left (e^2 \left (16 x^6+8 x^7+x^8\right )+e \left (-200 x^3-50 x^4\right ) \log (4)+625 \log ^2(4)\right ) \log ^2(x)} \, dx=\frac {10\,x^2\,\mathrm {e}\,\ln \left (2\right )}{\ln \left (x\right )\,\left (\mathrm {e}\,x^4+4\,\mathrm {e}\,x^3-50\,\ln \left (2\right )\right )} \]