Integrand size = 84, antiderivative size = 32 \[ \int \frac {e^{52+10 x^2+8 \log ^2(x)-2 \log ^4(x)} \left (5 x^2+4 \log (x)-2 \log ^3(x)\right )+e^{26+5 x^2+4 \log ^2(x)-\log ^4(x)} \left (-20 x^2-16 \log (x)+8 \log ^3(x)\right )}{4 x} \, dx=\frac {1}{16} \left (4-e^{5 \left (6+x^2\right )-\left (2-\log ^2(x)\right )^2}\right )^2 \] Output:
1/4*(4-exp(5*x^2+30-(2-ln(x)^2)^2))*(1-1/4*exp(5*x^2+30-(2-ln(x)^2)^2))
Time = 1.73 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {e^{52+10 x^2+8 \log ^2(x)-2 \log ^4(x)} \left (5 x^2+4 \log (x)-2 \log ^3(x)\right )+e^{26+5 x^2+4 \log ^2(x)-\log ^4(x)} \left (-20 x^2-16 \log (x)+8 \log ^3(x)\right )}{4 x} \, dx=\frac {1}{16} e^{-2 \log ^4(x)} \left (e^{52+10 x^2+8 \log ^2(x)}-8 e^{26+5 x^2+4 \log ^2(x)+\log ^4(x)}\right ) \] Input:
Integrate[(E^(52 + 10*x^2 + 8*Log[x]^2 - 2*Log[x]^4)*(5*x^2 + 4*Log[x] - 2 *Log[x]^3) + E^(26 + 5*x^2 + 4*Log[x]^2 - Log[x]^4)*(-20*x^2 - 16*Log[x] + 8*Log[x]^3))/(4*x),x]
Output:
(E^(52 + 10*x^2 + 8*Log[x]^2) - 8*E^(26 + 5*x^2 + 4*Log[x]^2 + Log[x]^4))/ (16*E^(2*Log[x]^4))
Time = 0.66 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.72, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {27, 2010, 2009}
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.
\(\displaystyle \int \frac {e^{10 x^2-2 \log ^4(x)+8 \log ^2(x)+52} \left (5 x^2-2 \log ^3(x)+4 \log (x)\right )+e^{5 x^2-\log ^4(x)+4 \log ^2(x)+26} \left (-20 x^2+8 \log ^3(x)-16 \log (x)\right )}{4 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {e^{-2 \log ^4(x)+8 \log ^2(x)+10 x^2+52} \left (-2 \log ^3(x)+4 \log (x)+5 x^2\right )-4 e^{-\log ^4(x)+4 \log ^2(x)+5 x^2+26} \left (-2 \log ^3(x)+4 \log (x)+5 x^2\right )}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{4} \int \left (\frac {e^{2 \left (-\log ^4(x)+4 \log ^2(x)+5 x^2+26\right )} \left (-2 \log ^3(x)+4 \log (x)+5 x^2\right )}{x}-\frac {4 e^{-\log ^4(x)+4 \log ^2(x)+5 x^2+26} \left (-2 \log ^3(x)+4 \log (x)+5 x^2\right )}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{4} e^{2 \left (5 x^2-\log ^4(x)+4 \log ^2(x)+26\right )}-2 e^{5 x^2-\log ^4(x)+4 \log ^2(x)+26}\right )\) |
Input:
Int[(E^(52 + 10*x^2 + 8*Log[x]^2 - 2*Log[x]^4)*(5*x^2 + 4*Log[x] - 2*Log[x ]^3) + E^(26 + 5*x^2 + 4*Log[x]^2 - Log[x]^4)*(-20*x^2 - 16*Log[x] + 8*Log [x]^3))/(4*x),x]
Output:
(-2*E^(26 + 5*x^2 + 4*Log[x]^2 - Log[x]^4) + E^(2*(26 + 5*x^2 + 4*Log[x]^2 - Log[x]^4))/4)/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.38 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44
method | result | size |
risch | \(\frac {{\mathrm e}^{-2 \ln \left (x \right )^{4}+8 \ln \left (x \right )^{2}+10 x^{2}+52}}{16}-\frac {{\mathrm e}^{-\ln \left (x \right )^{4}+4 \ln \left (x \right )^{2}+5 x^{2}+26}}{2}\) | \(46\) |
parallelrisch | \(\frac {{\mathrm e}^{-2 \ln \left (x \right )^{4}+8 \ln \left (x \right )^{2}+10 x^{2}+52}}{16}-\frac {{\mathrm e}^{-\ln \left (x \right )^{4}+4 \ln \left (x \right )^{2}+5 x^{2}+26}}{2}\) | \(48\) |
Input:
int(1/4*((-2*ln(x)^3+4*ln(x)+5*x^2)*exp(-ln(x)^4+4*ln(x)^2+5*x^2+26)^2+(8* ln(x)^3-16*ln(x)-20*x^2)*exp(-ln(x)^4+4*ln(x)^2+5*x^2+26))/x,x,method=_RET URNVERBOSE)
Output:
1/16*exp(-2*ln(x)^4+8*ln(x)^2+10*x^2+52)-1/2*exp(-ln(x)^4+4*ln(x)^2+5*x^2+ 26)
Time = 0.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {e^{52+10 x^2+8 \log ^2(x)-2 \log ^4(x)} \left (5 x^2+4 \log (x)-2 \log ^3(x)\right )+e^{26+5 x^2+4 \log ^2(x)-\log ^4(x)} \left (-20 x^2-16 \log (x)+8 \log ^3(x)\right )}{4 x} \, dx=-\frac {1}{2} \, e^{\left (-\log \left (x\right )^{4} + 5 \, x^{2} + 4 \, \log \left (x\right )^{2} + 26\right )} + \frac {1}{16} \, e^{\left (-2 \, \log \left (x\right )^{4} + 10 \, x^{2} + 8 \, \log \left (x\right )^{2} + 52\right )} \] Input:
integrate(1/4*((-2*log(x)^3+4*log(x)+5*x^2)*exp(-log(x)^4+4*log(x)^2+5*x^2 +26)^2+(8*log(x)^3-16*log(x)-20*x^2)*exp(-log(x)^4+4*log(x)^2+5*x^2+26))/x ,x, algorithm="fricas")
Output:
-1/2*e^(-log(x)^4 + 5*x^2 + 4*log(x)^2 + 26) + 1/16*e^(-2*log(x)^4 + 10*x^ 2 + 8*log(x)^2 + 52)
Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (20) = 40\).
Time = 0.14 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {e^{52+10 x^2+8 \log ^2(x)-2 \log ^4(x)} \left (5 x^2+4 \log (x)-2 \log ^3(x)\right )+e^{26+5 x^2+4 \log ^2(x)-\log ^4(x)} \left (-20 x^2-16 \log (x)+8 \log ^3(x)\right )}{4 x} \, dx=- \frac {e^{5 x^{2} - \log {\left (x \right )}^{4} + 4 \log {\left (x \right )}^{2} + 26}}{2} + \frac {e^{10 x^{2} - 2 \log {\left (x \right )}^{4} + 8 \log {\left (x \right )}^{2} + 52}}{16} \] Input:
integrate(1/4*((-2*ln(x)**3+4*ln(x)+5*x**2)*exp(-ln(x)**4+4*ln(x)**2+5*x** 2+26)**2+(8*ln(x)**3-16*ln(x)-20*x**2)*exp(-ln(x)**4+4*ln(x)**2+5*x**2+26) )/x,x)
Output:
-exp(5*x**2 - log(x)**4 + 4*log(x)**2 + 26)/2 + exp(10*x**2 - 2*log(x)**4 + 8*log(x)**2 + 52)/16
Time = 0.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {e^{52+10 x^2+8 \log ^2(x)-2 \log ^4(x)} \left (5 x^2+4 \log (x)-2 \log ^3(x)\right )+e^{26+5 x^2+4 \log ^2(x)-\log ^4(x)} \left (-20 x^2-16 \log (x)+8 \log ^3(x)\right )}{4 x} \, dx=-\frac {1}{2} \, e^{\left (-\log \left (x\right )^{4} + 5 \, x^{2} + 4 \, \log \left (x\right )^{2} + 26\right )} + \frac {1}{16} \, e^{\left (-2 \, \log \left (x\right )^{4} + 10 \, x^{2} + 8 \, \log \left (x\right )^{2} + 52\right )} \] Input:
integrate(1/4*((-2*log(x)^3+4*log(x)+5*x^2)*exp(-log(x)^4+4*log(x)^2+5*x^2 +26)^2+(8*log(x)^3-16*log(x)-20*x^2)*exp(-log(x)^4+4*log(x)^2+5*x^2+26))/x ,x, algorithm="maxima")
Output:
-1/2*e^(-log(x)^4 + 5*x^2 + 4*log(x)^2 + 26) + 1/16*e^(-2*log(x)^4 + 10*x^ 2 + 8*log(x)^2 + 52)
Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {e^{52+10 x^2+8 \log ^2(x)-2 \log ^4(x)} \left (5 x^2+4 \log (x)-2 \log ^3(x)\right )+e^{26+5 x^2+4 \log ^2(x)-\log ^4(x)} \left (-20 x^2-16 \log (x)+8 \log ^3(x)\right )}{4 x} \, dx=-\frac {1}{2} \, e^{\left (-\log \left (x\right )^{4} + 5 \, x^{2} + 4 \, \log \left (x\right )^{2} + 26\right )} + \frac {1}{16} \, e^{\left (-2 \, \log \left (x\right )^{4} + 10 \, x^{2} + 8 \, \log \left (x\right )^{2} + 52\right )} \] Input:
integrate(1/4*((-2*log(x)^3+4*log(x)+5*x^2)*exp(-log(x)^4+4*log(x)^2+5*x^2 +26)^2+(8*log(x)^3-16*log(x)-20*x^2)*exp(-log(x)^4+4*log(x)^2+5*x^2+26))/x ,x, algorithm="giac")
Output:
-1/2*e^(-log(x)^4 + 5*x^2 + 4*log(x)^2 + 26) + 1/16*e^(-2*log(x)^4 + 10*x^ 2 + 8*log(x)^2 + 52)
Time = 4.46 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {e^{52+10 x^2+8 \log ^2(x)-2 \log ^4(x)} \left (5 x^2+4 \log (x)-2 \log ^3(x)\right )+e^{26+5 x^2+4 \log ^2(x)-\log ^4(x)} \left (-20 x^2-16 \log (x)+8 \log ^3(x)\right )}{4 x} \, dx=-\frac {{\mathrm {e}}^{4\,{\ln \left (x\right )}^2}\,{\mathrm {e}}^{-2\,{\ln \left (x\right )}^4}\,{\mathrm {e}}^{26}\,{\mathrm {e}}^{5\,x^2}\,\left (8\,{\mathrm {e}}^{{\ln \left (x\right )}^4}-{\mathrm {e}}^{4\,{\ln \left (x\right )}^2}\,{\mathrm {e}}^{26}\,{\mathrm {e}}^{5\,x^2}\right )}{16} \] Input:
int(-((exp(4*log(x)^2 - log(x)^4 + 5*x^2 + 26)*(16*log(x) - 8*log(x)^3 + 2 0*x^2))/4 - (exp(8*log(x)^2 - 2*log(x)^4 + 10*x^2 + 52)*(4*log(x) - 2*log( x)^3 + 5*x^2))/4)/x,x)
Output:
-(exp(4*log(x)^2)*exp(-2*log(x)^4)*exp(26)*exp(5*x^2)*(8*exp(log(x)^4) - e xp(4*log(x)^2)*exp(26)*exp(5*x^2)))/16
Time = 0.16 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.75 \[ \int \frac {e^{52+10 x^2+8 \log ^2(x)-2 \log ^4(x)} \left (5 x^2+4 \log (x)-2 \log ^3(x)\right )+e^{26+5 x^2+4 \log ^2(x)-\log ^4(x)} \left (-20 x^2-16 \log (x)+8 \log ^3(x)\right )}{4 x} \, dx=\frac {e^{4 \mathrm {log}\left (x \right )^{2}+5 x^{2}} e^{26} \left (-8 e^{\mathrm {log}\left (x \right )^{4}}+e^{4 \mathrm {log}\left (x \right )^{2}+5 x^{2}} e^{26}\right )}{16 e^{2 \mathrm {log}\left (x \right )^{4}}} \] Input:
int(1/4*((-2*log(x)^3+4*log(x)+5*x^2)*exp(-log(x)^4+4*log(x)^2+5*x^2+26)^2 +(8*log(x)^3-16*log(x)-20*x^2)*exp(-log(x)^4+4*log(x)^2+5*x^2+26))/x,x)
Output:
(e**(4*log(x)**2 + 5*x**2)*e**26*( - 8*e**(log(x)**4) + e**(4*log(x)**2 + 5*x**2)*e**26))/(16*e**(2*log(x)**4))