Integrand size = 46, antiderivative size = 21 \[ \int \frac {e^{-\frac {x}{\log (3)}} \left (-x^4+4 x^3 \log (3)+e^{\frac {x}{\log (3)}} (15+6 x) \log (3)\right )}{3 \log (3)} \, dx=x \left (5+x+\frac {1}{3} e^{-\frac {x}{\log (3)}} x^3\right ) \] Output:
(x+5+1/3*x^3/exp(x/ln(3)))*x
Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-\frac {x}{\log (3)}} \left (-x^4+4 x^3 \log (3)+e^{\frac {x}{\log (3)}} (15+6 x) \log (3)\right )}{3 \log (3)} \, dx=5 x+x^2+\frac {1}{3} e^{-\frac {x}{\log (3)}} x^4 \] Input:
Integrate[(-x^4 + 4*x^3*Log[3] + E^(x/Log[3])*(15 + 6*x)*Log[3])/(3*E^(x/L og[3])*Log[3]),x]
Output:
5*x + x^2 + x^4/(3*E^(x/Log[3]))
Time = 0.41 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.81, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {27, 25, 7293, 2009}
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.
\(\displaystyle \int \frac {e^{-\frac {x}{\log (3)}} \left (-x^4+4 x^3 \log (3)+(6 x+15) \log (3) e^{\frac {x}{\log (3)}}\right )}{3 \log (3)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -e^{-\frac {x}{\log (3)}} \left (x^4-4 \log (3) x^3-3 e^{\frac {x}{\log (3)}} (2 x+5) \log (3)\right )dx}{3 \log (3)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int e^{-\frac {x}{\log (3)}} \left (x^4-4 \log (3) x^3-3 e^{\frac {x}{\log (3)}} (2 x+5) \log (3)\right )dx}{3 \log (3)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (e^{-\frac {x}{\log (3)}} x^4-4 e^{-\frac {x}{\log (3)}} \log (3) x^3-3 (2 x+5) \log (3)\right )dx}{3 \log (3)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {x^4 \log (3) \left (-e^{-\frac {x}{\log (3)}}\right )-\frac {3}{4} (2 x+5)^2 \log (3)}{3 \log (3)}\) |
Input:
Int[(-x^4 + 4*x^3*Log[3] + E^(x/Log[3])*(15 + 6*x)*Log[3])/(3*E^(x/Log[3]) *Log[3]),x]
Output:
-1/3*(-((x^4*Log[3])/E^(x/Log[3])) - (3*(5 + 2*x)^2*Log[3])/4)/Log[3]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00
method | result | size |
risch | \(x^{2}+5 x +\frac {x^{4} {\mathrm e}^{-\frac {x}{\ln \left (3\right )}}}{3}\) | \(21\) |
parts | \(x^{2}+5 x +\frac {x^{4} {\mathrm e}^{-\frac {x}{\ln \left (3\right )}}}{3}\) | \(22\) |
norman | \(\left (x^{2} {\mathrm e}^{\frac {x}{\ln \left (3\right )}}+\frac {x^{4}}{3}+5 x \,{\mathrm e}^{\frac {x}{\ln \left (3\right )}}\right ) {\mathrm e}^{-\frac {x}{\ln \left (3\right )}}\) | \(38\) |
parallelrisch | \(\frac {\left (x^{4} \ln \left (3\right )+15 \ln \left (3\right )^{2} \ln \left ({\mathrm e}^{\frac {x}{\ln \left (3\right )}}\right ) {\mathrm e}^{\frac {x}{\ln \left (3\right )}}+3 \,{\mathrm e}^{\frac {x}{\ln \left (3\right )}} x^{2} \ln \left (3\right )\right ) {\mathrm e}^{-\frac {x}{\ln \left (3\right )}}}{3 \ln \left (3\right )}\) | \(58\) |
derivativedivides | \(x^{2}+\frac {4 \ln \left (3\right )^{4} \left (-\frac {{\mathrm e}^{-\frac {x}{\ln \left (3\right )}} x^{3}}{\ln \left (3\right )^{3}}-\frac {3 x^{2} {\mathrm e}^{-\frac {x}{\ln \left (3\right )}}}{\ln \left (3\right )^{2}}-\frac {6 x \,{\mathrm e}^{-\frac {x}{\ln \left (3\right )}}}{\ln \left (3\right )}-6 \,{\mathrm e}^{-\frac {x}{\ln \left (3\right )}}\right )}{3}-\frac {\ln \left (3\right )^{4} \left (-\frac {{\mathrm e}^{-\frac {x}{\ln \left (3\right )}} x^{4}}{\ln \left (3\right )^{4}}-\frac {4 \,{\mathrm e}^{-\frac {x}{\ln \left (3\right )}} x^{3}}{\ln \left (3\right )^{3}}-\frac {12 x^{2} {\mathrm e}^{-\frac {x}{\ln \left (3\right )}}}{\ln \left (3\right )^{2}}-\frac {24 x \,{\mathrm e}^{-\frac {x}{\ln \left (3\right )}}}{\ln \left (3\right )}-24 \,{\mathrm e}^{-\frac {x}{\ln \left (3\right )}}\right )}{3}+5 x\) | \(166\) |
default | \(x^{2}+\frac {4 \ln \left (3\right )^{4} \left (-\frac {{\mathrm e}^{-\frac {x}{\ln \left (3\right )}} x^{3}}{\ln \left (3\right )^{3}}-\frac {3 x^{2} {\mathrm e}^{-\frac {x}{\ln \left (3\right )}}}{\ln \left (3\right )^{2}}-\frac {6 x \,{\mathrm e}^{-\frac {x}{\ln \left (3\right )}}}{\ln \left (3\right )}-6 \,{\mathrm e}^{-\frac {x}{\ln \left (3\right )}}\right )}{3}-\frac {\ln \left (3\right )^{4} \left (-\frac {{\mathrm e}^{-\frac {x}{\ln \left (3\right )}} x^{4}}{\ln \left (3\right )^{4}}-\frac {4 \,{\mathrm e}^{-\frac {x}{\ln \left (3\right )}} x^{3}}{\ln \left (3\right )^{3}}-\frac {12 x^{2} {\mathrm e}^{-\frac {x}{\ln \left (3\right )}}}{\ln \left (3\right )^{2}}-\frac {24 x \,{\mathrm e}^{-\frac {x}{\ln \left (3\right )}}}{\ln \left (3\right )}-24 \,{\mathrm e}^{-\frac {x}{\ln \left (3\right )}}\right )}{3}+5 x\) | \(166\) |
Input:
int(1/3*((6*x+15)*ln(3)*exp(x/ln(3))+4*x^3*ln(3)-x^4)/ln(3)/exp(x/ln(3)),x ,method=_RETURNVERBOSE)
Output:
x^2+5*x+1/3*x^4*exp(-x/ln(3))
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {e^{-\frac {x}{\log (3)}} \left (-x^4+4 x^3 \log (3)+e^{\frac {x}{\log (3)}} (15+6 x) \log (3)\right )}{3 \log (3)} \, dx=\frac {1}{3} \, {\left (x^{4} + 3 \, {\left (x^{2} + 5 \, x\right )} e^{\frac {x}{\log \left (3\right )}}\right )} e^{\left (-\frac {x}{\log \left (3\right )}\right )} \] Input:
integrate(1/3*((6*x+15)*log(3)*exp(x/log(3))+4*x^3*log(3)-x^4)/log(3)/exp( x/log(3)),x, algorithm="fricas")
Output:
1/3*(x^4 + 3*(x^2 + 5*x)*e^(x/log(3)))*e^(-x/log(3))
Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-\frac {x}{\log (3)}} \left (-x^4+4 x^3 \log (3)+e^{\frac {x}{\log (3)}} (15+6 x) \log (3)\right )}{3 \log (3)} \, dx=\frac {x^{4} e^{- \frac {x}{\log {\left (3 \right )}}}}{3} + x^{2} + 5 x \] Input:
integrate(1/3*((6*x+15)*ln(3)*exp(x/ln(3))+4*x**3*ln(3)-x**4)/ln(3)/exp(x/ ln(3)),x)
Output:
x**4*exp(-x/log(3))/3 + x**2 + 5*x
Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (20) = 40\).
Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 5.10 \[ \int \frac {e^{-\frac {x}{\log (3)}} \left (-x^4+4 x^3 \log (3)+e^{\frac {x}{\log (3)}} (15+6 x) \log (3)\right )}{3 \log (3)} \, dx=\frac {3 \, x^{2} \log \left (3\right ) - 4 \, {\left (x^{3} \log \left (3\right ) + 3 \, x^{2} \log \left (3\right )^{2} + 6 \, x \log \left (3\right )^{3} + 6 \, \log \left (3\right )^{4}\right )} e^{\left (-\frac {x}{\log \left (3\right )}\right )} \log \left (3\right ) + {\left (x^{4} \log \left (3\right ) + 4 \, x^{3} \log \left (3\right )^{2} + 12 \, x^{2} \log \left (3\right )^{3} + 24 \, x \log \left (3\right )^{4} + 24 \, \log \left (3\right )^{5}\right )} e^{\left (-\frac {x}{\log \left (3\right )}\right )} + 15 \, x \log \left (3\right )}{3 \, \log \left (3\right )} \] Input:
integrate(1/3*((6*x+15)*log(3)*exp(x/log(3))+4*x^3*log(3)-x^4)/log(3)/exp( x/log(3)),x, algorithm="maxima")
Output:
1/3*(3*x^2*log(3) - 4*(x^3*log(3) + 3*x^2*log(3)^2 + 6*x*log(3)^3 + 6*log( 3)^4)*e^(-x/log(3))*log(3) + (x^4*log(3) + 4*x^3*log(3)^2 + 12*x^2*log(3)^ 3 + 24*x*log(3)^4 + 24*log(3)^5)*e^(-x/log(3)) + 15*x*log(3))/log(3)
Time = 0.12 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {e^{-\frac {x}{\log (3)}} \left (-x^4+4 x^3 \log (3)+e^{\frac {x}{\log (3)}} (15+6 x) \log (3)\right )}{3 \log (3)} \, dx=\frac {x^{4} e^{\left (-\frac {x}{\log \left (3\right )}\right )} \log \left (3\right ) + 3 \, x^{2} \log \left (3\right ) + 15 \, x \log \left (3\right )}{3 \, \log \left (3\right )} \] Input:
integrate(1/3*((6*x+15)*log(3)*exp(x/log(3))+4*x^3*log(3)-x^4)/log(3)/exp( x/log(3)),x, algorithm="giac")
Output:
1/3*(x^4*e^(-x/log(3))*log(3) + 3*x^2*log(3) + 15*x*log(3))/log(3)
Time = 1.87 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-\frac {x}{\log (3)}} \left (-x^4+4 x^3 \log (3)+e^{\frac {x}{\log (3)}} (15+6 x) \log (3)\right )}{3 \log (3)} \, dx=5\,x+\frac {x^4\,{\mathrm {e}}^{-\frac {x}{\ln \left (3\right )}}}{3}+x^2 \] Input:
int((exp(-x/log(3))*((4*x^3*log(3))/3 - x^4/3 + (exp(x/log(3))*log(3)*(6*x + 15))/3))/log(3),x)
Output:
5*x + (x^4*exp(-x/log(3)))/3 + x^2
Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.81 \[ \int \frac {e^{-\frac {x}{\log (3)}} \left (-x^4+4 x^3 \log (3)+e^{\frac {x}{\log (3)}} (15+6 x) \log (3)\right )}{3 \log (3)} \, dx=\frac {x \left (3 e^{\frac {x}{\mathrm {log}\left (3\right )}} x +15 e^{\frac {x}{\mathrm {log}\left (3\right )}}+x^{3}\right )}{3 e^{\frac {x}{\mathrm {log}\left (3\right )}}} \] Input:
int(1/3*((6*x+15)*log(3)*exp(x/log(3))+4*x^3*log(3)-x^4)/log(3)/exp(x/log( 3)),x)
Output:
(x*(3*e**(x/log(3))*x + 15*e**(x/log(3)) + x**3))/(3*e**(x/log(3)))