Integrand size = 91, antiderivative size = 33 \[ \int \frac {e^{2-x} \left (e^{-5+x} x-3 \log (3)\right )^4 \left (8 e^{-5+x} x-24 \log (3)+\left (e^{-5+x} \left (4 x+3 x^2\right )+3 x \log (3)\right ) \log \left (\frac {2}{x^2}\right )\right )}{81 \left (e^{-5+x} x^2-3 x \log (3)\right ) \log ^5\left (\frac {2}{x^2}\right )} \, dx=\frac {e^{2-x} \left (\frac {1}{3} e^{-5+x} x-\log (3)\right )^4}{\log ^4\left (\frac {2}{x^2}\right )} \] Output:
exp(2-x+4*ln((1/3*x*exp(-5+x)-ln(3))/ln(2/x^2)))
Time = 5.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{2-x} \left (e^{-5+x} x-3 \log (3)\right )^4 \left (8 e^{-5+x} x-24 \log (3)+\left (e^{-5+x} \left (4 x+3 x^2\right )+3 x \log (3)\right ) \log \left (\frac {2}{x^2}\right )\right )}{81 \left (e^{-5+x} x^2-3 x \log (3)\right ) \log ^5\left (\frac {2}{x^2}\right )} \, dx=\frac {e^{-18-x} \left (e^x x-3 e^5 \log (3)\right )^4}{81 \log ^4\left (\frac {2}{x^2}\right )} \] Input:
Integrate[(E^(2 - x)*(E^(-5 + x)*x - 3*Log[3])^4*(8*E^(-5 + x)*x - 24*Log[ 3] + (E^(-5 + x)*(4*x + 3*x^2) + 3*x*Log[3])*Log[2/x^2]))/(81*(E^(-5 + x)* x^2 - 3*x*Log[3])*Log[2/x^2]^5),x]
Output:
(E^(-18 - x)*(E^x*x - 3*E^5*Log[3])^4)/(81*Log[2/x^2]^4)
Leaf count is larger than twice the leaf count of optimal. \(107\) vs. \(2(33)=66\).
Time = 2.94 (sec) , antiderivative size = 107, normalized size of antiderivative = 3.24, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {27, 7293, 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.
| \(\displaystyle \int \frac {e^{2-x} \left (e^{x-5} x-3 \log (3)\right )^4 \left (\left (e^{x-5} \left (3 x^2+4 x\right )+3 x \log (3)\right ) \log \left (\frac {2}{x^2}\right )+8 e^{x-5} x-24 \log (3)\right )}{81 \left (e^{x-5} x^2-3 x \log (3)\right ) \log ^5\left (\frac {2}{x^2}\right )} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{81} \int \frac {e^{2-x} \left (e^{x-5} x-3 \log (3)\right )^4 \left (8 e^{x-5} x+\left (3 \log (3) x+e^{x-5} \left (3 x^2+4 x\right )\right ) \log \left (\frac {2}{x^2}\right )-24 \log (3)\right )}{\left (e^{x-5} x^2-3 x \log (3)\right ) \log ^5\left (\frac {2}{x^2}\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{81} \int \left (\frac {e^{3 x-18} \left (3 x \log \left (\frac {2}{x^2}\right )+4 \log \left (\frac {2}{x^2}\right )+8\right ) x^3}{\log ^5\left (\frac {2}{x^2}\right )}-\frac {12 e^{2 x-13} \log (3) \left (2 x \log \left (\frac {2}{x^2}\right )+3 \log \left (\frac {2}{x^2}\right )+8\right ) x^2}{\log ^5\left (\frac {2}{x^2}\right )}+\frac {54 e^{x-8} \log ^2(3) \left (x \log \left (\frac {2}{x^2}\right )+2 \log \left (\frac {2}{x^2}\right )+8\right ) x}{\log ^5\left (\frac {2}{x^2}\right )}-\frac {108 \log ^3(3) \left (\log \left (\frac {2}{x^2}\right )+8\right )}{e^3 \log ^5\left (\frac {2}{x^2}\right )}-\frac {81 e^{2-x} \log ^4(3) \left (x \log \left (\frac {2}{x^2}\right )-8\right )}{\log ^5\left (\frac {2}{x^2}\right ) x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{81} \left (\frac {81 e^{2-x} \log ^4(3)}{\log ^4\left (\frac {2}{x^2}\right )}-\frac {108 x \log ^3(3)}{e^3 \log ^4\left (\frac {2}{x^2}\right )}+\frac {54 e^{x-8} x^2 \log ^2(3)}{\log ^4\left (\frac {2}{x^2}\right )}+\frac {e^{3 x-18} x^4}{\log ^4\left (\frac {2}{x^2}\right )}-\frac {12 e^{2 x-13} x^3 \log (3)}{\log ^4\left (\frac {2}{x^2}\right )}\right )\) |
Input:
Int[(E^(2 - x)*(E^(-5 + x)*x - 3*Log[3])^4*(8*E^(-5 + x)*x - 24*Log[3] + ( E^(-5 + x)*(4*x + 3*x^2) + 3*x*Log[3])*Log[2/x^2]))/(81*(E^(-5 + x)*x^2 - 3*x*Log[3])*Log[2/x^2]^5),x]
Output:
((E^(-18 + 3*x)*x^4)/Log[2/x^2]^4 - (12*E^(-13 + 2*x)*x^3*Log[3])/Log[2/x^ 2]^4 + (54*E^(-8 + x)*x^2*Log[3]^2)/Log[2/x^2]^4 - (108*x*Log[3]^3)/(E^3*L og[2/x^2]^4) + (81*E^(2 - x)*Log[3]^4)/Log[2/x^2]^4)/81
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 263.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94
| method | result | size |
| parallelrisch | \({\mathrm e}^{4 \ln \left (\frac {x \,{\mathrm e}^{-5+x}-3 \ln \left (3\right )}{3 \ln \left (\frac {2}{x^{2}}\right )}\right )+2-x}\) | \(31\) |
Input:
int((((3*x^2+4*x)*exp(-5+x)+3*x*ln(3))*ln(2/x^2)+8*x*exp(-5+x)-24*ln(3))*e xp(4*ln(1/3*(x*exp(-5+x)-3*ln(3))/ln(2/x^2))+2-x)/(x^2*exp(-5+x)-3*x*ln(3) )/ln(2/x^2),x,method=_RETURNVERBOSE)
Output:
exp(4*ln(1/3*(x*exp(-5+x)-3*ln(3))/ln(2/x^2))+2-x)
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (30) = 60\).
Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.18 \[ \int \frac {e^{2-x} \left (e^{-5+x} x-3 \log (3)\right )^4 \left (8 e^{-5+x} x-24 \log (3)+\left (e^{-5+x} \left (4 x+3 x^2\right )+3 x \log (3)\right ) \log \left (\frac {2}{x^2}\right )\right )}{81 \left (e^{-5+x} x^2-3 x \log (3)\right ) \log ^5\left (\frac {2}{x^2}\right )} \, dx=\frac {{\left (x^{4} e^{\left (4 \, x - 20\right )} - 12 \, x^{3} e^{\left (3 \, x - 15\right )} \log \left (3\right ) + 54 \, x^{2} e^{\left (2 \, x - 10\right )} \log \left (3\right )^{2} - 108 \, x e^{\left (x - 5\right )} \log \left (3\right )^{3} + 81 \, \log \left (3\right )^{4}\right )} e^{\left (-x + 2\right )}}{81 \, \log \left (\frac {2}{x^{2}}\right )^{4}} \] Input:
integrate((((3*x^2+4*x)*exp(-5+x)+3*x*log(3))*log(2/x^2)+8*x*exp(-5+x)-24* log(3))*exp(4*log(1/3*(x*exp(-5+x)-3*log(3))/log(2/x^2))+2-x)/(x^2*exp(-5+ x)-3*x*log(3))/log(2/x^2),x, algorithm="fricas")
Output:
1/81*(x^4*e^(4*x - 20) - 12*x^3*e^(3*x - 15)*log(3) + 54*x^2*e^(2*x - 10)* log(3)^2 - 108*x*e^(x - 5)*log(3)^3 + 81*log(3)^4)*e^(-x + 2)/log(2/x^2)^4
Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (26) = 52\).
Time = 0.36 (sec) , antiderivative size = 136, normalized size of antiderivative = 4.12 \[ \int \frac {e^{2-x} \left (e^{-5+x} x-3 \log (3)\right )^4 \left (8 e^{-5+x} x-24 \log (3)+\left (e^{-5+x} \left (4 x+3 x^2\right )+3 x \log (3)\right ) \log \left (\frac {2}{x^2}\right )\right )}{81 \left (e^{-5+x} x^2-3 x \log (3)\right ) \log ^5\left (\frac {2}{x^2}\right )} \, dx=- \frac {4 x \log {\left (3 \right )}^{3}}{3 e^{3} \log {\left (\frac {2}{x^{2}} \right )}^{4}} + \frac {81 x^{4} e^{15} e^{3 x - 6} \log {\left (\frac {2}{x^{2}} \right )}^{12} - 972 x^{3} e^{18} e^{2 x - 4} \log {\left (3 \right )} \log {\left (\frac {2}{x^{2}} \right )}^{12} + 4374 x^{2} e^{21} e^{x - 2} \log {\left (3 \right )}^{2} \log {\left (\frac {2}{x^{2}} \right )}^{12} + 6561 e^{27} e^{2 - x} \log {\left (3 \right )}^{4} \log {\left (\frac {2}{x^{2}} \right )}^{12}}{6561 e^{27} \log {\left (\frac {2}{x^{2}} \right )}^{16}} \] Input:
integrate((((3*x**2+4*x)*exp(-5+x)+3*x*ln(3))*ln(2/x**2)+8*x*exp(-5+x)-24* ln(3))*exp(4*ln(1/3*(x*exp(-5+x)-3*ln(3))/ln(2/x**2))+2-x)/(x**2*exp(-5+x) -3*x*ln(3))/ln(2/x**2),x)
Output:
-4*x*exp(-3)*log(3)**3/(3*log(2/x**2)**4) + (81*x**4*exp(15)*exp(3*x - 6)* log(2/x**2)**12 - 972*x**3*exp(18)*exp(2*x - 4)*log(3)*log(2/x**2)**12 + 4 374*x**2*exp(21)*exp(x - 2)*log(3)**2*log(2/x**2)**12 + 6561*exp(27)*exp(2 - x)*log(3)**4*log(2/x**2)**12)*exp(-27)/(6561*log(2/x**2)**16)
Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (30) = 60\).
Time = 0.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 3.27 \[ \int \frac {e^{2-x} \left (e^{-5+x} x-3 \log (3)\right )^4 \left (8 e^{-5+x} x-24 \log (3)+\left (e^{-5+x} \left (4 x+3 x^2\right )+3 x \log (3)\right ) \log \left (\frac {2}{x^2}\right )\right )}{81 \left (e^{-5+x} x^2-3 x \log (3)\right ) \log ^5\left (\frac {2}{x^2}\right )} \, dx=\frac {x^{4} e^{\left (3 \, x\right )} - 12 \, x^{3} e^{\left (2 \, x + 5\right )} \log \left (3\right ) + 54 \, x^{2} e^{\left (x + 10\right )} \log \left (3\right )^{2} - 108 \, x e^{15} \log \left (3\right )^{3} + 81 \, e^{\left (-x + 20\right )} \log \left (3\right )^{4}}{81 \, {\left (e^{18} \log \left (2\right )^{4} - 8 \, e^{18} \log \left (2\right )^{3} \log \left (x\right ) + 24 \, e^{18} \log \left (2\right )^{2} \log \left (x\right )^{2} - 32 \, e^{18} \log \left (2\right ) \log \left (x\right )^{3} + 16 \, e^{18} \log \left (x\right )^{4}\right )}} \] Input:
integrate((((3*x^2+4*x)*exp(-5+x)+3*x*log(3))*log(2/x^2)+8*x*exp(-5+x)-24* log(3))*exp(4*log(1/3*(x*exp(-5+x)-3*log(3))/log(2/x^2))+2-x)/(x^2*exp(-5+ x)-3*x*log(3))/log(2/x^2),x, algorithm="maxima")
Output:
1/81*(x^4*e^(3*x) - 12*x^3*e^(2*x + 5)*log(3) + 54*x^2*e^(x + 10)*log(3)^2 - 108*x*e^15*log(3)^3 + 81*e^(-x + 20)*log(3)^4)/(e^18*log(2)^4 - 8*e^18* log(2)^3*log(x) + 24*e^18*log(2)^2*log(x)^2 - 32*e^18*log(2)*log(x)^3 + 16 *e^18*log(x)^4)
Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (30) = 60\).
Time = 1.76 (sec) , antiderivative size = 256, normalized size of antiderivative = 7.76 \[ \int \frac {e^{2-x} \left (e^{-5+x} x-3 \log (3)\right )^4 \left (8 e^{-5+x} x-24 \log (3)+\left (e^{-5+x} \left (4 x+3 x^2\right )+3 x \log (3)\right ) \log \left (\frac {2}{x^2}\right )\right )}{81 \left (e^{-5+x} x^2-3 x \log (3)\right ) \log ^5\left (\frac {2}{x^2}\right )} \, dx=\frac {x^{4} e^{\left (4 \, x\right )} - 12 \, x^{3} e^{\left (3 \, x + 5\right )} \log \left (3\right ) + 54 \, x^{2} e^{\left (2 \, x + 10\right )} \log \left (3\right )^{2} - 108 \, x e^{\left (x + 15\right )} \log \left (3\right )^{3} + 81 \, e^{20} \log \left (3\right )^{4}}{81 \, {\left (e^{\left (x + 18\right )} \log \left (2\right )^{4} - 8 \, e^{\left (x + 18\right )} \log \left (2\right )^{3} \log \left (x\right ) + 24 \, e^{\left (x + 18\right )} \log \left (2\right )^{2} \log \left (x\right )^{2} - 32 \, e^{\left (x + 18\right )} \log \left (2\right ) \log \left (x\right )^{3} + 16 \, e^{\left (x + 18\right )} \log \left (x\right )^{4} - 8 \, e^{\left (x + 18\right )} \log \left (2\right )^{3} \log \left (\mathrm {sgn}\left (x\right )\right ) + 48 \, e^{\left (x + 18\right )} \log \left (2\right )^{2} \log \left (x\right ) \log \left (\mathrm {sgn}\left (x\right )\right ) - 96 \, e^{\left (x + 18\right )} \log \left (2\right ) \log \left (x\right )^{2} \log \left (\mathrm {sgn}\left (x\right )\right ) + 64 \, e^{\left (x + 18\right )} \log \left (x\right )^{3} \log \left (\mathrm {sgn}\left (x\right )\right ) + 24 \, e^{\left (x + 18\right )} \log \left (2\right )^{2} \log \left (\mathrm {sgn}\left (x\right )\right )^{2} - 96 \, e^{\left (x + 18\right )} \log \left (2\right ) \log \left (x\right ) \log \left (\mathrm {sgn}\left (x\right )\right )^{2} + 96 \, e^{\left (x + 18\right )} \log \left (x\right )^{2} \log \left (\mathrm {sgn}\left (x\right )\right )^{2} - 32 \, e^{\left (x + 18\right )} \log \left (2\right ) \log \left (\mathrm {sgn}\left (x\right )\right )^{3} + 64 \, e^{\left (x + 18\right )} \log \left (x\right ) \log \left (\mathrm {sgn}\left (x\right )\right )^{3} + 16 \, e^{\left (x + 18\right )} \log \left (\mathrm {sgn}\left (x\right )\right )^{4}\right )}} \] Input:
integrate((((3*x^2+4*x)*exp(-5+x)+3*x*log(3))*log(2/x^2)+8*x*exp(-5+x)-24* log(3))*exp(4*log(1/3*(x*exp(-5+x)-3*log(3))/log(2/x^2))+2-x)/(x^2*exp(-5+ x)-3*x*log(3))/log(2/x^2),x, algorithm="giac")
Output:
1/81*(x^4*e^(4*x) - 12*x^3*e^(3*x + 5)*log(3) + 54*x^2*e^(2*x + 10)*log(3) ^2 - 108*x*e^(x + 15)*log(3)^3 + 81*e^20*log(3)^4)/(e^(x + 18)*log(2)^4 - 8*e^(x + 18)*log(2)^3*log(x) + 24*e^(x + 18)*log(2)^2*log(x)^2 - 32*e^(x + 18)*log(2)*log(x)^3 + 16*e^(x + 18)*log(x)^4 - 8*e^(x + 18)*log(2)^3*log( sgn(x)) + 48*e^(x + 18)*log(2)^2*log(x)*log(sgn(x)) - 96*e^(x + 18)*log(2) *log(x)^2*log(sgn(x)) + 64*e^(x + 18)*log(x)^3*log(sgn(x)) + 24*e^(x + 18) *log(2)^2*log(sgn(x))^2 - 96*e^(x + 18)*log(2)*log(x)*log(sgn(x))^2 + 96*e ^(x + 18)*log(x)^2*log(sgn(x))^2 - 32*e^(x + 18)*log(2)*log(sgn(x))^3 + 64 *e^(x + 18)*log(x)*log(sgn(x))^3 + 16*e^(x + 18)*log(sgn(x))^4)
Time = 4.69 (sec) , antiderivative size = 1729, normalized size of antiderivative = 52.39 \[ \int \frac {e^{2-x} \left (e^{-5+x} x-3 \log (3)\right )^4 \left (8 e^{-5+x} x-24 \log (3)+\left (e^{-5+x} \left (4 x+3 x^2\right )+3 x \log (3)\right ) \log \left (\frac {2}{x^2}\right )\right )}{81 \left (e^{-5+x} x^2-3 x \log (3)\right ) \log ^5\left (\frac {2}{x^2}\right )} \, dx=\text {Too large to display} \] Input:
int(-(exp(4*log(-(log(3) - (x*exp(x - 5))/3)/log(2/x^2)) - x + 2)*(8*x*exp (x - 5) - 24*log(3) + log(2/x^2)*(exp(x - 5)*(4*x + 3*x^2) + 3*x*log(3)))) /(log(2/x^2)*(3*x*log(3) - x^2*exp(x - 5))),x)
Output:
exp(x - 5)*((x^2*exp(-3)*log(3)^2)/36 + (65*x^3*exp(-3)*log(3)^2)/576 + (5 5*x^4*exp(-3)*log(3)^2)/576 + (7*x^5*exp(-3)*log(3)^2)/288 + (x^6*exp(-3)* log(3)^2)/576) - ((x*exp(2 - x)*(4*exp(x - 5)*log(27)^3 - 432*exp(x - 5)*l og(3)^3 - 324*log(3)^4 + log(27)^4 + 12*x^3*exp(4*x - 20) + 9*x^4*exp(4*x - 20) + 432*x*exp(2*x - 10)*log(3)^2 - 144*x^2*exp(3*x - 15)*log(3) - 96*x ^3*exp(3*x - 15)*log(3) - 12*x*exp(2*x - 10)*log(27)^2 + 12*x^2*exp(3*x - 15)*log(27) + 8*x^3*exp(3*x - 15)*log(27) + 216*x^2*exp(2*x - 10)*log(3)^2 - 6*x^2*exp(2*x - 10)*log(27)^2))/1944 + (x*exp(2 - x)*log(2/x^2)*(x*log( 27)^4 - 4*exp(x - 5)*log(27)^3 - log(27)^4 + 16*x^3*exp(4*x - 20) + 27*x^4 *exp(4*x - 20) + 9*x^5*exp(4*x - 20) + 24*x*exp(2*x - 10)*log(27)^2 - 36*x ^2*exp(3*x - 15)*log(27) - 56*x^3*exp(3*x - 15)*log(27) - 16*x^4*exp(3*x - 15)*log(27) + 30*x^2*exp(2*x - 10)*log(27)^2 + 6*x^3*exp(2*x - 10)*log(27 )^2))/3888)/log(2/x^2)^3 + (exp(2 - x)*(log(3)^4 + (x^4*exp(4*x - 20))/81 - (4*x^3*exp(3*x - 15)*log(3))/27 + (2*x^2*exp(2*x - 10)*log(3)^2)/3 - (4* x*exp(x - 5)*log(3)^3)/3) - (x*exp(2 - x)*log(2/x^2)*(log(27) - x*exp(x - 5))^3*(4*exp(x - 5) + log(27) + 3*x*exp(x - 5)))/648)/log(2/x^2)^4 - ((x*e xp(2 - x)*(3*x^2*log(27)^4 - 324*x^2*log(3)^4 - 432*exp(x - 5)*log(3)^3 + 12*exp(x - 5)*log(27)^3 + 972*x*log(3)^4 - 9*x*log(27)^4 - 324*log(3)^4 + 3*log(27)^4 + 64*x^3*exp(4*x - 20) + 183*x^4*exp(4*x - 20) + 135*x^5*exp(4 *x - 20) + 27*x^6*exp(4*x - 20) + 1728*x*exp(2*x - 10)*log(3)^2 - 1296*...
Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.48 \[ \int \frac {e^{2-x} \left (e^{-5+x} x-3 \log (3)\right )^4 \left (8 e^{-5+x} x-24 \log (3)+\left (e^{-5+x} \left (4 x+3 x^2\right )+3 x \log (3)\right ) \log \left (\frac {2}{x^2}\right )\right )}{81 \left (e^{-5+x} x^2-3 x \log (3)\right ) \log ^5\left (\frac {2}{x^2}\right )} \, dx=\frac {e^{4 x} x^{4}-12 e^{3 x} \mathrm {log}\left (3\right ) e^{5} x^{3}+54 e^{2 x} \mathrm {log}\left (3\right )^{2} e^{10} x^{2}-108 e^{x} \mathrm {log}\left (3\right )^{3} e^{15} x +81 \mathrm {log}\left (3\right )^{4} e^{20}}{81 e^{x} \mathrm {log}\left (\frac {2}{x^{2}}\right )^{4} e^{18}} \] Input:
int((((3*x^2+4*x)*exp(-5+x)+3*x*log(3))*log(2/x^2)+8*x*exp(-5+x)-24*log(3) )*exp(4*log(1/3*(x*exp(-5+x)-3*log(3))/log(2/x^2))+2-x)/(x^2*exp(-5+x)-3*x *log(3))/log(2/x^2),x)
Output:
(e**(4*x)*x**4 - 12*e**(3*x)*log(3)*e**5*x**3 + 54*e**(2*x)*log(3)**2*e**1 0*x**2 - 108*e**x*log(3)**3*e**15*x + 81*log(3)**4*e**20)/(81*e**x*log(2/x **2)**4*e**18)