Integrand size = 108, antiderivative size = 33 \[ \int \frac {32-512 x-240 x^2+128 x^3+64 x^4+e^{4 x} \left (2-40 x+53 x^2+12 x^3-44 x^4-16 x^5\right )}{32 x-272 x^2-16 x^3+192 x^4+64 x^5+e^{4 x} \left (2 x-17 x^2-x^3+12 x^4+4 x^5\right )} \, dx=\log \left (\frac {5 \left (4 x^2-\frac {x}{2+x}\right )}{\left (16+e^{4 x}\right ) (1-x)}\right ) \] Output:
ln((4*x^2-x/(2+x))/(1-x)/(1/5*exp(4*x)+16/5))
Time = 1.64 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {32-512 x-240 x^2+128 x^3+64 x^4+e^{4 x} \left (2-40 x+53 x^2+12 x^3-44 x^4-16 x^5\right )}{32 x-272 x^2-16 x^3+192 x^4+64 x^5+e^{4 x} \left (2 x-17 x^2-x^3+12 x^4+4 x^5\right )} \, dx=-4 x-2 \text {arctanh}\left (1+\frac {e^{4 x}}{8}\right )+\log (x)+\log \left (1-8 x-4 x^2\right )-\log \left (2-x-x^2\right ) \] Input:
Integrate[(32 - 512*x - 240*x^2 + 128*x^3 + 64*x^4 + E^(4*x)*(2 - 40*x + 5 3*x^2 + 12*x^3 - 44*x^4 - 16*x^5))/(32*x - 272*x^2 - 16*x^3 + 192*x^4 + 64 *x^5 + E^(4*x)*(2*x - 17*x^2 - x^3 + 12*x^4 + 4*x^5)),x]
Output:
-4*x - 2*ArcTanh[1 + E^(4*x)/8] + Log[x] + Log[1 - 8*x - 4*x^2] - Log[2 - x - x^2]
Time = 3.59 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {7292, 2463, 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.
\(\displaystyle \int \frac {64 x^4+128 x^3-240 x^2+e^{4 x} \left (-16 x^5-44 x^4+12 x^3+53 x^2-40 x+2\right )-512 x+32}{64 x^5+192 x^4-16 x^3-272 x^2+e^{4 x} \left (4 x^5+12 x^4-x^3-17 x^2+2 x\right )+32 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {64 x^4+128 x^3-240 x^2+e^{4 x} \left (-16 x^5-44 x^4+12 x^3+53 x^2-40 x+2\right )-512 x+32}{\left (e^{4 x}+16\right ) x \left (4 x^4+12 x^3-x^2-17 x+2\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {64 x^4+128 x^3-240 x^2+e^{4 x} \left (-16 x^5-44 x^4+12 x^3+53 x^2-40 x+2\right )-512 x+32}{33 \left (e^{4 x}+16\right ) (x-1) x}+\frac {64 x^4+128 x^3-240 x^2+e^{4 x} \left (-16 x^5-44 x^4+12 x^3+53 x^2-40 x+2\right )-512 x+32}{3 \left (e^{4 x}+16\right ) x (x+2)}-\frac {4 (4 x+1) \left (64 x^4+128 x^3-240 x^2+e^{4 x} \left (-16 x^5-44 x^4+12 x^3+53 x^2-40 x+2\right )-512 x+32\right )}{11 \left (e^{4 x}+16\right ) x \left (4 x^2+8 x-1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log \left (-4 x^2-8 x+1\right )-\log \left (e^{4 x}+16\right )-\log (1-x)+\log (x)-\log (x+2)\) |
Input:
Int[(32 - 512*x - 240*x^2 + 128*x^3 + 64*x^4 + E^(4*x)*(2 - 40*x + 53*x^2 + 12*x^3 - 44*x^4 - 16*x^5))/(32*x - 272*x^2 - 16*x^3 + 192*x^4 + 64*x^5 + E^(4*x)*(2*x - 17*x^2 - x^3 + 12*x^4 + 4*x^5)),x]
Output:
-Log[16 + E^(4*x)] - Log[1 - x] + Log[x] - Log[2 + x] + Log[1 - 8*x - 4*x^ 2]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03
method | result | size |
parallelrisch | \(\ln \left (x \right )-\ln \left (-1+x \right )-\ln \left (2+x \right )-\ln \left ({\mathrm e}^{4 x}+16\right )+\ln \left (x^{2}+2 x -\frac {1}{4}\right )\) | \(34\) |
risch | \(-\ln \left (x^{2}+x -2\right )+\ln \left (4 x^{3}+8 x^{2}-x \right )-\ln \left ({\mathrm e}^{4 x}+16\right )\) | \(35\) |
norman | \(-\ln \left (-1+x \right )-\ln \left (2+x \right )-\ln \left ({\mathrm e}^{4 x}+16\right )+\ln \left (x \right )+\ln \left (4 x^{2}+8 x -1\right )\) | \(36\) |
Input:
int(((-16*x^5-44*x^4+12*x^3+53*x^2-40*x+2)*exp(4*x)+64*x^4+128*x^3-240*x^2 -512*x+32)/((4*x^5+12*x^4-x^3-17*x^2+2*x)*exp(4*x)+64*x^5+192*x^4-16*x^3-2 72*x^2+32*x),x,method=_RETURNVERBOSE)
Output:
ln(x)-ln(-1+x)-ln(2+x)-ln(exp(4*x)+16)+ln(x^2+2*x-1/4)
Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {32-512 x-240 x^2+128 x^3+64 x^4+e^{4 x} \left (2-40 x+53 x^2+12 x^3-44 x^4-16 x^5\right )}{32 x-272 x^2-16 x^3+192 x^4+64 x^5+e^{4 x} \left (2 x-17 x^2-x^3+12 x^4+4 x^5\right )} \, dx=\log \left (4 \, x^{3} + 8 \, x^{2} - x\right ) - \log \left (x^{2} + x - 2\right ) - \log \left (e^{\left (4 \, x\right )} + 16\right ) \] Input:
integrate(((-16*x^5-44*x^4+12*x^3+53*x^2-40*x+2)*exp(4*x)+64*x^4+128*x^3-2 40*x^2-512*x+32)/((4*x^5+12*x^4-x^3-17*x^2+2*x)*exp(4*x)+64*x^5+192*x^4-16 *x^3-272*x^2+32*x),x, algorithm="fricas")
Output:
log(4*x^3 + 8*x^2 - x) - log(x^2 + x - 2) - log(e^(4*x) + 16)
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {32-512 x-240 x^2+128 x^3+64 x^4+e^{4 x} \left (2-40 x+53 x^2+12 x^3-44 x^4-16 x^5\right )}{32 x-272 x^2-16 x^3+192 x^4+64 x^5+e^{4 x} \left (2 x-17 x^2-x^3+12 x^4+4 x^5\right )} \, dx=- \log {\left (e^{4 x} + 16 \right )} - \log {\left (x^{2} + x - 2 \right )} + \log {\left (4 x^{3} + 8 x^{2} - x \right )} \] Input:
integrate(((-16*x**5-44*x**4+12*x**3+53*x**2-40*x+2)*exp(4*x)+64*x**4+128* x**3-240*x**2-512*x+32)/((4*x**5+12*x**4-x**3-17*x**2+2*x)*exp(4*x)+64*x** 5+192*x**4-16*x**3-272*x**2+32*x),x)
Output:
-log(exp(4*x) + 16) - log(x**2 + x - 2) + log(4*x**3 + 8*x**2 - x)
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {32-512 x-240 x^2+128 x^3+64 x^4+e^{4 x} \left (2-40 x+53 x^2+12 x^3-44 x^4-16 x^5\right )}{32 x-272 x^2-16 x^3+192 x^4+64 x^5+e^{4 x} \left (2 x-17 x^2-x^3+12 x^4+4 x^5\right )} \, dx=\log \left (4 \, x^{2} + 8 \, x - 1\right ) - \log \left (x + 2\right ) - \log \left (x - 1\right ) + \log \left (x\right ) - \log \left (e^{\left (4 \, x\right )} + 16\right ) \] Input:
integrate(((-16*x^5-44*x^4+12*x^3+53*x^2-40*x+2)*exp(4*x)+64*x^4+128*x^3-2 40*x^2-512*x+32)/((4*x^5+12*x^4-x^3-17*x^2+2*x)*exp(4*x)+64*x^5+192*x^4-16 *x^3-272*x^2+32*x),x, algorithm="maxima")
Output:
log(4*x^2 + 8*x - 1) - log(x + 2) - log(x - 1) + log(x) - log(e^(4*x) + 16 )
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {32-512 x-240 x^2+128 x^3+64 x^4+e^{4 x} \left (2-40 x+53 x^2+12 x^3-44 x^4-16 x^5\right )}{32 x-272 x^2-16 x^3+192 x^4+64 x^5+e^{4 x} \left (2 x-17 x^2-x^3+12 x^4+4 x^5\right )} \, dx=\log \left (4 \, x^{2} + 8 \, x - 1\right ) - \log \left (x^{2} + x - 2\right ) + \log \left (x\right ) - \log \left (e^{\left (4 \, x\right )} + 16\right ) \] Input:
integrate(((-16*x^5-44*x^4+12*x^3+53*x^2-40*x+2)*exp(4*x)+64*x^4+128*x^3-2 40*x^2-512*x+32)/((4*x^5+12*x^4-x^3-17*x^2+2*x)*exp(4*x)+64*x^5+192*x^4-16 *x^3-272*x^2+32*x),x, algorithm="giac")
Output:
log(4*x^2 + 8*x - 1) - log(x^2 + x - 2) + log(x) - log(e^(4*x) + 16)
Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {32-512 x-240 x^2+128 x^3+64 x^4+e^{4 x} \left (2-40 x+53 x^2+12 x^3-44 x^4-16 x^5\right )}{32 x-272 x^2-16 x^3+192 x^4+64 x^5+e^{4 x} \left (2 x-17 x^2-x^3+12 x^4+4 x^5\right )} \, dx=\ln \left (x\,\left (4\,x^2+8\,x-1\right )\right )-\ln \left (x^2+x-2\right )-\ln \left ({\mathrm {e}}^{4\,x}+16\right ) \] Input:
int(-(512*x + exp(4*x)*(40*x - 53*x^2 - 12*x^3 + 44*x^4 + 16*x^5 - 2) + 24 0*x^2 - 128*x^3 - 64*x^4 - 32)/(32*x + exp(4*x)*(2*x - 17*x^2 - x^3 + 12*x ^4 + 4*x^5) - 272*x^2 - 16*x^3 + 192*x^4 + 64*x^5),x)
Output:
log(x*(8*x + 4*x^2 - 1)) - log(x + x^2 - 2) - log(exp(4*x) + 16)
Time = 0.17 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.03 \[ \int \frac {32-512 x-240 x^2+128 x^3+64 x^4+e^{4 x} \left (2-40 x+53 x^2+12 x^3-44 x^4-16 x^5\right )}{32 x-272 x^2-16 x^3+192 x^4+64 x^5+e^{4 x} \left (2 x-17 x^2-x^3+12 x^4+4 x^5\right )} \, dx=-\mathrm {log}\left (e^{2 x}-2 e^{x} \sqrt {2}+4\right )-\mathrm {log}\left (e^{2 x}+2 e^{x} \sqrt {2}+4\right )+\mathrm {log}\left (-\sqrt {5}+2 x +2\right )+\mathrm {log}\left (\sqrt {5}+2 x +2\right )-\mathrm {log}\left (x -1\right )-\mathrm {log}\left (x +2\right )+\mathrm {log}\left (x \right ) \] Input:
int(((-16*x^5-44*x^4+12*x^3+53*x^2-40*x+2)*exp(4*x)+64*x^4+128*x^3-240*x^2 -512*x+32)/((4*x^5+12*x^4-x^3-17*x^2+2*x)*exp(4*x)+64*x^5+192*x^4-16*x^3-2 72*x^2+32*x),x)
Output:
- log(e**(2*x) - 2*e**x*sqrt(2) + 4) - log(e**(2*x) + 2*e**x*sqrt(2) + 4) + log( - sqrt(5) + 2*x + 2) + log(sqrt(5) + 2*x + 2) - log(x - 1) - log(x + 2) + log(x)