Integrand size = 81, antiderivative size = 25 \[ \int \frac {-9+7 x+27 x^2+e^{-x} \left (5 x-3 x^2\right )+\left (-6-2 x+18 x^2+e^{-x} \left (2 x-x^2\right )\right ) \log (x)+\left (-1-2 x+3 x^2\right ) \log ^2(x)}{9+6 \log (x)+\log ^2(x)} \, dx=x \left (-1-x+x^2+\frac {\left (5+e^{-x}\right ) x}{3+\log (x)}\right ) \] Output:
(x^2-x-1+x*(exp(-x)+5)/(3+ln(x)))*x
Time = 5.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-9+7 x+27 x^2+e^{-x} \left (5 x-3 x^2\right )+\left (-6-2 x+18 x^2+e^{-x} \left (2 x-x^2\right )\right ) \log (x)+\left (-1-2 x+3 x^2\right ) \log ^2(x)}{9+6 \log (x)+\log ^2(x)} \, dx=x \left (-1-x+x^2+\frac {\left (5+e^{-x}\right ) x}{3+\log (x)}\right ) \] Input:
Integrate[(-9 + 7*x + 27*x^2 + (5*x - 3*x^2)/E^x + (-6 - 2*x + 18*x^2 + (2 *x - x^2)/E^x)*Log[x] + (-1 - 2*x + 3*x^2)*Log[x]^2)/(9 + 6*Log[x] + Log[x ]^2),x]
Output:
x*(-1 - x + x^2 + ((5 + E^(-x))*x)/(3 + Log[x]))
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 2.23 (sec) , antiderivative size = 167, normalized size of antiderivative = 6.68, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {7292, 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.
\(\displaystyle \int \frac {27 x^2+e^{-x} \left (5 x-3 x^2\right )+\left (3 x^2-2 x-1\right ) \log ^2(x)+\left (18 x^2+e^{-x} \left (2 x-x^2\right )-2 x-6\right ) \log (x)+7 x-9}{\log ^2(x)+6 \log (x)+9} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {27 x^2+e^{-x} \left (5 x-3 x^2\right )+\left (3 x^2-2 x-1\right ) \log ^2(x)+\left (18 x^2+e^{-x} \left (2 x-x^2\right )-2 x-6\right ) \log (x)+7 x-9}{(\log (x)+3)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {18 x^2 \log (x)}{(\log (x)+3)^2}+\frac {27 x^2}{(\log (x)+3)^2}+\frac {(x-1) (3 x+1) \log ^2(x)}{(\log (x)+3)^2}-\frac {e^{-x} x (3 x+x \log (x)-2 \log (x)-5)}{(\log (x)+3)^2}-\frac {2 x \log (x)}{(\log (x)+3)^2}+\frac {7 x}{(\log (x)+3)^2}-\frac {6 \log (x)}{(\log (x)+3)^2}-\frac {9}{(\log (x)+3)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {12 \operatorname {ExpIntegralEi}(2 (\log (x)+3))}{e^6}+\frac {162 \operatorname {ExpIntegralEi}(3 (\log (x)+3))}{e^9}-\frac {4 \log (x) \operatorname {ExpIntegralEi}(2 (\log (x)+3))}{e^6}+\frac {54 \log (x) \operatorname {ExpIntegralEi}(3 (\log (x)+3))}{e^9}+\frac {4 (\log (x)+3) \operatorname {ExpIntegralEi}(2 (\log (x)+3))}{e^6}-\frac {54 (\log (x)+3) \operatorname {ExpIntegralEi}(3 (\log (x)+3))}{e^9}+19 x^3-\frac {18 x^3 \log (x)}{\log (x)+3}-\frac {54 x^3}{\log (x)+3}-3 x^2+\frac {2 x^2 \log (x)}{\log (x)+3}+\frac {11 x^2}{\log (x)+3}-x+\frac {e^{-x} x (3 x+x \log (x))}{(\log (x)+3)^2}\) |
Input:
Int[(-9 + 7*x + 27*x^2 + (5*x - 3*x^2)/E^x + (-6 - 2*x + 18*x^2 + (2*x - x ^2)/E^x)*Log[x] + (-1 - 2*x + 3*x^2)*Log[x]^2)/(9 + 6*Log[x] + Log[x]^2),x ]
Output:
-x - 3*x^2 + 19*x^3 - (12*ExpIntegralEi[2*(3 + Log[x])])/E^6 + (162*ExpInt egralEi[3*(3 + Log[x])])/E^9 - (4*ExpIntegralEi[2*(3 + Log[x])]*Log[x])/E^ 6 + (54*ExpIntegralEi[3*(3 + Log[x])]*Log[x])/E^9 + (11*x^2)/(3 + Log[x]) - (54*x^3)/(3 + Log[x]) + (2*x^2*Log[x])/(3 + Log[x]) - (18*x^3*Log[x])/(3 + Log[x]) + (4*ExpIntegralEi[2*(3 + Log[x])]*(3 + Log[x]))/E^6 - (54*ExpI ntegralEi[3*(3 + Log[x])]*(3 + Log[x]))/E^9 + (x*(3*x + x*Log[x]))/(E^x*(3 + Log[x])^2)
Time = 9.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16
method | result | size |
risch | \(x^{3}-x^{2}-x +\frac {x^{2} \left ({\mathrm e}^{-x}+5\right )}{3+\ln \left (x \right )}\) | \(29\) |
default | \(\frac {5 x^{2}}{3+\ln \left (x \right )}-x -x^{2}+x^{3}+\frac {x^{2} {\mathrm e}^{-x}}{3+\ln \left (x \right )}\) | \(38\) |
parts | \(\frac {5 x^{2}}{3+\ln \left (x \right )}-x -x^{2}+x^{3}+\frac {x^{2} {\mathrm e}^{-x}}{3+\ln \left (x \right )}\) | \(38\) |
norman | \(\frac {x^{2} {\mathrm e}^{-x}+x^{3} \ln \left (x \right )-3 x +2 x^{2}+3 x^{3}-x \ln \left (x \right )-x^{2} \ln \left (x \right )}{3+\ln \left (x \right )}\) | \(48\) |
parallelrisch | \(-\frac {-x^{3} \ln \left (x \right )-3 x^{3}+x^{2} \ln \left (x \right )-x^{2} {\mathrm e}^{-x}-2 x^{2}+x \ln \left (x \right )+3 x}{3+\ln \left (x \right )}\) | \(49\) |
Input:
int(((3*x^2-2*x-1)*ln(x)^2+((-x^2+2*x)*exp(-x)+18*x^2-2*x-6)*ln(x)+(-3*x^2 +5*x)*exp(-x)+27*x^2+7*x-9)/(ln(x)^2+6*ln(x)+9),x,method=_RETURNVERBOSE)
Output:
x^3-x^2-x+x^2*(exp(-x)+5)/(3+ln(x))
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \[ \int \frac {-9+7 x+27 x^2+e^{-x} \left (5 x-3 x^2\right )+\left (-6-2 x+18 x^2+e^{-x} \left (2 x-x^2\right )\right ) \log (x)+\left (-1-2 x+3 x^2\right ) \log ^2(x)}{9+6 \log (x)+\log ^2(x)} \, dx=\frac {3 \, x^{3} + x^{2} e^{\left (-x\right )} + 2 \, x^{2} + {\left (x^{3} - x^{2} - x\right )} \log \left (x\right ) - 3 \, x}{\log \left (x\right ) + 3} \] Input:
integrate(((3*x^2-2*x-1)*log(x)^2+((-x^2+2*x)*exp(-x)+18*x^2-2*x-6)*log(x) +(-3*x^2+5*x)*exp(-x)+27*x^2+7*x-9)/(log(x)^2+6*log(x)+9),x, algorithm="fr icas")
Output:
(3*x^3 + x^2*e^(-x) + 2*x^2 + (x^3 - x^2 - x)*log(x) - 3*x)/(log(x) + 3)
Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-9+7 x+27 x^2+e^{-x} \left (5 x-3 x^2\right )+\left (-6-2 x+18 x^2+e^{-x} \left (2 x-x^2\right )\right ) \log (x)+\left (-1-2 x+3 x^2\right ) \log ^2(x)}{9+6 \log (x)+\log ^2(x)} \, dx=x^{3} - x^{2} + \frac {5 x^{2}}{\log {\left (x \right )} + 3} + \frac {x^{2} e^{- x}}{\log {\left (x \right )} + 3} - x \] Input:
integrate(((3*x**2-2*x-1)*ln(x)**2+((-x**2+2*x)*exp(-x)+18*x**2-2*x-6)*ln( x)+(-3*x**2+5*x)*exp(-x)+27*x**2+7*x-9)/(ln(x)**2+6*ln(x)+9),x)
Output:
x**3 - x**2 + 5*x**2/(log(x) + 3) + x**2*exp(-x)/(log(x) + 3) - x
\[ \int \frac {-9+7 x+27 x^2+e^{-x} \left (5 x-3 x^2\right )+\left (-6-2 x+18 x^2+e^{-x} \left (2 x-x^2\right )\right ) \log (x)+\left (-1-2 x+3 x^2\right ) \log ^2(x)}{9+6 \log (x)+\log ^2(x)} \, dx=\int { \frac {{\left (3 \, x^{2} - 2 \, x - 1\right )} \log \left (x\right )^{2} + 27 \, x^{2} - {\left (3 \, x^{2} - 5 \, x\right )} e^{\left (-x\right )} + {\left (18 \, x^{2} - {\left (x^{2} - 2 \, x\right )} e^{\left (-x\right )} - 2 \, x - 6\right )} \log \left (x\right ) + 7 \, x - 9}{\log \left (x\right )^{2} + 6 \, \log \left (x\right ) + 9} \,d x } \] Input:
integrate(((3*x^2-2*x-1)*log(x)^2+((-x^2+2*x)*exp(-x)+18*x^2-2*x-6)*log(x) +(-3*x^2+5*x)*exp(-x)+27*x^2+7*x-9)/(log(x)^2+6*log(x)+9),x, algorithm="ma xima")
Output:
9*e^(-3)*exp_integral_e(2, -log(x) - 3)/(log(x) + 3) - 7*e^(-6)*exp_integr al_e(2, -2*log(x) - 6)/(log(x) + 3) - 27*e^(-9)*exp_integral_e(2, -3*log(x ) - 9)/(log(x) + 3) + (30*x^3 + x^2*e^(-x) + 9*x^2 + (x^3 - x^2 - x)*log(x ) - 12*x)/(log(x) + 3) - integrate((81*x^2 + 14*x - 9)/(log(x) + 3), x)
Time = 0.12 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {-9+7 x+27 x^2+e^{-x} \left (5 x-3 x^2\right )+\left (-6-2 x+18 x^2+e^{-x} \left (2 x-x^2\right )\right ) \log (x)+\left (-1-2 x+3 x^2\right ) \log ^2(x)}{9+6 \log (x)+\log ^2(x)} \, dx=\frac {x^{3} \log \left (x\right ) + 3 \, x^{3} + x^{2} e^{\left (-x\right )} - x^{2} \log \left (x\right ) + 2 \, x^{2} - x \log \left (x\right ) - 3 \, x}{\log \left (x\right ) + 3} \] Input:
integrate(((3*x^2-2*x-1)*log(x)^2+((-x^2+2*x)*exp(-x)+18*x^2-2*x-6)*log(x) +(-3*x^2+5*x)*exp(-x)+27*x^2+7*x-9)/(log(x)^2+6*log(x)+9),x, algorithm="gi ac")
Output:
(x^3*log(x) + 3*x^3 + x^2*e^(-x) - x^2*log(x) + 2*x^2 - x*log(x) - 3*x)/(l og(x) + 3)
Time = 0.47 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-9+7 x+27 x^2+e^{-x} \left (5 x-3 x^2\right )+\left (-6-2 x+18 x^2+e^{-x} \left (2 x-x^2\right )\right ) \log (x)+\left (-1-2 x+3 x^2\right ) \log ^2(x)}{9+6 \log (x)+\log ^2(x)} \, dx=x^3-x+\frac {x^2\,\left ({\mathrm {e}}^{-x}-\ln \left (x\right )+2\right )}{\ln \left (x\right )+3} \] Input:
int((7*x + exp(-x)*(5*x - 3*x^2) - log(x)^2*(2*x - 3*x^2 + 1) - log(x)*(2* x - exp(-x)*(2*x - x^2) - 18*x^2 + 6) + 27*x^2 - 9)/(6*log(x) + log(x)^2 + 9),x)
Output:
x^3 - x + (x^2*(exp(-x) - log(x) + 2))/(log(x) + 3)
Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.32 \[ \int \frac {-9+7 x+27 x^2+e^{-x} \left (5 x-3 x^2\right )+\left (-6-2 x+18 x^2+e^{-x} \left (2 x-x^2\right )\right ) \log (x)+\left (-1-2 x+3 x^2\right ) \log ^2(x)}{9+6 \log (x)+\log ^2(x)} \, dx=\frac {x \left (e^{x} \mathrm {log}\left (x \right ) x^{2}-e^{x} \mathrm {log}\left (x \right ) x -e^{x} \mathrm {log}\left (x \right )+3 e^{x} x^{2}+2 e^{x} x -3 e^{x}+x \right )}{e^{x} \left (\mathrm {log}\left (x \right )+3\right )} \] Input:
int(((3*x^2-2*x-1)*log(x)^2+((-x^2+2*x)*exp(-x)+18*x^2-2*x-6)*log(x)+(-3*x ^2+5*x)*exp(-x)+27*x^2+7*x-9)/(log(x)^2+6*log(x)+9),x)
Output:
(x*(e**x*log(x)*x**2 - e**x*log(x)*x - e**x*log(x) + 3*e**x*x**2 + 2*e**x* x - 3*e**x + x))/(e**x*(log(x) + 3))