Integrand size = 62, antiderivative size = 28 \begin {dmath*} \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx=e^{\frac {e^{3+2 x} (3+(-4+x) x)}{x \log (2)}}+5 x \end {dmath*}
Time = 1.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \begin {dmath*} \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx=e^{\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}}+5 x \end {dmath*}
Integrate[(E^(3 + 2*x + (E^(3 + 2*x)*(3 - 4*x + x^2))/(x*Log[2]))*(-3 + 6* x - 7*x^2 + 2*x^3) + 5*x^2*Log[2])/(x^2*Log[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.
\(\displaystyle \int \frac {\left (2 x^3-7 x^2+6 x-3\right ) \exp \left (\frac {e^{2 x+3} \left (x^2-4 x+3\right )}{x \log (2)}+2 x+3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {\exp \left (2 x+3+\frac {e^{2 x+3} \left (x^2-4 x+3\right )}{\log (2) x}\right ) \left (-2 x^3+7 x^2-6 x+3\right )-5 x^2 \log (2)}{x^2}dx}{\log (2)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\exp \left (2 x+3+\frac {e^{2 x+3} \left (x^2-4 x+3\right )}{\log (2) x}\right ) \left (-2 x^3+7 x^2-6 x+3\right )-5 x^2 \log (2)}{x^2}dx}{\log (2)}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {\int \left (-\frac {\exp \left (2 x+3+\frac {e^{2 x+3} \left (x^2-4 x+3\right )}{\log (2) x}\right ) \left (2 x^3-7 x^2+6 x-3\right )}{x^2}-5 \log (2)\right )dx}{\log (2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {7 \int \exp \left (2 x+3+\frac {e^{2 x+3} \left (x^2-4 x+3\right )}{\log (2) x}\right )dx+3 \int \frac {\exp \left (2 x+3+\frac {e^{2 x+3} \left (x^2-4 x+3\right )}{\log (2) x}\right )}{x^2}dx-6 \int \frac {\exp \left (2 x+3+\frac {e^{2 x+3} \left (x^2-4 x+3\right )}{\log (2) x}\right )}{x}dx-2 \int \exp \left (2 x+3+\frac {e^{2 x+3} \left (x^2-4 x+3\right )}{\log (2) x}\right ) xdx-5 x \log (2)}{\log (2)}\) |
Int[(E^(3 + 2*x + (E^(3 + 2*x)*(3 - 4*x + x^2))/(x*Log[2]))*(-3 + 6*x - 7* x^2 + 2*x^3) + 5*x^2*Log[2])/(x^2*Log[2]),x]
3.3.36.3.1 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_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.91 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93
method | result | size |
risch | \(5 x +{\mathrm e}^{\frac {\left (-1+x \right ) \left (-3+x \right ) {\mathrm e}^{3+2 x}}{x \ln \left (2\right )}}\) | \(26\) |
parts | \(5 x +{\mathrm e}^{\frac {\left (x^{2}-4 x +3\right ) {\mathrm e}^{3+2 x}}{x \ln \left (2\right )}}\) | \(28\) |
norman | \(\frac {x \,{\mathrm e}^{\frac {\left (x^{2}-4 x +3\right ) {\mathrm e}^{3+2 x}}{x \ln \left (2\right )}}+5 x^{2}}{x}\) | \(36\) |
parallelrisch | \(\frac {5 x \ln \left (2\right )+\ln \left (2\right ) {\mathrm e}^{\frac {\left (x^{2}-4 x +3\right ) {\mathrm e}^{3+2 x}}{x \ln \left (2\right )}}}{\ln \left (2\right )}\) | \(38\) |
int(((2*x^3-7*x^2+6*x-3)*exp(3+2*x)*exp((x^2-4*x+3)*exp(3+2*x)/x/ln(2))+5* x^2*ln(2))/x^2/ln(2),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (26) = 52\).
Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.93 \begin {dmath*} \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx={\left (5 \, x e^{\left (2 \, x + 3\right )} + e^{\left (\frac {{\left (x^{2} - 4 \, x + 3\right )} e^{\left (2 \, x + 3\right )} + {\left (2 \, x^{2} + 3 \, x\right )} \log \left (2\right )}{x \log \left (2\right )}\right )}\right )} e^{\left (-2 \, x - 3\right )} \end {dmath*}
integrate(((2*x^3-7*x^2+6*x-3)*exp(3+2*x)*exp((x^2-4*x+3)*exp(3+2*x)/x/log (2))+5*x^2*log(2))/x^2/log(2),x, algorithm=\
(5*x*e^(2*x + 3) + e^(((x^2 - 4*x + 3)*e^(2*x + 3) + (2*x^2 + 3*x)*log(2)) /(x*log(2))))*e^(-2*x - 3)
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \begin {dmath*} \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx=5 x + e^{\frac {\left (x^{2} - 4 x + 3\right ) e^{2 x + 3}}{x \log {\left (2 \right )}}} \end {dmath*}
integrate(((2*x**3-7*x**2+6*x-3)*exp(3+2*x)*exp((x**2-4*x+3)*exp(3+2*x)/x/ ln(2))+5*x**2*ln(2))/x**2/ln(2),x)
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (26) = 52\).
Time = 0.44 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.96 \begin {dmath*} \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx=\frac {5 \, x \log \left (2\right ) + e^{\left (\frac {x e^{\left (2 \, x + 3\right )}}{\log \left (2\right )} - \frac {4 \, e^{\left (2 \, x + 3\right )}}{\log \left (2\right )} + \frac {3 \, e^{\left (2 \, x + 3\right )}}{x \log \left (2\right )}\right )} \log \left (2\right )}{\log \left (2\right )} \end {dmath*}
integrate(((2*x^3-7*x^2+6*x-3)*exp(3+2*x)*exp((x^2-4*x+3)*exp(3+2*x)/x/log (2))+5*x^2*log(2))/x^2/log(2),x, algorithm=\
(5*x*log(2) + e^(x*e^(2*x + 3)/log(2) - 4*e^(2*x + 3)/log(2) + 3*e^(2*x + 3)/(x*log(2)))*log(2))/log(2)
\begin {dmath*} \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx=\int { \frac {5 \, x^{2} \log \left (2\right ) + {\left (2 \, x^{3} - 7 \, x^{2} + 6 \, x - 3\right )} e^{\left (2 \, x + \frac {{\left (x^{2} - 4 \, x + 3\right )} e^{\left (2 \, x + 3\right )}}{x \log \left (2\right )} + 3\right )}}{x^{2} \log \left (2\right )} \,d x } \end {dmath*}
integrate(((2*x^3-7*x^2+6*x-3)*exp(3+2*x)*exp((x^2-4*x+3)*exp(3+2*x)/x/log (2))+5*x^2*log(2))/x^2/log(2),x, algorithm=\
integrate((5*x^2*log(2) + (2*x^3 - 7*x^2 + 6*x - 3)*e^(2*x + (x^2 - 4*x + 3)*e^(2*x + 3)/(x*log(2)) + 3))/(x^2*log(2)), x)
Time = 14.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.68 \begin {dmath*} \int \frac {e^{3+2 x+\frac {e^{3+2 x} \left (3-4 x+x^2\right )}{x \log (2)}} \left (-3+6 x-7 x^2+2 x^3\right )+5 x^2 \log (2)}{x^2 \log (2)} \, dx=5\,x+{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3}{x\,\ln \left (2\right )}}\,{\mathrm {e}}^{-\frac {4\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3}{\ln \left (2\right )}}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^3}{\ln \left (2\right )}} \end {dmath*}
int((5*x^2*log(2) + exp((exp(2*x + 3)*(x^2 - 4*x + 3))/(x*log(2)))*exp(2*x + 3)*(6*x - 7*x^2 + 2*x^3 - 3))/(x^2*log(2)),x)