Integrand size = 60, antiderivative size = 28 \[ \int \frac {-39-83 x+24 x^2+43 x^3-4 x^5+\left (18 x+3 x^2-9 x^3-3 x^4\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=\frac {1}{5} (3+x) \left (10+x+x^2 \left (5-x-\log \left (-3+x+x^2\right )\right )\right ) \] Output:
1/5*(10+x^2*(5-x-ln(x^2+x-3))+x)*(3+x)
Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.68 \[ \int \frac {-39-83 x+24 x^2+43 x^3-4 x^5+\left (18 x+3 x^2-9 x^3-3 x^4\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=\frac {1}{5} \left (13 x+16 x^2+2 x^3-x^4-3 x^2 \log \left (-3+x+x^2\right )-x^3 \log \left (-3+x+x^2\right )\right ) \] Input:
Integrate[(-39 - 83*x + 24*x^2 + 43*x^3 - 4*x^5 + (18*x + 3*x^2 - 9*x^3 - 3*x^4)*Log[-3 + x + x^2])/(-15 + 5*x + 5*x^2),x]
Output:
(13*x + 16*x^2 + 2*x^3 - x^4 - 3*x^2*Log[-3 + x + x^2] - x^3*Log[-3 + x + x^2])/5
Leaf count is larger than twice the leaf count of optimal. \(201\) vs. \(2(28)=56\).
Time = 0.59 (sec) , antiderivative size = 201, normalized size of antiderivative = 7.18, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {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.
\(\displaystyle \int \frac {-4 x^5+43 x^3+24 x^2+\left (-3 x^4-9 x^3+3 x^2+18 x\right ) \log \left (x^2+x-3\right )-83 x-39}{5 x^2+5 x-15} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {-4 x^5+43 x^3+24 x^2-83 x-39}{5 \left (x^2+x-3\right )}-\frac {3}{5} x (x+2) \log \left (x^2+x-3\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {x^4}{5}+\frac {2 x^3}{5}+\frac {16 x^2}{5}-\frac {3}{5} x^2 \log \left (x^2+x-3\right )-\frac {1}{5} x^3 \log \left (x^2+x-3\right )+\frac {13 x}{5}-\frac {1}{10} \left (11+\sqrt {13}\right ) \log \left (2 x-\sqrt {13}+1\right )+\frac {3}{10} \left (7-\sqrt {13}\right ) \log \left (2 x-\sqrt {13}+1\right )-\frac {1}{5} \left (5-2 \sqrt {13}\right ) \log \left (2 x-\sqrt {13}+1\right )-\frac {1}{5} \left (5+2 \sqrt {13}\right ) \log \left (2 x+\sqrt {13}+1\right )+\frac {3}{10} \left (7+\sqrt {13}\right ) \log \left (2 x+\sqrt {13}+1\right )-\frac {1}{10} \left (11-\sqrt {13}\right ) \log \left (2 x+\sqrt {13}+1\right )\) |
Input:
Int[(-39 - 83*x + 24*x^2 + 43*x^3 - 4*x^5 + (18*x + 3*x^2 - 9*x^3 - 3*x^4) *Log[-3 + x + x^2])/(-15 + 5*x + 5*x^2),x]
Output:
(13*x)/5 + (16*x^2)/5 + (2*x^3)/5 - x^4/5 - ((5 - 2*Sqrt[13])*Log[1 - Sqrt [13] + 2*x])/5 + (3*(7 - Sqrt[13])*Log[1 - Sqrt[13] + 2*x])/10 - ((11 + Sq rt[13])*Log[1 - Sqrt[13] + 2*x])/10 - ((11 - Sqrt[13])*Log[1 + Sqrt[13] + 2*x])/10 + (3*(7 + Sqrt[13])*Log[1 + Sqrt[13] + 2*x])/10 - ((5 + 2*Sqrt[13 ])*Log[1 + Sqrt[13] + 2*x])/5 - (3*x^2*Log[-3 + x + x^2])/5 - (x^3*Log[-3 + x + x^2])/5
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 = 0.78 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39
method | result | size |
risch | \(\left (-\frac {1}{5} x^{3}-\frac {3}{5} x^{2}\right ) \ln \left (x^{2}+x -3\right )-\frac {x^{4}}{5}+\frac {2 x^{3}}{5}+\frac {16 x^{2}}{5}+\frac {13 x}{5}\) | \(39\) |
default | \(-\frac {x^{4}}{5}+\frac {2 x^{3}}{5}+\frac {16 x^{2}}{5}+\frac {13 x}{5}-\frac {\ln \left (x^{2}+x -3\right ) x^{3}}{5}-\frac {3 \ln \left (x^{2}+x -3\right ) x^{2}}{5}\) | \(44\) |
norman | \(-\frac {x^{4}}{5}+\frac {2 x^{3}}{5}+\frac {16 x^{2}}{5}+\frac {13 x}{5}-\frac {\ln \left (x^{2}+x -3\right ) x^{3}}{5}-\frac {3 \ln \left (x^{2}+x -3\right ) x^{2}}{5}\) | \(44\) |
parts | \(-\frac {x^{4}}{5}+\frac {2 x^{3}}{5}+\frac {16 x^{2}}{5}+\frac {13 x}{5}-\frac {\ln \left (x^{2}+x -3\right ) x^{3}}{5}-\frac {3 \ln \left (x^{2}+x -3\right ) x^{2}}{5}\) | \(44\) |
parallelrisch | \(-\frac {x^{4}}{5}-\frac {\ln \left (x^{2}+x -3\right ) x^{3}}{5}+\frac {2 x^{3}}{5}-\frac {3 \ln \left (x^{2}+x -3\right ) x^{2}}{5}+\frac {16 x^{2}}{5}+\frac {13 x}{5}+15\) | \(45\) |
orering | \(\frac {\left (6 x^{9}+52 x^{8}+134 x^{7}+67 x^{6}-300 x^{5}-801 x^{4}-342 x^{3}+1202 x^{2}+438 x -792\right ) \left (\left (-3 x^{4}-9 x^{3}+3 x^{2}+18 x \right ) \ln \left (x^{2}+x -3\right )-4 x^{5}+43 x^{3}+24 x^{2}-83 x -39\right )}{3 \left (4 x^{8}+30 x^{7}+64 x^{6}-3 x^{5}-167 x^{4}-234 x^{3}-160 x^{2}+78 x +234\right ) \left (5 x^{2}+5 x -15\right )}-\frac {x \left (x^{7}+9 x^{6}+26 x^{5}+42 x^{4}-171 x^{2}-73 x +264\right ) \left (x^{2}+x -3\right ) \left (\frac {\left (-12 x^{3}-27 x^{2}+6 x +18\right ) \ln \left (x^{2}+x -3\right )+\frac {\left (-3 x^{4}-9 x^{3}+3 x^{2}+18 x \right ) \left (1+2 x \right )}{x^{2}+x -3}-20 x^{4}+129 x^{2}+48 x -83}{5 x^{2}+5 x -15}-\frac {\left (\left (-3 x^{4}-9 x^{3}+3 x^{2}+18 x \right ) \ln \left (x^{2}+x -3\right )-4 x^{5}+43 x^{3}+24 x^{2}-83 x -39\right ) \left (10 x +5\right )}{\left (5 x^{2}+5 x -15\right )^{2}}\right )}{3 \left (4 x^{8}+30 x^{7}+64 x^{6}-3 x^{5}-167 x^{4}-234 x^{3}-160 x^{2}+78 x +234\right )}\) | \(380\) |
Input:
int(((-3*x^4-9*x^3+3*x^2+18*x)*ln(x^2+x-3)-4*x^5+43*x^3+24*x^2-83*x-39)/(5 *x^2+5*x-15),x,method=_RETURNVERBOSE)
Output:
(-1/5*x^3-3/5*x^2)*ln(x^2+x-3)-1/5*x^4+2/5*x^3+16/5*x^2+13/5*x
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-39-83 x+24 x^2+43 x^3-4 x^5+\left (18 x+3 x^2-9 x^3-3 x^4\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=-\frac {1}{5} \, x^{4} + \frac {2}{5} \, x^{3} + \frac {16}{5} \, x^{2} - \frac {1}{5} \, {\left (x^{3} + 3 \, x^{2}\right )} \log \left (x^{2} + x - 3\right ) + \frac {13}{5} \, x \] Input:
integrate(((-3*x^4-9*x^3+3*x^2+18*x)*log(x^2+x-3)-4*x^5+43*x^3+24*x^2-83*x -39)/(5*x^2+5*x-15),x, algorithm="fricas")
Output:
-1/5*x^4 + 2/5*x^3 + 16/5*x^2 - 1/5*(x^3 + 3*x^2)*log(x^2 + x - 3) + 13/5* x
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {-39-83 x+24 x^2+43 x^3-4 x^5+\left (18 x+3 x^2-9 x^3-3 x^4\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=- \frac {x^{4}}{5} + \frac {2 x^{3}}{5} + \frac {16 x^{2}}{5} + \frac {13 x}{5} + \left (- \frac {x^{3}}{5} - \frac {3 x^{2}}{5}\right ) \log {\left (x^{2} + x - 3 \right )} \] Input:
integrate(((-3*x**4-9*x**3+3*x**2+18*x)*ln(x**2+x-3)-4*x**5+43*x**3+24*x** 2-83*x-39)/(5*x**2+5*x-15),x)
Output:
-x**4/5 + 2*x**3/5 + 16*x**2/5 + 13*x/5 + (-x**3/5 - 3*x**2/5)*log(x**2 + x - 3)
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (24) = 48\).
Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {-39-83 x+24 x^2+43 x^3-4 x^5+\left (18 x+3 x^2-9 x^3-3 x^4\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=-\frac {1}{5} \, x^{4} + \frac {2}{5} \, x^{3} + \frac {16}{5} \, x^{2} - \frac {1}{10} \, {\left (2 \, x^{3} + 6 \, x^{2} - 11\right )} \log \left (x^{2} + x - 3\right ) + \frac {13}{5} \, x - \frac {11}{10} \, \log \left (x^{2} + x - 3\right ) \] Input:
integrate(((-3*x^4-9*x^3+3*x^2+18*x)*log(x^2+x-3)-4*x^5+43*x^3+24*x^2-83*x -39)/(5*x^2+5*x-15),x, algorithm="maxima")
Output:
-1/5*x^4 + 2/5*x^3 + 16/5*x^2 - 1/10*(2*x^3 + 6*x^2 - 11)*log(x^2 + x - 3) + 13/5*x - 11/10*log(x^2 + x - 3)
Time = 0.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-39-83 x+24 x^2+43 x^3-4 x^5+\left (18 x+3 x^2-9 x^3-3 x^4\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=-\frac {1}{5} \, x^{4} + \frac {2}{5} \, x^{3} + \frac {16}{5} \, x^{2} - \frac {1}{5} \, {\left (x^{3} + 3 \, x^{2}\right )} \log \left (x^{2} + x - 3\right ) + \frac {13}{5} \, x \] Input:
integrate(((-3*x^4-9*x^3+3*x^2+18*x)*log(x^2+x-3)-4*x^5+43*x^3+24*x^2-83*x -39)/(5*x^2+5*x-15),x, algorithm="giac")
Output:
-1/5*x^4 + 2/5*x^3 + 16/5*x^2 - 1/5*(x^3 + 3*x^2)*log(x^2 + x - 3) + 13/5* x
Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {-39-83 x+24 x^2+43 x^3-4 x^5+\left (18 x+3 x^2-9 x^3-3 x^4\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=\frac {13\,x}{5}-x^3\,\left (\frac {\ln \left (x^2+x-3\right )}{5}-\frac {2}{5}\right )-x^2\,\left (\frac {3\,\ln \left (x^2+x-3\right )}{5}-\frac {16}{5}\right )-\frac {x^4}{5} \] Input:
int(-(83*x - log(x + x^2 - 3)*(18*x + 3*x^2 - 9*x^3 - 3*x^4) - 24*x^2 - 43 *x^3 + 4*x^5 + 39)/(5*x + 5*x^2 - 15),x)
Output:
(13*x)/5 - x^3*(log(x + x^2 - 3)/5 - 2/5) - x^2*((3*log(x + x^2 - 3))/5 - 16/5) - x^4/5
Time = 0.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 3.96 \[ \int \frac {-39-83 x+24 x^2+43 x^3-4 x^5+\left (18 x+3 x^2-9 x^3-3 x^4\right ) \log \left (-3+x+x^2\right )}{-15+5 x+5 x^2} \, dx=-\frac {\sqrt {13}\, \mathrm {log}\left (x^{2}+x -3\right )}{10}+\frac {\sqrt {13}\, \mathrm {log}\left (-\sqrt {13}+2 x +1\right )}{10}+\frac {\sqrt {13}\, \mathrm {log}\left (\sqrt {13}+2 x +1\right )}{10}-\frac {\mathrm {log}\left (x^{2}+x -3\right ) x^{3}}{5}-\frac {3 \,\mathrm {log}\left (x^{2}+x -3\right ) x^{2}}{5}+\frac {11 \,\mathrm {log}\left (x^{2}+x -3\right )}{10}-\frac {11 \,\mathrm {log}\left (-\sqrt {13}+2 x +1\right )}{10}-\frac {11 \,\mathrm {log}\left (\sqrt {13}+2 x +1\right )}{10}-\frac {x^{4}}{5}+\frac {2 x^{3}}{5}+\frac {16 x^{2}}{5}+\frac {13 x}{5} \] Input:
int(((-3*x^4-9*x^3+3*x^2+18*x)*log(x^2+x-3)-4*x^5+43*x^3+24*x^2-83*x-39)/( 5*x^2+5*x-15),x)
Output:
( - sqrt(13)*log(x**2 + x - 3) + sqrt(13)*log( - sqrt(13) + 2*x + 1) + sqr t(13)*log(sqrt(13) + 2*x + 1) - 2*log(x**2 + x - 3)*x**3 - 6*log(x**2 + x - 3)*x**2 + 11*log(x**2 + x - 3) - 11*log( - sqrt(13) + 2*x + 1) - 11*log( sqrt(13) + 2*x + 1) - 2*x**4 + 4*x**3 + 32*x**2 + 26*x)/10