Integrand size = 72, antiderivative size = 30 \[ \int \frac {-32+e^x (-72+8 x) \log (x)+e^x \left (64 x-8 x^2\right ) \log ^2(x)+\left (-e^{30}+e^{2 x} \left (144 x-34 x^2+2 x^3\right )\right ) \log ^3(x)}{e^{30} x \log ^3(x)} \, dx=3+\frac {\left (e^x (9-x)+\frac {4}{\log (x)}\right )^2}{e^{30}}-\log (3 x) \] Output:
(exp(x)*(9-x)+4/ln(x))^2/exp(15)^2+3-ln(3*x)
Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {-32+e^x (-72+8 x) \log (x)+e^x \left (64 x-8 x^2\right ) \log ^2(x)+\left (-e^{30}+e^{2 x} \left (144 x-34 x^2+2 x^3\right )\right ) \log ^3(x)}{e^{30} x \log ^3(x)} \, dx=-\frac {-e^{2 x} (-9+x)^2-\frac {16}{\log ^2(x)}+\frac {8 e^x (-9+x)}{\log (x)}+e^{30} \log (x)}{e^{30}} \] Input:
Integrate[(-32 + E^x*(-72 + 8*x)*Log[x] + E^x*(64*x - 8*x^2)*Log[x]^2 + (- E^30 + E^(2*x)*(144*x - 34*x^2 + 2*x^3))*Log[x]^3)/(E^30*x*Log[x]^3),x]
Output:
-((-(E^(2*x)*(-9 + x)^2) - 16/Log[x]^2 + (8*E^x*(-9 + x))/Log[x] + E^30*Lo g[x])/E^30)
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.
\(\displaystyle \int \frac {e^x \left (64 x-8 x^2\right ) \log ^2(x)+\left (e^{2 x} \left (2 x^3-34 x^2+144 x\right )-e^{30}\right ) \log ^3(x)+e^x (8 x-72) \log (x)-32}{e^{30} x \log ^3(x)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {\left (e^{30}-2 e^{2 x} \left (x^3-17 x^2+72 x\right )\right ) \log ^3(x)-8 e^x \left (8 x-x^2\right ) \log ^2(x)+8 e^x (9-x) \log (x)+32}{x \log ^3(x)}dx}{e^{30}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\left (e^{30}-2 e^{2 x} \left (x^3-17 x^2+72 x\right )\right ) \log ^3(x)-8 e^x \left (8 x-x^2\right ) \log ^2(x)+8 e^x (9-x) \log (x)+32}{x \log ^3(x)}dx}{e^{30}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (-2 e^{2 x} (x-9) (x-8)+\frac {8 e^x \left (\log (x) x^2-8 \log (x) x-x+9\right )}{x \log ^2(x)}+\frac {e^{30} \log ^3(x)+32}{x \log ^3(x)}\right )dx}{e^{30}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-8 \int \frac {e^x}{\log ^2(x)}dx+72 \int \frac {e^x}{x \log ^2(x)}dx-64 \int \frac {e^x}{\log (x)}dx+8 \int \frac {e^x x}{\log (x)}dx-e^{2 x} x^2+18 e^{2 x} x-81 e^{2 x}-\frac {16}{\log ^2(x)}+e^{30} \log (x)}{e^{30}}\) |
Input:
Int[(-32 + E^x*(-72 + 8*x)*Log[x] + E^x*(64*x - 8*x^2)*Log[x]^2 + (-E^30 + E^(2*x)*(144*x - 34*x^2 + 2*x^3))*Log[x]^3)/(E^30*x*Log[x]^3),x]
Output:
$Aborted
Time = 125.14 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83
method | result | size |
risch | \(x^{2} {\mathrm e}^{-30+2 x}-\ln \left (x \right )-18 x \,{\mathrm e}^{-30+2 x}+81 \,{\mathrm e}^{-30+2 x}-\frac {8 \,{\mathrm e}^{-30} \left (x \,{\mathrm e}^{x} \ln \left (x \right )-9 \,{\mathrm e}^{x} \ln \left (x \right )-2\right )}{\ln \left (x \right )^{2}}\) | \(55\) |
default | \({\mathrm e}^{-30} \left (\frac {-8 \,{\mathrm e}^{x} x +72 \,{\mathrm e}^{x}}{\ln \left (x \right )}+{\mathrm e}^{2 x} x^{2}-18 x \,{\mathrm e}^{2 x}+81 \,{\mathrm e}^{2 x}+\frac {16}{\ln \left (x \right )^{2}}-{\mathrm e}^{30} \ln \left (x \right )\right )\) | \(57\) |
parallelrisch | \(-\frac {{\mathrm e}^{-30} \left (-\ln \left (x \right )^{2} {\mathrm e}^{2 x} x^{2}+\ln \left (x \right )^{3} {\mathrm e}^{30}+18 \ln \left (x \right )^{2} {\mathrm e}^{2 x} x -81 \,{\mathrm e}^{2 x} \ln \left (x \right )^{2}+8 x \,{\mathrm e}^{x} \ln \left (x \right )-16-72 \,{\mathrm e}^{x} \ln \left (x \right )\right )}{\ln \left (x \right )^{2}}\) | \(69\) |
parts | \(\frac {\left (72 \,{\mathrm e}^{-15} {\mathrm e}^{x}-8 \,{\mathrm e}^{-15} x \,{\mathrm e}^{x}\right ) {\mathrm e}^{-15}}{\ln \left (x \right )}-\ln \left (x \right )+\frac {16 \,{\mathrm e}^{-30}}{\ln \left (x \right )^{2}}+2 \,{\mathrm e}^{-30} \left (\frac {{\mathrm e}^{2 x} x^{2}}{2}-9 x \,{\mathrm e}^{2 x}+\frac {81 \,{\mathrm e}^{2 x}}{2}\right )\) | \(72\) |
Input:
int((((2*x^3-34*x^2+144*x)*exp(x)^2-exp(15)^2)*ln(x)^3+(-8*x^2+64*x)*exp(x )*ln(x)^2+(8*x-72)*exp(x)*ln(x)-32)/x/exp(15)^2/ln(x)^3,x,method=_RETURNVE RBOSE)
Output:
x^2*exp(-30+2*x)-ln(x)-18*x*exp(-30+2*x)+81*exp(-30+2*x)-8*exp(-30)*(x*exp (x)*ln(x)-9*exp(x)*ln(x)-2)/ln(x)^2
Time = 0.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \frac {-32+e^x (-72+8 x) \log (x)+e^x \left (64 x-8 x^2\right ) \log ^2(x)+\left (-e^{30}+e^{2 x} \left (144 x-34 x^2+2 x^3\right )\right ) \log ^3(x)}{e^{30} x \log ^3(x)} \, dx=\frac {{\left ({\left (x^{2} - 18 \, x + 81\right )} e^{\left (2 \, x\right )} \log \left (x\right )^{2} - e^{30} \log \left (x\right )^{3} - 8 \, {\left (x - 9\right )} e^{x} \log \left (x\right ) + 16\right )} e^{\left (-30\right )}}{\log \left (x\right )^{2}} \] Input:
integrate((((2*x^3-34*x^2+144*x)*exp(x)^2-exp(15)^2)*log(x)^3+(-8*x^2+64*x )*exp(x)*log(x)^2+(8*x-72)*exp(x)*log(x)-32)/x/exp(15)^2/log(x)^3,x, algor ithm="fricas")
Output:
((x^2 - 18*x + 81)*e^(2*x)*log(x)^2 - e^30*log(x)^3 - 8*(x - 9)*e^x*log(x) + 16)*e^(-30)/log(x)^2
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (22) = 44\).
Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.27 \[ \int \frac {-32+e^x (-72+8 x) \log (x)+e^x \left (64 x-8 x^2\right ) \log ^2(x)+\left (-e^{30}+e^{2 x} \left (144 x-34 x^2+2 x^3\right )\right ) \log ^3(x)}{e^{30} x \log ^3(x)} \, dx=\frac {\left (- 8 x e^{30} + 72 e^{30}\right ) e^{x} + \left (x^{2} e^{30} \log {\left (x \right )} - 18 x e^{30} \log {\left (x \right )} + 81 e^{30} \log {\left (x \right )}\right ) e^{2 x}}{e^{60} \log {\left (x \right )}} - \log {\left (x \right )} + \frac {16}{e^{30} \log {\left (x \right )}^{2}} \] Input:
integrate((((2*x**3-34*x**2+144*x)*exp(x)**2-exp(15)**2)*ln(x)**3+(-8*x**2 +64*x)*exp(x)*ln(x)**2+(8*x-72)*exp(x)*ln(x)-32)/x/exp(15)**2/ln(x)**3,x)
Output:
((-8*x*exp(30) + 72*exp(30))*exp(x) + (x**2*exp(30)*log(x) - 18*x*exp(30)* log(x) + 81*exp(30)*log(x))*exp(2*x))*exp(-60)/log(x) - log(x) + 16*exp(-3 0)/log(x)**2
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (26) = 52\).
Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.00 \[ \int \frac {-32+e^x (-72+8 x) \log (x)+e^x \left (64 x-8 x^2\right ) \log ^2(x)+\left (-e^{30}+e^{2 x} \left (144 x-34 x^2+2 x^3\right )\right ) \log ^3(x)}{e^{30} x \log ^3(x)} \, dx=\frac {1}{2} \, {\left ({\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} - 17 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - 2 \, e^{30} \log \left (x\right ) - \frac {16 \, {\left (x - 9\right )} e^{x}}{\log \left (x\right )} + \frac {32}{\log \left (x\right )^{2}} + 144 \, e^{\left (2 \, x\right )}\right )} e^{\left (-30\right )} \] Input:
integrate((((2*x^3-34*x^2+144*x)*exp(x)^2-exp(15)^2)*log(x)^3+(-8*x^2+64*x )*exp(x)*log(x)^2+(8*x-72)*exp(x)*log(x)-32)/x/exp(15)^2/log(x)^3,x, algor ithm="maxima")
Output:
1/2*((2*x^2 - 2*x + 1)*e^(2*x) - 17*(2*x - 1)*e^(2*x) - 2*e^30*log(x) - 16 *(x - 9)*e^x/log(x) + 32/log(x)^2 + 144*e^(2*x))*e^(-30)
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (26) = 52\).
Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.10 \[ \int \frac {-32+e^x (-72+8 x) \log (x)+e^x \left (64 x-8 x^2\right ) \log ^2(x)+\left (-e^{30}+e^{2 x} \left (144 x-34 x^2+2 x^3\right )\right ) \log ^3(x)}{e^{30} x \log ^3(x)} \, dx=\frac {{\left (x^{2} e^{\left (2 \, x\right )} \log \left (x\right )^{2} - 18 \, x e^{\left (2 \, x\right )} \log \left (x\right )^{2} - e^{30} \log \left (x\right )^{3} - 8 \, x e^{x} \log \left (x\right ) + 81 \, e^{\left (2 \, x\right )} \log \left (x\right )^{2} + 72 \, e^{x} \log \left (x\right ) + 16\right )} e^{\left (-30\right )}}{\log \left (x\right )^{2}} \] Input:
integrate((((2*x^3-34*x^2+144*x)*exp(x)^2-exp(15)^2)*log(x)^3+(-8*x^2+64*x )*exp(x)*log(x)^2+(8*x-72)*exp(x)*log(x)-32)/x/exp(15)^2/log(x)^3,x, algor ithm="giac")
Output:
(x^2*e^(2*x)*log(x)^2 - 18*x*e^(2*x)*log(x)^2 - e^30*log(x)^3 - 8*x*e^x*lo g(x) + 81*e^(2*x)*log(x)^2 + 72*e^x*log(x) + 16)*e^(-30)/log(x)^2
Time = 2.97 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.03 \[ \int \frac {-32+e^x (-72+8 x) \log (x)+e^x \left (64 x-8 x^2\right ) \log ^2(x)+\left (-e^{30}+e^{2 x} \left (144 x-34 x^2+2 x^3\right )\right ) \log ^3(x)}{e^{30} x \log ^3(x)} \, dx=81\,{\mathrm {e}}^{2\,x-30}-\ln \left (x\right )+\frac {16\,{\mathrm {e}}^{-30}}{{\ln \left (x\right )}^2}-18\,x\,{\mathrm {e}}^{2\,x-30}+x^2\,{\mathrm {e}}^{2\,x-30}+\frac {72\,{\mathrm {e}}^{x-30}}{\ln \left (x\right )}-\frac {8\,x\,{\mathrm {e}}^{x-30}}{\ln \left (x\right )} \] Input:
int(-(exp(-30)*(log(x)^3*(exp(30) - exp(2*x)*(144*x - 34*x^2 + 2*x^3)) - e xp(x)*log(x)*(8*x - 72) - exp(x)*log(x)^2*(64*x - 8*x^2) + 32))/(x*log(x)^ 3),x)
Output:
81*exp(2*x - 30) - log(x) + (16*exp(-30))/log(x)^2 - 18*x*exp(2*x - 30) + x^2*exp(2*x - 30) + (72*exp(x - 30))/log(x) - (8*x*exp(x - 30))/log(x)
Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.33 \[ \int \frac {-32+e^x (-72+8 x) \log (x)+e^x \left (64 x-8 x^2\right ) \log ^2(x)+\left (-e^{30}+e^{2 x} \left (144 x-34 x^2+2 x^3\right )\right ) \log ^3(x)}{e^{30} x \log ^3(x)} \, dx=\frac {e^{2 x} \mathrm {log}\left (x \right )^{2} x^{2}-18 e^{2 x} \mathrm {log}\left (x \right )^{2} x +81 e^{2 x} \mathrm {log}\left (x \right )^{2}-8 e^{x} \mathrm {log}\left (x \right ) x +72 e^{x} \mathrm {log}\left (x \right )-\mathrm {log}\left (x \right )^{3} e^{30}+16}{\mathrm {log}\left (x \right )^{2} e^{30}} \] Input:
int((((2*x^3-34*x^2+144*x)*exp(x)^2-exp(15)^2)*log(x)^3+(-8*x^2+64*x)*exp( x)*log(x)^2+(8*x-72)*exp(x)*log(x)-32)/x/exp(15)^2/log(x)^3,x)
Output:
(e**(2*x)*log(x)**2*x**2 - 18*e**(2*x)*log(x)**2*x + 81*e**(2*x)*log(x)**2 - 8*e**x*log(x)*x + 72*e**x*log(x) - log(x)**3*e**30 + 16)/(log(x)**2*e** 30)