Integrand size = 51, antiderivative size = 28 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=2+\frac {1}{15} \left (-e^2-x\right )-e^x \log \left (-1+\frac {4}{x^2}\right ) \] Output:
2-exp(x)*ln(4/x^2-1)-1/15*x-1/15*exp(2)
Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=\frac {1}{15} \left (-x-15 e^x \log \left (-1+\frac {4}{x^2}\right )\right ) \] Input:
Integrate[(-120*E^x + 4*x - x^3 + E^x*(60*x - 15*x^3)*Log[(4 - x^2)/x^2])/ (-60*x + 15*x^3),x]
Output:
(-x - 15*E^x*Log[-1 + 4/x^2])/15
Time = 0.62 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2026, 7276, 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.
\(\displaystyle \int \frac {-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )+4 x-120 e^x}{15 x^3-60 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )+4 x-120 e^x}{x \left (15 x^2-60\right )}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (-\frac {e^x \left (-4 x \log \left (\frac {4}{x^2}-1\right )+x^3 \log \left (\frac {4}{x^2}-1\right )+8\right )}{x \left (x^2-4\right )}-\frac {1}{15}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^x \left (4 x \log \left (\frac {4}{x^2}-1\right )-x^3 \log \left (\frac {4}{x^2}-1\right )\right )}{\left (4-x^2\right ) x}-\frac {x}{15}\) |
Input:
Int[(-120*E^x + 4*x - x^3 + E^x*(60*x - 15*x^3)*Log[(4 - x^2)/x^2])/(-60*x + 15*x^3),x]
Output:
-1/15*x - (E^x*(4*x*Log[-1 + 4/x^2] - x^3*Log[-1 + 4/x^2]))/(x*(4 - x^2))
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])
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 0.64 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71
method | result | size |
parallelrisch | \(-{\mathrm e}^{x} \ln \left (-\frac {x^{2}-4}{x^{2}}\right )-\frac {x}{15}\) | \(20\) |
default | \(-\frac {x}{15}-\ln \left (\frac {-x^{2}+4}{x^{2}}\right ) {\mathrm e}^{x}\) | \(21\) |
norman | \(-\frac {x}{15}-\ln \left (\frac {-x^{2}+4}{x^{2}}\right ) {\mathrm e}^{x}\) | \(21\) |
parts | \(-\frac {x}{15}-\ln \left (\frac {-x^{2}+4}{x^{2}}\right ) {\mathrm e}^{x}\) | \(21\) |
risch | \(-\ln \left (x^{2}-4\right ) {\mathrm e}^{x}+2 \,{\mathrm e}^{x} \ln \left (x \right )-\frac {x}{15}-\frac {i {\mathrm e}^{x} \pi \,\operatorname {csgn}\left (i \left (x^{2}-4\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right )}^{2}}{2}+\frac {i {\mathrm e}^{x} \pi \,\operatorname {csgn}\left (i \left (x^{2}-4\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right )}{2}-\frac {i {\mathrm e}^{x} \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right )}^{3}}{2}+i {\mathrm e}^{x} \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right )}^{2}-\frac {i {\mathrm e}^{x} \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-4\right )}{x^{2}}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right )}{2}-\frac {i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}}{2}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}-\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}}{2}-i {\mathrm e}^{x} \pi \) | \(212\) |
orering | \(\frac {\left (5+x \right ) \left (\left (-15 x^{3}+60 x \right ) {\mathrm e}^{x} \ln \left (\frac {-x^{2}+4}{x^{2}}\right )-120 \,{\mathrm e}^{x}-x^{3}+4 x \right )}{15 x^{3}-60 x}-\frac {2 \left (x^{7}-x^{6}-20 x^{5}+20 x^{4}+100 x^{3}-16 x^{2}-80 x +64\right ) \left (\frac {\left (-45 x^{2}+60\right ) {\mathrm e}^{x} \ln \left (\frac {-x^{2}+4}{x^{2}}\right )+\left (-15 x^{3}+60 x \right ) {\mathrm e}^{x} \ln \left (\frac {-x^{2}+4}{x^{2}}\right )+\frac {\left (-15 x^{3}+60 x \right ) {\mathrm e}^{x} \left (-\frac {2}{x}-\frac {2 \left (-x^{2}+4\right )}{x^{3}}\right ) x^{2}}{-x^{2}+4}-120 \,{\mathrm e}^{x}-3 x^{2}+4}{15 x^{3}-60 x}-\frac {\left (\left (-15 x^{3}+60 x \right ) {\mathrm e}^{x} \ln \left (\frac {-x^{2}+4}{x^{2}}\right )-120 \,{\mathrm e}^{x}-x^{3}+4 x \right ) \left (45 x^{2}-60\right )}{\left (15 x^{3}-60 x \right )^{2}}\right )}{x^{6}-6 x^{5}+4 x^{4}+32 x^{3}-8 x^{2}-32 x +32}+\frac {\left (x^{4}-16 x^{2}-8 x +16\right ) \left (-2+x \right ) \left (2+x \right ) x \left (\frac {-90 \ln \left (\frac {-x^{2}+4}{x^{2}}\right ) {\mathrm e}^{x} x +2 \left (-45 x^{2}+60\right ) {\mathrm e}^{x} \ln \left (\frac {-x^{2}+4}{x^{2}}\right )+\frac {2 \left (-45 x^{2}+60\right ) {\mathrm e}^{x} \left (-\frac {2}{x}-\frac {2 \left (-x^{2}+4\right )}{x^{3}}\right ) x^{2}}{-x^{2}+4}+\left (-15 x^{3}+60 x \right ) {\mathrm e}^{x} \ln \left (\frac {-x^{2}+4}{x^{2}}\right )+\frac {2 \left (-15 x^{3}+60 x \right ) {\mathrm e}^{x} \left (-\frac {2}{x}-\frac {2 \left (-x^{2}+4\right )}{x^{3}}\right ) x^{2}}{-x^{2}+4}+\frac {\left (-15 x^{3}+60 x \right ) {\mathrm e}^{x} \left (\frac {6}{x^{2}}+\frac {-6 x^{2}+24}{x^{4}}\right ) x^{2}}{-x^{2}+4}+\frac {2 \left (-15 x^{3}+60 x \right ) {\mathrm e}^{x} \left (-\frac {2}{x}-\frac {2 \left (-x^{2}+4\right )}{x^{3}}\right ) x^{3}}{\left (-x^{2}+4\right )^{2}}+\frac {2 \left (-15 x^{3}+60 x \right ) {\mathrm e}^{x} \left (-\frac {2}{x}-\frac {2 \left (-x^{2}+4\right )}{x^{3}}\right ) x}{-x^{2}+4}-120 \,{\mathrm e}^{x}-6 x}{15 x^{3}-60 x}-\frac {2 \left (\left (-45 x^{2}+60\right ) {\mathrm e}^{x} \ln \left (\frac {-x^{2}+4}{x^{2}}\right )+\left (-15 x^{3}+60 x \right ) {\mathrm e}^{x} \ln \left (\frac {-x^{2}+4}{x^{2}}\right )+\frac {\left (-15 x^{3}+60 x \right ) {\mathrm e}^{x} \left (-\frac {2}{x}-\frac {2 \left (-x^{2}+4\right )}{x^{3}}\right ) x^{2}}{-x^{2}+4}-120 \,{\mathrm e}^{x}-3 x^{2}+4\right ) \left (45 x^{2}-60\right )}{\left (15 x^{3}-60 x \right )^{2}}+\frac {2 \left (\left (-15 x^{3}+60 x \right ) {\mathrm e}^{x} \ln \left (\frac {-x^{2}+4}{x^{2}}\right )-120 \,{\mathrm e}^{x}-x^{3}+4 x \right ) \left (45 x^{2}-60\right )^{2}}{\left (15 x^{3}-60 x \right )^{3}}-\frac {90 \left (\left (-15 x^{3}+60 x \right ) {\mathrm e}^{x} \ln \left (\frac {-x^{2}+4}{x^{2}}\right )-120 \,{\mathrm e}^{x}-x^{3}+4 x \right ) x}{\left (15 x^{3}-60 x \right )^{2}}\right )}{x^{6}-6 x^{5}+4 x^{4}+32 x^{3}-8 x^{2}-32 x +32}\) | \(863\) |
Input:
int(((-15*x^3+60*x)*exp(x)*ln((-x^2+4)/x^2)-120*exp(x)-x^3+4*x)/(15*x^3-60 *x),x,method=_RETURNVERBOSE)
Output:
-exp(x)*ln(-(x^2-4)/x^2)-1/15*x
Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=-e^{x} \log \left (-\frac {x^{2} - 4}{x^{2}}\right ) - \frac {1}{15} \, x \] Input:
integrate(((-15*x^3+60*x)*exp(x)*log((-x^2+4)/x^2)-120*exp(x)-x^3+4*x)/(15 *x^3-60*x),x, algorithm="fricas")
Output:
-e^x*log(-(x^2 - 4)/x^2) - 1/15*x
Time = 0.16 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=- \frac {x}{15} - e^{x} \log {\left (\frac {4 - x^{2}}{x^{2}} \right )} \] Input:
integrate(((-15*x**3+60*x)*exp(x)*ln((-x**2+4)/x**2)-120*exp(x)-x**3+4*x)/ (15*x**3-60*x),x)
Output:
-x/15 - exp(x)*log((4 - x**2)/x**2)
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=-e^{x} \log \left (x + 2\right ) + 2 \, e^{x} \log \left (x\right ) - e^{x} \log \left (-x + 2\right ) - \frac {1}{15} \, x \] Input:
integrate(((-15*x^3+60*x)*exp(x)*log((-x^2+4)/x^2)-120*exp(x)-x^3+4*x)/(15 *x^3-60*x),x, algorithm="maxima")
Output:
-e^x*log(x + 2) + 2*e^x*log(x) - e^x*log(-x + 2) - 1/15*x
Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=-e^{x} \log \left (-\frac {x^{2} - 4}{x^{2}}\right ) - \frac {1}{15} \, x \] Input:
integrate(((-15*x^3+60*x)*exp(x)*log((-x^2+4)/x^2)-120*exp(x)-x^3+4*x)/(15 *x^3-60*x),x, algorithm="giac")
Output:
-e^x*log(-(x^2 - 4)/x^2) - 1/15*x
Time = 3.81 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=-\frac {x}{15}-{\mathrm {e}}^x\,\ln \left (-\frac {x^2-4}{x^2}\right ) \] Input:
int(-(4*x - 120*exp(x) - x^3 + exp(x)*log(-(x^2 - 4)/x^2)*(60*x - 15*x^3)) /(60*x - 15*x^3),x)
Output:
- x/15 - exp(x)*log(-(x^2 - 4)/x^2)
Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-120 e^x+4 x-x^3+e^x \left (60 x-15 x^3\right ) \log \left (\frac {4-x^2}{x^2}\right )}{-60 x+15 x^3} \, dx=-e^{x} \mathrm {log}\left (\frac {-x^{2}+4}{x^{2}}\right )-\frac {x}{15} \] Input:
int(((-15*x^3+60*x)*exp(x)*log((-x^2+4)/x^2)-120*exp(x)-x^3+4*x)/(15*x^3-6 0*x),x)
Output:
( - 15*e**x*log(( - x**2 + 4)/x**2) - x)/15