Integrand size = 113, antiderivative size = 28 \[ \int \frac {-96 x+24 x^2+e^{x^2} (-96+24 x)+\left (-32 x+16 x^2+e^{x^2} \left (8 x-64 x^2+16 x^3\right )\right ) \log \left (\frac {1}{x^3}\right )}{16 x^3-8 x^4+x^5+e^{2 x^2} \left (16 x-8 x^2+x^3\right )+e^{x^2} \left (32 x^2-16 x^3+2 x^4\right )} \, dx=\frac {4 \log \left (\frac {1}{x^3}\right )}{\left (e^{x^2}+x\right ) \left (-x+\frac {4+x}{2}\right )} \]
Time = 0.39 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {-96 x+24 x^2+e^{x^2} (-96+24 x)+\left (-32 x+16 x^2+e^{x^2} \left (8 x-64 x^2+16 x^3\right )\right ) \log \left (\frac {1}{x^3}\right )}{16 x^3-8 x^4+x^5+e^{2 x^2} \left (16 x-8 x^2+x^3\right )+e^{x^2} \left (32 x^2-16 x^3+2 x^4\right )} \, dx=-\frac {8 \log \left (\frac {1}{x^3}\right )}{(-4+x) \left (e^{x^2}+x\right )} \]
Integrate[(-96*x + 24*x^2 + E^x^2*(-96 + 24*x) + (-32*x + 16*x^2 + E^x^2*( 8*x - 64*x^2 + 16*x^3))*Log[x^(-3)])/(16*x^3 - 8*x^4 + x^5 + E^(2*x^2)*(16 *x - 8*x^2 + x^3) + E^x^2*(32*x^2 - 16*x^3 + 2*x^4)),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.
\(\displaystyle \int \frac {24 x^2+e^{x^2} (24 x-96)+\left (16 x^2+e^{x^2} \left (16 x^3-64 x^2+8 x\right )-32 x\right ) \log \left (\frac {1}{x^3}\right )-96 x}{x^5-8 x^4+16 x^3+e^{2 x^2} \left (x^3-8 x^2+16 x\right )+e^{x^2} \left (2 x^4-16 x^3+32 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {8 \left (3 (x-4) \left (e^{x^2}+x\right )+x \left (e^{x^2} \left (2 x^2-8 x+1\right )+2 (x-2)\right ) \log \left (\frac {1}{x^3}\right )\right )}{(4-x)^2 x \left (e^{x^2}+x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 8 \int -\frac {3 (4-x) \left (x+e^{x^2}\right )+x \left (2 (2-x)-e^{x^2} \left (2 x^2-8 x+1\right )\right ) \log \left (\frac {1}{x^3}\right )}{(4-x)^2 x \left (x+e^{x^2}\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -8 \int \frac {3 (4-x) \left (x+e^{x^2}\right )+x \left (2 (2-x)-e^{x^2} \left (2 x^2-8 x+1\right )\right ) \log \left (\frac {1}{x^3}\right )}{(4-x)^2 x \left (x+e^{x^2}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -8 \int \left (\frac {\left (2 x^2-1\right ) \log \left (\frac {1}{x^3}\right )}{(x-4) \left (x+e^{x^2}\right )^2}-\frac {2 \log \left (\frac {1}{x^3}\right ) x^3-8 \log \left (\frac {1}{x^3}\right ) x^2+\log \left (\frac {1}{x^3}\right ) x+3 x-12}{(x-4)^2 x \left (x+e^{x^2}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \left (-\frac {3}{4} \int \frac {1}{(x-4) \left (x+e^{x^2}\right )}dx+\frac {3}{4} \int \frac {1}{x \left (x+e^{x^2}\right )}dx+24 \int \frac {\int \frac {1}{\left (x+e^{x^2}\right )^2}dx}{x}dx+93 \int \frac {\int \frac {1}{(x-4) \left (x+e^{x^2}\right )^2}dx}{x}dx+6 \int \frac {\int \frac {x}{\left (x+e^{x^2}\right )^2}dx}{x}dx-6 \int \frac {\int \frac {1}{x+e^{x^2}}dx}{x}dx-3 \int \frac {\int \frac {1}{(x-4)^2 \left (x+e^{x^2}\right )}dx}{x}dx-24 \int \frac {\int \frac {1}{(x-4) \left (x+e^{x^2}\right )}dx}{x}dx+8 \log \left (\frac {1}{x^3}\right ) \int \frac {1}{\left (x+e^{x^2}\right )^2}dx+31 \log \left (\frac {1}{x^3}\right ) \int \frac {1}{(x-4) \left (x+e^{x^2}\right )^2}dx+2 \log \left (\frac {1}{x^3}\right ) \int \frac {x}{\left (x+e^{x^2}\right )^2}dx-2 \log \left (\frac {1}{x^3}\right ) \int \frac {1}{x+e^{x^2}}dx-\log \left (\frac {1}{x^3}\right ) \int \frac {1}{(x-4)^2 \left (x+e^{x^2}\right )}dx-8 \log \left (\frac {1}{x^3}\right ) \int \frac {1}{(x-4) \left (x+e^{x^2}\right )}dx\right )\) |
Int[(-96*x + 24*x^2 + E^x^2*(-96 + 24*x) + (-32*x + 16*x^2 + E^x^2*(8*x - 64*x^2 + 16*x^3))*Log[x^(-3)])/(16*x^3 - 8*x^4 + x^5 + E^(2*x^2)*(16*x - 8 *x^2 + x^3) + E^x^2*(32*x^2 - 16*x^3 + 2*x^4)),x]
3.14.67.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.37 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(-\frac {8 \ln \left (\frac {1}{x^{3}}\right )}{{\mathrm e}^{x^{2}} x +x^{2}-4 \,{\mathrm e}^{x^{2}}-4 x}\) | \(28\) |
risch | \(\frac {24 \ln \left (x \right )}{\left (x -4\right ) \left ({\mathrm e}^{x^{2}}+x \right )}-\frac {4 i \pi \left (\operatorname {csgn}\left (i x^{3}\right )^{3}-\operatorname {csgn}\left (i x^{3}\right )^{2} \operatorname {csgn}\left (i x \right )-\operatorname {csgn}\left (i x^{3}\right )^{2} \operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\operatorname {csgn}\left (i x^{2}\right )^{3}\right )}{\left (x -4\right ) \left ({\mathrm e}^{x^{2}}+x \right )}\) | \(140\) |
int((((16*x^3-64*x^2+8*x)*exp(x^2)+16*x^2-32*x)*ln(1/x^3)+(24*x-96)*exp(x^ 2)+24*x^2-96*x)/((x^3-8*x^2+16*x)*exp(x^2)^2+(2*x^4-16*x^3+32*x^2)*exp(x^2 )+x^5-8*x^4+16*x^3),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {-96 x+24 x^2+e^{x^2} (-96+24 x)+\left (-32 x+16 x^2+e^{x^2} \left (8 x-64 x^2+16 x^3\right )\right ) \log \left (\frac {1}{x^3}\right )}{16 x^3-8 x^4+x^5+e^{2 x^2} \left (16 x-8 x^2+x^3\right )+e^{x^2} \left (32 x^2-16 x^3+2 x^4\right )} \, dx=-\frac {8 \, \log \left (\frac {1}{x^{3}}\right )}{x^{2} + {\left (x - 4\right )} e^{\left (x^{2}\right )} - 4 \, x} \]
integrate((((16*x^3-64*x^2+8*x)*exp(x^2)+16*x^2-32*x)*log(1/x^3)+(24*x-96) *exp(x^2)+24*x^2-96*x)/((x^3-8*x^2+16*x)*exp(x^2)^2+(2*x^4-16*x^3+32*x^2)* exp(x^2)+x^5-8*x^4+16*x^3),x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-96 x+24 x^2+e^{x^2} (-96+24 x)+\left (-32 x+16 x^2+e^{x^2} \left (8 x-64 x^2+16 x^3\right )\right ) \log \left (\frac {1}{x^3}\right )}{16 x^3-8 x^4+x^5+e^{2 x^2} \left (16 x-8 x^2+x^3\right )+e^{x^2} \left (32 x^2-16 x^3+2 x^4\right )} \, dx=- \frac {8 \log {\left (\frac {1}{x^{3}} \right )}}{x^{2} - 4 x + \left (x - 4\right ) e^{x^{2}}} \]
integrate((((16*x**3-64*x**2+8*x)*exp(x**2)+16*x**2-32*x)*ln(1/x**3)+(24*x -96)*exp(x**2)+24*x**2-96*x)/((x**3-8*x**2+16*x)*exp(x**2)**2+(2*x**4-16*x **3+32*x**2)*exp(x**2)+x**5-8*x**4+16*x**3),x)
Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-96 x+24 x^2+e^{x^2} (-96+24 x)+\left (-32 x+16 x^2+e^{x^2} \left (8 x-64 x^2+16 x^3\right )\right ) \log \left (\frac {1}{x^3}\right )}{16 x^3-8 x^4+x^5+e^{2 x^2} \left (16 x-8 x^2+x^3\right )+e^{x^2} \left (32 x^2-16 x^3+2 x^4\right )} \, dx=\frac {24 \, \log \left (x\right )}{x^{2} + {\left (x - 4\right )} e^{\left (x^{2}\right )} - 4 \, x} \]
integrate((((16*x^3-64*x^2+8*x)*exp(x^2)+16*x^2-32*x)*log(1/x^3)+(24*x-96) *exp(x^2)+24*x^2-96*x)/((x^3-8*x^2+16*x)*exp(x^2)^2+(2*x^4-16*x^3+32*x^2)* exp(x^2)+x^5-8*x^4+16*x^3),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-96 x+24 x^2+e^{x^2} (-96+24 x)+\left (-32 x+16 x^2+e^{x^2} \left (8 x-64 x^2+16 x^3\right )\right ) \log \left (\frac {1}{x^3}\right )}{16 x^3-8 x^4+x^5+e^{2 x^2} \left (16 x-8 x^2+x^3\right )+e^{x^2} \left (32 x^2-16 x^3+2 x^4\right )} \, dx=\frac {8 \, \log \left (x^{3}\right )}{x^{2} + x e^{\left (x^{2}\right )} - 4 \, x - 4 \, e^{\left (x^{2}\right )}} \]
integrate((((16*x^3-64*x^2+8*x)*exp(x^2)+16*x^2-32*x)*log(1/x^3)+(24*x-96) *exp(x^2)+24*x^2-96*x)/((x^3-8*x^2+16*x)*exp(x^2)^2+(2*x^4-16*x^3+32*x^2)* exp(x^2)+x^5-8*x^4+16*x^3),x, algorithm=\
Time = 9.57 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-96 x+24 x^2+e^{x^2} (-96+24 x)+\left (-32 x+16 x^2+e^{x^2} \left (8 x-64 x^2+16 x^3\right )\right ) \log \left (\frac {1}{x^3}\right )}{16 x^3-8 x^4+x^5+e^{2 x^2} \left (16 x-8 x^2+x^3\right )+e^{x^2} \left (32 x^2-16 x^3+2 x^4\right )} \, dx=-\frac {8\,\ln \left (\frac {1}{x^3}\right )}{\left (x+{\mathrm {e}}^{x^2}\right )\,\left (x-4\right )} \]