Integrand size = 115, antiderivative size = 28 \[ \int \frac {e^{\frac {-x^2+x \log \left (x^2\right )+\log \left (\frac {16-\log (3+x)}{e}\right )}{x}} \left (95 x-16 x^2-16 x^3+\left (-6 x+x^2+x^3\right ) \log (3+x)+(-48-16 x+(3+x) \log (3+x)) \log \left (\frac {16-\log (3+x)}{e}\right )\right )}{-48 x^2-16 x^3+\left (3 x^2+x^3\right ) \log (3+x)} \, dx=-e^{-x+\frac {\log \left (\frac {16-\log (3+x)}{e}\right )}{x}} x^2 \]
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-x^2+x \log \left (x^2\right )+\log \left (\frac {16-\log (3+x)}{e}\right )}{x}} \left (95 x-16 x^2-16 x^3+\left (-6 x+x^2+x^3\right ) \log (3+x)+(-48-16 x+(3+x) \log (3+x)) \log \left (\frac {16-\log (3+x)}{e}\right )\right )}{-48 x^2-16 x^3+\left (3 x^2+x^3\right ) \log (3+x)} \, dx=-e^{-\frac {1}{x}-x} x^2 (16-\log (3+x))^{\frac {1}{x}} \]
Integrate[(E^((-x^2 + x*Log[x^2] + Log[(16 - Log[3 + x])/E])/x)*(95*x - 16 *x^2 - 16*x^3 + (-6*x + x^2 + x^3)*Log[3 + x] + (-48 - 16*x + (3 + x)*Log[ 3 + x])*Log[(16 - Log[3 + x])/E]))/(-48*x^2 - 16*x^3 + (3*x^2 + x^3)*Log[3 + x]),x]
Time = 0.62 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {7257}
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.
\(\displaystyle \int \frac {e^{\frac {-x^2+x \log \left (x^2\right )+\log \left (\frac {16-\log (x+3)}{e}\right )}{x}} \left (-16 x^3-16 x^2+\left (x^3+x^2-6 x\right ) \log (x+3)+95 x+(-16 x+(x+3) \log (x+3)-48) \log \left (\frac {16-\log (x+3)}{e}\right )\right )}{-16 x^3-48 x^2+\left (x^3+3 x^2\right ) \log (x+3)} \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle -e^{-x-\frac {1}{x}} x^2 (16-\log (x+3))^{\frac {1}{x}}\) |
Int[(E^((-x^2 + x*Log[x^2] + Log[(16 - Log[3 + x])/E])/x)*(95*x - 16*x^2 - 16*x^3 + (-6*x + x^2 + x^3)*Log[3 + x] + (-48 - 16*x + (3 + x)*Log[3 + x] )*Log[(16 - Log[3 + x])/E]))/(-48*x^2 - 16*x^3 + (3*x^2 + x^3)*Log[3 + x]) ,x]
3.9.56.3.1 Defintions of rubi rules used
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 38.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18
method | result | size |
parallelrisch | \(-{\mathrm e}^{\frac {\ln \left (-\left (\ln \left (3+x \right )-16\right ) {\mathrm e}^{-1}\right )+x \ln \left (x^{2}\right )-x^{2}}{x}}\) | \(33\) |
risch | \(-x^{2} \left (\left (-\ln \left (3+x \right )+16\right ) {\mathrm e}^{-1}\right )^{\frac {1}{x}} {\mathrm e}^{-\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}+i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}}{2}-x}\) | \(75\) |
int((((3+x)*ln(3+x)-16*x-48)*ln((-ln(3+x)+16)/exp(1))+(x^3+x^2-6*x)*ln(3+x )-16*x^3-16*x^2+95*x)*exp((ln((-ln(3+x)+16)/exp(1))+x*ln(x^2)-x^2)/x)/((x^ 3+3*x^2)*ln(3+x)-16*x^3-48*x^2),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {-x^2+x \log \left (x^2\right )+\log \left (\frac {16-\log (3+x)}{e}\right )}{x}} \left (95 x-16 x^2-16 x^3+\left (-6 x+x^2+x^3\right ) \log (3+x)+(-48-16 x+(3+x) \log (3+x)) \log \left (\frac {16-\log (3+x)}{e}\right )\right )}{-48 x^2-16 x^3+\left (3 x^2+x^3\right ) \log (3+x)} \, dx=-e^{\left (-\frac {x^{2} - x \log \left (x^{2}\right ) - \log \left (-{\left (\log \left (x + 3\right ) - 16\right )} e^{\left (-1\right )}\right )}{x}\right )} \]
integrate((((3+x)*log(3+x)-16*x-48)*log((-log(3+x)+16)/exp(1))+(x^3+x^2-6* x)*log(3+x)-16*x^3-16*x^2+95*x)*exp((log((-log(3+x)+16)/exp(1))+x*log(x^2) -x^2)/x)/((x^3+3*x^2)*log(3+x)-16*x^3-48*x^2),x, algorithm=\
Time = 1.48 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {-x^2+x \log \left (x^2\right )+\log \left (\frac {16-\log (3+x)}{e}\right )}{x}} \left (95 x-16 x^2-16 x^3+\left (-6 x+x^2+x^3\right ) \log (3+x)+(-48-16 x+(3+x) \log (3+x)) \log \left (\frac {16-\log (3+x)}{e}\right )\right )}{-48 x^2-16 x^3+\left (3 x^2+x^3\right ) \log (3+x)} \, dx=- e^{\frac {- x^{2} + x \log {\left (x^{2} \right )} + \log {\left (\frac {16 - \log {\left (x + 3 \right )}}{e} \right )}}{x}} \]
integrate((((3+x)*ln(3+x)-16*x-48)*ln((-ln(3+x)+16)/exp(1))+(x**3+x**2-6*x )*ln(3+x)-16*x**3-16*x**2+95*x)*exp((ln((-ln(3+x)+16)/exp(1))+x*ln(x**2)-x **2)/x)/((x**3+3*x**2)*ln(3+x)-16*x**3-48*x**2),x)
Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-x^2+x \log \left (x^2\right )+\log \left (\frac {16-\log (3+x)}{e}\right )}{x}} \left (95 x-16 x^2-16 x^3+\left (-6 x+x^2+x^3\right ) \log (3+x)+(-48-16 x+(3+x) \log (3+x)) \log \left (\frac {16-\log (3+x)}{e}\right )\right )}{-48 x^2-16 x^3+\left (3 x^2+x^3\right ) \log (3+x)} \, dx=-x^{2} e^{\left (-x + \frac {\log \left (-\log \left (x + 3\right ) + 16\right )}{x} - \frac {1}{x}\right )} \]
integrate((((3+x)*log(3+x)-16*x-48)*log((-log(3+x)+16)/exp(1))+(x^3+x^2-6* x)*log(3+x)-16*x^3-16*x^2+95*x)*exp((log((-log(3+x)+16)/exp(1))+x*log(x^2) -x^2)/x)/((x^3+3*x^2)*log(3+x)-16*x^3-48*x^2),x, algorithm=\
\[ \int \frac {e^{\frac {-x^2+x \log \left (x^2\right )+\log \left (\frac {16-\log (3+x)}{e}\right )}{x}} \left (95 x-16 x^2-16 x^3+\left (-6 x+x^2+x^3\right ) \log (3+x)+(-48-16 x+(3+x) \log (3+x)) \log \left (\frac {16-\log (3+x)}{e}\right )\right )}{-48 x^2-16 x^3+\left (3 x^2+x^3\right ) \log (3+x)} \, dx=\int { \frac {{\left (16 \, x^{3} + 16 \, x^{2} - {\left ({\left (x + 3\right )} \log \left (x + 3\right ) - 16 \, x - 48\right )} \log \left (-{\left (\log \left (x + 3\right ) - 16\right )} e^{\left (-1\right )}\right ) - {\left (x^{3} + x^{2} - 6 \, x\right )} \log \left (x + 3\right ) - 95 \, x\right )} e^{\left (-\frac {x^{2} - x \log \left (x^{2}\right ) - \log \left (-{\left (\log \left (x + 3\right ) - 16\right )} e^{\left (-1\right )}\right )}{x}\right )}}{16 \, x^{3} + 48 \, x^{2} - {\left (x^{3} + 3 \, x^{2}\right )} \log \left (x + 3\right )} \,d x } \]
integrate((((3+x)*log(3+x)-16*x-48)*log((-log(3+x)+16)/exp(1))+(x^3+x^2-6* x)*log(3+x)-16*x^3-16*x^2+95*x)*exp((log((-log(3+x)+16)/exp(1))+x*log(x^2) -x^2)/x)/((x^3+3*x^2)*log(3+x)-16*x^3-48*x^2),x, algorithm=\
integrate((16*x^3 + 16*x^2 - ((x + 3)*log(x + 3) - 16*x - 48)*log(-(log(x + 3) - 16)*e^(-1)) - (x^3 + x^2 - 6*x)*log(x + 3) - 95*x)*e^(-(x^2 - x*log (x^2) - log(-(log(x + 3) - 16)*e^(-1)))/x)/(16*x^3 + 48*x^2 - (x^3 + 3*x^2 )*log(x + 3)), x)
Time = 9.75 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {-x^2+x \log \left (x^2\right )+\log \left (\frac {16-\log (3+x)}{e}\right )}{x}} \left (95 x-16 x^2-16 x^3+\left (-6 x+x^2+x^3\right ) \log (3+x)+(-48-16 x+(3+x) \log (3+x)) \log \left (\frac {16-\log (3+x)}{e}\right )\right )}{-48 x^2-16 x^3+\left (3 x^2+x^3\right ) \log (3+x)} \, dx=-x^2\,{\mathrm {e}}^{-x}\,{\left (16\,{\mathrm {e}}^{-1}-\ln \left (x+3\right )\,{\mathrm {e}}^{-1}\right )}^{1/x} \]