Integrand size = 47, antiderivative size = 24 \[ \int \frac {2 e^5 \left (120-60 x-4 x^3\right )}{\left (-3 x+x^2\right ) \left (-3 x^2+x^3+\left (-3 x^2+x^3\right ) \log (4)\right )} \, dx=\frac {8 e^5 \left (5+x^2\right )}{(-3+x) x^2 (1+\log (4))} \] Output:
4*(x^2+5)*exp(ln(2/x^2/(1+2*ln(2))/(-3+x))+5)
Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {2 e^5 \left (120-60 x-4 x^3\right )}{\left (-3 x+x^2\right ) \left (-3 x^2+x^3+\left (-3 x^2+x^3\right ) \log (4)\right )} \, dx=\frac {8 e^5 \left (5+x^2\right )}{(-3+x) x^2 (1+\log (4))} \] Input:
Integrate[(2*E^5*(120 - 60*x - 4*x^3))/((-3*x + x^2)*(-3*x^2 + x^3 + (-3*x ^2 + x^3)*Log[4])),x]
Output:
(8*E^5*(5 + x^2))/((-3 + x)*x^2*(1 + Log[4]))
Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {27, 27, 2026, 9, 35, 2123, 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.
\(\displaystyle \int \frac {2 e^5 \left (-4 x^3-60 x+120\right )}{\left (x^2-3 x\right ) \left (x^3-3 x^2+\left (x^3-3 x^2\right ) \log (4)\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 e^5 \int \frac {4 \left (-x^3-15 x+30\right )}{\left (3 x-x^2\right ) \left (-x^3+3 x^2+\left (3 x^2-x^3\right ) \log (4)\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 8 e^5 \int \frac {-x^3-15 x+30}{\left (3 x-x^2\right ) \left (-x^3+3 x^2+\left (3 x^2-x^3\right ) \log (4)\right )}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle 8 e^5 \int \frac {-x^3-15 x+30}{(3-x) x \left (-x^3+3 x^2+\left (3 x^2-x^3\right ) \log (4)\right )}dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle 8 e^5 \int \frac {-x^3-15 x+30}{(3-x) x^3 (-((1+\log (4)) x)+\log (64)+3)}dx\) |
\(\Big \downarrow \) 35 |
\(\displaystyle \frac {8 e^5 \int \frac {-x^3-15 x+30}{(3-x)^2 x^3}dx}{1+\log (4)}\) |
\(\Big \downarrow \) 2123 |
\(\displaystyle \frac {8 e^5 \int \left (\frac {5}{9 x^2}+\frac {10}{3 x^3}-\frac {14}{9 (x-3)^2}\right )dx}{1+\log (4)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {8 e^5 \left (-\frac {5}{3 x^2}-\frac {5}{9 x}-\frac {14}{9 (3-x)}\right )}{1+\log (4)}\) |
Input:
Int[(2*E^5*(120 - 60*x - 4*x^3))/((-3*x + x^2)*(-3*x^2 + x^3 + (-3*x^2 + x ^3)*Log[4])),x]
Output:
(8*E^5*(-14/(9*(3 - x)) - 5/(3*x^2) - 5/(9*x)))/(1 + Log[4])
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} , x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && !(IntegerQ[n] && SimplerQ[a + b*x, c + d*x])
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[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17
method | result | size |
risch | \(\frac {4 \,{\mathrm e}^{5} \left (2 x^{2}+10\right )}{x^{2} \left (-3+x \right ) \left (1+2 \ln \left (2\right )\right )}\) | \(28\) |
gosper | \(4 \left (x^{2}+5\right ) {\mathrm e}^{\ln \left (\frac {2}{x^{2} \left (2 x \ln \left (2\right )-6 \ln \left (2\right )+x -3\right )}\right )+5}\) | \(31\) |
norman | \(\frac {\frac {8 \,{\mathrm e}^{5} x^{2}}{1+2 \ln \left (2\right )}+\frac {40 \,{\mathrm e}^{5}}{1+2 \ln \left (2\right )}}{x^{2} \left (-3+x \right )}\) | \(38\) |
parallelrisch | \(4 x^{2} {\mathrm e}^{\ln \left (\frac {2}{x^{2} \left (2 x \ln \left (2\right )-6 \ln \left (2\right )+x -3\right )}\right )+5}+20 \,{\mathrm e}^{\ln \left (\frac {2}{x^{2} \left (2 x \ln \left (2\right )-6 \ln \left (2\right )+x -3\right )}\right )+5}\) | \(55\) |
orering | \(-\frac {\left (x^{2}+5\right ) x \left (-3+x \right ) \left (-4 x^{3}-60 x +120\right ) {\mathrm e}^{\ln \left (\frac {2}{2 \left (x^{3}-3 x^{2}\right ) \ln \left (2\right )+x^{3}-3 x^{2}}\right )+5}}{\left (x^{3}+15 x -30\right ) \left (x^{2}-3 x \right )}\) | \(71\) |
default | \(\frac {4 \,{\mathrm e}^{5+\ln \left (\frac {2}{2 \left (x^{3}-3 x^{2}\right ) \ln \left (2\right )+x^{3}-3 x^{2}}\right )+\ln \left (2 \left (x^{3}-3 x^{2}\right ) \ln \left (2\right )+x^{3}-3 x^{2}\right )} \left (\frac {14}{9 \left (-3+x \right )}-\frac {5}{3 x^{2}}-\frac {5}{9 x}\right )}{1+2 \ln \left (2\right )}\) | \(82\) |
Input:
int((-4*x^3-60*x+120)*exp(ln(2/(2*(x^3-3*x^2)*ln(2)+x^3-3*x^2))+5)/(x^2-3* x),x,method=_RETURNVERBOSE)
Output:
4*exp(5)*(2*x^2+10)/x^2/(-3+x)/(1+2*ln(2))
Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {2 e^5 \left (120-60 x-4 x^3\right )}{\left (-3 x+x^2\right ) \left (-3 x^2+x^3+\left (-3 x^2+x^3\right ) \log (4)\right )} \, dx=\frac {8 \, {\left (x^{2} + 5\right )} e^{5}}{x^{3} - 3 \, x^{2} + 2 \, {\left (x^{3} - 3 \, x^{2}\right )} \log \left (2\right )} \] Input:
integrate((-4*x^3-60*x+120)*exp(log(2/(2*(x^3-3*x^2)*log(2)+x^3-3*x^2))+5) /(x^2-3*x),x, algorithm="fricas")
Output:
8*(x^2 + 5)*e^5/(x^3 - 3*x^2 + 2*(x^3 - 3*x^2)*log(2))
Time = 0.35 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {2 e^5 \left (120-60 x-4 x^3\right )}{\left (-3 x+x^2\right ) \left (-3 x^2+x^3+\left (-3 x^2+x^3\right ) \log (4)\right )} \, dx=- \frac {- 8 x^{2} e^{5} - 40 e^{5}}{x^{3} \cdot \left (1 + 2 \log {\left (2 \right )}\right ) + x^{2} \left (- 6 \log {\left (2 \right )} - 3\right )} \] Input:
integrate((-4*x**3-60*x+120)*exp(ln(2/(2*(x**3-3*x**2)*ln(2)+x**3-3*x**2)) +5)/(x**2-3*x),x)
Output:
-(-8*x**2*exp(5) - 40*exp(5))/(x**3*(1 + 2*log(2)) + x**2*(-6*log(2) - 3))
Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {2 e^5 \left (120-60 x-4 x^3\right )}{\left (-3 x+x^2\right ) \left (-3 x^2+x^3+\left (-3 x^2+x^3\right ) \log (4)\right )} \, dx=\frac {8 \, {\left (x^{2} + 5\right )} e^{5}}{x^{3} {\left (2 \, \log \left (2\right ) + 1\right )} - 3 \, x^{2} {\left (2 \, \log \left (2\right ) + 1\right )}} \] Input:
integrate((-4*x^3-60*x+120)*exp(log(2/(2*(x^3-3*x^2)*log(2)+x^3-3*x^2))+5) /(x^2-3*x),x, algorithm="maxima")
Output:
8*(x^2 + 5)*e^5/(x^3*(2*log(2) + 1) - 3*x^2*(2*log(2) + 1))
Time = 0.11 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {2 e^5 \left (120-60 x-4 x^3\right )}{\left (-3 x+x^2\right ) \left (-3 x^2+x^3+\left (-3 x^2+x^3\right ) \log (4)\right )} \, dx=\frac {112 \, e^{5}}{9 \, {\left (x - 3\right )} {\left (2 \, \log \left (2\right ) + 1\right )}} - \frac {40 \, {\left (x e^{5} + 3 \, e^{5}\right )}}{9 \, x^{2} {\left (2 \, \log \left (2\right ) + 1\right )}} \] Input:
integrate((-4*x^3-60*x+120)*exp(log(2/(2*(x^3-3*x^2)*log(2)+x^3-3*x^2))+5) /(x^2-3*x),x, algorithm="giac")
Output:
112/9*e^5/((x - 3)*(2*log(2) + 1)) - 40/9*(x*e^5 + 3*e^5)/(x^2*(2*log(2) + 1))
Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {2 e^5 \left (120-60 x-4 x^3\right )}{\left (-3 x+x^2\right ) \left (-3 x^2+x^3+\left (-3 x^2+x^3\right ) \log (4)\right )} \, dx=-\frac {\frac {8\,{\mathrm {e}}^5\,x^3}{3}+40\,{\mathrm {e}}^5}{6\,x^2\,\ln \left (2\right )-2\,x^3\,\ln \left (2\right )+3\,x^2-x^3} \] Input:
int((exp(log(-2/(2*log(2)*(3*x^2 - x^3) + 3*x^2 - x^3)) + 5)*(60*x + 4*x^3 - 120))/(3*x - x^2),x)
Output:
-(40*exp(5) + (8*x^3*exp(5))/3)/(6*x^2*log(2) - 2*x^3*log(2) + 3*x^2 - x^3 )
Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {2 e^5 \left (120-60 x-4 x^3\right )}{\left (-3 x+x^2\right ) \left (-3 x^2+x^3+\left (-3 x^2+x^3\right ) \log (4)\right )} \, dx=\frac {8 e^{5} \left (x^{3}+15\right )}{3 x^{2} \left (2 \,\mathrm {log}\left (2\right ) x -6 \,\mathrm {log}\left (2\right )+x -3\right )} \] Input:
int((-4*x^3-60*x+120)*exp(log(2/(2*(x^3-3*x^2)*log(2)+x^3-3*x^2))+5)/(x^2- 3*x),x)
Output:
(8*e**5*(x**3 + 15))/(3*x**2*(2*log(2)*x - 6*log(2) + x - 3))