Integrand size = 67, antiderivative size = 25 \[ \int \frac {-312+72 x+\left (108 x-24 x^2\right ) \log (5)+\left (-9 x^2+2 x^3\right ) \log ^2(5)+e^x \left (6+6 x-x^2 \log (5)\right )}{36-12 x \log (5)+x^2 \log ^2(5)} \, dx=x+(-10+x) x+\frac {2+e^x}{\frac {6}{x}-\log (5)} \]
Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {-312+72 x+\left (108 x-24 x^2\right ) \log (5)+\left (-9 x^2+2 x^3\right ) \log ^2(5)+e^x \left (6+6 x-x^2 \log (5)\right )}{36-12 x \log (5)+x^2 \log ^2(5)} \, dx=\frac {-12-e^x x \log (5)+x^3 \log ^2(5)+18 x \log (125)-3 x^2 \log (5) (2+\log (125))}{\log (5) (-6+x \log (5))} \]
Integrate[(-312 + 72*x + (108*x - 24*x^2)*Log[5] + (-9*x^2 + 2*x^3)*Log[5] ^2 + E^x*(6 + 6*x - x^2*Log[5]))/(36 - 12*x*Log[5] + x^2*Log[5]^2),x]
(-12 - E^x*x*Log[5] + x^3*Log[5]^2 + 18*x*Log[125] - 3*x^2*Log[5]*(2 + Log [125]))/(Log[5]*(-6 + x*Log[5]))
Leaf count is larger than twice the leaf count of optimal. \(156\) vs. \(2(25)=50\).
Time = 0.83 (sec) , antiderivative size = 156, normalized size of antiderivative = 6.24, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {7277, 27, 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.
\(\displaystyle \int \frac {e^x \left (x^2 (-\log (5))+6 x+6\right )+\left (108 x-24 x^2\right ) \log (5)+\left (2 x^3-9 x^2\right ) \log ^2(5)+72 x-312}{x^2 \log ^2(5)-12 x \log (5)+36} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \log ^2(5) \int -\frac {-72 x-e^x \left (-\log (5) x^2+6 x+6\right )+\left (9 x^2-2 x^3\right ) \log ^2(5)-12 \left (9 x-2 x^2\right ) \log (5)+312}{4 \log ^2(5) (6-x \log (5))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {-72 x-e^x \left (-\log (5) x^2+6 x+6\right )+\left (9 x^2-2 x^3\right ) \log ^2(5)-12 \left (9 x-2 x^2\right ) \log (5)+312}{(6-x \log (5))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (-\frac {(2 x-9) \log ^2(5) x^2}{(x \log (5)-6)^2}+\frac {12 (2 x-9) \log (5) x}{(x \log (5)-6)^2}-\frac {72 x}{(x \log (5)-6)^2}+\frac {e^x \left (\log (5) x^2-6 x-6\right )}{(x \log (5)-6)^2}+\frac {312}{(x \log (5)-6)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^2-\frac {36 (8-\log (125)) \log (6-x \log (5))}{\log ^2(5)}+\frac {108 (2-\log (5)) \log (6-x \log (5))}{\log ^2(5)}+\frac {72 \log (6-x \log (5))}{\log ^2(5)}+\frac {432}{\log ^2(5) (6-x \log (5))}-\frac {108 (4-\log (125))}{\log ^2(5) (6-x \log (5))}+\frac {3 x (8-\log (125))}{\log (5)}-\frac {24 x}{\log (5)}+\frac {6 e^x}{\log (5) (6-x \log (5))}-\frac {312}{\log (5) (6-x \log (5))}-\frac {e^x}{\log (5)}\) |
Int[(-312 + 72*x + (108*x - 24*x^2)*Log[5] + (-9*x^2 + 2*x^3)*Log[5]^2 + E ^x*(6 + 6*x - x^2*Log[5]))/(36 - 12*x*Log[5] + x^2*Log[5]^2),x]
x^2 - E^x/Log[5] - (24*x)/Log[5] + 432/(Log[5]^2*(6 - x*Log[5])) - 312/(Lo g[5]*(6 - x*Log[5])) + (6*E^x)/(Log[5]*(6 - x*Log[5])) - (108*(4 - Log[125 ]))/(Log[5]^2*(6 - x*Log[5])) + (3*x*(8 - Log[125]))/Log[5] + (72*Log[6 - x*Log[5]])/Log[5]^2 + (108*(2 - Log[5])*Log[6 - x*Log[5]])/Log[5]^2 - (36* (8 - Log[125])*Log[6 - x*Log[5]])/Log[5]^2
3.7.56.3.1 Defintions of rubi rules used
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_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40
method | result | size |
norman | \(\frac {x^{3} \ln \left (5\right )+52 x +\left (-9 \ln \left (5\right )-6\right ) x^{2}-{\mathrm e}^{x} x}{x \ln \left (5\right )-6}\) | \(35\) |
risch | \(x^{2}-9 x -\frac {12}{\ln \left (5\right ) \left (x \ln \left (5\right )-6\right )}-\frac {x \,{\mathrm e}^{x}}{x \ln \left (5\right )-6}\) | \(35\) |
parallelrisch | \(\frac {6 x^{3} \ln \left (5\right )-54 x^{2} \ln \left (5\right )-36 x^{2}-6 \,{\mathrm e}^{x} x +312 x}{6 x \ln \left (5\right )-36}\) | \(39\) |
parts | \(-9 x +x^{2}-\frac {12}{\ln \left (5\right ) \left (x \ln \left (5\right )-6\right )}-\frac {{\mathrm e}^{x}}{\ln \left (5\right )}-\frac {6 \,{\mathrm e}^{x}}{\ln \left (5\right )^{2} \left (x -\frac {6}{\ln \left (5\right )}\right )}\) | \(48\) |
default | \(-\frac {12}{\ln \left (5\right ) \left (x \ln \left (5\right )-6\right )}+\frac {-\frac {6 \,{\mathrm e}^{x}}{x -\frac {6}{\ln \left (5\right )}}-6 \,{\mathrm e}^{\frac {6}{\ln \left (5\right )}} \operatorname {Ei}_{1}\left (-x +\frac {6}{\ln \left (5\right )}\right )}{\ln \left (5\right )^{2}}-9 x +x^{2}+\frac {6 \,{\mathrm e}^{\frac {6}{\ln \left (5\right )}} \operatorname {Ei}_{1}\left (-x +\frac {6}{\ln \left (5\right )}\right )}{\ln \left (5\right )^{2}}+\frac {-\frac {36 \,{\mathrm e}^{x}}{x -\frac {6}{\ln \left (5\right )}}-36 \,{\mathrm e}^{\frac {6}{\ln \left (5\right )}} \operatorname {Ei}_{1}\left (-x +\frac {6}{\ln \left (5\right )}\right )}{\ln \left (5\right )^{3}}-\frac {{\mathrm e}^{x}}{\ln \left (5\right )}+\frac {36 \,{\mathrm e}^{x}}{\ln \left (5\right )^{3} \left (x -\frac {6}{\ln \left (5\right )}\right )}+\frac {36 \,{\mathrm e}^{\frac {6}{\ln \left (5\right )}} \operatorname {Ei}_{1}\left (-x +\frac {6}{\ln \left (5\right )}\right )}{\ln \left (5\right )^{3}}\) | \(182\) |
int(((-x^2*ln(5)+6*x+6)*exp(x)+(2*x^3-9*x^2)*ln(5)^2+(-24*x^2+108*x)*ln(5) +72*x-312)/(x^2*ln(5)^2-12*x*ln(5)+36),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (24) = 48\).
Time = 0.24 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int \frac {-312+72 x+\left (108 x-24 x^2\right ) \log (5)+\left (-9 x^2+2 x^3\right ) \log ^2(5)+e^x \left (6+6 x-x^2 \log (5)\right )}{36-12 x \log (5)+x^2 \log ^2(5)} \, dx=-\frac {x e^{x} \log \left (5\right ) - {\left (x^{3} - 9 \, x^{2}\right )} \log \left (5\right )^{2} + 6 \, {\left (x^{2} - 9 \, x\right )} \log \left (5\right ) + 12}{x \log \left (5\right )^{2} - 6 \, \log \left (5\right )} \]
integrate(((-x^2*log(5)+6*x+6)*exp(x)+(2*x^3-9*x^2)*log(5)^2+(-24*x^2+108* x)*log(5)+72*x-312)/(x^2*log(5)^2-12*x*log(5)+36),x, algorithm=\
-(x*e^x*log(5) - (x^3 - 9*x^2)*log(5)^2 + 6*(x^2 - 9*x)*log(5) + 12)/(x*lo g(5)^2 - 6*log(5))
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {-312+72 x+\left (108 x-24 x^2\right ) \log (5)+\left (-9 x^2+2 x^3\right ) \log ^2(5)+e^x \left (6+6 x-x^2 \log (5)\right )}{36-12 x \log (5)+x^2 \log ^2(5)} \, dx=x^{2} - 9 x - \frac {x e^{x}}{x \log {\left (5 \right )} - 6} - \frac {12}{x \log {\left (5 \right )}^{2} - 6 \log {\left (5 \right )}} \]
integrate(((-x**2*ln(5)+6*x+6)*exp(x)+(2*x**3-9*x**2)*ln(5)**2+(-24*x**2+1 08*x)*ln(5)+72*x-312)/(x**2*ln(5)**2-12*x*ln(5)+36),x)
\[ \int \frac {-312+72 x+\left (108 x-24 x^2\right ) \log (5)+\left (-9 x^2+2 x^3\right ) \log ^2(5)+e^x \left (6+6 x-x^2 \log (5)\right )}{36-12 x \log (5)+x^2 \log ^2(5)} \, dx=\int { \frac {{\left (2 \, x^{3} - 9 \, x^{2}\right )} \log \left (5\right )^{2} - {\left (x^{2} \log \left (5\right ) - 6 \, x - 6\right )} e^{x} - 12 \, {\left (2 \, x^{2} - 9 \, x\right )} \log \left (5\right ) + 72 \, x - 312}{x^{2} \log \left (5\right )^{2} - 12 \, x \log \left (5\right ) + 36} \,d x } \]
integrate(((-x^2*log(5)+6*x+6)*exp(x)+(2*x^3-9*x^2)*log(5)^2+(-24*x^2+108* x)*log(5)+72*x-312)/(x^2*log(5)^2-12*x*log(5)+36),x, algorithm=\
-(432/(x*log(5)^5 - 6*log(5)^4) - (x^2*log(5) + 24*x)/log(5)^3 - 216*log(x *log(5) - 6)/log(5)^4)*log(5)^2 + 9*(36/(x*log(5)^4 - 6*log(5)^3) - x/log( 5)^2 - 12*log(x*log(5) - 6)/log(5)^3)*log(5)^2 + 24*(36/(x*log(5)^4 - 6*lo g(5)^3) - x/log(5)^2 - 12*log(x*log(5) - 6)/log(5)^3)*log(5) - 108*(6/(x*l og(5)^3 - 6*log(5)^2) - log(x*log(5) - 6)/log(5)^2)*log(5) - x*e^x/(x*log( 5) - 6) - 6*e^(6/log(5))*exp_integral_e(2, -(x*log(5) - 6)/log(5))/((x*log (5) - 6)*log(5)) - 432/(x*log(5)^3 - 6*log(5)^2) + 312/(x*log(5)^2 - 6*log (5)) + 72*log(x*log(5) - 6)/log(5)^2 - 6*integrate(e^x/(x^2*log(5)^2 - 12* x*log(5) + 36), x)
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (24) = 48\).
Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.08 \[ \int \frac {-312+72 x+\left (108 x-24 x^2\right ) \log (5)+\left (-9 x^2+2 x^3\right ) \log ^2(5)+e^x \left (6+6 x-x^2 \log (5)\right )}{36-12 x \log (5)+x^2 \log ^2(5)} \, dx=\frac {x^{3} \log \left (5\right )^{2} - 9 \, x^{2} \log \left (5\right )^{2} - 6 \, x^{2} \log \left (5\right ) - x e^{x} \log \left (5\right ) + 54 \, x \log \left (5\right ) - 12}{x \log \left (5\right )^{2} - 6 \, \log \left (5\right )} \]
integrate(((-x^2*log(5)+6*x+6)*exp(x)+(2*x^3-9*x^2)*log(5)^2+(-24*x^2+108* x)*log(5)+72*x-312)/(x^2*log(5)^2-12*x*log(5)+36),x, algorithm=\
(x^3*log(5)^2 - 9*x^2*log(5)^2 - 6*x^2*log(5) - x*e^x*log(5) + 54*x*log(5) - 12)/(x*log(5)^2 - 6*log(5))
Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-312+72 x+\left (108 x-24 x^2\right ) \log (5)+\left (-9 x^2+2 x^3\right ) \log ^2(5)+e^x \left (6+6 x-x^2 \log (5)\right )}{36-12 x \log (5)+x^2 \log ^2(5)} \, dx=-\frac {x\,\left (6\,x+{\mathrm {e}}^x+9\,x\,\ln \left (5\right )-x^2\,\ln \left (5\right )-52\right )}{x\,\ln \left (5\right )-6} \]