Integrand size = 83, antiderivative size = 24 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=3+x+\frac {1}{20} x \log \left (\left (\frac {9}{e^4}-x\right ) \left (x+x^2\right )\right ) \] Output:
3+1/20*x*ln((x^2+x)*(9/exp(2)^2-x))+x
Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {1}{20} \left (16 x+x \log \left (-x (1+x) \left (-9+e^4 x\right )\right )\right ) \] Input:
Integrate[(-189 - 198*x + E^4*(22*x + 23*x^2) + (-9 - 9*x + E^4*(x + x^2)) *Log[(9*x + 9*x^2 + E^4*(-x^2 - x^3))/E^4])/(-180 - 180*x + E^4*(20*x + 20 *x^2)),x]
Output:
(16*x + x*Log[-(x*(1 + x)*(-9 + E^4*x))])/20
Time = 0.61 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {7292, 7279, 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.
\(\displaystyle \int \frac {e^4 \left (23 x^2+22 x\right )+\left (e^4 \left (x^2+x\right )-9 x-9\right ) \log \left (\frac {9 x^2+e^4 \left (-x^3-x^2\right )+9 x}{e^4}\right )-198 x-189}{e^4 \left (20 x^2+20 x\right )-180 x-180} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-e^4 \left (23 x^2+22 x\right )-\left (e^4 \left (x^2+x\right )-9 x-9\right ) \log \left (\frac {9 x^2+e^4 \left (-x^3-x^2\right )+9 x}{e^4}\right )+198 x+189}{-20 e^4 x^2+20 \left (9-e^4\right ) x+180}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {-23 e^4 x^2+22 \left (9-e^4\right ) x+189}{20 (x+1) \left (9-e^4 x\right )}+\frac {1}{20} \left (\log \left (-x (x+1) \left (e^4 x-9\right )\right )-4\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 x}{5}+\frac {1}{20} x \log \left (x (x+1) \left (9-e^4 x\right )\right )\) |
Input:
Int[(-189 - 198*x + E^4*(22*x + 23*x^2) + (-9 - 9*x + E^4*(x + x^2))*Log[( 9*x + 9*x^2 + E^4*(-x^2 - x^3))/E^4])/(-180 - 180*x + E^4*(20*x + 20*x^2)) ,x]
Output:
(4*x)/5 + (x*Log[x*(1 + x)*(9 - E^4*x)])/20
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 2.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33
method | result | size |
default | \(\frac {4 x}{5}+\frac {x \ln \left (x \left (-x^{2} {\mathrm e}^{4}-x \,{\mathrm e}^{4}+9 x +9\right )\right )}{20}\) | \(32\) |
parts | \(\frac {4 x}{5}+\frac {x \ln \left (x \left (-x^{2} {\mathrm e}^{4}-x \,{\mathrm e}^{4}+9 x +9\right )\right )}{20}\) | \(32\) |
risch | \(x +\frac {x \ln \left (\left (\left (-x^{3}-x^{2}\right ) {\mathrm e}^{4}+9 x^{2}+9 x \right ) {\mathrm e}^{-4}\right )}{20}\) | \(33\) |
norman | \(x +\frac {x \ln \left (\left (\left (-x^{3}-x^{2}\right ) {\mathrm e}^{4}+9 x^{2}+9 x \right ) {\mathrm e}^{-4}\right )}{20}\) | \(37\) |
parallelrisch | \(\frac {\left (\ln \left (\left (\left (-x^{3}-x^{2}\right ) {\mathrm e}^{4}+9 x^{2}+9 x \right ) {\mathrm e}^{-4}\right ) x \,{\mathrm e}^{8}+20 x \,{\mathrm e}^{8}-40 \,{\mathrm e}^{8}+360 \,{\mathrm e}^{4}\right ) {\mathrm e}^{-8}}{20}\) | \(64\) |
Input:
int((((x^2+x)*exp(2)^2-9*x-9)*ln(((-x^3-x^2)*exp(2)^2+9*x^2+9*x)/exp(2)^2) +(23*x^2+22*x)*exp(2)^2-198*x-189)/((20*x^2+20*x)*exp(2)^2-180*x-180),x,me thod=_RETURNVERBOSE)
Output:
4/5*x+1/20*x*ln(x*(-x^2*exp(2)^2-x*exp(2)^2+9*x+9))
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {1}{20} \, x \log \left ({\left (9 \, x^{2} - {\left (x^{3} + x^{2}\right )} e^{4} + 9 \, x\right )} e^{\left (-4\right )}\right ) + x \] Input:
integrate((((x^2+x)*exp(2)^2-9*x-9)*log(((-x^3-x^2)*exp(2)^2+9*x^2+9*x)/ex p(2)^2)+(23*x^2+22*x)*exp(2)^2-198*x-189)/((20*x^2+20*x)*exp(2)^2-180*x-18 0),x, algorithm="fricas")
Output:
1/20*x*log((9*x^2 - (x^3 + x^2)*e^4 + 9*x)*e^(-4)) + x
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {x \log {\left (\frac {9 x^{2} + 9 x + \left (- x^{3} - x^{2}\right ) e^{4}}{e^{4}} \right )}}{20} + x \] Input:
integrate((((x**2+x)*exp(2)**2-9*x-9)*ln(((-x**3-x**2)*exp(2)**2+9*x**2+9* x)/exp(2)**2)+(23*x**2+22*x)*exp(2)**2-198*x-189)/((20*x**2+20*x)*exp(2)** 2-180*x-180),x)
Output:
x*log((9*x**2 + 9*x + (-x**3 - x**2)*exp(4))*exp(-4))/20 + x
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (20) = 40\).
Time = 0.19 (sec) , antiderivative size = 162, normalized size of antiderivative = 6.75 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {23}{20} \, {\left (x e^{\left (-4\right )} + \frac {81 \, \log \left (x e^{4} - 9\right )}{e^{12} + 9 \, e^{8}} - \frac {\log \left (x + 1\right )}{e^{4} + 9}\right )} e^{4} + \frac {11}{10} \, {\left (\frac {9 \, \log \left (x e^{4} - 9\right )}{e^{8} + 9 \, e^{4}} + \frac {\log \left (x + 1\right )}{e^{4} + 9}\right )} e^{4} + \frac {1}{20} \, {\left (x e^{4} \log \left (x\right ) - 7 \, x e^{4} + {\left (x e^{4} - 9\right )} \log \left (-x e^{4} + 9\right ) + {\left (x e^{4} + e^{4}\right )} \log \left (x + 1\right )\right )} e^{\left (-4\right )} - \frac {891 \, \log \left (x e^{4} - 9\right )}{10 \, {\left (e^{8} + 9 \, e^{4}\right )}} - \frac {189 \, \log \left (x e^{4} - 9\right )}{20 \, {\left (e^{4} + 9\right )}} - \frac {9 \, \log \left (x + 1\right )}{20 \, {\left (e^{4} + 9\right )}} \] Input:
integrate((((x^2+x)*exp(2)^2-9*x-9)*log(((-x^3-x^2)*exp(2)^2+9*x^2+9*x)/ex p(2)^2)+(23*x^2+22*x)*exp(2)^2-198*x-189)/((20*x^2+20*x)*exp(2)^2-180*x-18 0),x, algorithm="maxima")
Output:
23/20*(x*e^(-4) + 81*log(x*e^4 - 9)/(e^12 + 9*e^8) - log(x + 1)/(e^4 + 9)) *e^4 + 11/10*(9*log(x*e^4 - 9)/(e^8 + 9*e^4) + log(x + 1)/(e^4 + 9))*e^4 + 1/20*(x*e^4*log(x) - 7*x*e^4 + (x*e^4 - 9)*log(-x*e^4 + 9) + (x*e^4 + e^4 )*log(x + 1))*e^(-4) - 891/10*log(x*e^4 - 9)/(e^8 + 9*e^4) - 189/20*log(x* e^4 - 9)/(e^4 + 9) - 9/20*log(x + 1)/(e^4 + 9)
Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {1}{20} \, x \log \left (-x^{3} e^{4} - x^{2} e^{4} + 9 \, x^{2} + 9 \, x\right ) + \frac {4}{5} \, x \] Input:
integrate((((x^2+x)*exp(2)^2-9*x-9)*log(((-x^3-x^2)*exp(2)^2+9*x^2+9*x)/ex p(2)^2)+(23*x^2+22*x)*exp(2)^2-198*x-189)/((20*x^2+20*x)*exp(2)^2-180*x-18 0),x, algorithm="giac")
Output:
1/20*x*log(-x^3*e^4 - x^2*e^4 + 9*x^2 + 9*x) + 4/5*x
Time = 3.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {x\,\left (\ln \left ({\mathrm {e}}^{-4}\,\left (9\,x-{\mathrm {e}}^4\,\left (x^3+x^2\right )+9\,x^2\right )\right )+20\right )}{20} \] Input:
int((198*x - exp(4)*(22*x + 23*x^2) + log(exp(-4)*(9*x - exp(4)*(x^2 + x^3 ) + 9*x^2))*(9*x - exp(4)*(x + x^2) + 9) + 189)/(180*x - exp(4)*(20*x + 20 *x^2) + 180),x)
Output:
(x*(log(exp(-4)*(9*x - exp(4)*(x^2 + x^3) + 9*x^2)) + 20))/20
Time = 0.19 (sec) , antiderivative size = 99, normalized size of antiderivative = 4.12 \[ \int \frac {-189-198 x+e^4 \left (22 x+23 x^2\right )+\left (-9-9 x+e^4 \left (x+x^2\right )\right ) \log \left (\frac {9 x+9 x^2+e^4 \left (-x^2-x^3\right )}{e^4}\right )}{-180-180 x+e^4 \left (20 x+20 x^2\right )} \, dx=\frac {9 \,\mathrm {log}\left (e^{4} x -9\right )+9 \,\mathrm {log}\left (x +1\right )+\mathrm {log}\left (\frac {-e^{4} x^{3}-e^{4} x^{2}+9 x^{2}+9 x}{e^{4}}\right ) e^{4} x -9 \,\mathrm {log}\left (\frac {-e^{4} x^{3}-e^{4} x^{2}+9 x^{2}+9 x}{e^{4}}\right )+9 \,\mathrm {log}\left (x \right )+20 e^{4} x}{20 e^{4}} \] Input:
int((((x^2+x)*exp(2)^2-9*x-9)*log(((-x^3-x^2)*exp(2)^2+9*x^2+9*x)/exp(2)^2 )+(23*x^2+22*x)*exp(2)^2-198*x-189)/((20*x^2+20*x)*exp(2)^2-180*x-180),x)
Output:
(9*log(e**4*x - 9) + 9*log(x + 1) + log(( - e**4*x**3 - e**4*x**2 + 9*x**2 + 9*x)/e**4)*e**4*x - 9*log(( - e**4*x**3 - e**4*x**2 + 9*x**2 + 9*x)/e** 4) + 9*log(x) + 20*e**4*x)/(20*e**4)