Integrand size = 68, antiderivative size = 25 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=\frac {x \log \left (e^{6+x+x (3+x (5+x))}\right )}{-x+\log (5)} \] Output:
x/(ln(5)-x)*ln(exp(3)^2*exp(x+((5+x)*x+3)*x))
Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=-x \left (4+5 x+x^2\right )-\frac {\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x-\log (5)} \] Input:
Integrate[(-4*x^2 - 10*x^3 - 3*x^4 + (4*x + 10*x^2 + 3*x^3)*Log[5] + Log[5 ]*Log[E^(6 + 4*x + 5*x^2 + x^3)])/(x^2 - 2*x*Log[5] + Log[5]^2),x]
Output:
-(x*(4 + 5*x + x^2)) - (Log[5]*Log[E^(6 + 4*x + 5*x^2 + x^3)])/(x - Log[5] )
Leaf count is larger than twice the leaf count of optimal. \(234\) vs. \(2(25)=50\).
Time = 0.76 (sec) , antiderivative size = 234, normalized size of antiderivative = 9.36, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, 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.
\(\displaystyle \int \frac {-3 x^4-10 x^3-4 x^2+\log (5) \log \left (e^{x^3+5 x^2+4 x+6}\right )+\left (3 x^3+10 x^2+4 x\right ) \log (5)}{x^2-2 x \log (5)+\log ^2(5)} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int -\frac {3 x^4+10 x^3+4 x^2-\log (5) \log \left (e^{x^3+5 x^2+4 x+6}\right )-\left (3 x^3+10 x^2+4 x\right ) \log (5)}{4 (x-\log (5))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {3 x^4+10 x^3+4 x^2-\log (5) \log \left (e^{x^3+5 x^2+4 x+6}\right )-\left (3 x^3+10 x^2+4 x\right ) \log (5)}{(x-\log (5))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (\frac {3 x^4}{(x-\log (5))^2}+\frac {10 x^3}{(x-\log (5))^2}+\frac {4 x^2}{(x-\log (5))^2}-\frac {\left (3 x^2+10 x+4\right ) \log (5) x}{(x-\log (5))^2}-\frac {\log (5) \log \left (e^{x^3+5 x^2+4 x+6}\right )}{(x-\log (5))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x^3-5 x^2-\frac {3}{2} x^2 \log (25)+3 x^2 \log (5)-\frac {\log (5) \log \left (e^{x^3+5 x^2+4 x+6}\right )}{x-\log (5)}-4 x+\frac {3 \log ^4(5)}{x-\log (5)}-12 \log ^3(5) \log (x-\log (5))+\frac {10 \log ^3(5)}{x-\log (5)}-9 x \log ^2(5)+\log (5) \left (4+3 \log ^2(5)+10 \log (5)\right ) \log (x-\log (5))-30 \log ^2(5) \log (x-\log (5))-\frac {\log ^2(5) \left (4+3 \log ^2(5)+10 \log (5)\right )}{x-\log (5)}+\frac {4 \log ^2(5)}{x-\log (5)}+x \log (5) (10+\log (125))+2 x \log (5) (5+\log (125))-10 x \log (25)-4 \log (25) \log (x-\log (5))+\log (5) (2+\log (5)) (2+9 \log (5)) \log (x-\log (5))\) |
Input:
Int[(-4*x^2 - 10*x^3 - 3*x^4 + (4*x + 10*x^2 + 3*x^3)*Log[5] + Log[5]*Log[ E^(6 + 4*x + 5*x^2 + x^3)])/(x^2 - 2*x*Log[5] + Log[5]^2),x]
Output:
-4*x - 5*x^2 - x^3 + 3*x^2*Log[5] - 9*x*Log[5]^2 + (4*Log[5]^2)/(x - Log[5 ]) + (10*Log[5]^3)/(x - Log[5]) + (3*Log[5]^4)/(x - Log[5]) - (Log[5]^2*(4 + 10*Log[5] + 3*Log[5]^2))/(x - Log[5]) - 10*x*Log[25] - (3*x^2*Log[25])/ 2 + 2*x*Log[5]*(5 + Log[125]) + x*Log[5]*(10 + Log[125]) - (Log[5]*Log[E^( 6 + 4*x + 5*x^2 + x^3)])/(x - Log[5]) - 30*Log[5]^2*Log[x - Log[5]] - 12*L og[5]^3*Log[x - Log[5]] + Log[5]*(2 + Log[5])*(2 + 9*Log[5])*Log[x - Log[5 ]] + Log[5]*(4 + 10*Log[5] + 3*Log[5]^2)*Log[x - Log[5]] - 4*Log[25]*Log[x - Log[5]]
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 = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20
method | result | size |
norman | \(\frac {x \ln \left ({\mathrm e}^{6} {\mathrm e}^{x^{3}+5 x^{2}+4 x}\right )}{\ln \left (5\right )-x}\) | \(30\) |
parallelrisch | \(\frac {-x^{3} \ln \left (5\right )+x^{4}-5 x^{2} \ln \left (5\right )+5 x^{3}-4 \ln \left (5\right )^{2}+\ln \left (5\right ) \ln \left ({\mathrm e}^{6} {\mathrm e}^{x^{3}+5 x^{2}+4 x}\right )+4 x^{2}}{\ln \left (5\right )-x}\) | \(66\) |
risch | \(\frac {\ln \left (5\right ) \ln \left ({\mathrm e}^{x \left (4+x \right ) \left (1+x \right )}\right )}{\ln \left (5\right )-x}-\frac {-2 x^{3} \ln \left (5\right )+2 x^{4}-10 x^{2} \ln \left (5\right )+10 x^{3}-8 x \ln \left (5\right )+8 x^{2}+12 \ln \left (5\right )}{2 \left (-\ln \left (5\right )+x \right )}\) | \(72\) |
default | \(-3 x \ln \left (5\right )^{2}-\frac {3 x^{2} \ln \left (5\right )}{2}-x^{3}-10 x \ln \left (5\right )-5 x^{2}-4 x -\left (3 \ln \left (5\right )^{2}+10 \ln \left (5\right )+4\right ) \ln \left (5\right ) \ln \left (-\ln \left (5\right )+x \right )+\ln \left (5\right ) \left (-\frac {\ln \left ({\mathrm e}^{6} {\mathrm e}^{x^{3}+5 x^{2}+4 x}\right )}{-\ln \left (5\right )+x}+\frac {3 x^{2}}{2}+3 x \ln \left (5\right )+10 x +\left (3 \ln \left (5\right )^{2}+10 \ln \left (5\right )+4\right ) \ln \left (-\ln \left (5\right )+x \right )\right )\) | \(123\) |
parts | \(-3 x \ln \left (5\right )^{2}-\frac {3 x^{2} \ln \left (5\right )}{2}-x^{3}-10 x \ln \left (5\right )-5 x^{2}-4 x -\left (3 \ln \left (5\right )^{2}+10 \ln \left (5\right )+4\right ) \ln \left (5\right ) \ln \left (-\ln \left (5\right )+x \right )+\ln \left (5\right ) \left (-\frac {\ln \left ({\mathrm e}^{6} {\mathrm e}^{x^{3}+5 x^{2}+4 x}\right )}{-\ln \left (5\right )+x}+\frac {3 x^{2}}{2}+3 x \ln \left (5\right )+10 x +\left (3 \ln \left (5\right )^{2}+10 \ln \left (5\right )+4\right ) \ln \left (-\ln \left (5\right )+x \right )\right )\) | \(123\) |
Input:
int((ln(5)*ln(exp(3)^2*exp(x^3+5*x^2+4*x))+(3*x^3+10*x^2+4*x)*ln(5)-3*x^4- 10*x^3-4*x^2)/(ln(5)^2-2*x*ln(5)+x^2),x,method=_RETURNVERBOSE)
Output:
x*ln(exp(3)^2*exp(x^3+5*x^2+4*x))/(ln(5)-x)
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (25) = 50\).
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.28 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=-\frac {x^{4} - {\left (x - 5\right )} \log \left (5\right )^{3} + \log \left (5\right )^{4} + 5 \, x^{3} - {\left (5 \, x - 4\right )} \log \left (5\right )^{2} + 4 \, x^{2} - 2 \, {\left (2 \, x - 3\right )} \log \left (5\right )}{x - \log \left (5\right )} \] Input:
integrate((log(5)*log(exp(3)^2*exp(x^3+5*x^2+4*x))+(3*x^3+10*x^2+4*x)*log( 5)-3*x^4-10*x^3-4*x^2)/(log(5)^2-2*x*log(5)+x^2),x, algorithm="fricas")
Output:
-(x^4 - (x - 5)*log(5)^3 + log(5)^4 + 5*x^3 - (5*x - 4)*log(5)^2 + 4*x^2 - 2*(2*x - 3)*log(5))/(x - log(5))
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).
Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=- x^{3} - x^{2} \left (\log {\left (5 \right )} + 5\right ) - x \left (\log {\left (5 \right )}^{2} + 4 + 5 \log {\left (5 \right )}\right ) - \frac {\log {\left (5 \right )}^{4} + 6 \log {\left (5 \right )} + 4 \log {\left (5 \right )}^{2} + 5 \log {\left (5 \right )}^{3}}{x - \log {\left (5 \right )}} \] Input:
integrate((ln(5)*ln(exp(3)**2*exp(x**3+5*x**2+4*x))+(3*x**3+10*x**2+4*x)*l n(5)-3*x**4-10*x**3-4*x**2)/(ln(5)**2-2*x*ln(5)+x**2),x)
Output:
-x**3 - x**2*(log(5) + 5) - x*(log(5)**2 + 4 + 5*log(5)) - (log(5)**4 + 6* log(5) + 4*log(5)**2 + 5*log(5)**3)/(x - log(5))
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (25) = 50\).
Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.36 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=-x^{3} - x^{2} {\left (\log \left (5\right ) + 5\right )} - {\left (\log \left (5\right )^{2} + 5 \, \log \left (5\right ) + 4\right )} x - \frac {\log \left (5\right )^{4} + 5 \, \log \left (5\right )^{3} + 4 \, \log \left (5\right )^{2} + 6 \, \log \left (5\right )}{x - \log \left (5\right )} \] Input:
integrate((log(5)*log(exp(3)^2*exp(x^3+5*x^2+4*x))+(3*x^3+10*x^2+4*x)*log( 5)-3*x^4-10*x^3-4*x^2)/(log(5)^2-2*x*log(5)+x^2),x, algorithm="maxima")
Output:
-x^3 - x^2*(log(5) + 5) - (log(5)^2 + 5*log(5) + 4)*x - (log(5)^4 + 5*log( 5)^3 + 4*log(5)^2 + 6*log(5))/(x - log(5))
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (25) = 50\).
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.56 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=-x^{3} - x^{2} \log \left (5\right ) - x \log \left (5\right )^{2} - 5 \, x^{2} - 5 \, x \log \left (5\right ) - 4 \, x - \frac {\log \left (5\right )^{4} + 5 \, \log \left (5\right )^{3} + 4 \, \log \left (5\right )^{2} + 6 \, \log \left (5\right )}{x - \log \left (5\right )} \] Input:
integrate((log(5)*log(exp(3)^2*exp(x^3+5*x^2+4*x))+(3*x^3+10*x^2+4*x)*log( 5)-3*x^4-10*x^3-4*x^2)/(log(5)^2-2*x*log(5)+x^2),x, algorithm="giac")
Output:
-x^3 - x^2*log(5) - x*log(5)^2 - 5*x^2 - 5*x*log(5) - 4*x - (log(5)^4 + 5* log(5)^3 + 4*log(5)^2 + 6*log(5))/(x - log(5))
Time = 0.17 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.16 \[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=x\,\left (15\,\ln \left (5\right )-2\,\ln \left (5\right )\,\left (6\,\ln \left (5\right )-\ln \left (625\right )+10\right )+3\,{\ln \left (5\right )}^2-4\right )-\frac {6\,\ln \left (5\right )+\ln \left (5\right )\,\ln \left (625\right )+5\,{\ln \left (5\right )}^3+{\ln \left (5\right )}^4}{x-\ln \left (5\right )}-x^3-x^2\,\left (3\,\ln \left (5\right )-\frac {\ln \left (625\right )}{2}+5\right ) \] Input:
int(-(4*x^2 - log(5)*log(exp(6)*exp(4*x + 5*x^2 + x^3)) - log(5)*(4*x + 10 *x^2 + 3*x^3) + 10*x^3 + 3*x^4)/(log(5)^2 - 2*x*log(5) + x^2),x)
Output:
x*(15*log(5) - 2*log(5)*(6*log(5) - log(625) + 10) + 3*log(5)^2 - 4) - (6* log(5) + log(5)*log(625) + 5*log(5)^3 + log(5)^4)/(x - log(5)) - x^3 - x^2 *(3*log(5) - log(625)/2 + 5)
\[ \int \frac {-4 x^2-10 x^3-3 x^4+\left (4 x+10 x^2+3 x^3\right ) \log (5)+\log (5) \log \left (e^{6+4 x+5 x^2+x^3}\right )}{x^2-2 x \log (5)+\log ^2(5)} \, dx=\left (\int \frac {\mathrm {log}\left (e^{x^{3}+5 x^{2}+4 x} e^{6}\right )}{\mathrm {log}\left (5\right )^{2}-2 \,\mathrm {log}\left (5\right ) x +x^{2}}d x \right ) \mathrm {log}\left (5\right )-3 \,\mathrm {log}\left (\mathrm {log}\left (5\right )-x \right ) \mathrm {log}\left (5\right )^{3}-10 \,\mathrm {log}\left (\mathrm {log}\left (5\right )-x \right ) \mathrm {log}\left (5\right )^{2}-4 \,\mathrm {log}\left (\mathrm {log}\left (5\right )-x \right ) \mathrm {log}\left (5\right )-3 x \mathrm {log}\left (5\right )^{2}-\frac {3 \,\mathrm {log}\left (5\right ) x^{2}}{2}-10 \,\mathrm {log}\left (5\right ) x -x^{3}-5 x^{2}-4 x \] Input:
int((log(5)*log(exp(3)^2*exp(x^3+5*x^2+4*x))+(3*x^3+10*x^2+4*x)*log(5)-3*x ^4-10*x^3-4*x^2)/(log(5)^2-2*x*log(5)+x^2),x)
Output:
(2*int(log(e**(x**3 + 5*x**2 + 4*x)*e**6)/(log(5)**2 - 2*log(5)*x + x**2), x)*log(5) - 6*log(log(5) - x)*log(5)**3 - 20*log(log(5) - x)*log(5)**2 - 8 *log(log(5) - x)*log(5) - 6*log(5)**2*x - 3*log(5)*x**2 - 20*log(5)*x - 2* x**3 - 10*x**2 - 8*x)/2