Integrand size = 65, antiderivative size = 31 \[ \int \frac {16-32 x^2-44 x^4-12 x^6+e^{\frac {9 x}{2+x^2}} \left (-18+9 x^2\right )+\left (32+32 x^2+8 x^4\right ) \log (x)}{4+4 x^2+x^4} \, dx=2-e^{\frac {9}{\frac {2}{x}+x}}+4 x \left (-1-x^2+2 \log (x)\right ) \] Output:
4*(2*ln(x)-x^2-1)*x-exp(9/(x+2/x))+2
Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {16-32 x^2-44 x^4-12 x^6+e^{\frac {9 x}{2+x^2}} \left (-18+9 x^2\right )+\left (32+32 x^2+8 x^4\right ) \log (x)}{4+4 x^2+x^4} \, dx=-e^{\frac {9 x}{2+x^2}}-4 x-4 x^3+8 x \log (x) \] Input:
Integrate[(16 - 32*x^2 - 44*x^4 - 12*x^6 + E^((9*x)/(2 + x^2))*(-18 + 9*x^ 2) + (32 + 32*x^2 + 8*x^4)*Log[x])/(4 + 4*x^2 + x^4),x]
Output:
-E^((9*x)/(2 + x^2)) - 4*x - 4*x^3 + 8*x*Log[x]
Time = 0.95 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.046, Rules used = {1380, 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.
\(\displaystyle \int \frac {-12 x^6-44 x^4-32 x^2+e^{\frac {9 x}{x^2+2}} \left (9 x^2-18\right )+\left (8 x^4+32 x^2+32\right ) \log (x)+16}{x^4+4 x^2+4} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \int \frac {-12 x^6-44 x^4-32 x^2-9 e^{\frac {9 x}{x^2+2}} \left (2-x^2\right )+8 \left (x^4+4 x^2+4\right ) \log (x)+16}{\left (x^2+2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {32 x^2}{\left (x^2+2\right )^2}+\frac {9 e^{\frac {9 x}{x^2+2}} \left (x^2-2\right )}{\left (x^2+2\right )^2}+\frac {16}{\left (x^2+2\right )^2}-\frac {12 x^6}{\left (x^2+2\right )^2}-\frac {44 x^4}{\left (x^2+2\right )^2}+8 \log (x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -10 x^3+\frac {20 x}{x^2+2}-e^{\frac {9 x}{x^2+2}}+\frac {6 x^5}{x^2+2}+\frac {22 x^3}{x^2+2}-14 x+8 x \log (x)\) |
Input:
Int[(16 - 32*x^2 - 44*x^4 - 12*x^6 + E^((9*x)/(2 + x^2))*(-18 + 9*x^2) + ( 32 + 32*x^2 + 8*x^4)*Log[x])/(4 + 4*x^2 + x^4),x]
Output:
-E^((9*x)/(2 + x^2)) - 14*x - 10*x^3 + (20*x)/(2 + x^2) + (22*x^3)/(2 + x^ 2) + (6*x^5)/(2 + x^2) + 8*x*Log[x]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 1.43 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90
method | result | size |
risch | \(-4 x^{3}+8 x \ln \left (x \right )-4 x -{\mathrm e}^{\frac {9 x}{x^{2}+2}}\) | \(28\) |
parallelrisch | \(-4 x^{3}+8 x \ln \left (x \right )-4 x -{\mathrm e}^{\frac {9 x}{x^{2}+2}}\) | \(28\) |
default | \(-4 x^{3}-4 x +\frac {-x^{2} {\mathrm e}^{\frac {9 x}{x^{2}+2}}-2 \,{\mathrm e}^{\frac {9 x}{x^{2}+2}}}{x^{2}+2}+8 x \ln \left (x \right )\) | \(53\) |
parts | \(-4 x^{3}-4 x +\frac {-x^{2} {\mathrm e}^{\frac {9 x}{x^{2}+2}}-2 \,{\mathrm e}^{\frac {9 x}{x^{2}+2}}}{x^{2}+2}+8 x \ln \left (x \right )\) | \(53\) |
orering | \(\frac {\left (\left (8 x^{4}+32 x^{2}+32\right ) \ln \left (x \right )+\left (9 x^{2}-18\right ) {\mathrm e}^{\frac {9 x}{x^{2}+2}}-12 x^{6}-44 x^{4}-32 x^{2}+16\right ) x}{x^{4}+4 x^{2}+4}+\frac {\left (6 x^{14}-54 x^{13}-898 x^{12}-144 x^{11}+5142 x^{10}-4212 x^{9}-12117 x^{8}+82656 x^{7}+222110 x^{6}+94752 x^{5}-241020 x^{4}-454464 x^{3}+189000 x^{2}-576 x +128\right ) \left (\frac {\left (32 x^{3}+64 x \right ) \ln \left (x \right )+\frac {8 x^{4}+32 x^{2}+32}{x}+18 x \,{\mathrm e}^{\frac {9 x}{x^{2}+2}}+\left (9 x^{2}-18\right ) \left (\frac {9}{x^{2}+2}-\frac {18 x^{2}}{\left (x^{2}+2\right )^{2}}\right ) {\mathrm e}^{\frac {9 x}{x^{2}+2}}-72 x^{5}-176 x^{3}-64 x}{x^{4}+4 x^{2}+4}-\frac {\left (\left (8 x^{4}+32 x^{2}+32\right ) \ln \left (x \right )+\left (9 x^{2}-18\right ) {\mathrm e}^{\frac {9 x}{x^{2}+2}}-12 x^{6}-44 x^{4}-32 x^{2}+16\right ) \left (4 x^{3}+8 x \right )}{\left (x^{4}+4 x^{2}+4\right )^{2}}\right )}{432 x^{11}+3402 x^{10}+1710 x^{9}-18306 x^{8}-42390 x^{7}-9720 x^{6}+47412 x^{5}+72576 x^{4}-51336 x^{3}-15552 x^{2}+8208 x +2592}-\frac {\left (6 x^{10}+90 x^{9}+208 x^{8}-90 x^{7}+2529 x^{5}-1288 x^{4}-12204 x^{3}-8800 x^{2}+12420 x -32\right ) \left (x^{2}+2\right )^{2} x \left (\frac {\left (96 x^{2}+64\right ) \ln \left (x \right )+\frac {64 x^{3}+128 x}{x}-\frac {8 x^{4}+32 x^{2}+32}{x^{2}}+18 \,{\mathrm e}^{\frac {9 x}{x^{2}+2}}+36 x \left (\frac {9}{x^{2}+2}-\frac {18 x^{2}}{\left (x^{2}+2\right )^{2}}\right ) {\mathrm e}^{\frac {9 x}{x^{2}+2}}+\left (9 x^{2}-18\right ) \left (-\frac {54 x}{\left (x^{2}+2\right )^{2}}+\frac {72 x^{3}}{\left (x^{2}+2\right )^{3}}\right ) {\mathrm e}^{\frac {9 x}{x^{2}+2}}+\left (9 x^{2}-18\right ) \left (\frac {9}{x^{2}+2}-\frac {18 x^{2}}{\left (x^{2}+2\right )^{2}}\right )^{2} {\mathrm e}^{\frac {9 x}{x^{2}+2}}-360 x^{4}-528 x^{2}-64}{x^{4}+4 x^{2}+4}-\frac {2 \left (\left (32 x^{3}+64 x \right ) \ln \left (x \right )+\frac {8 x^{4}+32 x^{2}+32}{x}+18 x \,{\mathrm e}^{\frac {9 x}{x^{2}+2}}+\left (9 x^{2}-18\right ) \left (\frac {9}{x^{2}+2}-\frac {18 x^{2}}{\left (x^{2}+2\right )^{2}}\right ) {\mathrm e}^{\frac {9 x}{x^{2}+2}}-72 x^{5}-176 x^{3}-64 x \right ) \left (4 x^{3}+8 x \right )}{\left (x^{4}+4 x^{2}+4\right )^{2}}+\frac {2 \left (\left (8 x^{4}+32 x^{2}+32\right ) \ln \left (x \right )+\left (9 x^{2}-18\right ) {\mathrm e}^{\frac {9 x}{x^{2}+2}}-12 x^{6}-44 x^{4}-32 x^{2}+16\right ) \left (4 x^{3}+8 x \right )^{2}}{\left (x^{4}+4 x^{2}+4\right )^{3}}-\frac {\left (\left (8 x^{4}+32 x^{2}+32\right ) \ln \left (x \right )+\left (9 x^{2}-18\right ) {\mathrm e}^{\frac {9 x}{x^{2}+2}}-12 x^{6}-44 x^{4}-32 x^{2}+16\right ) \left (12 x^{2}+8\right )}{\left (x^{4}+4 x^{2}+4\right )^{2}}\right )}{18 \left (24 x^{11}+189 x^{10}+95 x^{9}-1017 x^{8}-2355 x^{7}-540 x^{6}+2634 x^{5}+4032 x^{4}-2852 x^{3}-864 x^{2}+456 x +144\right )}\) | \(962\) |
Input:
int(((8*x^4+32*x^2+32)*ln(x)+(9*x^2-18)*exp(9*x/(x^2+2))-12*x^6-44*x^4-32* x^2+16)/(x^4+4*x^2+4),x,method=_RETURNVERBOSE)
Output:
-4*x^3+8*x*ln(x)-4*x-exp(9*x/(x^2+2))
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {16-32 x^2-44 x^4-12 x^6+e^{\frac {9 x}{2+x^2}} \left (-18+9 x^2\right )+\left (32+32 x^2+8 x^4\right ) \log (x)}{4+4 x^2+x^4} \, dx=-4 \, x^{3} + 8 \, x \log \left (x\right ) - 4 \, x - e^{\left (\frac {9 \, x}{x^{2} + 2}\right )} \] Input:
integrate(((8*x^4+32*x^2+32)*log(x)+(9*x^2-18)*exp(9*x/(x^2+2))-12*x^6-44* x^4-32*x^2+16)/(x^4+4*x^2+4),x, algorithm="fricas")
Output:
-4*x^3 + 8*x*log(x) - 4*x - e^(9*x/(x^2 + 2))
Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {16-32 x^2-44 x^4-12 x^6+e^{\frac {9 x}{2+x^2}} \left (-18+9 x^2\right )+\left (32+32 x^2+8 x^4\right ) \log (x)}{4+4 x^2+x^4} \, dx=- 4 x^{3} + 8 x \log {\left (x \right )} - 4 x - e^{\frac {9 x}{x^{2} + 2}} \] Input:
integrate(((8*x**4+32*x**2+32)*ln(x)+(9*x**2-18)*exp(9*x/(x**2+2))-12*x**6 -44*x**4-32*x**2+16)/(x**4+4*x**2+4),x)
Output:
-4*x**3 + 8*x*log(x) - 4*x - exp(9*x/(x**2 + 2))
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {16-32 x^2-44 x^4-12 x^6+e^{\frac {9 x}{2+x^2}} \left (-18+9 x^2\right )+\left (32+32 x^2+8 x^4\right ) \log (x)}{4+4 x^2+x^4} \, dx=-4 \, x^{3} + 8 \, x \log \left (x\right ) - 4 \, x - e^{\left (\frac {9 \, x}{x^{2} + 2}\right )} \] Input:
integrate(((8*x^4+32*x^2+32)*log(x)+(9*x^2-18)*exp(9*x/(x^2+2))-12*x^6-44* x^4-32*x^2+16)/(x^4+4*x^2+4),x, algorithm="maxima")
Output:
-4*x^3 + 8*x*log(x) - 4*x - e^(9*x/(x^2 + 2))
Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {16-32 x^2-44 x^4-12 x^6+e^{\frac {9 x}{2+x^2}} \left (-18+9 x^2\right )+\left (32+32 x^2+8 x^4\right ) \log (x)}{4+4 x^2+x^4} \, dx=-4 \, x^{3} + 8 \, x \log \left (x\right ) - 4 \, x - e^{\left (\frac {9 \, x}{x^{2} + 2}\right )} \] Input:
integrate(((8*x^4+32*x^2+32)*log(x)+(9*x^2-18)*exp(9*x/(x^2+2))-12*x^6-44* x^4-32*x^2+16)/(x^4+4*x^2+4),x, algorithm="giac")
Output:
-4*x^3 + 8*x*log(x) - 4*x - e^(9*x/(x^2 + 2))
Time = 2.82 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {16-32 x^2-44 x^4-12 x^6+e^{\frac {9 x}{2+x^2}} \left (-18+9 x^2\right )+\left (32+32 x^2+8 x^4\right ) \log (x)}{4+4 x^2+x^4} \, dx=4\,x-{\mathrm {e}}^{\frac {9\,x}{x^2+2}}+8\,x\,\left (\ln \left (x\right )-1\right )-4\,x^3 \] Input:
int((log(x)*(32*x^2 + 8*x^4 + 32) + exp((9*x)/(x^2 + 2))*(9*x^2 - 18) - 32 *x^2 - 44*x^4 - 12*x^6 + 16)/(4*x^2 + x^4 + 4),x)
Output:
4*x - exp((9*x)/(x^2 + 2)) + 8*x*(log(x) - 1) - 4*x^3
Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {16-32 x^2-44 x^4-12 x^6+e^{\frac {9 x}{2+x^2}} \left (-18+9 x^2\right )+\left (32+32 x^2+8 x^4\right ) \log (x)}{4+4 x^2+x^4} \, dx=-e^{\frac {9 x}{x^{2}+2}}+8 \,\mathrm {log}\left (x \right ) x -4 x^{3}-4 x \] Input:
int(((8*x^4+32*x^2+32)*log(x)+(9*x^2-18)*exp(9*x/(x^2+2))-12*x^6-44*x^4-32 *x^2+16)/(x^4+4*x^2+4),x)
Output:
- e**((9*x)/(x**2 + 2)) + 8*log(x)*x - 4*x**3 - 4*x