Integrand size = 94, antiderivative size = 20 \[ \int \frac {e^{e^{2 x^2}} \left (16+8 x-7 x^2-2 x^3+x^4\right )^{e^{e^{2 x^2}}} \left (-2+4 x+e^{2 x^2} \left (-16 x-4 x^2+4 x^3\right ) \log \left (16+8 x-7 x^2-2 x^3+x^4\right )\right )}{-4-x+x^2} \, dx=\left (\left (4+x-x^2\right )^2\right )^{e^{e^{2 x^2}}} \]
Time = 4.51 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{2 x^2}} \left (16+8 x-7 x^2-2 x^3+x^4\right )^{e^{e^{2 x^2}}} \left (-2+4 x+e^{2 x^2} \left (-16 x-4 x^2+4 x^3\right ) \log \left (16+8 x-7 x^2-2 x^3+x^4\right )\right )}{-4-x+x^2} \, dx=\left (\left (4+x-x^2\right )^2\right )^{e^{e^{2 x^2}}} \]
Integrate[(E^E^(2*x^2)*(16 + 8*x - 7*x^2 - 2*x^3 + x^4)^E^E^(2*x^2)*(-2 + 4*x + E^(2*x^2)*(-16*x - 4*x^2 + 4*x^3)*Log[16 + 8*x - 7*x^2 - 2*x^3 + x^4 ]))/(-4 - x + x^2),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.
\(\displaystyle \int \frac {e^{e^{2 x^2}} \left (x^4-2 x^3-7 x^2+8 x+16\right )^{e^{e^{2 x^2}}} \left (e^{2 x^2} \left (4 x^3-4 x^2-16 x\right ) \log \left (x^4-2 x^3-7 x^2+8 x+16\right )+4 x-2\right )}{x^2-x-4} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {2 e^{e^{2 x^2}} (2 x-1) \left (x^4-2 x^3-7 x^2+8 x+16\right )^{e^{e^{2 x^2}}}}{x^2-x-4}+4 e^{2 x^2+e^{2 x^2}} x \left (x^4-2 x^3-7 x^2+8 x+16\right )^{e^{e^{2 x^2}}} \log \left (\left (-x^2+x+4\right )^2\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {16 \int \frac {\int e^{2 x^2+e^{2 x^2}} x \left (\left (-x^2+x+4\right )^2\right )^{e^{e^{2 x^2}}}dx}{-2 x+\sqrt {17}+1}dx}{\sqrt {17}}-\frac {16}{17} \left (17+\sqrt {17}\right ) \int \frac {\int e^{2 x^2+e^{2 x^2}} x \left (\left (-x^2+x+4\right )^2\right )^{e^{e^{2 x^2}}}dx}{2 x-\sqrt {17}-1}dx-\frac {16}{17} \left (17-\sqrt {17}\right ) \int \frac {\int e^{2 x^2+e^{2 x^2}} x \left (\left (-x^2+x+4\right )^2\right )^{e^{e^{2 x^2}}}dx}{2 x+\sqrt {17}-1}dx-\frac {16 \int \frac {\int e^{2 x^2+e^{2 x^2}} x \left (\left (-x^2+x+4\right )^2\right )^{e^{e^{2 x^2}}}dx}{2 x+\sqrt {17}-1}dx}{\sqrt {17}}+4 \int \frac {e^{e^{2 x^2}} \left (x^4-2 x^3-7 x^2+8 x+16\right )^{e^{e^{2 x^2}}}}{2 x-\sqrt {17}-1}dx+4 \int \frac {e^{e^{2 x^2}} \left (x^4-2 x^3-7 x^2+8 x+16\right )^{e^{e^{2 x^2}}}}{2 x+\sqrt {17}-1}dx+4 \log \left (\left (-x^2+x+4\right )^2\right ) \int e^{2 x^2+e^{2 x^2}} x \left (x^4-2 x^3-7 x^2+8 x+16\right )^{e^{e^{2 x^2}}}dx\) |
Int[(E^E^(2*x^2)*(16 + 8*x - 7*x^2 - 2*x^3 + x^4)^E^E^(2*x^2)*(-2 + 4*x + E^(2*x^2)*(-16*x - 4*x^2 + 4*x^3)*Log[16 + 8*x - 7*x^2 - 2*x^3 + x^4]))/(- 4 - x + x^2),x]
3.2.61.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 26.94 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45
method | result | size |
parallelrisch | \({\mathrm e}^{\ln \left (x^{4}-2 x^{3}-7 x^{2}+8 x +16\right ) {\mathrm e}^{{\mathrm e}^{2 x^{2}}}}\) | \(29\) |
risch | \({\mathrm e}^{\frac {\left (-i \pi {\operatorname {csgn}\left (i \left (x^{2}-x -4\right )^{2}\right )}^{3}+2 i \pi {\operatorname {csgn}\left (i \left (x^{2}-x -4\right )^{2}\right )}^{2} \operatorname {csgn}\left (i \left (x^{2}-x -4\right )\right )-i \pi \,\operatorname {csgn}\left (i \left (x^{2}-x -4\right )^{2}\right ) {\operatorname {csgn}\left (i \left (x^{2}-x -4\right )\right )}^{2}+4 \ln \left (x^{2}-x -4\right )\right ) {\mathrm e}^{{\mathrm e}^{2 x^{2}}}}{2}}\) | \(107\) |
int(((4*x^3-4*x^2-16*x)*exp(x^2)^2*ln(x^4-2*x^3-7*x^2+8*x+16)+4*x-2)*exp(e xp(x^2)^2)*exp(ln(x^4-2*x^3-7*x^2+8*x+16)*exp(exp(x^2)^2))/(x^2-x-4),x,met hod=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {e^{e^{2 x^2}} \left (16+8 x-7 x^2-2 x^3+x^4\right )^{e^{e^{2 x^2}}} \left (-2+4 x+e^{2 x^2} \left (-16 x-4 x^2+4 x^3\right ) \log \left (16+8 x-7 x^2-2 x^3+x^4\right )\right )}{-4-x+x^2} \, dx={\left (x^{4} - 2 \, x^{3} - 7 \, x^{2} + 8 \, x + 16\right )}^{e^{\left (e^{\left (2 \, x^{2}\right )}\right )}} \]
integrate(((4*x^3-4*x^2-16*x)*exp(x^2)^2*log(x^4-2*x^3-7*x^2+8*x+16)+4*x-2 )*exp(exp(x^2)^2)*exp(log(x^4-2*x^3-7*x^2+8*x+16)*exp(exp(x^2)^2))/(x^2-x- 4),x, algorithm=\
Timed out. \[ \int \frac {e^{e^{2 x^2}} \left (16+8 x-7 x^2-2 x^3+x^4\right )^{e^{e^{2 x^2}}} \left (-2+4 x+e^{2 x^2} \left (-16 x-4 x^2+4 x^3\right ) \log \left (16+8 x-7 x^2-2 x^3+x^4\right )\right )}{-4-x+x^2} \, dx=\text {Timed out} \]
integrate(((4*x**3-4*x**2-16*x)*exp(x**2)**2*ln(x**4-2*x**3-7*x**2+8*x+16) +4*x-2)*exp(exp(x**2)**2)*exp(ln(x**4-2*x**3-7*x**2+8*x+16)*exp(exp(x**2)* *2))/(x**2-x-4),x)
Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{e^{2 x^2}} \left (16+8 x-7 x^2-2 x^3+x^4\right )^{e^{e^{2 x^2}}} \left (-2+4 x+e^{2 x^2} \left (-16 x-4 x^2+4 x^3\right ) \log \left (16+8 x-7 x^2-2 x^3+x^4\right )\right )}{-4-x+x^2} \, dx={\left (x^{2} - x - 4\right )}^{2 \, e^{\left (e^{\left (2 \, x^{2}\right )}\right )}} \]
integrate(((4*x^3-4*x^2-16*x)*exp(x^2)^2*log(x^4-2*x^3-7*x^2+8*x+16)+4*x-2 )*exp(exp(x^2)^2)*exp(log(x^4-2*x^3-7*x^2+8*x+16)*exp(exp(x^2)^2))/(x^2-x- 4),x, algorithm=\
\[ \int \frac {e^{e^{2 x^2}} \left (16+8 x-7 x^2-2 x^3+x^4\right )^{e^{e^{2 x^2}}} \left (-2+4 x+e^{2 x^2} \left (-16 x-4 x^2+4 x^3\right ) \log \left (16+8 x-7 x^2-2 x^3+x^4\right )\right )}{-4-x+x^2} \, dx=\int { \frac {2 \, {\left (2 \, {\left (x^{3} - x^{2} - 4 \, x\right )} e^{\left (2 \, x^{2}\right )} \log \left (x^{4} - 2 \, x^{3} - 7 \, x^{2} + 8 \, x + 16\right ) + 2 \, x - 1\right )} {\left (x^{4} - 2 \, x^{3} - 7 \, x^{2} + 8 \, x + 16\right )}^{e^{\left (e^{\left (2 \, x^{2}\right )}\right )}} e^{\left (e^{\left (2 \, x^{2}\right )}\right )}}{x^{2} - x - 4} \,d x } \]
integrate(((4*x^3-4*x^2-16*x)*exp(x^2)^2*log(x^4-2*x^3-7*x^2+8*x+16)+4*x-2 )*exp(exp(x^2)^2)*exp(log(x^4-2*x^3-7*x^2+8*x+16)*exp(exp(x^2)^2))/(x^2-x- 4),x, algorithm=\
integrate(2*(2*(x^3 - x^2 - 4*x)*e^(2*x^2)*log(x^4 - 2*x^3 - 7*x^2 + 8*x + 16) + 2*x - 1)*(x^4 - 2*x^3 - 7*x^2 + 8*x + 16)^e^(e^(2*x^2))*e^(e^(2*x^2 ))/(x^2 - x - 4), x)
Time = 8.63 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {e^{e^{2 x^2}} \left (16+8 x-7 x^2-2 x^3+x^4\right )^{e^{e^{2 x^2}}} \left (-2+4 x+e^{2 x^2} \left (-16 x-4 x^2+4 x^3\right ) \log \left (16+8 x-7 x^2-2 x^3+x^4\right )\right )}{-4-x+x^2} \, dx={\left (x^4-2\,x^3-7\,x^2+8\,x+16\right )}^{{\mathrm {e}}^{{\mathrm {e}}^{2\,x^2}}} \]