Integrand size = 50, antiderivative size = 19 \[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=e^{\frac {1}{25} x^2 (x+\log (3)-25 \log (5 x))} \]
\[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=\int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx \]
Integrate[(E^((x^3 + x^2*Log[3] - 25*x^2*Log[5*x])/25)*(-25*x + 3*x^2 + 2* x*Log[3] - 50*x*Log[5*x]))/25,x]
Integrate[E^((x^3 + x^2*Log[3] - 25*x^2*Log[5*x])/25)*(-25*x + 3*x^2 + 2*x *Log[3] - 50*x*Log[5*x]), x]/25
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.
\(\displaystyle \int \frac {1}{25} e^{\frac {1}{25} \left (x^3-25 x^2 \log (5 x)+x^2 \log (3)\right )} \left (3 x^2-25 x-50 x \log (5 x)+2 x \log (3)\right ) \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {1}{25} e^{\frac {1}{25} \left (x^3-25 x^2 \log (5 x)+x^2 \log (3)\right )} \left (3 x^2-50 x \log (5 x)+x (2 \log (3)-25)\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{25} \int 3^{\frac {x^2}{25}} 5^{-x^2} e^{\frac {x^3}{25}} x^{-x^2} \left (3 x^2-50 \log (5 x) x-(25-\log (9)) x\right )dx\) |
\(\Big \downarrow \) 2725 |
\(\displaystyle \frac {1}{25} \int e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} x^{-x^2} \left (3 x^2-50 \log (5 x) x-(25-\log (9)) x\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{25} \int \left (-50 e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} \log (5 x) x^{1-x^2}+e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} (-25+\log (9)) x^{1-x^2}+3 e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} x^{2-x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{25} \left (50 \int \frac {\int e^{\frac {1}{25} x^2 (x-25 \log (5)+\log (3))} x^{1-x^2}dx}{x}dx-50 \log (5 x) \int e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} x^{1-x^2}dx-(25-\log (9)) \int e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} x^{1-x^2}dx+3 \int e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} x^{2-x^2}dx\right )\) |
Int[(E^((x^3 + x^2*Log[3] - 25*x^2*Log[5*x])/25)*(-25*x + 3*x^2 + 2*x*Log[ 3] - 50*x*Log[5*x]))/25,x]
3.15.24.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x], x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]
Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11
method | result | size |
parallelrisch | \({\mathrm e}^{-\frac {x^{2} \left (25 \ln \left (5 x \right )-\ln \left (3\right )-x \right )}{25}}\) | \(21\) |
norman | \({\mathrm e}^{-x^{2} \ln \left (5 x \right )+\frac {x^{2} \ln \left (3\right )}{25}+\frac {x^{3}}{25}}\) | \(24\) |
risch | \(\left (5 x \right )^{-x^{2}} 3^{\frac {x^{2}}{25}} {\mathrm e}^{\frac {x^{3}}{25}}\) | \(24\) |
int(1/25*(-50*x*ln(5*x)+2*x*ln(3)+3*x^2-25*x)*exp(-x^2*ln(5*x)+1/25*x^2*ln (3)+1/25*x^3),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=e^{\left (\frac {1}{25} \, x^{3} + \frac {1}{25} \, x^{2} \log \left (3\right ) - x^{2} \log \left (5 \, x\right )\right )} \]
integrate(1/25*(-50*x*log(5*x)+2*x*log(3)+3*x^2-25*x)*exp(-x^2*log(5*x)+1/ 25*x^2*log(3)+1/25*x^3),x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=e^{\frac {x^{3}}{25} - x^{2} \log {\left (5 x \right )} + \frac {x^{2} \log {\left (3 \right )}}{25}} \]
integrate(1/25*(-50*x*ln(5*x)+2*x*ln(3)+3*x**2-25*x)*exp(-x**2*ln(5*x)+1/2 5*x**2*ln(3)+1/25*x**3),x)
Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=e^{\left (\frac {1}{25} \, x^{3} + \frac {1}{25} \, x^{2} \log \left (3\right ) - x^{2} \log \left (5 \, x\right )\right )} \]
integrate(1/25*(-50*x*log(5*x)+2*x*log(3)+3*x^2-25*x)*exp(-x^2*log(5*x)+1/ 25*x^2*log(3)+1/25*x^3),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=e^{\left (\frac {1}{25} \, x^{3} + \frac {1}{25} \, x^{2} \log \left (3\right ) - x^{2} \log \left (5 \, x\right )\right )} \]
integrate(1/25*(-50*x*log(5*x)+2*x*log(3)+3*x^2-25*x)*exp(-x^2*log(5*x)+1/ 25*x^2*log(3)+1/25*x^3),x, algorithm=\
Time = 11.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=\frac {{\left (\frac {1}{5}\right )}^{x^2}\,3^{\frac {x^2}{25}}\,{\mathrm {e}}^{\frac {x^3}{25}}}{x^{x^2}} \]