Integrand size = 99, antiderivative size = 30 \[ \int \frac {-e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} x+e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} \left (6 x^3+3 x^4+e^5 \left (-6-2 x^2\right )\right ) \log (2 x)}{x^2 \log ^2(2 x)} \, dx=\frac {e^{(3+x) \left (x^2-\frac {e^5 (-2+2 x)}{x}\right )}}{\log (2 x)} \] Output:
exp((x^2-(-2+2*x)*exp(5)/x)*(3+x))/ln(2*x)
Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} x+e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} \left (6 x^3+3 x^4+e^5 \left (-6-2 x^2\right )\right ) \log (2 x)}{x^2 \log ^2(2 x)} \, dx=\frac {e^{\frac {(3+x) \left (-2 e^5 (-1+x)+x^3\right )}{x}}}{\log (2 x)} \] Input:
Integrate[(-(E^((3*x^3 + x^4 + E^5*(6 - 4*x - 2*x^2))/x)*x) + E^((3*x^3 + x^4 + E^5*(6 - 4*x - 2*x^2))/x)*(6*x^3 + 3*x^4 + E^5*(-6 - 2*x^2))*Log[2*x ])/(x^2*Log[2*x]^2),x]
Output:
E^(((3 + x)*(-2*E^5*(-1 + x) + x^3))/x)/Log[2*x]
Leaf count is larger than twice the leaf count of optimal. \(117\) vs. \(2(30)=60\).
Time = 1.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 3.90, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {7239, 2726}
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.
\(\displaystyle \int \frac {e^{\frac {x^4+3 x^3+e^5 \left (-2 x^2-4 x+6\right )}{x}} \left (3 x^4+6 x^3+e^5 \left (-2 x^2-6\right )\right ) \log (2 x)-e^{\frac {x^4+3 x^3+e^5 \left (-2 x^2-4 x+6\right )}{x}} x}{x^2 \log ^2(2 x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{\frac {(x+3) \left (x^3-2 e^5 x+2 e^5\right )}{x}} \left (\left (3 x^3 (x+2)-2 e^5 \left (x^2+3\right )\right ) \log (2 x)-x\right )}{x^2 \log ^2(2 x)}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle -\frac {e^{\frac {(x+3) \left (x^3-2 e^5 x+2 e^5\right )}{x}} \left (3 x^3 (x+2)-2 e^5 \left (x^2+3\right )\right )}{x^2 \left (-\frac {x^3-2 e^5 x+2 e^5}{x}+\frac {(x+3) \left (2 e^5-3 x^2\right )}{x}+\frac {(x+3) \left (x^3-2 e^5 x+2 e^5\right )}{x^2}\right ) \log (2 x)}\) |
Input:
Int[(-(E^((3*x^3 + x^4 + E^5*(6 - 4*x - 2*x^2))/x)*x) + E^((3*x^3 + x^4 + E^5*(6 - 4*x - 2*x^2))/x)*(6*x^3 + 3*x^4 + E^5*(-6 - 2*x^2))*Log[2*x])/(x^ 2*Log[2*x]^2),x]
Output:
-((E^(((3 + x)*(2*E^5 - 2*E^5*x + x^3))/x)*(3*x^3*(2 + x) - 2*E^5*(3 + x^2 )))/(x^2*(((3 + x)*(2*E^5 - 3*x^2))/x - (2*E^5 - 2*E^5*x + x^3)/x + ((3 + x)*(2*E^5 - 2*E^5*x + x^3))/x^2)*Log[2*x]))
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.71 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07
method | result | size |
risch | \(\frac {{\mathrm e}^{-\frac {\left (3+x \right ) \left (-x^{3}+2 x \,{\mathrm e}^{5}-2 \,{\mathrm e}^{5}\right )}{x}}}{\ln \left (2 x \right )}\) | \(32\) |
parallelrisch | \(\frac {{\mathrm e}^{-\frac {-x^{4}+2 x^{2} {\mathrm e}^{5}-3 x^{3}+4 x \,{\mathrm e}^{5}-6 \,{\mathrm e}^{5}}{x}}}{\ln \left (2 x \right )}\) | \(41\) |
Input:
int((((-2*x^2-6)*exp(5)+3*x^4+6*x^3)*exp(((-2*x^2-4*x+6)*exp(5)+x^4+3*x^3) /x)*ln(2*x)-x*exp(((-2*x^2-4*x+6)*exp(5)+x^4+3*x^3)/x))/x^2/ln(2*x)^2,x,me thod=_RETURNVERBOSE)
Output:
exp(-(3+x)*(-x^3+2*x*exp(5)-2*exp(5))/x)/ln(2*x)
Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {-e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} x+e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} \left (6 x^3+3 x^4+e^5 \left (-6-2 x^2\right )\right ) \log (2 x)}{x^2 \log ^2(2 x)} \, dx=\frac {e^{\left (\frac {x^{4} + 3 \, x^{3} - 2 \, {\left (x^{2} + 2 \, x - 3\right )} e^{5}}{x}\right )}}{\log \left (2 \, x\right )} \] Input:
integrate((((-2*x^2-6)*exp(5)+3*x^4+6*x^3)*exp(((-2*x^2-4*x+6)*exp(5)+x^4+ 3*x^3)/x)*log(2*x)-x*exp(((-2*x^2-4*x+6)*exp(5)+x^4+3*x^3)/x))/x^2/log(2*x )^2,x, algorithm="fricas")
Output:
e^((x^4 + 3*x^3 - 2*(x^2 + 2*x - 3)*e^5)/x)/log(2*x)
Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {-e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} x+e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} \left (6 x^3+3 x^4+e^5 \left (-6-2 x^2\right )\right ) \log (2 x)}{x^2 \log ^2(2 x)} \, dx=\frac {e^{\frac {x^{4} + 3 x^{3} + \left (- 2 x^{2} - 4 x + 6\right ) e^{5}}{x}}}{\log {\left (2 x \right )}} \] Input:
integrate((((-2*x**2-6)*exp(5)+3*x**4+6*x**3)*exp(((-2*x**2-4*x+6)*exp(5)+ x**4+3*x**3)/x)*ln(2*x)-x*exp(((-2*x**2-4*x+6)*exp(5)+x**4+3*x**3)/x))/x** 2/ln(2*x)**2,x)
Output:
exp((x**4 + 3*x**3 + (-2*x**2 - 4*x + 6)*exp(5))/x)/log(2*x)
Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {-e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} x+e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} \left (6 x^3+3 x^4+e^5 \left (-6-2 x^2\right )\right ) \log (2 x)}{x^2 \log ^2(2 x)} \, dx=\frac {e^{\left (x^{3} + 3 \, x^{2} - 2 \, x e^{5} + \frac {6 \, e^{5}}{x}\right )}}{e^{\left (4 \, e^{5}\right )} \log \left (2\right ) + e^{\left (4 \, e^{5}\right )} \log \left (x\right )} \] Input:
integrate((((-2*x^2-6)*exp(5)+3*x^4+6*x^3)*exp(((-2*x^2-4*x+6)*exp(5)+x^4+ 3*x^3)/x)*log(2*x)-x*exp(((-2*x^2-4*x+6)*exp(5)+x^4+3*x^3)/x))/x^2/log(2*x )^2,x, algorithm="maxima")
Output:
e^(x^3 + 3*x^2 - 2*x*e^5 + 6*e^5/x)/(e^(4*e^5)*log(2) + e^(4*e^5)*log(x))
Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {-e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} x+e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} \left (6 x^3+3 x^4+e^5 \left (-6-2 x^2\right )\right ) \log (2 x)}{x^2 \log ^2(2 x)} \, dx=\frac {e^{\left (\frac {x^{4} + 3 \, x^{3} - 2 \, x^{2} e^{5} - 4 \, x e^{5} + 6 \, e^{5}}{x}\right )}}{\log \left (2 \, x\right )} \] Input:
integrate((((-2*x^2-6)*exp(5)+3*x^4+6*x^3)*exp(((-2*x^2-4*x+6)*exp(5)+x^4+ 3*x^3)/x)*log(2*x)-x*exp(((-2*x^2-4*x+6)*exp(5)+x^4+3*x^3)/x))/x^2/log(2*x )^2,x, algorithm="giac")
Output:
e^((x^4 + 3*x^3 - 2*x^2*e^5 - 4*x*e^5 + 6*e^5)/x)/log(2*x)
Time = 2.49 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {-e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} x+e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} \left (6 x^3+3 x^4+e^5 \left (-6-2 x^2\right )\right ) \log (2 x)}{x^2 \log ^2(2 x)} \, dx=\frac {{\mathrm {e}}^{\frac {6\,{\mathrm {e}}^5}{x}}\,{\mathrm {e}}^{-4\,{\mathrm {e}}^5}\,{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{3\,x^2}\,{\mathrm {e}}^{-2\,x\,{\mathrm {e}}^5}}{\ln \left (2\right )+\ln \left (x\right )} \] Input:
int(-(x*exp((3*x^3 - exp(5)*(4*x + 2*x^2 - 6) + x^4)/x) - log(2*x)*exp((3* x^3 - exp(5)*(4*x + 2*x^2 - 6) + x^4)/x)*(6*x^3 - exp(5)*(2*x^2 + 6) + 3*x ^4))/(x^2*log(2*x)^2),x)
Output:
(exp((6*exp(5))/x)*exp(-4*exp(5))*exp(x^3)*exp(3*x^2)*exp(-2*x*exp(5)))/(l og(2) + log(x))
\[ \int \frac {-e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} x+e^{\frac {3 x^3+x^4+e^5 \left (6-4 x-2 x^2\right )}{x}} \left (6 x^3+3 x^4+e^5 \left (-6-2 x^2\right )\right ) \log (2 x)}{x^2 \log ^2(2 x)} \, dx=\int \frac {\left (\left (-2 x^{2}-6\right ) {\mathrm e}^{5}+3 x^{4}+6 x^{3}\right ) {\mathrm e}^{\frac {\left (-2 x^{2}-4 x +6\right ) {\mathrm e}^{5}+x^{4}+3 x^{3}}{x}} \mathrm {log}\left (2 x \right )-x \,{\mathrm e}^{\frac {\left (-2 x^{2}-4 x +6\right ) {\mathrm e}^{5}+x^{4}+3 x^{3}}{x}}}{x^{2} \mathrm {log}\left (2 x \right )^{2}}d x \] Input:
int((((-2*x^2-6)*exp(5)+3*x^4+6*x^3)*exp(((-2*x^2-4*x+6)*exp(5)+x^4+3*x^3) /x)*log(2*x)-x*exp(((-2*x^2-4*x+6)*exp(5)+x^4+3*x^3)/x))/x^2/log(2*x)^2,x)
Output:
int((((-2*x^2-6)*exp(5)+3*x^4+6*x^3)*exp(((-2*x^2-4*x+6)*exp(5)+x^4+3*x^3) /x)*log(2*x)-x*exp(((-2*x^2-4*x+6)*exp(5)+x^4+3*x^3)/x))/x^2/log(2*x)^2,x)