Integrand size = 75, antiderivative size = 29 \[ \int \frac {126-54 \log (x)+6 \log ^2(x)+e^{\frac {1}{6} \left (-30+35 x^2+5 x \log (x)\right )} \left (80+1120 x+(40-560 x) \log (x)+(-35+70 x) \log ^2(x)+5 \log ^3(x)\right )}{96-48 \log (x)+6 \log ^2(x)} \, dx=e^{-5+5 x \left (x+\frac {1}{6} (x+\log (x))\right )}+x+\frac {x}{4-\log (x)} \] Output:
exp(5*(7/6*x+1/6*ln(x))*x-5)+x/(-ln(x)+4)+x
Time = 3.78 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {126-54 \log (x)+6 \log ^2(x)+e^{\frac {1}{6} \left (-30+35 x^2+5 x \log (x)\right )} \left (80+1120 x+(40-560 x) \log (x)+(-35+70 x) \log ^2(x)+5 \log ^3(x)\right )}{96-48 \log (x)+6 \log ^2(x)} \, dx=\frac {1}{6} \left (6 x+6 e^{-5+\frac {35 x^2}{6}} x^{5 x/6}-\frac {6 x}{-4+\log (x)}\right ) \] Input:
Integrate[(126 - 54*Log[x] + 6*Log[x]^2 + E^((-30 + 35*x^2 + 5*x*Log[x])/6 )*(80 + 1120*x + (40 - 560*x)*Log[x] + (-35 + 70*x)*Log[x]^2 + 5*Log[x]^3) )/(96 - 48*Log[x] + 6*Log[x]^2),x]
Output:
(6*x + 6*E^(-5 + (35*x^2)/6)*x^((5*x)/6) - (6*x)/(-4 + Log[x]))/6
Time = 0.83 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {7292, 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.
\(\displaystyle \int \frac {e^{\frac {1}{6} \left (35 x^2+5 x \log (x)-30\right )} \left (1120 x+5 \log ^3(x)+(70 x-35) \log ^2(x)+(40-560 x) \log (x)+80\right )+6 \log ^2(x)-54 \log (x)+126}{6 \log ^2(x)-48 \log (x)+96} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {1}{6} \left (35 x^2+5 x \log (x)-30\right )} \left (1120 x+5 \log ^3(x)+(70 x-35) \log ^2(x)+(40-560 x) \log (x)+80\right )+6 \log ^2(x)-54 \log (x)+126}{6 (4-\log (x))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} \int \frac {5 e^{-\frac {5}{6} \left (6-7 x^2\right )} \left (\log ^3(x)-7 (1-2 x) \log ^2(x)+8 (1-14 x) \log (x)+224 x+16\right ) x^{5 x/6}+6 \log ^2(x)-54 \log (x)+126}{(4-\log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{6} \int \left (5 e^{\frac {35 x^2}{6}-5} (14 x+\log (x)+1) x^{5 x/6}+\frac {6 \left (\log ^2(x)-9 \log (x)+21\right )}{(\log (x)-4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{6} \left (6 e^{\frac {35 x^2}{6}-5} x^{5 x/6}+6 x+\frac {6 x}{4-\log (x)}\right )\) |
Input:
Int[(126 - 54*Log[x] + 6*Log[x]^2 + E^((-30 + 35*x^2 + 5*x*Log[x])/6)*(80 + 1120*x + (40 - 560*x)*Log[x] + (-35 + 70*x)*Log[x]^2 + 5*Log[x]^3))/(96 - 48*Log[x] + 6*Log[x]^2),x]
Output:
(6*x + 6*E^(-5 + (35*x^2)/6)*x^((5*x)/6) + (6*x)/(4 - Log[x]))/6
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.70 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90
method | result | size |
risch | \(x -\frac {x}{\ln \left (x \right )-4}+x^{\frac {5 x}{6}} {\mathrm e}^{-5+\frac {35 x^{2}}{6}}\) | \(26\) |
norman | \(\frac {x \ln \left (x \right )+\ln \left (x \right ) {\mathrm e}^{\frac {5 x \ln \left (x \right )}{6}+\frac {35 x^{2}}{6}-5}-5 x -4 \,{\mathrm e}^{\frac {5 x \ln \left (x \right )}{6}+\frac {35 x^{2}}{6}-5}}{\ln \left (x \right )-4}\) | \(47\) |
parallelrisch | \(\frac {6 x \ln \left (x \right )+6 \ln \left (x \right ) {\mathrm e}^{\frac {5 x \ln \left (x \right )}{6}+\frac {35 x^{2}}{6}-5}-30 x -24 \,{\mathrm e}^{\frac {5 x \ln \left (x \right )}{6}+\frac {35 x^{2}}{6}-5}}{6 \ln \left (x \right )-24}\) | \(50\) |
parts | \(\frac {x \left (\ln \left (x \right )-5\right )}{\ln \left (x \right )-4}+\frac {\ln \left (x \right ) {\mathrm e}^{\frac {5 x \ln \left (x \right )}{6}+\frac {35 x^{2}}{6}-5}-4 \,{\mathrm e}^{\frac {5 x \ln \left (x \right )}{6}+\frac {35 x^{2}}{6}-5}}{\ln \left (x \right )-4}\) | \(53\) |
default | \(\frac {x \left (\ln \left (x \right )-5\right )}{\ln \left (x \right )-4}+\frac {6 \ln \left (x \right ) {\mathrm e}^{\frac {5 x \ln \left (x \right )}{6}+\frac {35 x^{2}}{6}-5}-24 \,{\mathrm e}^{\frac {5 x \ln \left (x \right )}{6}+\frac {35 x^{2}}{6}-5}}{6 \ln \left (x \right )-24}\) | \(55\) |
Input:
int(((5*ln(x)^3+(70*x-35)*ln(x)^2+(-560*x+40)*ln(x)+1120*x+80)*exp(5/6*x*l n(x)+35/6*x^2-5)+6*ln(x)^2-54*ln(x)+126)/(6*ln(x)^2-48*ln(x)+96),x,method= _RETURNVERBOSE)
Output:
x-x/(ln(x)-4)+x^(5/6*x)*exp(-5+35/6*x^2)
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {126-54 \log (x)+6 \log ^2(x)+e^{\frac {1}{6} \left (-30+35 x^2+5 x \log (x)\right )} \left (80+1120 x+(40-560 x) \log (x)+(-35+70 x) \log ^2(x)+5 \log ^3(x)\right )}{96-48 \log (x)+6 \log ^2(x)} \, dx=\frac {{\left (\log \left (x\right ) - 4\right )} e^{\left (\frac {35}{6} \, x^{2} + \frac {5}{6} \, x \log \left (x\right ) - 5\right )} + x \log \left (x\right ) - 5 \, x}{\log \left (x\right ) - 4} \] Input:
integrate(((5*log(x)^3+(70*x-35)*log(x)^2+(-560*x+40)*log(x)+1120*x+80)*ex p(5/6*x*log(x)+35/6*x^2-5)+6*log(x)^2-54*log(x)+126)/(6*log(x)^2-48*log(x) +96),x, algorithm="fricas")
Output:
((log(x) - 4)*e^(35/6*x^2 + 5/6*x*log(x) - 5) + x*log(x) - 5*x)/(log(x) - 4)
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {126-54 \log (x)+6 \log ^2(x)+e^{\frac {1}{6} \left (-30+35 x^2+5 x \log (x)\right )} \left (80+1120 x+(40-560 x) \log (x)+(-35+70 x) \log ^2(x)+5 \log ^3(x)\right )}{96-48 \log (x)+6 \log ^2(x)} \, dx=x - \frac {x}{\log {\left (x \right )} - 4} + e^{\frac {35 x^{2}}{6} + \frac {5 x \log {\left (x \right )}}{6} - 5} \] Input:
integrate(((5*ln(x)**3+(70*x-35)*ln(x)**2+(-560*x+40)*ln(x)+1120*x+80)*exp (5/6*x*ln(x)+35/6*x**2-5)+6*ln(x)**2-54*ln(x)+126)/(6*ln(x)**2-48*ln(x)+96 ),x)
Output:
x - x/(log(x) - 4) + exp(35*x**2/6 + 5*x*log(x)/6 - 5)
\[ \int \frac {126-54 \log (x)+6 \log ^2(x)+e^{\frac {1}{6} \left (-30+35 x^2+5 x \log (x)\right )} \left (80+1120 x+(40-560 x) \log (x)+(-35+70 x) \log ^2(x)+5 \log ^3(x)\right )}{96-48 \log (x)+6 \log ^2(x)} \, dx=\int { \frac {5 \, {\left (7 \, {\left (2 \, x - 1\right )} \log \left (x\right )^{2} + \log \left (x\right )^{3} - 8 \, {\left (14 \, x - 1\right )} \log \left (x\right ) + 224 \, x + 16\right )} e^{\left (\frac {35}{6} \, x^{2} + \frac {5}{6} \, x \log \left (x\right ) - 5\right )} + 6 \, \log \left (x\right )^{2} - 54 \, \log \left (x\right ) + 126}{6 \, {\left (\log \left (x\right )^{2} - 8 \, \log \left (x\right ) + 16\right )}} \,d x } \] Input:
integrate(((5*log(x)^3+(70*x-35)*log(x)^2+(-560*x+40)*log(x)+1120*x+80)*ex p(5/6*x*log(x)+35/6*x^2-5)+6*log(x)^2-54*log(x)+126)/(6*log(x)^2-48*log(x) +96),x, algorithm="maxima")
Output:
-21*e^4*exp_integral_e(2, -log(x) + 4)/(log(x) - 4) + (x*e^5*log(x) + 16*x *e^5 + (log(x) - 4)*e^(35/6*x^2 + 5/6*x*log(x)))/(e^5*log(x) - 4*e^5) - 21 *integrate(1/(log(x) - 4), x)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (23) = 46\).
Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.86 \[ \int \frac {126-54 \log (x)+6 \log ^2(x)+e^{\frac {1}{6} \left (-30+35 x^2+5 x \log (x)\right )} \left (80+1120 x+(40-560 x) \log (x)+(-35+70 x) \log ^2(x)+5 \log ^3(x)\right )}{96-48 \log (x)+6 \log ^2(x)} \, dx=\frac {x e^{5} \log \left (x\right ) - 5 \, x e^{5} + e^{\left (\frac {35}{6} \, x^{2} + \frac {5}{6} \, x \log \left (x\right )\right )} \log \left (x\right ) - 4 \, e^{\left (\frac {35}{6} \, x^{2} + \frac {5}{6} \, x \log \left (x\right )\right )}}{e^{5} \log \left (x\right ) - 4 \, e^{5}} \] Input:
integrate(((5*log(x)^3+(70*x-35)*log(x)^2+(-560*x+40)*log(x)+1120*x+80)*ex p(5/6*x*log(x)+35/6*x^2-5)+6*log(x)^2-54*log(x)+126)/(6*log(x)^2-48*log(x) +96),x, algorithm="giac")
Output:
(x*e^5*log(x) - 5*x*e^5 + e^(35/6*x^2 + 5/6*x*log(x))*log(x) - 4*e^(35/6*x ^2 + 5/6*x*log(x)))/(e^5*log(x) - 4*e^5)
Time = 4.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {126-54 \log (x)+6 \log ^2(x)+e^{\frac {1}{6} \left (-30+35 x^2+5 x \log (x)\right )} \left (80+1120 x+(40-560 x) \log (x)+(-35+70 x) \log ^2(x)+5 \log ^3(x)\right )}{96-48 \log (x)+6 \log ^2(x)} \, dx=x^{\frac {5\,x}{6}}\,{\mathrm {e}}^{\frac {35\,x^2}{6}-5}-\frac {5\,x-x\,\ln \left (x\right )}{\ln \left (x\right )-4} \] Input:
int((6*log(x)^2 - 54*log(x) + exp((5*x*log(x))/6 + (35*x^2)/6 - 5)*(1120*x + 5*log(x)^3 - log(x)*(560*x - 40) + log(x)^2*(70*x - 35) + 80) + 126)/(6 *log(x)^2 - 48*log(x) + 96),x)
Output:
x^((5*x)/6)*exp((35*x^2)/6 - 5) - (5*x - x*log(x))/(log(x) - 4)
Time = 0.17 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {126-54 \log (x)+6 \log ^2(x)+e^{\frac {1}{6} \left (-30+35 x^2+5 x \log (x)\right )} \left (80+1120 x+(40-560 x) \log (x)+(-35+70 x) \log ^2(x)+5 \log ^3(x)\right )}{96-48 \log (x)+6 \log ^2(x)} \, dx=\frac {x^{\frac {5 x}{6}} e^{\frac {35 x^{2}}{6}} \mathrm {log}\left (x \right )-4 x^{\frac {5 x}{6}} e^{\frac {35 x^{2}}{6}}+\mathrm {log}\left (x \right ) e^{5} x -5 e^{5} x}{e^{5} \left (\mathrm {log}\left (x \right )-4\right )} \] Input:
int(((5*log(x)^3+(70*x-35)*log(x)^2+(-560*x+40)*log(x)+1120*x+80)*exp(5/6* x*log(x)+35/6*x^2-5)+6*log(x)^2-54*log(x)+126)/(6*log(x)^2-48*log(x)+96),x )
Output:
(x**((5*x)/6)*e**((35*x**2)/6)*log(x) - 4*x**((5*x)/6)*e**((35*x**2)/6) + log(x)*e**5*x - 5*e**5*x)/(e**5*(log(x) - 4))