Integrand size = 72, antiderivative size = 24 \[ \int \frac {\left (324-162 e^4\right ) \log (4) \log (x)+\left (162-81 e^4\right ) \log (4) \log ^2(x)}{164025+(4050 x-810 x \log (4)) \log ^2(x)+\left (25 x^2-10 x^2 \log (4)+x^2 \log ^2(4)\right ) \log ^4(x)} \, dx=\frac {-2+e^4}{-5+\frac {x \log (4)}{x+\frac {81}{\log ^2(x)}}} \] Output:
(exp(4)-2)/(2/(x+81/ln(x)^2)*x*ln(2)-5)
Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\left (324-162 e^4\right ) \log (4) \log (x)+\left (162-81 e^4\right ) \log (4) \log ^2(x)}{164025+(4050 x-810 x \log (4)) \log ^2(x)+\left (25 x^2-10 x^2 \log (4)+x^2 \log ^2(4)\right ) \log ^4(x)} \, dx=\frac {81 \left (-2+e^4\right ) \log (4)}{(-5+\log (4)) \left (-405+x (-5+\log (4)) \log ^2(x)\right )} \] Input:
Integrate[((324 - 162*E^4)*Log[4]*Log[x] + (162 - 81*E^4)*Log[4]*Log[x]^2) /(164025 + (4050*x - 810*x*Log[4])*Log[x]^2 + (25*x^2 - 10*x^2*Log[4] + x^ 2*Log[4]^2)*Log[x]^4),x]
Output:
(81*(-2 + E^4)*Log[4])/((-5 + Log[4])*(-405 + x*(-5 + Log[4])*Log[x]^2))
Time = 0.82 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {7239, 27, 7237}
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.
\(\displaystyle \int \frac {\left (162-81 e^4\right ) \log (4) \log ^2(x)+\left (324-162 e^4\right ) \log (4) \log (x)}{\left (25 x^2+x^2 \log ^2(4)-10 x^2 \log (4)\right ) \log ^4(x)+(4050 x-810 x \log (4)) \log ^2(x)+164025} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {81 \left (2-e^4\right ) \log (4) \log (x) (\log (x)+2)}{\left (405-x (\log (4)-5) \log ^2(x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 81 \left (2-e^4\right ) \log (4) \int \frac {\log (x) (\log (x)+2)}{\left (x (5-\log (4)) \log ^2(x)+405\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle -\frac {81 \left (2-e^4\right ) \log (4)}{(5-\log (4)) \left (x (5-\log (4)) \log ^2(x)+405\right )}\) |
Input:
Int[((324 - 162*E^4)*Log[4]*Log[x] + (162 - 81*E^4)*Log[4]*Log[x]^2)/(1640 25 + (4050*x - 810*x*Log[4])*Log[x]^2 + (25*x^2 - 10*x^2*Log[4] + x^2*Log[ 4]^2)*Log[x]^4),x]
Output:
(-81*(2 - E^4)*Log[4])/((5 - Log[4])*(405 + x*(5 - Log[4])*Log[x]^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.56 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50
method | result | size |
default | \(-\frac {162 \ln \left (2\right ) \left (\frac {2}{405}-\frac {{\mathrm e}^{4}}{405}\right ) \ln \left (x \right )^{2} x}{2 \ln \left (x \right )^{2} \ln \left (2\right ) x -5 x \ln \left (x \right )^{2}-405}\) | \(36\) |
norman | \(\frac {\left (\frac {2 \,{\mathrm e}^{4} \ln \left (2\right )}{5}-\frac {4 \ln \left (2\right )}{5}\right ) x \ln \left (x \right )^{2}}{2 \ln \left (x \right )^{2} \ln \left (2\right ) x -5 x \ln \left (x \right )^{2}-405}\) | \(38\) |
parallelrisch | \(\frac {162 x \ln \left (x \right )^{2} {\mathrm e}^{4} \ln \left (2\right )-324 \ln \left (x \right )^{2} \ln \left (2\right ) x}{810 \ln \left (x \right )^{2} \ln \left (2\right ) x -2025 x \ln \left (x \right )^{2}-164025}\) | \(44\) |
risch | \(\frac {162 \ln \left (2\right ) {\mathrm e}^{4}}{\left (2 \ln \left (2\right )-5\right ) \left (2 \ln \left (x \right )^{2} \ln \left (2\right ) x -5 x \ln \left (x \right )^{2}-405\right )}-\frac {324 \ln \left (2\right )}{\left (2 \ln \left (2\right )-5\right ) \left (2 \ln \left (x \right )^{2} \ln \left (2\right ) x -5 x \ln \left (x \right )^{2}-405\right )}\) | \(68\) |
Input:
int((2*(-81*exp(4)+162)*ln(2)*ln(x)^2+2*(-162*exp(4)+324)*ln(2)*ln(x))/((4 *x^2*ln(2)^2-20*x^2*ln(2)+25*x^2)*ln(x)^4+(-1620*x*ln(2)+4050*x)*ln(x)^2+1 64025),x,method=_RETURNVERBOSE)
Output:
-162*ln(2)*(2/405-1/405*exp(4))*ln(x)^2*x/(2*ln(x)^2*ln(2)*x-5*x*ln(x)^2-4 05)
Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {\left (324-162 e^4\right ) \log (4) \log (x)+\left (162-81 e^4\right ) \log (4) \log ^2(x)}{164025+(4050 x-810 x \log (4)) \log ^2(x)+\left (25 x^2-10 x^2 \log (4)+x^2 \log ^2(4)\right ) \log ^4(x)} \, dx=\frac {162 \, {\left (e^{4} - 2\right )} \log \left (2\right )}{{\left (4 \, x \log \left (2\right )^{2} - 20 \, x \log \left (2\right ) + 25 \, x\right )} \log \left (x\right )^{2} - 810 \, \log \left (2\right ) + 2025} \] Input:
integrate((2*(-81*exp(4)+162)*log(2)*log(x)^2+2*(-162*exp(4)+324)*log(2)*l og(x))/((4*x^2*log(2)^2-20*x^2*log(2)+25*x^2)*log(x)^4+(-1620*x*log(2)+405 0*x)*log(x)^2+164025),x, algorithm="fricas")
Output:
162*(e^4 - 2)*log(2)/((4*x*log(2)^2 - 20*x*log(2) + 25*x)*log(x)^2 - 810*l og(2) + 2025)
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {\left (324-162 e^4\right ) \log (4) \log (x)+\left (162-81 e^4\right ) \log (4) \log ^2(x)}{164025+(4050 x-810 x \log (4)) \log ^2(x)+\left (25 x^2-10 x^2 \log (4)+x^2 \log ^2(4)\right ) \log ^4(x)} \, dx=\frac {- 324 \log {\left (2 \right )} + 162 e^{4} \log {\left (2 \right )}}{\left (- 20 x \log {\left (2 \right )} + 4 x \log {\left (2 \right )}^{2} + 25 x\right ) \log {\left (x \right )}^{2} - 810 \log {\left (2 \right )} + 2025} \] Input:
integrate((2*(-81*exp(4)+162)*ln(2)*ln(x)**2+2*(-162*exp(4)+324)*ln(2)*ln( x))/((4*x**2*ln(2)**2-20*x**2*ln(2)+25*x**2)*ln(x)**4+(-1620*x*ln(2)+4050* x)*ln(x)**2+164025),x)
Output:
(-324*log(2) + 162*exp(4)*log(2))/((-20*x*log(2) + 4*x*log(2)**2 + 25*x)*l og(x)**2 - 810*log(2) + 2025)
Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {\left (324-162 e^4\right ) \log (4) \log (x)+\left (162-81 e^4\right ) \log (4) \log ^2(x)}{164025+(4050 x-810 x \log (4)) \log ^2(x)+\left (25 x^2-10 x^2 \log (4)+x^2 \log ^2(4)\right ) \log ^4(x)} \, dx=\frac {162 \, {\left (e^{4} \log \left (2\right ) - 2 \, \log \left (2\right )\right )}}{{\left (4 \, \log \left (2\right )^{2} - 20 \, \log \left (2\right ) + 25\right )} x \log \left (x\right )^{2} - 810 \, \log \left (2\right ) + 2025} \] Input:
integrate((2*(-81*exp(4)+162)*log(2)*log(x)^2+2*(-162*exp(4)+324)*log(2)*l og(x))/((4*x^2*log(2)^2-20*x^2*log(2)+25*x^2)*log(x)^4+(-1620*x*log(2)+405 0*x)*log(x)^2+164025),x, algorithm="maxima")
Output:
162*(e^4*log(2) - 2*log(2))/((4*log(2)^2 - 20*log(2) + 25)*x*log(x)^2 - 81 0*log(2) + 2025)
Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {\left (324-162 e^4\right ) \log (4) \log (x)+\left (162-81 e^4\right ) \log (4) \log ^2(x)}{164025+(4050 x-810 x \log (4)) \log ^2(x)+\left (25 x^2-10 x^2 \log (4)+x^2 \log ^2(4)\right ) \log ^4(x)} \, dx=\frac {162 \, {\left (e^{4} \log \left (2\right ) - 2 \, \log \left (2\right )\right )}}{4 \, x \log \left (2\right )^{2} \log \left (x\right )^{2} - 20 \, x \log \left (2\right ) \log \left (x\right )^{2} + 25 \, x \log \left (x\right )^{2} - 810 \, \log \left (2\right ) + 2025} \] Input:
integrate((2*(-81*exp(4)+162)*log(2)*log(x)^2+2*(-162*exp(4)+324)*log(2)*l og(x))/((4*x^2*log(2)^2-20*x^2*log(2)+25*x^2)*log(x)^4+(-1620*x*log(2)+405 0*x)*log(x)^2+164025),x, algorithm="giac")
Output:
162*(e^4*log(2) - 2*log(2))/(4*x*log(2)^2*log(x)^2 - 20*x*log(2)*log(x)^2 + 25*x*log(x)^2 - 810*log(2) + 2025)
Time = 3.56 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88 \[ \int \frac {\left (324-162 e^4\right ) \log (4) \log (x)+\left (162-81 e^4\right ) \log (4) \log ^2(x)}{164025+(4050 x-810 x \log (4)) \log ^2(x)+\left (25 x^2-10 x^2 \log (4)+x^2 \log ^2(4)\right ) \log ^4(x)} \, dx=\frac {x^2\,{\ln \left (x\right )}^2\,\left (\frac {4\,\ln \left (2\right )}{5}-\frac {2\,{\mathrm {e}}^4\,\ln \left (2\right )}{5}\right )}{405\,x+5\,x^2\,{\ln \left (x\right )}^2-2\,x^2\,\ln \left (2\right )\,{\ln \left (x\right )}^2} \] Input:
int(-(2*log(2)*log(x)^2*(81*exp(4) - 162) + 2*log(2)*log(x)*(162*exp(4) - 324))/(log(x)^2*(4050*x - 1620*x*log(2)) + log(x)^4*(4*x^2*log(2)^2 - 20*x ^2*log(2) + 25*x^2) + 164025),x)
Output:
(x^2*log(x)^2*((4*log(2))/5 - (2*exp(4)*log(2))/5))/(405*x + 5*x^2*log(x)^ 2 - 2*x^2*log(2)*log(x)^2)
Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {\left (324-162 e^4\right ) \log (4) \log (x)+\left (162-81 e^4\right ) \log (4) \log ^2(x)}{164025+(4050 x-810 x \log (4)) \log ^2(x)+\left (25 x^2-10 x^2 \log (4)+x^2 \log ^2(4)\right ) \log ^4(x)} \, dx=\frac {2 \mathrm {log}\left (x \right )^{2} \mathrm {log}\left (2\right ) x \left (e^{4}-2\right )}{10 \mathrm {log}\left (x \right )^{2} \mathrm {log}\left (2\right ) x -25 \mathrm {log}\left (x \right )^{2} x -2025} \] Input:
int((2*(-81*exp(4)+162)*log(2)*log(x)^2+2*(-162*exp(4)+324)*log(2)*log(x)) /((4*x^2*log(2)^2-20*x^2*log(2)+25*x^2)*log(x)^4+(-1620*x*log(2)+4050*x)*l og(x)^2+164025),x)
Output:
(2*log(x)**2*log(2)*x*(e**4 - 2))/(5*(2*log(x)**2*log(2)*x - 5*log(x)**2*x - 405))