Integrand size = 88, antiderivative size = 27 \[ \int \frac {e^{\frac {16 \log ^4(2)+24 \log ^2(2) \log ^2(x)+9 \log ^4(x)}{\log ^4(x)}} \left (-320 x \log ^4(2)-240 x \log ^2(2) \log ^2(x)+5 x \log ^5(x)+\left (-64 \log ^4(2)-48 \log ^2(2) \log ^2(x)\right ) \log (\log (5))\right )}{2 x \log ^5(x)} \, dx=\frac {1}{2} e^{\left (3+\frac {4 \log ^2(2)}{\log ^2(x)}\right )^2} (5 x+\log (\log (5))) \] Output:
1/2*(5*x+ln(ln(5)))*exp((4*ln(2)^2/ln(x)^2+3)^2)
Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {e^{\frac {16 \log ^4(2)+24 \log ^2(2) \log ^2(x)+9 \log ^4(x)}{\log ^4(x)}} \left (-320 x \log ^4(2)-240 x \log ^2(2) \log ^2(x)+5 x \log ^5(x)+\left (-64 \log ^4(2)-48 \log ^2(2) \log ^2(x)\right ) \log (\log (5))\right )}{2 x \log ^5(x)} \, dx=\frac {1}{2} e^{\frac {\left (4 \log ^2(2)+3 \log ^2(x)\right )^2}{\log ^4(x)}} (5 x+\log (\log (5))) \] Input:
Integrate[(E^((16*Log[2]^4 + 24*Log[2]^2*Log[x]^2 + 9*Log[x]^4)/Log[x]^4)* (-320*x*Log[2]^4 - 240*x*Log[2]^2*Log[x]^2 + 5*x*Log[x]^5 + (-64*Log[2]^4 - 48*Log[2]^2*Log[x]^2)*Log[Log[5]]))/(2*x*Log[x]^5),x]
Output:
(E^((4*Log[2]^2 + 3*Log[x]^2)^2/Log[x]^4)*(5*x + Log[Log[5]]))/2
Leaf count is larger than twice the leaf count of optimal. \(141\) vs. \(2(27)=54\).
Time = 0.67 (sec) , antiderivative size = 141, normalized size of antiderivative = 5.22, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {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.
| \(\displaystyle \int \frac {\left (5 x \log ^5(x)-320 x \log ^4(2)-240 x \log ^2(2) \log ^2(x)+\log (\log (5)) \left (-48 \log ^2(2) \log ^2(x)-64 \log ^4(2)\right )\right ) \exp \left (\frac {9 \log ^4(x)+24 \log ^2(2) \log ^2(x)+16 \log ^4(2)}{\log ^4(x)}\right )}{2 x \log ^5(x)} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int -\frac {\exp \left (\frac {9 \log ^4(x)+24 \log ^2(2) \log ^2(x)+16 \log ^4(2)}{\log ^4(x)}\right ) \left (-5 x \log ^5(x)+240 x \log ^2(2) \log ^2(x)+16 \left (3 \log ^2(2) \log ^2(x)+4 \log ^4(2)\right ) \log (\log (5))+320 x \log ^4(2)\right )}{x \log ^5(x)}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{2} \int \frac {\exp \left (\frac {9 \log ^4(x)+24 \log ^2(2) \log ^2(x)+16 \log ^4(2)}{\log ^4(x)}\right ) \left (-5 x \log ^5(x)+240 x \log ^2(2) \log ^2(x)+16 \left (3 \log ^2(2) \log ^2(x)+4 \log ^4(2)\right ) \log (\log (5))+320 x \log ^4(2)\right )}{x \log ^5(x)}dx\) |
| \(\Big \downarrow \) 2726 |
| \(\displaystyle -\frac {2 \left (20 x \log ^4(2)+15 x \log ^2(2) \log ^2(x)+\log (\log (5)) \left (3 \log ^2(2) \log ^2(x)+4 \log ^4(2)\right )\right ) \exp \left (\frac {9 \log ^4(x)+24 \log ^2(2) \log ^2(x)+16 \log ^4(2)}{\log ^4(x)}\right )}{x \log ^5(x) \left (\frac {3 \left (\frac {3 \log ^3(x)}{x}+\frac {4 \log ^2(2) \log (x)}{x}\right )}{\log ^4(x)}-\frac {9 \log ^4(x)+24 \log ^2(2) \log ^2(x)+16 \log ^4(2)}{x \log ^5(x)}\right )}\) |
Input:
Int[(E^((16*Log[2]^4 + 24*Log[2]^2*Log[x]^2 + 9*Log[x]^4)/Log[x]^4)*(-320* x*Log[2]^4 - 240*x*Log[2]^2*Log[x]^2 + 5*x*Log[x]^5 + (-64*Log[2]^4 - 48*L og[2]^2*Log[x]^2)*Log[Log[5]]))/(2*x*Log[x]^5),x]
Output:
(-2*E^((16*Log[2]^4 + 24*Log[2]^2*Log[x]^2 + 9*Log[x]^4)/Log[x]^4)*(20*x*L og[2]^4 + 15*x*Log[2]^2*Log[x]^2 + (4*Log[2]^4 + 3*Log[2]^2*Log[x]^2)*Log[ Log[5]]))/(x*Log[x]^5*((3*((4*Log[2]^2*Log[x])/x + (3*Log[x]^3)/x))/Log[x] ^4 - (16*Log[2]^4 + 24*Log[2]^2*Log[x]^2 + 9*Log[x]^4)/(x*Log[x]^5)))
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 = 6.86 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15
| method | result | size |
| risch | \(\frac {\left (5 x +\ln \left (\ln \left (5\right )\right )\right ) {\mathrm e}^{\frac {\left (4 \ln \left (2\right )^{2}+3 \ln \left (x \right )^{2}\right )^{2}}{\ln \left (x \right )^{4}}}}{2}\) | \(31\) |
| parallelrisch | \(\frac {\ln \left (\ln \left (5\right )\right ) {\mathrm e}^{\frac {9 \ln \left (x \right )^{4}+24 \ln \left (2\right )^{2} \ln \left (x \right )^{2}+16 \ln \left (2\right )^{4}}{\ln \left (x \right )^{4}}}}{2}+\frac {5 \,{\mathrm e}^{\frac {9 \ln \left (x \right )^{4}+24 \ln \left (2\right )^{2} \ln \left (x \right )^{2}+16 \ln \left (2\right )^{4}}{\ln \left (x \right )^{4}}} x}{2}\) | \(68\) |
Input:
int(1/2*((-48*ln(2)^2*ln(x)^2-64*ln(2)^4)*ln(ln(5))+5*x*ln(x)^5-240*x*ln(2 )^2*ln(x)^2-320*x*ln(2)^4)*exp((9*ln(x)^4+24*ln(2)^2*ln(x)^2+16*ln(2)^4)/l n(x)^4)/x/ln(x)^5,x,method=_RETURNVERBOSE)
Output:
1/2*(5*x+ln(ln(5)))*exp((4*ln(2)^2+3*ln(x)^2)^2/ln(x)^4)
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {e^{\frac {16 \log ^4(2)+24 \log ^2(2) \log ^2(x)+9 \log ^4(x)}{\log ^4(x)}} \left (-320 x \log ^4(2)-240 x \log ^2(2) \log ^2(x)+5 x \log ^5(x)+\left (-64 \log ^4(2)-48 \log ^2(2) \log ^2(x)\right ) \log (\log (5))\right )}{2 x \log ^5(x)} \, dx=\frac {1}{2} \, {\left (5 \, x + \log \left (\log \left (5\right )\right )\right )} e^{\left (\frac {16 \, \log \left (2\right )^{4} + 24 \, \log \left (2\right )^{2} \log \left (x\right )^{2} + 9 \, \log \left (x\right )^{4}}{\log \left (x\right )^{4}}\right )} \] Input:
integrate(1/2*((-48*log(2)^2*log(x)^2-64*log(2)^4)*log(log(5))+5*x*log(x)^ 5-240*x*log(2)^2*log(x)^2-320*x*log(2)^4)*exp((9*log(x)^4+24*log(2)^2*log( x)^2+16*log(2)^4)/log(x)^4)/x/log(x)^5,x, algorithm="fricas")
Output:
1/2*(5*x + log(log(5)))*e^((16*log(2)^4 + 24*log(2)^2*log(x)^2 + 9*log(x)^ 4)/log(x)^4)
Time = 0.98 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {e^{\frac {16 \log ^4(2)+24 \log ^2(2) \log ^2(x)+9 \log ^4(x)}{\log ^4(x)}} \left (-320 x \log ^4(2)-240 x \log ^2(2) \log ^2(x)+5 x \log ^5(x)+\left (-64 \log ^4(2)-48 \log ^2(2) \log ^2(x)\right ) \log (\log (5))\right )}{2 x \log ^5(x)} \, dx=\frac {\left (5 x + \log {\left (\log {\left (5 \right )} \right )}\right ) e^{\frac {9 \log {\left (x \right )}^{4} + 24 \log {\left (2 \right )}^{2} \log {\left (x \right )}^{2} + 16 \log {\left (2 \right )}^{4}}{\log {\left (x \right )}^{4}}}}{2} \] Input:
integrate(1/2*((-48*ln(2)**2*ln(x)**2-64*ln(2)**4)*ln(ln(5))+5*x*ln(x)**5- 240*x*ln(2)**2*ln(x)**2-320*x*ln(2)**4)*exp((9*ln(x)**4+24*ln(2)**2*ln(x)* *2+16*ln(2)**4)/ln(x)**4)/x/ln(x)**5,x)
Output:
(5*x + log(log(5)))*exp((9*log(x)**4 + 24*log(2)**2*log(x)**2 + 16*log(2)* *4)/log(x)**4)/2
\[ \int \frac {e^{\frac {16 \log ^4(2)+24 \log ^2(2) \log ^2(x)+9 \log ^4(x)}{\log ^4(x)}} \left (-320 x \log ^4(2)-240 x \log ^2(2) \log ^2(x)+5 x \log ^5(x)+\left (-64 \log ^4(2)-48 \log ^2(2) \log ^2(x)\right ) \log (\log (5))\right )}{2 x \log ^5(x)} \, dx=\int { \frac {{\left (5 \, x \log \left (x\right )^{5} - 320 \, x \log \left (2\right )^{4} - 240 \, x \log \left (2\right )^{2} \log \left (x\right )^{2} - 16 \, {\left (4 \, \log \left (2\right )^{4} + 3 \, \log \left (2\right )^{2} \log \left (x\right )^{2}\right )} \log \left (\log \left (5\right )\right )\right )} e^{\left (\frac {16 \, \log \left (2\right )^{4} + 24 \, \log \left (2\right )^{2} \log \left (x\right )^{2} + 9 \, \log \left (x\right )^{4}}{\log \left (x\right )^{4}}\right )}}{2 \, x \log \left (x\right )^{5}} \,d x } \] Input:
integrate(1/2*((-48*log(2)^2*log(x)^2-64*log(2)^4)*log(log(5))+5*x*log(x)^ 5-240*x*log(2)^2*log(x)^2-320*x*log(2)^4)*exp((9*log(x)^4+24*log(2)^2*log( x)^2+16*log(2)^4)/log(x)^4)/x/log(x)^5,x, algorithm="maxima")
Output:
1/2*integrate((5*x*log(x)^5 - 320*x*log(2)^4 - 240*x*log(2)^2*log(x)^2 - 1 6*(4*log(2)^4 + 3*log(2)^2*log(x)^2)*log(log(5)))*e^((16*log(2)^4 + 24*log (2)^2*log(x)^2 + 9*log(x)^4)/log(x)^4)/(x*log(x)^5), x)
\[ \int \frac {e^{\frac {16 \log ^4(2)+24 \log ^2(2) \log ^2(x)+9 \log ^4(x)}{\log ^4(x)}} \left (-320 x \log ^4(2)-240 x \log ^2(2) \log ^2(x)+5 x \log ^5(x)+\left (-64 \log ^4(2)-48 \log ^2(2) \log ^2(x)\right ) \log (\log (5))\right )}{2 x \log ^5(x)} \, dx=\int { \frac {{\left (5 \, x \log \left (x\right )^{5} - 320 \, x \log \left (2\right )^{4} - 240 \, x \log \left (2\right )^{2} \log \left (x\right )^{2} - 16 \, {\left (4 \, \log \left (2\right )^{4} + 3 \, \log \left (2\right )^{2} \log \left (x\right )^{2}\right )} \log \left (\log \left (5\right )\right )\right )} e^{\left (\frac {16 \, \log \left (2\right )^{4} + 24 \, \log \left (2\right )^{2} \log \left (x\right )^{2} + 9 \, \log \left (x\right )^{4}}{\log \left (x\right )^{4}}\right )}}{2 \, x \log \left (x\right )^{5}} \,d x } \] Input:
integrate(1/2*((-48*log(2)^2*log(x)^2-64*log(2)^4)*log(log(5))+5*x*log(x)^ 5-240*x*log(2)^2*log(x)^2-320*x*log(2)^4)*exp((9*log(x)^4+24*log(2)^2*log( x)^2+16*log(2)^4)/log(x)^4)/x/log(x)^5,x, algorithm="giac")
Output:
undef
Time = 7.59 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11 \[ \int \frac {e^{\frac {16 \log ^4(2)+24 \log ^2(2) \log ^2(x)+9 \log ^4(x)}{\log ^4(x)}} \left (-320 x \log ^4(2)-240 x \log ^2(2) \log ^2(x)+5 x \log ^5(x)+\left (-64 \log ^4(2)-48 \log ^2(2) \log ^2(x)\right ) \log (\log (5))\right )}{2 x \log ^5(x)} \, dx=\frac {{\mathrm {e}}^9\,{\mathrm {e}}^{\frac {16\,{\ln \left (2\right )}^4}{{\ln \left (x\right )}^4}}\,{\mathrm {e}}^{\frac {24\,{\ln \left (2\right )}^2}{{\ln \left (x\right )}^2}}\,\ln \left (\ln \left (5\right )\right )}{2}+\frac {5\,x\,{\mathrm {e}}^9\,{\mathrm {e}}^{\frac {16\,{\ln \left (2\right )}^4}{{\ln \left (x\right )}^4}}\,{\mathrm {e}}^{\frac {24\,{\ln \left (2\right )}^2}{{\ln \left (x\right )}^2}}}{2} \] Input:
int(-(exp((9*log(x)^4 + 24*log(2)^2*log(x)^2 + 16*log(2)^4)/log(x)^4)*(log (log(5))*(48*log(2)^2*log(x)^2 + 64*log(2)^4) - 5*x*log(x)^5 + 320*x*log(2 )^4 + 240*x*log(2)^2*log(x)^2))/(2*x*log(x)^5),x)
Output:
(exp(9)*exp((16*log(2)^4)/log(x)^4)*exp((24*log(2)^2)/log(x)^2)*log(log(5) ))/2 + (5*x*exp(9)*exp((16*log(2)^4)/log(x)^4)*exp((24*log(2)^2)/log(x)^2) )/2
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{\frac {16 \log ^4(2)+24 \log ^2(2) \log ^2(x)+9 \log ^4(x)}{\log ^4(x)}} \left (-320 x \log ^4(2)-240 x \log ^2(2) \log ^2(x)+5 x \log ^5(x)+\left (-64 \log ^4(2)-48 \log ^2(2) \log ^2(x)\right ) \log (\log (5))\right )}{2 x \log ^5(x)} \, dx=\frac {e^{\frac {24 \mathrm {log}\left (x \right )^{2} \mathrm {log}\left (2\right )^{2}+16 \mathrm {log}\left (2\right )^{4}}{\mathrm {log}\left (x \right )^{4}}} e^{9} \left (\mathrm {log}\left (\mathrm {log}\left (5\right )\right )+5 x \right )}{2} \] Input:
int(1/2*((-48*log(2)^2*log(x)^2-64*log(2)^4)*log(log(5))+5*x*log(x)^5-240* x*log(2)^2*log(x)^2-320*x*log(2)^4)*exp((9*log(x)^4+24*log(2)^2*log(x)^2+1 6*log(2)^4)/log(x)^4)/x/log(x)^5,x)
Output:
(e**((24*log(x)**2*log(2)**2 + 16*log(2)**4)/log(x)**4)*e**9*(log(log(5)) + 5*x))/2