Integrand size = 60, antiderivative size = 29 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {-x+\frac {e^{-x-8 x^2} x^2}{\log ^4(9)}}{5+x} \] Output:
(1/16*x^2/ln(3)^4/exp(2*x^2)^4/exp(x)-x)/(5+x)
Time = 1.55 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {e^{-x (1+8 x)} x^2+5 \log ^4(9)}{(5+x) \log ^4(9)} \] Input:
Integrate[(E^(-x - 8*x^2)*(10*x - 4*x^2 - 81*x^3 - 16*x^4 - 5*E^(x + 8*x^2 )*Log[9]^4))/((25 + 10*x + x^2)*Log[9]^4),x]
Output:
(x^2/E^(x*(1 + 8*x)) + 5*Log[9]^4)/((5 + x)*Log[9]^4)
Leaf count is larger than twice the leaf count of optimal. \(61\) vs. \(2(29)=58\).
Time = 1.82 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.10, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {27, 2007, 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.
\(\displaystyle \int \frac {e^{-8 x^2-x} \left (-16 x^4-81 x^3-4 x^2-5 e^{8 x^2+x} \log ^4(9)+10 x\right )}{\left (x^2+10 x+25\right ) \log ^4(9)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {e^{-8 x^2-x} \left (-16 x^4-81 x^3-4 x^2+10 x-5 e^{8 x^2+x} \log ^4(9)\right )}{x^2+10 x+25}dx}{\log ^4(9)}\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \frac {\int \frac {e^{-8 x^2-x} \left (-16 x^4-81 x^3-4 x^2+10 x-5 e^{8 x^2+x} \log ^4(9)\right )}{(x+5)^2}dx}{\log ^4(9)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (-\frac {16 e^{-8 x^2-x} x^4}{(x+5)^2}-\frac {81 e^{-8 x^2-x} x^3}{(x+5)^2}-\frac {4 e^{-8 x^2-x} x^2}{(x+5)^2}+\frac {10 e^{-8 x^2-x} x}{(x+5)^2}-\frac {5 \log ^4(9)}{(x+5)^2}\right )dx}{\log ^4(9)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{-8 x^2-x} x-5 e^{-8 x^2-x}+\frac {25 e^{-8 x^2-x}}{x+5}+\frac {5 \log ^4(9)}{x+5}}{\log ^4(9)}\) |
Input:
Int[(E^(-x - 8*x^2)*(10*x - 4*x^2 - 81*x^3 - 16*x^4 - 5*E^(x + 8*x^2)*Log[ 9]^4))/((25 + 10*x + x^2)*Log[9]^4),x]
Output:
(-5*E^(-x - 8*x^2) + E^(-x - 8*x^2)*x + (25*E^(-x - 8*x^2))/(5 + x) + (5*L og[9]^4)/(5 + x))/Log[9]^4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Time = 0.71 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10
method | result | size |
risch | \(\frac {5}{5+x}+\frac {x^{2} {\mathrm e}^{-x \left (8 x +1\right )}}{16 \left (5+x \right ) \ln \left (3\right )^{4}}\) | \(32\) |
parallelrisch | \(\frac {\left (80 \ln \left (3\right )^{4} {\mathrm e}^{x} {\mathrm e}^{8 x^{2}}+x^{2}\right ) {\mathrm e}^{-x} {\mathrm e}^{-8 x^{2}}}{16 \ln \left (3\right )^{4} \left (5+x \right )}\) | \(44\) |
orering | \(-\frac {\left (256 x^{5}+1312 x^{4}+113 x^{3}-397 x^{2}-22 x +10\right ) \left (-80 \ln \left (3\right )^{4} {\mathrm e}^{x} {\mathrm e}^{8 x^{2}}-16 x^{4}-81 x^{3}-4 x^{2}+10 x \right ) {\mathrm e}^{-x} {\mathrm e}^{-8 x^{2}}}{16 \left (256 x^{4}+32 x^{3}-79 x^{2}-4 x +2\right ) \left (x^{2}+10 x +25\right ) \ln \left (3\right )^{4}}-\frac {x \left (16 x^{2}+x -2\right ) \left (5+x \right ) \left (\frac {\left (-80 \ln \left (3\right )^{4} {\mathrm e}^{x} {\mathrm e}^{8 x^{2}}-1280 \ln \left (3\right )^{4} {\mathrm e}^{x} {\mathrm e}^{8 x^{2}} x -64 x^{3}-243 x^{2}-8 x +10\right ) {\mathrm e}^{-x} {\mathrm e}^{-8 x^{2}}}{16 \left (x^{2}+10 x +25\right ) \ln \left (3\right )^{4}}-\frac {\left (-80 \ln \left (3\right )^{4} {\mathrm e}^{x} {\mathrm e}^{8 x^{2}}-16 x^{4}-81 x^{3}-4 x^{2}+10 x \right ) {\mathrm e}^{-x} {\mathrm e}^{-8 x^{2}} \left (2 x +10\right )}{16 \left (x^{2}+10 x +25\right )^{2} \ln \left (3\right )^{4}}-\frac {\left (-80 \ln \left (3\right )^{4} {\mathrm e}^{x} {\mathrm e}^{8 x^{2}}-16 x^{4}-81 x^{3}-4 x^{2}+10 x \right ) {\mathrm e}^{-x} {\mathrm e}^{-8 x^{2}}}{16 \left (x^{2}+10 x +25\right ) \ln \left (3\right )^{4}}-\frac {\left (-80 \ln \left (3\right )^{4} {\mathrm e}^{x} {\mathrm e}^{8 x^{2}}-16 x^{4}-81 x^{3}-4 x^{2}+10 x \right ) {\mathrm e}^{-x} {\mathrm e}^{-8 x^{2}} x}{\left (x^{2}+10 x +25\right ) \ln \left (3\right )^{4}}\right )}{256 x^{4}+32 x^{3}-79 x^{2}-4 x +2}\) | \(420\) |
Input:
int(1/16*(-80*ln(3)^4*exp(x)*exp(2*x^2)^4-16*x^4-81*x^3-4*x^2+10*x)/(x^2+1 0*x+25)/ln(3)^4/exp(x)/exp(2*x^2)^4,x,method=_RETURNVERBOSE)
Output:
5/(5+x)+1/16/(5+x)*x^2/ln(3)^4*exp(-x*(8*x+1))
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {{\left (80 \, e^{\left (8 \, x^{2} + x\right )} \log \left (3\right )^{4} + x^{2}\right )} e^{\left (-8 \, x^{2} - x\right )}}{16 \, {\left (x + 5\right )} \log \left (3\right )^{4}} \] Input:
integrate(1/16*(-80*log(3)^4*exp(x)*exp(2*x^2)^4-16*x^4-81*x^3-4*x^2+10*x) /(x^2+10*x+25)/log(3)^4/exp(x)/exp(2*x^2)^4,x, algorithm="fricas")
Output:
1/16*(80*e^(8*x^2 + x)*log(3)^4 + x^2)*e^(-8*x^2 - x)/((x + 5)*log(3)^4)
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {x^{2} e^{- x} e^{- 8 x^{2}}}{16 x \log {\left (3 \right )}^{4} + 80 \log {\left (3 \right )}^{4}} + \frac {5}{x + 5} \] Input:
integrate(1/16*(-80*ln(3)**4*exp(x)*exp(2*x**2)**4-16*x**4-81*x**3-4*x**2+ 10*x)/(x**2+10*x+25)/ln(3)**4/exp(x)/exp(2*x**2)**4,x)
Output:
x**2*exp(-x)*exp(-8*x**2)/(16*x*log(3)**4 + 80*log(3)**4) + 5/(x + 5)
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {\frac {80 \, \log \left (3\right )^{4}}{x + 5} + \frac {x^{2} e^{\left (-8 \, x^{2} - x\right )}}{x + 5}}{16 \, \log \left (3\right )^{4}} \] Input:
integrate(1/16*(-80*log(3)^4*exp(x)*exp(2*x^2)^4-16*x^4-81*x^3-4*x^2+10*x) /(x^2+10*x+25)/log(3)^4/exp(x)/exp(2*x^2)^4,x, algorithm="maxima")
Output:
1/16*(80*log(3)^4/(x + 5) + x^2*e^(-8*x^2 - x)/(x + 5))/log(3)^4
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {80 \, \log \left (3\right )^{4} + x^{2} e^{\left (-8 \, x^{2} - x\right )}}{16 \, {\left (x + 5\right )} \log \left (3\right )^{4}} \] Input:
integrate(1/16*(-80*log(3)^4*exp(x)*exp(2*x^2)^4-16*x^4-81*x^3-4*x^2+10*x) /(x^2+10*x+25)/log(3)^4/exp(x)/exp(2*x^2)^4,x, algorithm="giac")
Output:
1/16*(80*log(3)^4 + x^2*e^(-8*x^2 - x))/((x + 5)*log(3)^4)
Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.79 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {x^2-16\,x\,{\mathrm {e}}^{8\,x^2+x}\,{\ln \left (3\right )}^4}{80\,{\mathrm {e}}^{8\,x^2+x}\,{\ln \left (3\right )}^4+16\,x\,{\mathrm {e}}^{8\,x^2+x}\,{\ln \left (3\right )}^4} \] Input:
int(-(exp(-x)*exp(-8*x^2)*(x^2/4 - (5*x)/8 + (81*x^3)/16 + x^4 + 5*exp(8*x ^2)*exp(x)*log(3)^4))/(log(3)^4*(10*x + x^2 + 25)),x)
Output:
(x^2 - 16*x*exp(x + 8*x^2)*log(3)^4)/(80*exp(x + 8*x^2)*log(3)^4 + 16*x*ex p(x + 8*x^2)*log(3)^4)
Time = 0.17 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {x \left (-16 e^{8 x^{2}+x} \mathrm {log}\left (3\right )^{4}+x \right )}{16 e^{8 x^{2}+x} \mathrm {log}\left (3\right )^{4} \left (x +5\right )} \] Input:
int(1/16*(-80*log(3)^4*exp(x)*exp(2*x^2)^4-16*x^4-81*x^3-4*x^2+10*x)/(x^2+ 10*x+25)/log(3)^4/exp(x)/exp(2*x^2)^4,x)
Output:
(x*( - 16*e**(8*x**2 + x)*log(3)**4 + x))/(16*e**(8*x**2 + x)*log(3)**4*(x + 5))