Integrand size = 93, antiderivative size = 25 \[ \int \frac {e^x \left (54+108 x-162 x^2-135 x^3+54 x^4\right )-27 e^x x \log \left (x^2\right )}{16 x^3-8 x^4-15 x^5+4 x^6+4 x^7+\left (8 x^2-2 x^3-4 x^4\right ) \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx=\frac {27 e^x}{-4 x+x^2+2 x^3-\log \left (x^2\right )} \] Output:
27*exp(x)/(x^2-ln(x^2)-4*x+2*x^3)
Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (54+108 x-162 x^2-135 x^3+54 x^4\right )-27 e^x x \log \left (x^2\right )}{16 x^3-8 x^4-15 x^5+4 x^6+4 x^7+\left (8 x^2-2 x^3-4 x^4\right ) \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx=-\frac {27 e^x}{4 x-x^2-2 x^3+\log \left (x^2\right )} \] Input:
Integrate[(E^x*(54 + 108*x - 162*x^2 - 135*x^3 + 54*x^4) - 27*E^x*x*Log[x^ 2])/(16*x^3 - 8*x^4 - 15*x^5 + 4*x^6 + 4*x^7 + (8*x^2 - 2*x^3 - 4*x^4)*Log [x^2] + x*Log[x^2]^2),x]
Output:
(-27*E^x)/(4*x - x^2 - 2*x^3 + Log[x^2])
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.
\(\displaystyle \int \frac {e^x \left (54 x^4-135 x^3-162 x^2+108 x+54\right )-27 e^x x \log \left (x^2\right )}{4 x^7+4 x^6-15 x^5-8 x^4+16 x^3+x \log ^2\left (x^2\right )+\left (-4 x^4-2 x^3+8 x^2\right ) \log \left (x^2\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {27 e^x \left (2 x^4-5 x^3-6 x^2-x \log \left (x^2\right )+4 x+2\right )}{x \left (x \left (2 x^2+x-4\right )-\log \left (x^2\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 27 \int \frac {e^x \left (2 x^4-5 x^3-6 x^2-\log \left (x^2\right ) x+4 x+2\right )}{x \left (x \left (-2 x^2-x+4\right )+\log \left (x^2\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 27 \int \left (\frac {e^x}{2 x^3+x^2-4 x-\log \left (x^2\right )}-\frac {2 e^x \left (3 x^3+x^2-2 x-1\right )}{x \left (2 x^3+x^2-4 x-\log \left (x^2\right )\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 27 \left (4 \int \frac {e^x}{\left (2 x^3+x^2-4 x-\log \left (x^2\right )\right )^2}dx+2 \int \frac {e^x}{x \left (2 x^3+x^2-4 x-\log \left (x^2\right )\right )^2}dx-2 \int \frac {e^x x}{\left (2 x^3+x^2-4 x-\log \left (x^2\right )\right )^2}dx-6 \int \frac {e^x x^2}{\left (2 x^3+x^2-4 x-\log \left (x^2\right )\right )^2}dx+\int \frac {e^x}{2 x^3+x^2-4 x-\log \left (x^2\right )}dx\right )\) |
Input:
Int[(E^x*(54 + 108*x - 162*x^2 - 135*x^3 + 54*x^4) - 27*E^x*x*Log[x^2])/(1 6*x^3 - 8*x^4 - 15*x^5 + 4*x^6 + 4*x^7 + (8*x^2 - 2*x^3 - 4*x^4)*Log[x^2] + x*Log[x^2]^2),x]
Output:
$Aborted
Time = 0.34 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {27 \,{\mathrm e}^{x}}{x^{2}-\ln \left (x^{2}\right )-4 x +2 x^{3}}\) | \(25\) |
risch | \(-\frac {54 i {\mathrm e}^{x}}{\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-4 i x^{3}-2 i x^{2}+8 i x +4 i \ln \left (x \right )}\) | \(74\) |
Input:
int((-27*x*exp(x)*ln(x^2)+(54*x^4-135*x^3-162*x^2+108*x+54)*exp(x))/(x*ln( x^2)^2+(-4*x^4-2*x^3+8*x^2)*ln(x^2)+4*x^7+4*x^6-15*x^5-8*x^4+16*x^3),x,met hod=_RETURNVERBOSE)
Output:
27*exp(x)/(x^2-ln(x^2)-4*x+2*x^3)
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^x \left (54+108 x-162 x^2-135 x^3+54 x^4\right )-27 e^x x \log \left (x^2\right )}{16 x^3-8 x^4-15 x^5+4 x^6+4 x^7+\left (8 x^2-2 x^3-4 x^4\right ) \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx=\frac {27 \, e^{x}}{2 \, x^{3} + x^{2} - 4 \, x - \log \left (x^{2}\right )} \] Input:
integrate((-27*x*exp(x)*log(x^2)+(54*x^4-135*x^3-162*x^2+108*x+54)*exp(x)) /(x*log(x^2)^2+(-4*x^4-2*x^3+8*x^2)*log(x^2)+4*x^7+4*x^6-15*x^5-8*x^4+16*x ^3),x, algorithm="fricas")
Output:
27*e^x/(2*x^3 + x^2 - 4*x - log(x^2))
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^x \left (54+108 x-162 x^2-135 x^3+54 x^4\right )-27 e^x x \log \left (x^2\right )}{16 x^3-8 x^4-15 x^5+4 x^6+4 x^7+\left (8 x^2-2 x^3-4 x^4\right ) \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx=\frac {27 e^{x}}{2 x^{3} + x^{2} - 4 x - \log {\left (x^{2} \right )}} \] Input:
integrate((-27*x*exp(x)*ln(x**2)+(54*x**4-135*x**3-162*x**2+108*x+54)*exp( x))/(x*ln(x**2)**2+(-4*x**4-2*x**3+8*x**2)*ln(x**2)+4*x**7+4*x**6-15*x**5- 8*x**4+16*x**3),x)
Output:
27*exp(x)/(2*x**3 + x**2 - 4*x - log(x**2))
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^x \left (54+108 x-162 x^2-135 x^3+54 x^4\right )-27 e^x x \log \left (x^2\right )}{16 x^3-8 x^4-15 x^5+4 x^6+4 x^7+\left (8 x^2-2 x^3-4 x^4\right ) \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx=\frac {27 \, e^{x}}{2 \, x^{3} + x^{2} - 4 \, x - 2 \, \log \left (x\right )} \] Input:
integrate((-27*x*exp(x)*log(x^2)+(54*x^4-135*x^3-162*x^2+108*x+54)*exp(x)) /(x*log(x^2)^2+(-4*x^4-2*x^3+8*x^2)*log(x^2)+4*x^7+4*x^6-15*x^5-8*x^4+16*x ^3),x, algorithm="maxima")
Output:
27*e^x/(2*x^3 + x^2 - 4*x - 2*log(x))
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^x \left (54+108 x-162 x^2-135 x^3+54 x^4\right )-27 e^x x \log \left (x^2\right )}{16 x^3-8 x^4-15 x^5+4 x^6+4 x^7+\left (8 x^2-2 x^3-4 x^4\right ) \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx=\frac {27 \, e^{x}}{2 \, x^{3} + x^{2} - 4 \, x - \log \left (x^{2}\right )} \] Input:
integrate((-27*x*exp(x)*log(x^2)+(54*x^4-135*x^3-162*x^2+108*x+54)*exp(x)) /(x*log(x^2)^2+(-4*x^4-2*x^3+8*x^2)*log(x^2)+4*x^7+4*x^6-15*x^5-8*x^4+16*x ^3),x, algorithm="giac")
Output:
27*e^x/(2*x^3 + x^2 - 4*x - log(x^2))
Timed out. \[ \int \frac {e^x \left (54+108 x-162 x^2-135 x^3+54 x^4\right )-27 e^x x \log \left (x^2\right )}{16 x^3-8 x^4-15 x^5+4 x^6+4 x^7+\left (8 x^2-2 x^3-4 x^4\right ) \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx=\int \frac {{\mathrm {e}}^x\,\left (54\,x^4-135\,x^3-162\,x^2+108\,x+54\right )-27\,x\,\ln \left (x^2\right )\,{\mathrm {e}}^x}{x\,{\ln \left (x^2\right )}^2-\ln \left (x^2\right )\,\left (4\,x^4+2\,x^3-8\,x^2\right )+16\,x^3-8\,x^4-15\,x^5+4\,x^6+4\,x^7} \,d x \] Input:
int((exp(x)*(108*x - 162*x^2 - 135*x^3 + 54*x^4 + 54) - 27*x*log(x^2)*exp( x))/(x*log(x^2)^2 - log(x^2)*(2*x^3 - 8*x^2 + 4*x^4) + 16*x^3 - 8*x^4 - 15 *x^5 + 4*x^6 + 4*x^7),x)
Output:
int((exp(x)*(108*x - 162*x^2 - 135*x^3 + 54*x^4 + 54) - 27*x*log(x^2)*exp( x))/(x*log(x^2)^2 - log(x^2)*(2*x^3 - 8*x^2 + 4*x^4) + 16*x^3 - 8*x^4 - 15 *x^5 + 4*x^6 + 4*x^7), x)
Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (54+108 x-162 x^2-135 x^3+54 x^4\right )-27 e^x x \log \left (x^2\right )}{16 x^3-8 x^4-15 x^5+4 x^6+4 x^7+\left (8 x^2-2 x^3-4 x^4\right ) \log \left (x^2\right )+x \log ^2\left (x^2\right )} \, dx=-\frac {27 e^{x}}{\mathrm {log}\left (x^{2}\right )-2 x^{3}-x^{2}+4 x} \] Input:
int((-27*x*exp(x)*log(x^2)+(54*x^4-135*x^3-162*x^2+108*x+54)*exp(x))/(x*lo g(x^2)^2+(-4*x^4-2*x^3+8*x^2)*log(x^2)+4*x^7+4*x^6-15*x^5-8*x^4+16*x^3),x)
Output:
( - 27*e**x)/(log(x**2) - 2*x**3 - x**2 + 4*x)