Integrand size = 111, antiderivative size = 31 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=-x+9 x^2+\frac {3 e^{3-x}}{x (4 x-2 \log (x))} \] Output:
9*x^2+3/exp(x)*exp(3)/(4*x-2*ln(x))/x-x
Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=\frac {1}{2} \left (-2 x+18 x^2-\frac {3 e^{3-x}}{x (-2 x+\log (x))}\right ) \] Input:
Integrate[(E^3*(3 - 12*x - 6*x^2) + E^x*(-8*x^4 + 144*x^5) + (E^3*(3 + 3*x ) + E^x*(8*x^3 - 144*x^4))*Log[x] + E^x*(-2*x^2 + 36*x^3)*Log[x]^2)/(8*E^x *x^4 - 8*E^x*x^3*Log[x] + 2*E^x*x^2*Log[x]^2),x]
Output:
(-2*x + 18*x^2 - (3*E^(3 - x))/(x*(-2*x + Log[x])))/2
Time = 1.54 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {7292, 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.
\(\displaystyle \int \frac {e^3 \left (-6 x^2-12 x+3\right )+e^x \left (144 x^5-8 x^4\right )+\left (e^x \left (8 x^3-144 x^4\right )+e^3 (3 x+3)\right ) \log (x)+e^x \left (36 x^3-2 x^2\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-x} \left (e^3 \left (-6 x^2-12 x+3\right )+e^x \left (144 x^5-8 x^4\right )+\left (e^x \left (8 x^3-144 x^4\right )+e^3 (3 x+3)\right ) \log (x)+e^x \left (36 x^3-2 x^2\right ) \log ^2(x)\right )}{2 x^2 (2 x-\log (x))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {e^{-x} \left (-2 e^x \left (x^2-18 x^3\right ) \log ^2(x)+\left (3 e^3 (x+1)+8 e^x \left (x^3-18 x^4\right )\right ) \log (x)+3 e^3 \left (-2 x^2-4 x+1\right )-8 e^x \left (x^4-18 x^5\right )\right )}{x^2 (2 x-\log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{2} \int \left (2 (18 x-1)-\frac {3 e^{3-x} \left (2 x^2-\log (x) x+4 x-\log (x)-1\right )}{x^2 (2 x-\log (x))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {3 e^{3-x} \left (2 x^2-x \log (x)\right )}{x^2 (2 x-\log (x))^2}+\frac {1}{18} (1-18 x)^2\right )\) |
Input:
Int[(E^3*(3 - 12*x - 6*x^2) + E^x*(-8*x^4 + 144*x^5) + (E^3*(3 + 3*x) + E^ x*(8*x^3 - 144*x^4))*Log[x] + E^x*(-2*x^2 + 36*x^3)*Log[x]^2)/(8*E^x*x^4 - 8*E^x*x^3*Log[x] + 2*E^x*x^2*Log[x]^2),x]
Output:
((1 - 18*x)^2/18 + (3*E^(3 - x)*(2*x^2 - x*Log[x]))/(x^2*(2*x - Log[x])^2) )/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 2.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00
method | result | size |
risch | \(9 x^{2}-x +\frac {3 \,{\mathrm e}^{-x +3}}{2 \left (2 x -\ln \left (x \right )\right ) x}\) | \(31\) |
parallelrisch | \(\frac {\left (36 \,{\mathrm e}^{x} x^{4}-18 x^{3} {\mathrm e}^{x} \ln \left (x \right )-4 \,{\mathrm e}^{x} x^{3}+2 x^{2} {\mathrm e}^{x} \ln \left (x \right )+3 \,{\mathrm e}^{3}\right ) {\mathrm e}^{-x}}{2 x \left (2 x -\ln \left (x \right )\right )}\) | \(57\) |
Input:
int(((36*x^3-2*x^2)*exp(x)*ln(x)^2+((-144*x^4+8*x^3)*exp(x)+(3*x+3)*exp(3) )*ln(x)+(144*x^5-8*x^4)*exp(x)+(-6*x^2-12*x+3)*exp(3))/(2*x^2*exp(x)*ln(x) ^2-8*x^3*exp(x)*ln(x)+8*exp(x)*x^4),x,method=_RETURNVERBOSE)
Output:
9*x^2-x+3/2/(2*x-ln(x))/x*exp(-x+3)
Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.81 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=-\frac {2 \, {\left (9 \, x^{3} - x^{2}\right )} e^{x} \log \left (x\right ) - 4 \, {\left (9 \, x^{4} - x^{3}\right )} e^{x} - 3 \, e^{3}}{2 \, {\left (2 \, x^{2} e^{x} - x e^{x} \log \left (x\right )\right )}} \] Input:
integrate(((36*x^3-2*x^2)*exp(x)*log(x)^2+((-144*x^4+8*x^3)*exp(x)+(3*x+3) *exp(3))*log(x)+(144*x^5-8*x^4)*exp(x)+(-6*x^2-12*x+3)*exp(3))/(2*x^2*exp( x)*log(x)^2-8*x^3*exp(x)*log(x)+8*exp(x)*x^4),x, algorithm="fricas")
Output:
-1/2*(2*(9*x^3 - x^2)*e^x*log(x) - 4*(9*x^4 - x^3)*e^x - 3*e^3)/(2*x^2*e^x - x*e^x*log(x))
Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=9 x^{2} - x + \frac {3 e^{3} e^{- x}}{4 x^{2} - 2 x \log {\left (x \right )}} \] Input:
integrate(((36*x**3-2*x**2)*exp(x)*ln(x)**2+((-144*x**4+8*x**3)*exp(x)+(3* x+3)*exp(3))*ln(x)+(144*x**5-8*x**4)*exp(x)+(-6*x**2-12*x+3)*exp(3))/(2*x* *2*exp(x)*ln(x)**2-8*x**3*exp(x)*ln(x)+8*exp(x)*x**4),x)
Output:
9*x**2 - x + 3*exp(3)*exp(-x)/(4*x**2 - 2*x*log(x))
Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=\frac {36 \, x^{4} - 4 \, x^{3} - 2 \, {\left (9 \, x^{3} - x^{2}\right )} \log \left (x\right ) + 3 \, e^{\left (-x + 3\right )}}{2 \, {\left (2 \, x^{2} - x \log \left (x\right )\right )}} \] Input:
integrate(((36*x^3-2*x^2)*exp(x)*log(x)^2+((-144*x^4+8*x^3)*exp(x)+(3*x+3) *exp(3))*log(x)+(144*x^5-8*x^4)*exp(x)+(-6*x^2-12*x+3)*exp(3))/(2*x^2*exp( x)*log(x)^2-8*x^3*exp(x)*log(x)+8*exp(x)*x^4),x, algorithm="maxima")
Output:
1/2*(36*x^4 - 4*x^3 - 2*(9*x^3 - x^2)*log(x) + 3*e^(-x + 3))/(2*x^2 - x*lo g(x))
Time = 0.13 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=\frac {36 \, x^{4} - 18 \, x^{3} \log \left (x\right ) - 4 \, x^{3} + 2 \, x^{2} \log \left (x\right ) + 3 \, e^{\left (-x + 3\right )}}{2 \, {\left (2 \, x^{2} - x \log \left (x\right )\right )}} \] Input:
integrate(((36*x^3-2*x^2)*exp(x)*log(x)^2+((-144*x^4+8*x^3)*exp(x)+(3*x+3) *exp(3))*log(x)+(144*x^5-8*x^4)*exp(x)+(-6*x^2-12*x+3)*exp(3))/(2*x^2*exp( x)*log(x)^2-8*x^3*exp(x)*log(x)+8*exp(x)*x^4),x, algorithm="giac")
Output:
1/2*(36*x^4 - 18*x^3*log(x) - 4*x^3 + 2*x^2*log(x) + 3*e^(-x + 3))/(2*x^2 - x*log(x))
Time = 2.99 (sec) , antiderivative size = 105, normalized size of antiderivative = 3.39 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=\frac {\frac {3\,{\mathrm {e}}^{-x}\,\left (2\,{\mathrm {e}}^3\,x^2+4\,{\mathrm {e}}^3\,x-{\mathrm {e}}^3\right )}{2\,x\,\left (2\,x-1\right )}-\frac {3\,{\mathrm {e}}^{-x}\,\ln \left (x\right )\,\left ({\mathrm {e}}^3+x\,{\mathrm {e}}^3\right )}{2\,x\,\left (2\,x-1\right )}}{2\,x-\ln \left (x\right )}-x+9\,x^2+\frac {{\mathrm {e}}^{-x}\,\left (\frac {3\,{\mathrm {e}}^3}{4}+\frac {3\,x\,{\mathrm {e}}^3}{4}\right )}{\frac {x}{2}-x^2} \] Input:
int(-(exp(x)*(8*x^4 - 144*x^5) + exp(3)*(12*x + 6*x^2 - 3) - log(x)*(exp(x )*(8*x^3 - 144*x^4) + exp(3)*(3*x + 3)) + exp(x)*log(x)^2*(2*x^2 - 36*x^3) )/(8*x^4*exp(x) + 2*x^2*exp(x)*log(x)^2 - 8*x^3*exp(x)*log(x)),x)
Output:
((3*exp(-x)*(4*x*exp(3) - exp(3) + 2*x^2*exp(3)))/(2*x*(2*x - 1)) - (3*exp (-x)*log(x)*(exp(3) + x*exp(3)))/(2*x*(2*x - 1)))/(2*x - log(x)) - x + 9*x ^2 + (exp(-x)*((3*exp(3))/4 + (3*x*exp(3))/4))/(x/2 - x^2)
Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.94 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=\frac {18 e^{x} \mathrm {log}\left (x \right ) x^{3}-2 e^{x} \mathrm {log}\left (x \right ) x^{2}-36 e^{x} x^{4}+4 e^{x} x^{3}-3 e^{3}}{2 e^{x} x \left (\mathrm {log}\left (x \right )-2 x \right )} \] Input:
int(((36*x^3-2*x^2)*exp(x)*log(x)^2+((-144*x^4+8*x^3)*exp(x)+(3*x+3)*exp(3 ))*log(x)+(144*x^5-8*x^4)*exp(x)+(-6*x^2-12*x+3)*exp(3))/(2*x^2*exp(x)*log (x)^2-8*x^3*exp(x)*log(x)+8*exp(x)*x^4),x)
Output:
(18*e**x*log(x)*x**3 - 2*e**x*log(x)*x**2 - 36*e**x*x**4 + 4*e**x*x**3 - 3 *e**3)/(2*e**x*x*(log(x) - 2*x))