Integrand size = 66, antiderivative size = 27 \[ \int \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{64 x+32 x^2+4 x^3} \, dx=x-\frac {-x+e^{2 x} (10+\log (-2 x))}{16+4 x} \] Output:
x-(exp(2*x)*(ln(-2*x)+10)-x)/(4*x+16)
Time = 0.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{64 x+32 x^2+4 x^3} \, dx=-\frac {4+10 e^{2 x}-16 x-4 x^2+e^{2 x} \log (-2 x)}{4 (4+x)} \] Input:
Integrate[(68*x + 32*x^2 + 4*x^3 + E^(2*x)*(-4 - 71*x - 20*x^2) + E^(2*x)* (-7*x - 2*x^2)*Log[-2*x])/(64*x + 32*x^2 + 4*x^3),x]
Output:
-1/4*(4 + 10*E^(2*x) - 16*x - 4*x^2 + E^(2*x)*Log[-2*x])/(4 + x)
Time = 1.48 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2026, 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.
\(\displaystyle \int \frac {4 x^3+32 x^2+e^{2 x} \left (-20 x^2-71 x-4\right )+e^{2 x} \left (-2 x^2-7 x\right ) \log (-2 x)+68 x}{4 x^3+32 x^2+64 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {4 x^3+32 x^2+e^{2 x} \left (-20 x^2-71 x-4\right )+e^{2 x} \left (-2 x^2-7 x\right ) \log (-2 x)+68 x}{x \left (4 x^2+32 x+64\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {4 x^3+32 x^2+e^{2 x} \left (-20 x^2-71 x-4\right )+e^{2 x} \left (-2 x^2-7 x\right ) \log (-2 x)+68 x}{x (2 x+8)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2+8 x+17}{(x+4)^2}-\frac {e^{2 x} \left (20 x^2+2 x^2 \log (-2 x)+71 x+7 x \log (-2 x)+4\right )}{4 x (x+4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x-\frac {5 e^{2 x}}{2 (x+4)}-\frac {1}{x+4}-\frac {e^{2 x} \log (-2 x)}{4 (x+4)}\) |
Input:
Int[(68*x + 32*x^2 + 4*x^3 + E^(2*x)*(-4 - 71*x - 20*x^2) + E^(2*x)*(-7*x - 2*x^2)*Log[-2*x])/(64*x + 32*x^2 + 4*x^3),x]
Output:
x - (4 + x)^(-1) - (5*E^(2*x))/(2*(4 + x)) - (E^(2*x)*Log[-2*x])/(4*(4 + 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]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 0.68 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
norman | \(\frac {x^{2}-17-\frac {\ln \left (-2 x \right ) {\mathrm e}^{2 x}}{4}-\frac {5 \,{\mathrm e}^{2 x}}{2}}{4+x}\) | \(28\) |
parallelrisch | \(-\frac {272-16 x^{2}+4 \ln \left (-2 x \right ) {\mathrm e}^{2 x}+40 \,{\mathrm e}^{2 x}}{16 \left (4+x \right )}\) | \(31\) |
parts | \(\frac {-\frac {\ln \left (-2 x \right ) {\mathrm e}^{2 x}}{4}-\frac {5 \,{\mathrm e}^{2 x}}{2}}{4+x}+x -\frac {1}{4+x}\) | \(33\) |
default | \(\frac {-\ln \left (-2 x \right ) {\mathrm e}^{2 x}-10 \,{\mathrm e}^{2 x}}{4 x +16}+x -\frac {1}{4+x}\) | \(34\) |
risch | \(-\frac {{\mathrm e}^{2 x} \ln \left (-2 x \right )}{4 \left (4+x \right )}+\frac {2 x^{2}+8 x -5 \,{\mathrm e}^{2 x}-2}{2 x +8}\) | \(40\) |
orering | \(\frac {\left (8 x^{5}+28 x^{4}-40 x^{3}-108 x^{2}+153 x -87\right ) \left (\left (-2 x^{2}-7 x \right ) {\mathrm e}^{2 x} \ln \left (-2 x \right )+\left (-20 x^{2}-71 x -4\right ) {\mathrm e}^{2 x}+4 x^{3}+32 x^{2}+68 x \right )}{2 \left (4 x^{4}+20 x^{3}+18 x^{2}-33 x +27\right ) \left (4 x^{3}+32 x^{2}+64 x \right )}-\frac {\left (32 x^{5}+112 x^{4}-280 x^{3}-804 x^{2}+675 x -264\right ) \left (\frac {\left (-4 x -7\right ) {\mathrm e}^{2 x} \ln \left (-2 x \right )+2 \left (-2 x^{2}-7 x \right ) {\mathrm e}^{2 x} \ln \left (-2 x \right )+\frac {\left (-2 x^{2}-7 x \right ) {\mathrm e}^{2 x}}{x}+\left (-40 x -71\right ) {\mathrm e}^{2 x}+2 \left (-20 x^{2}-71 x -4\right ) {\mathrm e}^{2 x}+12 x^{2}+64 x +68}{4 x^{3}+32 x^{2}+64 x}-\frac {\left (\left (-2 x^{2}-7 x \right ) {\mathrm e}^{2 x} \ln \left (-2 x \right )+\left (-20 x^{2}-71 x -4\right ) {\mathrm e}^{2 x}+4 x^{3}+32 x^{2}+68 x \right ) \left (12 x^{2}+64 x +64\right )}{\left (4 x^{3}+32 x^{2}+64 x \right )^{2}}\right )}{8 \left (4 x^{4}+20 x^{3}+18 x^{2}-33 x +27\right )}+\frac {\left (8 x^{3}-78 x +33\right ) \left (4+x \right ) x \left (\frac {-4 \ln \left (-2 x \right ) {\mathrm e}^{2 x}+4 \left (-4 x -7\right ) {\mathrm e}^{2 x} \ln \left (-2 x \right )+\frac {2 \left (-4 x -7\right ) {\mathrm e}^{2 x}}{x}+4 \left (-2 x^{2}-7 x \right ) {\mathrm e}^{2 x} \ln \left (-2 x \right )+\frac {4 \left (-2 x^{2}-7 x \right ) {\mathrm e}^{2 x}}{x}-\frac {\left (-2 x^{2}-7 x \right ) {\mathrm e}^{2 x}}{x^{2}}-40 \,{\mathrm e}^{2 x}+4 \left (-40 x -71\right ) {\mathrm e}^{2 x}+4 \left (-20 x^{2}-71 x -4\right ) {\mathrm e}^{2 x}+24 x +64}{4 x^{3}+32 x^{2}+64 x}-\frac {2 \left (\left (-4 x -7\right ) {\mathrm e}^{2 x} \ln \left (-2 x \right )+2 \left (-2 x^{2}-7 x \right ) {\mathrm e}^{2 x} \ln \left (-2 x \right )+\frac {\left (-2 x^{2}-7 x \right ) {\mathrm e}^{2 x}}{x}+\left (-40 x -71\right ) {\mathrm e}^{2 x}+2 \left (-20 x^{2}-71 x -4\right ) {\mathrm e}^{2 x}+12 x^{2}+64 x +68\right ) \left (12 x^{2}+64 x +64\right )}{\left (4 x^{3}+32 x^{2}+64 x \right )^{2}}+\frac {2 \left (\left (-2 x^{2}-7 x \right ) {\mathrm e}^{2 x} \ln \left (-2 x \right )+\left (-20 x^{2}-71 x -4\right ) {\mathrm e}^{2 x}+4 x^{3}+32 x^{2}+68 x \right ) \left (12 x^{2}+64 x +64\right )^{2}}{\left (4 x^{3}+32 x^{2}+64 x \right )^{3}}-\frac {\left (\left (-2 x^{2}-7 x \right ) {\mathrm e}^{2 x} \ln \left (-2 x \right )+\left (-20 x^{2}-71 x -4\right ) {\mathrm e}^{2 x}+4 x^{3}+32 x^{2}+68 x \right ) \left (24 x +64\right )}{\left (4 x^{3}+32 x^{2}+64 x \right )^{2}}\right )}{32 x^{4}+160 x^{3}+144 x^{2}-264 x +216}\) | \(791\) |
Input:
int(((-2*x^2-7*x)*exp(2*x)*ln(-2*x)+(-20*x^2-71*x-4)*exp(2*x)+4*x^3+32*x^2 +68*x)/(4*x^3+32*x^2+64*x),x,method=_RETURNVERBOSE)
Output:
(x^2-17-1/4*ln(-2*x)*exp(2*x)-5/2*exp(2*x))/(4+x)
Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{64 x+32 x^2+4 x^3} \, dx=\frac {4 \, x^{2} - e^{\left (2 \, x\right )} \log \left (-2 \, x\right ) + 16 \, x - 10 \, e^{\left (2 \, x\right )} - 4}{4 \, {\left (x + 4\right )}} \] Input:
integrate(((-2*x^2-7*x)*exp(2*x)*log(-2*x)+(-20*x^2-71*x-4)*exp(2*x)+4*x^3 +32*x^2+68*x)/(4*x^3+32*x^2+64*x),x, algorithm="fricas")
Output:
1/4*(4*x^2 - e^(2*x)*log(-2*x) + 16*x - 10*e^(2*x) - 4)/(x + 4)
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{64 x+32 x^2+4 x^3} \, dx=x + \frac {\left (- \log {\left (- 2 x \right )} - 10\right ) e^{2 x}}{4 x + 16} - \frac {1}{x + 4} \] Input:
integrate(((-2*x**2-7*x)*exp(2*x)*ln(-2*x)+(-20*x**2-71*x-4)*exp(2*x)+4*x* *3+32*x**2+68*x)/(4*x**3+32*x**2+64*x),x)
Output:
x + (-log(-2*x) - 10)*exp(2*x)/(4*x + 16) - 1/(x + 4)
Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{64 x+32 x^2+4 x^3} \, dx=x - \frac {{\left (\log \left (2\right ) + 10\right )} e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )} \log \left (-x\right )}{4 \, {\left (x + 4\right )}} - \frac {1}{x + 4} \] Input:
integrate(((-2*x^2-7*x)*exp(2*x)*log(-2*x)+(-20*x^2-71*x-4)*exp(2*x)+4*x^3 +32*x^2+68*x)/(4*x^3+32*x^2+64*x),x, algorithm="maxima")
Output:
x - 1/4*((log(2) + 10)*e^(2*x) + e^(2*x)*log(-x))/(x + 4) - 1/(x + 4)
Time = 0.12 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{64 x+32 x^2+4 x^3} \, dx=\frac {4 \, x^{2} - e^{\left (2 \, x\right )} \log \left (-2 \, x\right ) + 16 \, x - 10 \, e^{\left (2 \, x\right )} - 4}{4 \, {\left (x + 4\right )}} \] Input:
integrate(((-2*x^2-7*x)*exp(2*x)*log(-2*x)+(-20*x^2-71*x-4)*exp(2*x)+4*x^3 +32*x^2+68*x)/(4*x^3+32*x^2+64*x),x, algorithm="giac")
Output:
1/4*(4*x^2 - e^(2*x)*log(-2*x) + 16*x - 10*e^(2*x) - 4)/(x + 4)
Time = 1.69 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{64 x+32 x^2+4 x^3} \, dx=\frac {17\,x-10\,{\mathrm {e}}^{2\,x}+4\,x^2-\ln \left (-2\,x\right )\,{\mathrm {e}}^{2\,x}}{4\,\left (x+4\right )} \] Input:
int((68*x - exp(2*x)*(71*x + 20*x^2 + 4) + 32*x^2 + 4*x^3 - log(-2*x)*exp( 2*x)*(7*x + 2*x^2))/(64*x + 32*x^2 + 4*x^3),x)
Output:
(17*x - 10*exp(2*x) + 4*x^2 - log(-2*x)*exp(2*x))/(4*(x + 4))
Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {68 x+32 x^2+4 x^3+e^{2 x} \left (-4-71 x-20 x^2\right )+e^{2 x} \left (-7 x-2 x^2\right ) \log (-2 x)}{64 x+32 x^2+4 x^3} \, dx=\frac {-e^{2 x} \mathrm {log}\left (-2 x \right )-10 e^{2 x}+4 x^{2}+17 x}{4 x +16} \] Input:
int(((-2*x^2-7*x)*exp(2*x)*log(-2*x)+(-20*x^2-71*x-4)*exp(2*x)+4*x^3+32*x^ 2+68*x)/(4*x^3+32*x^2+64*x),x)
Output:
( - e**(2*x)*log( - 2*x) - 10*e**(2*x) + 4*x**2 + 17*x)/(4*(x + 4))