Integrand size = 55, antiderivative size = 30 \[ \int \frac {32-12 x-2 x^2+e^x \left (-320-400 x-85 x^2-5 x^3\right )+16 \log \left (5 e^{-x} x^2\right )}{320+80 x+5 x^2} \, dx=-e^x x+\frac {4 x \log \left (5 e^{-x} x^2\right )}{5 (16+2 x)} \] Output:
4/5/(2*x+16)*x*ln(5*x^2/exp(x))-exp(x)*x
Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {32-12 x-2 x^2+e^x \left (-320-400 x-85 x^2-5 x^3\right )+16 \log \left (5 e^{-x} x^2\right )}{320+80 x+5 x^2} \, dx=\frac {1}{5} \left (-2 x-5 e^x x+4 \log (x)-\frac {16 \log \left (5 e^{-x} x^2\right )}{8+x}\right ) \] Input:
Integrate[(32 - 12*x - 2*x^2 + E^x*(-320 - 400*x - 85*x^2 - 5*x^3) + 16*Lo g[(5*x^2)/E^x])/(320 + 80*x + 5*x^2),x]
Output:
(-2*x - 5*E^x*x + 4*Log[x] - (16*Log[(5*x^2)/E^x])/(8 + x))/5
Time = 0.51 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {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.
\(\displaystyle \int \frac {-2 x^2+16 \log \left (5 e^{-x} x^2\right )+e^x \left (-5 x^3-85 x^2-400 x-320\right )-12 x+32}{5 x^2+80 x+320} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {-2 x^2+16 \log \left (5 e^{-x} x^2\right )+e^x \left (-5 x^3-85 x^2-400 x-320\right )-12 x+32}{\left (\sqrt {5} x+8 \sqrt {5}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {2 x^2}{5 (x+8)^2}+\frac {16 \log \left (5 e^{-x} x^2\right )}{5 (x+8)^2}-\frac {12 x}{5 (x+8)^2}-e^x (x+1)+\frac {32}{5 (x+8)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {16 \log \left (5 e^{-x} x^2\right )}{5 (x+8)}-\frac {2 x}{5}+e^x-e^x (x+1)+\frac {4 \log (x)}{5}\) |
Input:
Int[(32 - 12*x - 2*x^2 + E^x*(-320 - 400*x - 85*x^2 - 5*x^3) + 16*Log[(5*x ^2)/E^x])/(320 + 80*x + 5*x^2),x]
Output:
E^x - (2*x)/5 - E^x*(1 + x) + (4*Log[x])/5 - (16*Log[(5*x^2)/E^x])/(5*(8 + 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 = 0.56 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03
method | result | size |
default | \(-{\mathrm e}^{x} x -\frac {2 x}{5}-\frac {16 \ln \left (5 x^{2} {\mathrm e}^{-x}\right )}{5 \left (x +8\right )}+\frac {4 \ln \left (x \right )}{5}\) | \(31\) |
parts | \(-{\mathrm e}^{x} x -\frac {2 x}{5}-\frac {16 \ln \left (5 x^{2} {\mathrm e}^{-x}\right )}{5 \left (x +8\right )}+\frac {4 \ln \left (x \right )}{5}\) | \(31\) |
norman | \(\frac {-8 \,{\mathrm e}^{x} x -{\mathrm e}^{x} x^{2}+\frac {2 \ln \left (5 x^{2} {\mathrm e}^{-x}\right ) x}{5}}{x +8}\) | \(33\) |
parallelrisch | \(\frac {-10 \,{\mathrm e}^{x} x^{2}-80 \,{\mathrm e}^{x} x +4 \ln \left (5 x^{2} {\mathrm e}^{-x}\right ) x}{10 x +80}\) | \(34\) |
risch | \(\frac {16 \ln \left ({\mathrm e}^{x}\right )}{5 \left (x +8\right )}-\frac {-8 i \pi \,\operatorname {csgn}\left (i x^{2} {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x^{2}\right )+8 i \pi \operatorname {csgn}\left (i x^{2} {\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right )+8 i \pi \operatorname {csgn}\left (i x^{2} {\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x^{2}\right )-8 i \pi \operatorname {csgn}\left (i x^{2} {\mathrm e}^{-x}\right )^{3}+16 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-8 i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-8 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+5 \,{\mathrm e}^{x} x^{2}-4 x \ln \left (x \right )+2 x^{2}+40 \,{\mathrm e}^{x} x +16 \ln \left (5\right )+16 x}{5 \left (x +8\right )}\) | \(194\) |
Input:
int((16*ln(5*x^2/exp(x))+(-5*x^3-85*x^2-400*x-320)*exp(x)-2*x^2-12*x+32)/( 5*x^2+80*x+320),x,method=_RETURNVERBOSE)
Output:
-exp(x)*x-2/5*x-16/5*ln(5*x^2/exp(x))/(x+8)+4/5*ln(x)
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {32-12 x-2 x^2+e^x \left (-320-400 x-85 x^2-5 x^3\right )+16 \log \left (5 e^{-x} x^2\right )}{320+80 x+5 x^2} \, dx=-\frac {5 \, {\left (x^{2} + 8 \, x\right )} e^{x} - 2 \, x \log \left (5 \, x^{2} e^{\left (-x\right )}\right )}{5 \, {\left (x + 8\right )}} \] Input:
integrate((16*log(5*x^2/exp(x))+(-5*x^3-85*x^2-400*x-320)*exp(x)-2*x^2-12* x+32)/(5*x^2+80*x+320),x, algorithm="fricas")
Output:
-1/5*(5*(x^2 + 8*x)*e^x - 2*x*log(5*x^2*e^(-x)))/(x + 8)
Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {32-12 x-2 x^2+e^x \left (-320-400 x-85 x^2-5 x^3\right )+16 \log \left (5 e^{-x} x^2\right )}{320+80 x+5 x^2} \, dx=- x e^{x} - \frac {2 x}{5} + \frac {4 \log {\left (x \right )}}{5} - \frac {16 \log {\left (5 x^{2} e^{- x} \right )}}{5 x + 40} \] Input:
integrate((16*ln(5*x**2/exp(x))+(-5*x**3-85*x**2-400*x-320)*exp(x)-2*x**2- 12*x+32)/(5*x**2+80*x+320),x)
Output:
-x*exp(x) - 2*x/5 + 4*log(x)/5 - 16*log(5*x**2*exp(-x))/(5*x + 40)
\[ \int \frac {32-12 x-2 x^2+e^x \left (-320-400 x-85 x^2-5 x^3\right )+16 \log \left (5 e^{-x} x^2\right )}{320+80 x+5 x^2} \, dx=\int { -\frac {2 \, x^{2} + 5 \, {\left (x^{3} + 17 \, x^{2} + 80 \, x + 64\right )} e^{x} + 12 \, x - 16 \, \log \left (5 \, x^{2} e^{\left (-x\right )}\right ) - 32}{5 \, {\left (x^{2} + 16 \, x + 64\right )}} \,d x } \] Input:
integrate((16*log(5*x^2/exp(x))+(-5*x^3-85*x^2-400*x-320)*exp(x)-2*x^2-12* x+32)/(5*x^2+80*x+320),x, algorithm="maxima")
Output:
-2/5*x + 64*e^(-8)*exp_integral_e(2, -x - 8)/(x + 8) + 4/5*(x*log(x) - 4*l og(5) - 32)/(x + 8) - 1/5*integrate(5*(x^3 + 17*x^2 + 80*x)*e^x/(x^2 + 16* x + 64), x)
Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {32-12 x-2 x^2+e^x \left (-320-400 x-85 x^2-5 x^3\right )+16 \log \left (5 e^{-x} x^2\right )}{320+80 x+5 x^2} \, dx=-\frac {5 \, x^{2} e^{x} + 2 \, x^{2} + 40 \, x e^{x} - 4 \, x \log \left (x\right ) + 16 \, x + 16 \, \log \left (5 \, x^{2}\right ) - 32 \, \log \left (x\right ) + 128}{5 \, {\left (x + 8\right )}} \] Input:
integrate((16*log(5*x^2/exp(x))+(-5*x^3-85*x^2-400*x-320)*exp(x)-2*x^2-12* x+32)/(5*x^2+80*x+320),x, algorithm="giac")
Output:
-1/5*(5*x^2*e^x + 2*x^2 + 40*x*e^x - 4*x*log(x) + 16*x + 16*log(5*x^2) - 3 2*log(x) + 128)/(x + 8)
Time = 1.84 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \frac {32-12 x-2 x^2+e^x \left (-320-400 x-85 x^2-5 x^3\right )+16 \log \left (5 e^{-x} x^2\right )}{320+80 x+5 x^2} \, dx=\frac {4\,\ln \left (x\right )}{5}-\frac {2\,x}{5}-\frac {16\,\ln \left (5\right )}{5\,\left (x+8\right )}+\frac {16\,x}{5\,\left (x+8\right )}-x\,{\mathrm {e}}^x-\frac {16\,\ln \left (x^2\right )}{5\,\left (x+8\right )} \] Input:
int(-(12*x - 16*log(5*x^2*exp(-x)) + 2*x^2 + exp(x)*(400*x + 85*x^2 + 5*x^ 3 + 320) - 32)/(80*x + 5*x^2 + 320),x)
Output:
(4*log(x))/5 - (2*x)/5 - (16*log(5))/(5*(x + 8)) + (16*x)/(5*(x + 8)) - x* exp(x) - (16*log(x^2))/(5*(x + 8))
Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77 \[ \int \frac {32-12 x-2 x^2+e^x \left (-320-400 x-85 x^2-5 x^3\right )+16 \log \left (5 e^{-x} x^2\right )}{320+80 x+5 x^2} \, dx=\frac {-5 e^{x} x^{2}-40 e^{x} x -16 \,\mathrm {log}\left (\frac {5 x^{2}}{e^{x}}\right )+4 \,\mathrm {log}\left (x \right ) x +32 \,\mathrm {log}\left (x \right )-2 x^{2}-16 x}{5 x +40} \] Input:
int((16*log(5*x^2/exp(x))+(-5*x^3-85*x^2-400*x-320)*exp(x)-2*x^2-12*x+32)/ (5*x^2+80*x+320),x)
Output:
( - 5*e**x*x**2 - 40*e**x*x - 16*log((5*x**2)/e**x) + 4*log(x)*x + 32*log( x) - 2*x**2 - 16*x)/(5*(x + 8))