Integrand size = 221, antiderivative size = 33 \[ \int \frac {e^{2+2 x} (36-9 x)+4 x-x^2+e^{1+x} \left (8 x-6 x^2+x^3\right )+\left (-8 x+2 x^2+e^{1+x} (-72+18 x)\right ) \log (x)+(36-9 x) \log ^2(x)+\left (e^{2+2 x} (-48-9 x)-e^{1+x} x^2+\left (x^2+e^{1+x} (96+18 x)\right ) \log (x)+(-48-9 x) \log ^2(x)\right ) \log \left (\frac {x^2+e^{1+x} (48+9 x)+(-48-9 x) \log (x)}{-12 e^{1+x}+12 \log (x)}\right )}{e^{1+x} x^2+e^{2+2 x} (48+9 x)+\left (e^{1+x} (-96-18 x)-x^2\right ) \log (x)+(48+9 x) \log ^2(x)} \, dx=(4-x) \log \left (-4+\frac {1}{4} x \left (-3+\frac {x}{3 \left (-e^{1+x}+\log (x)\right )}\right )\right ) \] Output:
ln(1/4*(x/(3*ln(x)-3*exp(1+x))-3)*x-4)*(4-x)
Leaf count is larger than twice the leaf count of optimal. \(89\) vs. \(2(33)=66\).
Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.70 \[ \int \frac {e^{2+2 x} (36-9 x)+4 x-x^2+e^{1+x} \left (8 x-6 x^2+x^3\right )+\left (-8 x+2 x^2+e^{1+x} (-72+18 x)\right ) \log (x)+(36-9 x) \log ^2(x)+\left (e^{2+2 x} (-48-9 x)-e^{1+x} x^2+\left (x^2+e^{1+x} (96+18 x)\right ) \log (x)+(-48-9 x) \log ^2(x)\right ) \log \left (\frac {x^2+e^{1+x} (48+9 x)+(-48-9 x) \log (x)}{-12 e^{1+x}+12 \log (x)}\right )}{e^{1+x} x^2+e^{2+2 x} (48+9 x)+\left (e^{1+x} (-96-18 x)-x^2\right ) \log (x)+(48+9 x) \log ^2(x)} \, dx=-4 \log \left (e^{1+x}-\log (x)\right )+4 \log \left (48 e^{1+x}+9 e^{1+x} x+x^2-48 \log (x)-9 x \log (x)\right )-x \log \left (\frac {x^2+e^{1+x} (48+9 x)-3 (16+3 x) \log (x)}{12 \left (-e^{1+x}+\log (x)\right )}\right ) \] Input:
Integrate[(E^(2 + 2*x)*(36 - 9*x) + 4*x - x^2 + E^(1 + x)*(8*x - 6*x^2 + x ^3) + (-8*x + 2*x^2 + E^(1 + x)*(-72 + 18*x))*Log[x] + (36 - 9*x)*Log[x]^2 + (E^(2 + 2*x)*(-48 - 9*x) - E^(1 + x)*x^2 + (x^2 + E^(1 + x)*(96 + 18*x) )*Log[x] + (-48 - 9*x)*Log[x]^2)*Log[(x^2 + E^(1 + x)*(48 + 9*x) + (-48 - 9*x)*Log[x])/(-12*E^(1 + x) + 12*Log[x])])/(E^(1 + x)*x^2 + E^(2 + 2*x)*(4 8 + 9*x) + (E^(1 + x)*(-96 - 18*x) - x^2)*Log[x] + (48 + 9*x)*Log[x]^2),x]
Output:
-4*Log[E^(1 + x) - Log[x]] + 4*Log[48*E^(1 + x) + 9*E^(1 + x)*x + x^2 - 48 *Log[x] - 9*x*Log[x]] - x*Log[(x^2 + E^(1 + x)*(48 + 9*x) - 3*(16 + 3*x)*L og[x])/(12*(-E^(1 + x) + Log[x]))]
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.
| \(\displaystyle \int \frac {-x^2+\left (-e^{x+1} x^2+\left (x^2+e^{x+1} (18 x+96)\right ) \log (x)+e^{2 x+2} (-9 x-48)+(-9 x-48) \log ^2(x)\right ) \log \left (\frac {x^2+e^{x+1} (9 x+48)+(-9 x-48) \log (x)}{12 \log (x)-12 e^{x+1}}\right )+\left (2 x^2-8 x+e^{x+1} (18 x-72)\right ) \log (x)+e^{x+1} \left (x^3-6 x^2+8 x\right )+4 x+e^{2 x+2} (36-9 x)+(36-9 x) \log ^2(x)}{e^{x+1} x^2+\left (e^{x+1} (-18 x-96)-x^2\right ) \log (x)+e^{2 x+2} (9 x+48)+(9 x+48) \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-x^2+\left (-e^{x+1} x^2+\left (x^2+e^{x+1} (18 x+96)\right ) \log (x)+e^{2 x+2} (-9 x-48)+(-9 x-48) \log ^2(x)\right ) \log \left (\frac {x^2+e^{x+1} (9 x+48)+(-9 x-48) \log (x)}{12 \log (x)-12 e^{x+1}}\right )+\left (2 x^2-8 x+e^{x+1} (18 x-72)\right ) \log (x)+e^{x+1} \left (x^3-6 x^2+8 x\right )+4 x+e^{2 x+2} (36-9 x)+(36-9 x) \log ^2(x)}{\left (e^{x+1}-\log (x)\right ) \left (x^2+9 e^{x+1} x+48 e^{x+1}-9 x \log (x)-48 \log (x)\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {-3 x \log \left (\frac {x^2+e^{x+1} (9 x+48)-3 (3 x+16) \log (x)}{12 \left (\log (x)-e^{x+1}\right )}\right )-16 \log \left (\frac {x^2+e^{x+1} (9 x+48)-3 (3 x+16) \log (x)}{12 \left (\log (x)-e^{x+1}\right )}\right )-3 x+12}{3 x+16}+\frac {(x-4) \left (3 x^4+13 x^3-27 x^3 \log (x)-5 x^2-288 x^2 \log (x)+288 x-768 x \log (x)+768\right )}{x (3 x+16) \left (x^2+9 e^{x+1} x+48 e^{x+1}-9 x \log (x)-48 \log (x)\right )}+\frac {(x-4) (x \log (x)-1)}{x \left (e^{x+1}-\log (x)\right )}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \left (\frac {-3 x \log \left (\frac {x^2+e^{x+1} (9 x+48)-3 (3 x+16) \log (x)}{12 \left (\log (x)-e^{x+1}\right )}\right )-16 \log \left (\frac {x^2+e^{x+1} (9 x+48)-3 (3 x+16) \log (x)}{12 \left (\log (x)-e^{x+1}\right )}\right )-3 x+12}{3 x+16}+\frac {(x-4) \left (3 x^4+13 x^3-27 x^3 \log (x)-5 x^2-288 x^2 \log (x)+288 x-768 x \log (x)+768\right )}{x (3 x+16) \left (x^2+9 e^{x+1} x+48 e^{x+1}-9 x \log (x)-48 \log (x)\right )}+\frac {(x-4) (x \log (x)-1)}{x \left (e^{x+1}-\log (x)\right )}\right )dx\) |
Input:
Int[(E^(2 + 2*x)*(36 - 9*x) + 4*x - x^2 + E^(1 + x)*(8*x - 6*x^2 + x^3) + (-8*x + 2*x^2 + E^(1 + x)*(-72 + 18*x))*Log[x] + (36 - 9*x)*Log[x]^2 + (E^ (2 + 2*x)*(-48 - 9*x) - E^(1 + x)*x^2 + (x^2 + E^(1 + x)*(96 + 18*x))*Log[ x] + (-48 - 9*x)*Log[x]^2)*Log[(x^2 + E^(1 + x)*(48 + 9*x) + (-48 - 9*x)*L og[x])/(-12*E^(1 + x) + 12*Log[x])])/(E^(1 + x)*x^2 + E^(2 + 2*x)*(48 + 9* x) + (E^(1 + x)*(-96 - 18*x) - x^2)*Log[x] + (48 + 9*x)*Log[x]^2),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(78\) vs. \(2(29)=58\).
Time = 9.57 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.39
| method | result | size |
| parallelrisch | \(-\ln \left (\frac {\left (-9 x -48\right ) \ln \left (x \right )+\left (9 x +48\right ) {\mathrm e}^{1+x}+x^{2}}{12 \ln \left (x \right )-12 \,{\mathrm e}^{1+x}}\right ) x +4 \ln \left (\frac {\left (-9 x -48\right ) \ln \left (x \right )+\left (9 x +48\right ) {\mathrm e}^{1+x}+x^{2}}{12 \ln \left (x \right )-12 \,{\mathrm e}^{1+x}}\right )\) | \(79\) |
| risch | \(-x \ln \left (x^{2}+\left (-9 \ln \left (x \right )+9 \,{\mathrm e}^{1+x}\right ) x -48 \ln \left (x \right )+48 \,{\mathrm e}^{1+x}\right )+x \ln \left (-\ln \left (x \right )+{\mathrm e}^{1+x}\right )+\frac {i \pi x \,\operatorname {csgn}\left (i \left (-x^{2}+\left (9 \ln \left (x \right )-9 \,{\mathrm e}^{1+x}\right ) x +48 \ln \left (x \right )-48 \,{\mathrm e}^{1+x}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+\left (9 \ln \left (x \right )-9 \,{\mathrm e}^{1+x}\right ) x +48 \ln \left (x \right )-48 \,{\mathrm e}^{1+x}\right )}{\ln \left (x \right )-{\mathrm e}^{1+x}}\right )}^{2}}{2}+i \pi x {\operatorname {csgn}\left (\frac {i \left (-x^{2}+\left (9 \ln \left (x \right )-9 \,{\mathrm e}^{1+x}\right ) x +48 \ln \left (x \right )-48 \,{\mathrm e}^{1+x}\right )}{\ln \left (x \right )-{\mathrm e}^{1+x}}\right )}^{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )-{\mathrm e}^{1+x}}\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+\left (9 \ln \left (x \right )-9 \,{\mathrm e}^{1+x}\right ) x +48 \ln \left (x \right )-48 \,{\mathrm e}^{1+x}\right )}{\ln \left (x \right )-{\mathrm e}^{1+x}}\right )}^{2}}{2}-\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (-x^{2}+\left (9 \ln \left (x \right )-9 \,{\mathrm e}^{1+x}\right ) x +48 \ln \left (x \right )-48 \,{\mathrm e}^{1+x}\right )}{\ln \left (x \right )-{\mathrm e}^{1+x}}\right )}^{3}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )-{\mathrm e}^{1+x}}\right ) \operatorname {csgn}\left (i \left (-x^{2}+\left (9 \ln \left (x \right )-9 \,{\mathrm e}^{1+x}\right ) x +48 \ln \left (x \right )-48 \,{\mathrm e}^{1+x}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-x^{2}+\left (9 \ln \left (x \right )-9 \,{\mathrm e}^{1+x}\right ) x +48 \ln \left (x \right )-48 \,{\mathrm e}^{1+x}\right )}{\ln \left (x \right )-{\mathrm e}^{1+x}}\right )}{2}-i \pi x +x \ln \left (3\right )+2 x \ln \left (2\right )+4 \ln \left (3 x +16\right )+4 \ln \left (\ln \left (x \right )-\frac {x^{2}+9 x \,{\mathrm e}^{1+x}+48 \,{\mathrm e}^{1+x}}{3 \left (3 x +16\right )}\right )-4 \ln \left (\ln \left (x \right )-{\mathrm e}^{1+x}\right )\) | \(460\) |
Input:
int((((-9*x-48)*ln(x)^2+((18*x+96)*exp(1+x)+x^2)*ln(x)+(-9*x-48)*exp(1+x)^ 2-x^2*exp(1+x))*ln(((-9*x-48)*ln(x)+(9*x+48)*exp(1+x)+x^2)/(12*ln(x)-12*ex p(1+x)))+(-9*x+36)*ln(x)^2+((18*x-72)*exp(1+x)+2*x^2-8*x)*ln(x)+(-9*x+36)* exp(1+x)^2+(x^3-6*x^2+8*x)*exp(1+x)-x^2+4*x)/((9*x+48)*ln(x)^2+((-18*x-96) *exp(1+x)-x^2)*ln(x)+(9*x+48)*exp(1+x)^2+x^2*exp(1+x)),x,method=_RETURNVER BOSE)
Output:
-ln(1/12/(ln(x)-exp(1+x))*((-9*x-48)*ln(x)+(9*x+48)*exp(1+x)+x^2))*x+4*ln( 1/12/(ln(x)-exp(1+x))*((-9*x-48)*ln(x)+(9*x+48)*exp(1+x)+x^2))
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {e^{2+2 x} (36-9 x)+4 x-x^2+e^{1+x} \left (8 x-6 x^2+x^3\right )+\left (-8 x+2 x^2+e^{1+x} (-72+18 x)\right ) \log (x)+(36-9 x) \log ^2(x)+\left (e^{2+2 x} (-48-9 x)-e^{1+x} x^2+\left (x^2+e^{1+x} (96+18 x)\right ) \log (x)+(-48-9 x) \log ^2(x)\right ) \log \left (\frac {x^2+e^{1+x} (48+9 x)+(-48-9 x) \log (x)}{-12 e^{1+x}+12 \log (x)}\right )}{e^{1+x} x^2+e^{2+2 x} (48+9 x)+\left (e^{1+x} (-96-18 x)-x^2\right ) \log (x)+(48+9 x) \log ^2(x)} \, dx=-{\left (x - 4\right )} \log \left (-\frac {x^{2} + 3 \, {\left (3 \, x + 16\right )} e^{\left (x + 1\right )} - 3 \, {\left (3 \, x + 16\right )} \log \left (x\right )}{12 \, {\left (e^{\left (x + 1\right )} - \log \left (x\right )\right )}}\right ) \] Input:
integrate((((-9*x-48)*log(x)^2+((18*x+96)*exp(1+x)+x^2)*log(x)+(-9*x-48)*e xp(1+x)^2-x^2*exp(1+x))*log(((-9*x-48)*log(x)+(9*x+48)*exp(1+x)+x^2)/(12*l og(x)-12*exp(1+x)))+(-9*x+36)*log(x)^2+((18*x-72)*exp(1+x)+2*x^2-8*x)*log( x)+(-9*x+36)*exp(1+x)^2+(x^3-6*x^2+8*x)*exp(1+x)-x^2+4*x)/((9*x+48)*log(x) ^2+((-18*x-96)*exp(1+x)-x^2)*log(x)+(9*x+48)*exp(1+x)^2+x^2*exp(1+x)),x, a lgorithm="fricas")
Output:
-(x - 4)*log(-1/12*(x^2 + 3*(3*x + 16)*e^(x + 1) - 3*(3*x + 16)*log(x))/(e ^(x + 1) - log(x)))
Exception generated. \[ \int \frac {e^{2+2 x} (36-9 x)+4 x-x^2+e^{1+x} \left (8 x-6 x^2+x^3\right )+\left (-8 x+2 x^2+e^{1+x} (-72+18 x)\right ) \log (x)+(36-9 x) \log ^2(x)+\left (e^{2+2 x} (-48-9 x)-e^{1+x} x^2+\left (x^2+e^{1+x} (96+18 x)\right ) \log (x)+(-48-9 x) \log ^2(x)\right ) \log \left (\frac {x^2+e^{1+x} (48+9 x)+(-48-9 x) \log (x)}{-12 e^{1+x}+12 \log (x)}\right )}{e^{1+x} x^2+e^{2+2 x} (48+9 x)+\left (e^{1+x} (-96-18 x)-x^2\right ) \log (x)+(48+9 x) \log ^2(x)} \, dx=\text {Exception raised: PolynomialError} \] Input:
integrate((((-9*x-48)*ln(x)**2+((18*x+96)*exp(1+x)+x**2)*ln(x)+(-9*x-48)*e xp(1+x)**2-x**2*exp(1+x))*ln(((-9*x-48)*ln(x)+(9*x+48)*exp(1+x)+x**2)/(12* ln(x)-12*exp(1+x)))+(-9*x+36)*ln(x)**2+((18*x-72)*exp(1+x)+2*x**2-8*x)*ln( x)+(-9*x+36)*exp(1+x)**2+(x**3-6*x**2+8*x)*exp(1+x)-x**2+4*x)/((9*x+48)*ln (x)**2+((-18*x-96)*exp(1+x)-x**2)*ln(x)+(9*x+48)*exp(1+x)**2+x**2*exp(1+x) ),x)
Output:
Exception raised: PolynomialError >> 1/(27*x**3 + 288*x**2 + 768*x) contai ns an element of the set of generators.
Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (26) = 52\).
Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 3.64 \[ \int \frac {e^{2+2 x} (36-9 x)+4 x-x^2+e^{1+x} \left (8 x-6 x^2+x^3\right )+\left (-8 x+2 x^2+e^{1+x} (-72+18 x)\right ) \log (x)+(36-9 x) \log ^2(x)+\left (e^{2+2 x} (-48-9 x)-e^{1+x} x^2+\left (x^2+e^{1+x} (96+18 x)\right ) \log (x)+(-48-9 x) \log ^2(x)\right ) \log \left (\frac {x^2+e^{1+x} (48+9 x)+(-48-9 x) \log (x)}{-12 e^{1+x}+12 \log (x)}\right )}{e^{1+x} x^2+e^{2+2 x} (48+9 x)+\left (e^{1+x} (-96-18 x)-x^2\right ) \log (x)+(48+9 x) \log ^2(x)} \, dx=x {\left (\log \left (3\right ) + 2 \, \log \left (2\right )\right )} - x \log \left (x^{2} + 3 \, {\left (3 \, x e + 16 \, e\right )} e^{x} - 3 \, {\left (3 \, x + 16\right )} \log \left (x\right )\right ) + x \log \left (-e^{\left (x + 1\right )} + \log \left (x\right )\right ) - 4 \, \log \left ({\left (e^{\left (x + 1\right )} - \log \left (x\right )\right )} e^{\left (-1\right )}\right ) + 4 \, \log \left (3 \, x + 16\right ) + 4 \, \log \left (\frac {x^{2} + 3 \, {\left (3 \, x e + 16 \, e\right )} e^{x} - 3 \, {\left (3 \, x + 16\right )} \log \left (x\right )}{3 \, {\left (3 \, x e + 16 \, e\right )}}\right ) \] Input:
integrate((((-9*x-48)*log(x)^2+((18*x+96)*exp(1+x)+x^2)*log(x)+(-9*x-48)*e xp(1+x)^2-x^2*exp(1+x))*log(((-9*x-48)*log(x)+(9*x+48)*exp(1+x)+x^2)/(12*l og(x)-12*exp(1+x)))+(-9*x+36)*log(x)^2+((18*x-72)*exp(1+x)+2*x^2-8*x)*log( x)+(-9*x+36)*exp(1+x)^2+(x^3-6*x^2+8*x)*exp(1+x)-x^2+4*x)/((9*x+48)*log(x) ^2+((-18*x-96)*exp(1+x)-x^2)*log(x)+(9*x+48)*exp(1+x)^2+x^2*exp(1+x)),x, a lgorithm="maxima")
Output:
x*(log(3) + 2*log(2)) - x*log(x^2 + 3*(3*x*e + 16*e)*e^x - 3*(3*x + 16)*lo g(x)) + x*log(-e^(x + 1) + log(x)) - 4*log((e^(x + 1) - log(x))*e^(-1)) + 4*log(3*x + 16) + 4*log(1/3*(x^2 + 3*(3*x*e + 16*e)*e^x - 3*(3*x + 16)*log (x))/(3*x*e + 16*e))
Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (26) = 52\).
Time = 2.33 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.73 \[ \int \frac {e^{2+2 x} (36-9 x)+4 x-x^2+e^{1+x} \left (8 x-6 x^2+x^3\right )+\left (-8 x+2 x^2+e^{1+x} (-72+18 x)\right ) \log (x)+(36-9 x) \log ^2(x)+\left (e^{2+2 x} (-48-9 x)-e^{1+x} x^2+\left (x^2+e^{1+x} (96+18 x)\right ) \log (x)+(-48-9 x) \log ^2(x)\right ) \log \left (\frac {x^2+e^{1+x} (48+9 x)+(-48-9 x) \log (x)}{-12 e^{1+x}+12 \log (x)}\right )}{e^{1+x} x^2+e^{2+2 x} (48+9 x)+\left (e^{1+x} (-96-18 x)-x^2\right ) \log (x)+(48+9 x) \log ^2(x)} \, dx=-x \log \left (-x^{2} - 9 \, x e^{\left (x + 1\right )} + 9 \, x \log \left (x\right ) - 48 \, e^{\left (x + 1\right )} + 48 \, \log \left (x\right )\right ) + x \log \left (12 \, e^{\left (x + 1\right )} - 12 \, \log \left (x\right )\right ) + 4 \, \log \left (-x^{2} - 9 \, x e^{\left (x + 1\right )} + 9 \, x \log \left (x\right ) - 48 \, e^{\left (x + 1\right )} + 48 \, \log \left (x\right )\right ) - 4 \, \log \left (-e^{\left (x + 1\right )} + \log \left (x\right )\right ) \] Input:
integrate((((-9*x-48)*log(x)^2+((18*x+96)*exp(1+x)+x^2)*log(x)+(-9*x-48)*e xp(1+x)^2-x^2*exp(1+x))*log(((-9*x-48)*log(x)+(9*x+48)*exp(1+x)+x^2)/(12*l og(x)-12*exp(1+x)))+(-9*x+36)*log(x)^2+((18*x-72)*exp(1+x)+2*x^2-8*x)*log( x)+(-9*x+36)*exp(1+x)^2+(x^3-6*x^2+8*x)*exp(1+x)-x^2+4*x)/((9*x+48)*log(x) ^2+((-18*x-96)*exp(1+x)-x^2)*log(x)+(9*x+48)*exp(1+x)^2+x^2*exp(1+x)),x, a lgorithm="giac")
Output:
-x*log(-x^2 - 9*x*e^(x + 1) + 9*x*log(x) - 48*e^(x + 1) + 48*log(x)) + x*l og(12*e^(x + 1) - 12*log(x)) + 4*log(-x^2 - 9*x*e^(x + 1) + 9*x*log(x) - 4 8*e^(x + 1) + 48*log(x)) - 4*log(-e^(x + 1) + log(x))
Timed out. \[ \int \frac {e^{2+2 x} (36-9 x)+4 x-x^2+e^{1+x} \left (8 x-6 x^2+x^3\right )+\left (-8 x+2 x^2+e^{1+x} (-72+18 x)\right ) \log (x)+(36-9 x) \log ^2(x)+\left (e^{2+2 x} (-48-9 x)-e^{1+x} x^2+\left (x^2+e^{1+x} (96+18 x)\right ) \log (x)+(-48-9 x) \log ^2(x)\right ) \log \left (\frac {x^2+e^{1+x} (48+9 x)+(-48-9 x) \log (x)}{-12 e^{1+x}+12 \log (x)}\right )}{e^{1+x} x^2+e^{2+2 x} (48+9 x)+\left (e^{1+x} (-96-18 x)-x^2\right ) \log (x)+(48+9 x) \log ^2(x)} \, dx=\int -\frac {{\mathrm {e}}^{2\,x+2}\,\left (9\,x-36\right )-4\,x+\ln \left (-\frac {{\mathrm {e}}^{x+1}\,\left (9\,x+48\right )-\ln \left (x\right )\,\left (9\,x+48\right )+x^2}{12\,{\mathrm {e}}^{x+1}-12\,\ln \left (x\right )}\right )\,\left ({\mathrm {e}}^{2\,x+2}\,\left (9\,x+48\right )-\ln \left (x\right )\,\left ({\mathrm {e}}^{x+1}\,\left (18\,x+96\right )+x^2\right )+x^2\,{\mathrm {e}}^{x+1}+{\ln \left (x\right )}^2\,\left (9\,x+48\right )\right )-\ln \left (x\right )\,\left ({\mathrm {e}}^{x+1}\,\left (18\,x-72\right )-8\,x+2\,x^2\right )+x^2+{\ln \left (x\right )}^2\,\left (9\,x-36\right )-{\mathrm {e}}^{x+1}\,\left (x^3-6\,x^2+8\,x\right )}{{\mathrm {e}}^{2\,x+2}\,\left (9\,x+48\right )-\ln \left (x\right )\,\left ({\mathrm {e}}^{x+1}\,\left (18\,x+96\right )+x^2\right )+x^2\,{\mathrm {e}}^{x+1}+{\ln \left (x\right )}^2\,\left (9\,x+48\right )} \,d x \] Input:
int(-(exp(2*x + 2)*(9*x - 36) - 4*x + log(-(exp(x + 1)*(9*x + 48) - log(x) *(9*x + 48) + x^2)/(12*exp(x + 1) - 12*log(x)))*(exp(2*x + 2)*(9*x + 48) - log(x)*(exp(x + 1)*(18*x + 96) + x^2) + x^2*exp(x + 1) + log(x)^2*(9*x + 48)) - log(x)*(exp(x + 1)*(18*x - 72) - 8*x + 2*x^2) + x^2 + log(x)^2*(9*x - 36) - exp(x + 1)*(8*x - 6*x^2 + x^3))/(exp(2*x + 2)*(9*x + 48) - log(x) *(exp(x + 1)*(18*x + 96) + x^2) + x^2*exp(x + 1) + log(x)^2*(9*x + 48)),x)
Output:
int(-(exp(2*x + 2)*(9*x - 36) - 4*x + log(-(exp(x + 1)*(9*x + 48) - log(x) *(9*x + 48) + x^2)/(12*exp(x + 1) - 12*log(x)))*(exp(2*x + 2)*(9*x + 48) - log(x)*(exp(x + 1)*(18*x + 96) + x^2) + x^2*exp(x + 1) + log(x)^2*(9*x + 48)) - log(x)*(exp(x + 1)*(18*x - 72) - 8*x + 2*x^2) + x^2 + log(x)^2*(9*x - 36) - exp(x + 1)*(8*x - 6*x^2 + x^3))/(exp(2*x + 2)*(9*x + 48) - log(x) *(exp(x + 1)*(18*x + 96) + x^2) + x^2*exp(x + 1) + log(x)^2*(9*x + 48)), x )
Time = 0.24 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int \frac {e^{2+2 x} (36-9 x)+4 x-x^2+e^{1+x} \left (8 x-6 x^2+x^3\right )+\left (-8 x+2 x^2+e^{1+x} (-72+18 x)\right ) \log (x)+(36-9 x) \log ^2(x)+\left (e^{2+2 x} (-48-9 x)-e^{1+x} x^2+\left (x^2+e^{1+x} (96+18 x)\right ) \log (x)+(-48-9 x) \log ^2(x)\right ) \log \left (\frac {x^2+e^{1+x} (48+9 x)+(-48-9 x) \log (x)}{-12 e^{1+x}+12 \log (x)}\right )}{e^{1+x} x^2+e^{2+2 x} (48+9 x)+\left (e^{1+x} (-96-18 x)-x^2\right ) \log (x)+(48+9 x) \log ^2(x)} \, dx=\mathrm {log}\left (\frac {-9 e^{x} e x -48 e^{x} e +9 \,\mathrm {log}\left (x \right ) x +48 \,\mathrm {log}\left (x \right )-x^{2}}{12 e^{x} e -12 \,\mathrm {log}\left (x \right )}\right ) \left (-x +4\right ) \] Input:
int((((-9*x-48)*log(x)^2+((18*x+96)*exp(1+x)+x^2)*log(x)+(-9*x-48)*exp(1+x )^2-x^2*exp(1+x))*log(((-9*x-48)*log(x)+(9*x+48)*exp(1+x)+x^2)/(12*log(x)- 12*exp(1+x)))+(-9*x+36)*log(x)^2+((18*x-72)*exp(1+x)+2*x^2-8*x)*log(x)+(-9 *x+36)*exp(1+x)^2+(x^3-6*x^2+8*x)*exp(1+x)-x^2+4*x)/((9*x+48)*log(x)^2+((- 18*x-96)*exp(1+x)-x^2)*log(x)+(9*x+48)*exp(1+x)^2+x^2*exp(1+x)),x)
Output:
log(( - 9*e**x*e*x - 48*e**x*e + 9*log(x)*x + 48*log(x) - x**2)/(12*e**x*e - 12*log(x)))*( - x + 4)