Integrand size = 109, antiderivative size = 21 \[ \int \frac {4 x+e^{2 x} x+x^2+x \log ^2\left (4 e^3 x\right )+\left (5 x+x^2+e^x (10+2 x)+e^{2 x} \left (10 x+2 x^2\right )\right ) \log (5+x)+\log \left (4 e^3 x\right ) \left (2 e^x x+\left (10+2 x+e^x \left (10 x+2 x^2\right )\right ) \log (5+x)\right )}{5 x+x^2} \, dx=\left (4+x+\left (e^x+\log \left (4 e^3 x\right )\right )^2\right ) \log (5+x) \] Output:
ln(5+x)*(x+4+(ln(4*x*exp(3))+exp(x))^2)
Leaf count is larger than twice the leaf count of optimal. \(45\) vs. \(2(21)=42\).
Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.14 \[ \int \frac {4 x+e^{2 x} x+x^2+x \log ^2\left (4 e^3 x\right )+\left (5 x+x^2+e^x (10+2 x)+e^{2 x} \left (10 x+2 x^2\right )\right ) \log (5+x)+\log \left (4 e^3 x\right ) \left (2 e^x x+\left (10+2 x+e^x \left (10 x+2 x^2\right )\right ) \log (5+x)\right )}{5 x+x^2} \, dx=\left (13+e^{2 x}+x+6 \log (4)+\log ^2(4)+e^x (6+\log (16))+\left (6+2 e^x+\log (16)\right ) \log (x)+\log ^2(x)\right ) \log (5+x) \] Input:
Integrate[(4*x + E^(2*x)*x + x^2 + x*Log[4*E^3*x]^2 + (5*x + x^2 + E^x*(10 + 2*x) + E^(2*x)*(10*x + 2*x^2))*Log[5 + x] + Log[4*E^3*x]*(2*E^x*x + (10 + 2*x + E^x*(10*x + 2*x^2))*Log[5 + x]))/(5*x + x^2),x]
Output:
(13 + E^(2*x) + x + 6*Log[4] + Log[4]^2 + E^x*(6 + Log[16]) + (6 + 2*E^x + Log[16])*Log[x] + Log[x]^2)*Log[5 + x]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 3.82 (sec) , antiderivative size = 122, normalized size of antiderivative = 5.81, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {2026, 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.
\(\displaystyle \int \frac {x^2+\left (x^2+e^{2 x} \left (2 x^2+10 x\right )+5 x+e^x (2 x+10)\right ) \log (x+5)+\log \left (4 e^3 x\right ) \left (\left (e^x \left (2 x^2+10 x\right )+2 x+10\right ) \log (x+5)+2 e^x x\right )+e^{2 x} x+4 x+x \log ^2\left (4 e^3 x\right )}{x^2+5 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {x^2+\left (x^2+e^{2 x} \left (2 x^2+10 x\right )+5 x+e^x (2 x+10)\right ) \log (x+5)+\log \left (4 e^3 x\right ) \left (\left (e^x \left (2 x^2+10 x\right )+2 x+10\right ) \log (x+5)+2 e^x x\right )+e^{2 x} x+4 x+x \log ^2\left (4 e^3 x\right )}{x (x+5)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 e^x \left (x^2 \log (x) \log (x+5)+3 x^2 \left (1+\frac {2 \log (2)}{3}\right ) \log (x+5)+x \log (x)+5 x \log (x) \log (x+5)+16 x \left (1+\frac {5 \log (2)}{8}\right ) \log (x+5)+3 x \left (1+\frac {2 \log (2)}{3}\right )+5 \log (x+5)\right )}{x (x+5)}+\frac {x}{x+5}+\frac {4}{x+5}+\frac {\left (\log (x)+3 \left (1+\frac {2 \log (2)}{3}\right )\right )^2}{x+5}+\frac {10 \log (x+5) \left (\log (x)+3 \left (1+\frac {2 \log (2)}{3}\right )\right )}{x (x+5)}+\frac {2 \log (x+5) \left (\log (x)+3 \left (1+\frac {2 \log (2)}{3}\right )\right )}{x+5}+\frac {x \log (x+5)}{x+5}+\frac {5 \log (x+5)}{x+5}+\frac {e^{2 x} (2 x \log (x+5)+10 \log (x+5)+1)}{x+5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {6 \operatorname {ExpIntegralEi}(x+5)}{e^5}-\frac {2 \operatorname {ExpIntegralEi}(x+5) \log (x)}{e^5}+\frac {2 \operatorname {ExpIntegralEi}(x+5) \log (4 x)}{e^5}-\frac {2 (3+\log (4)) \operatorname {ExpIntegralEi}(x+5)}{e^5}+\log (x+5) (\log (x)+3+\log (4))^2+(x+5) \log (x+5)+2 e^x \log (x) \log (x+5)+2 e^x (3+\log (4)) \log (x+5)-\log (x+5)+\frac {e^{2 x} (x \log (x+5)+5 \log (x+5))}{x+5}\) |
Input:
Int[(4*x + E^(2*x)*x + x^2 + x*Log[4*E^3*x]^2 + (5*x + x^2 + E^x*(10 + 2*x ) + E^(2*x)*(10*x + 2*x^2))*Log[5 + x] + Log[4*E^3*x]*(2*E^x*x + (10 + 2*x + E^x*(10*x + 2*x^2))*Log[5 + x]))/(5*x + x^2),x]
Output:
(6*ExpIntegralEi[5 + x])/E^5 - (2*ExpIntegralEi[5 + x]*(3 + Log[4]))/E^5 - (2*ExpIntegralEi[5 + x]*Log[x])/E^5 + (2*ExpIntegralEi[5 + x]*Log[4*x])/E ^5 - Log[5 + x] + (5 + x)*Log[5 + x] + 2*E^x*(3 + Log[4])*Log[5 + x] + 2*E ^x*Log[x]*Log[5 + x] + (3 + Log[4] + Log[x])^2*Log[5 + x] + (E^(2*x)*(5*Lo g[5 + x] + x*Log[5 + x]))/(5 + 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])
Leaf count of result is larger than twice the leaf count of optimal. \(49\) vs. \(2(19)=38\).
Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.38
\[\ln \left (5+x \right ) \ln \left (4 x \,{\mathrm e}^{3}\right )^{2}+2 \,{\mathrm e}^{x} \ln \left (5+x \right ) \ln \left (4 x \,{\mathrm e}^{3}\right )+\ln \left (5+x \right ) {\mathrm e}^{2 x}+x \ln \left (5+x \right )+4 \ln \left (5+x \right )\]
Input:
int((x*ln(4*x*exp(3))^2+(((2*x^2+10*x)*exp(x)+2*x+10)*ln(5+x)+2*exp(x)*x)* ln(4*x*exp(3))+((2*x^2+10*x)*exp(x)^2+(2*x+10)*exp(x)+x^2+5*x)*ln(5+x)+x*e xp(x)^2+x^2+4*x)/(x^2+5*x),x)
Output:
ln(5+x)*ln(4*x*exp(3))^2+2*exp(x)*ln(5+x)*ln(4*x*exp(3))+exp(x)^2*ln(5+x)+ x*ln(5+x)+4*ln(5+x)
Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (19) = 38\).
Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.90 \[ \int \frac {4 x+e^{2 x} x+x^2+x \log ^2\left (4 e^3 x\right )+\left (5 x+x^2+e^x (10+2 x)+e^{2 x} \left (10 x+2 x^2\right )\right ) \log (5+x)+\log \left (4 e^3 x\right ) \left (2 e^x x+\left (10+2 x+e^x \left (10 x+2 x^2\right )\right ) \log (5+x)\right )}{5 x+x^2} \, dx=2 \, e^{x} \log \left (4 \, x e^{3}\right ) \log \left (x + 5\right ) + \log \left (4 \, x e^{3}\right )^{2} \log \left (x + 5\right ) + {\left (x + e^{\left (2 \, x\right )} + 4\right )} \log \left (x + 5\right ) \] Input:
integrate((x*log(4*x*exp(3))^2+(((2*x^2+10*x)*exp(x)+2*x+10)*log(5+x)+2*ex p(x)*x)*log(4*x*exp(3))+((2*x^2+10*x)*exp(x)^2+(2*x+10)*exp(x)+x^2+5*x)*lo g(5+x)+x*exp(x)^2+x^2+4*x)/(x^2+5*x),x, algorithm="fricas")
Output:
2*e^x*log(4*x*e^3)*log(x + 5) + log(4*x*e^3)^2*log(x + 5) + (x + e^(2*x) + 4)*log(x + 5)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (20) = 40\).
Time = 4.77 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.67 \[ \int \frac {4 x+e^{2 x} x+x^2+x \log ^2\left (4 e^3 x\right )+\left (5 x+x^2+e^x (10+2 x)+e^{2 x} \left (10 x+2 x^2\right )\right ) \log (5+x)+\log \left (4 e^3 x\right ) \left (2 e^x x+\left (10+2 x+e^x \left (10 x+2 x^2\right )\right ) \log (5+x)\right )}{5 x+x^2} \, dx=x \log {\left (x + 5 \right )} + e^{2 x} \log {\left (x + 5 \right )} + 2 e^{x} \log {\left (4 x e^{3} \right )} \log {\left (x + 5 \right )} + \log {\left (4 x e^{3} \right )}^{2} \log {\left (x + 5 \right )} + 4 \log {\left (x + 5 \right )} \] Input:
integrate((x*ln(4*x*exp(3))**2+(((2*x**2+10*x)*exp(x)+2*x+10)*ln(5+x)+2*ex p(x)*x)*ln(4*x*exp(3))+((2*x**2+10*x)*exp(x)**2+(2*x+10)*exp(x)+x**2+5*x)* ln(5+x)+x*exp(x)**2+x**2+4*x)/(x**2+5*x),x)
Output:
x*log(x + 5) + exp(2*x)*log(x + 5) + 2*exp(x)*log(4*x*exp(3))*log(x + 5) + log(4*x*exp(3))**2*log(x + 5) + 4*log(x + 5)
\[ \int \frac {4 x+e^{2 x} x+x^2+x \log ^2\left (4 e^3 x\right )+\left (5 x+x^2+e^x (10+2 x)+e^{2 x} \left (10 x+2 x^2\right )\right ) \log (5+x)+\log \left (4 e^3 x\right ) \left (2 e^x x+\left (10+2 x+e^x \left (10 x+2 x^2\right )\right ) \log (5+x)\right )}{5 x+x^2} \, dx=\int { \frac {x \log \left (4 \, x e^{3}\right )^{2} + x^{2} + x e^{\left (2 \, x\right )} + 2 \, {\left (x e^{x} + {\left ({\left (x^{2} + 5 \, x\right )} e^{x} + x + 5\right )} \log \left (x + 5\right )\right )} \log \left (4 \, x e^{3}\right ) + {\left (x^{2} + 2 \, {\left (x^{2} + 5 \, x\right )} e^{\left (2 \, x\right )} + 2 \, {\left (x + 5\right )} e^{x} + 5 \, x\right )} \log \left (x + 5\right ) + 4 \, x}{x^{2} + 5 \, x} \,d x } \] Input:
integrate((x*log(4*x*exp(3))^2+(((2*x^2+10*x)*exp(x)+2*x+10)*log(5+x)+2*ex p(x)*x)*log(4*x*exp(3))+((2*x^2+10*x)*exp(x)^2+(2*x+10)*exp(x)+x^2+5*x)*lo g(5+x)+x*exp(x)^2+x^2+4*x)/(x^2+5*x),x, algorithm="maxima")
Output:
-e^(-10)*exp_integral_e(1, -2*x - 10) + (2*(2*log(2) + log(x) + 3)*e^x + 4 *log(2)^2 + 2*(2*log(2) + 3)*log(x) + log(x)^2 + x + e^(2*x) + 12*log(2) + 14)*log(x + 5) - integrate(e^(2*x)/(x + 5), x) - log(x + 5)
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (19) = 38\).
Time = 0.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 4.52 \[ \int \frac {4 x+e^{2 x} x+x^2+x \log ^2\left (4 e^3 x\right )+\left (5 x+x^2+e^x (10+2 x)+e^{2 x} \left (10 x+2 x^2\right )\right ) \log (5+x)+\log \left (4 e^3 x\right ) \left (2 e^x x+\left (10+2 x+e^x \left (10 x+2 x^2\right )\right ) \log (5+x)\right )}{5 x+x^2} \, dx=4 \, e^{x} \log \left (2\right ) \log \left (x + 5\right ) + 4 \, \log \left (2\right )^{2} \log \left (x + 5\right ) + 2 \, e^{x} \log \left (x + 5\right ) \log \left (x\right ) + 4 \, \log \left (2\right ) \log \left (x + 5\right ) \log \left (x\right ) + \log \left (x + 5\right ) \log \left (x\right )^{2} + x \log \left (x + 5\right ) + e^{\left (2 \, x\right )} \log \left (x + 5\right ) + 6 \, e^{x} \log \left (x + 5\right ) + 12 \, \log \left (2\right ) \log \left (x + 5\right ) + 6 \, \log \left (x + 5\right ) \log \left (x\right ) + 13 \, \log \left (x + 5\right ) \] Input:
integrate((x*log(4*x*exp(3))^2+(((2*x^2+10*x)*exp(x)+2*x+10)*log(5+x)+2*ex p(x)*x)*log(4*x*exp(3))+((2*x^2+10*x)*exp(x)^2+(2*x+10)*exp(x)+x^2+5*x)*lo g(5+x)+x*exp(x)^2+x^2+4*x)/(x^2+5*x),x, algorithm="giac")
Output:
4*e^x*log(2)*log(x + 5) + 4*log(2)^2*log(x + 5) + 2*e^x*log(x + 5)*log(x) + 4*log(2)*log(x + 5)*log(x) + log(x + 5)*log(x)^2 + x*log(x + 5) + e^(2*x )*log(x + 5) + 6*e^x*log(x + 5) + 12*log(2)*log(x + 5) + 6*log(x + 5)*log( x) + 13*log(x + 5)
Timed out. \[ \int \frac {4 x+e^{2 x} x+x^2+x \log ^2\left (4 e^3 x\right )+\left (5 x+x^2+e^x (10+2 x)+e^{2 x} \left (10 x+2 x^2\right )\right ) \log (5+x)+\log \left (4 e^3 x\right ) \left (2 e^x x+\left (10+2 x+e^x \left (10 x+2 x^2\right )\right ) \log (5+x)\right )}{5 x+x^2} \, dx=\int \frac {4\,x+x\,{\mathrm {e}}^{2\,x}+\ln \left (x+5\right )\,\left (5\,x+{\mathrm {e}}^{2\,x}\,\left (2\,x^2+10\,x\right )+{\mathrm {e}}^x\,\left (2\,x+10\right )+x^2\right )+\ln \left (4\,x\,{\mathrm {e}}^3\right )\,\left (\ln \left (x+5\right )\,\left (2\,x+{\mathrm {e}}^x\,\left (2\,x^2+10\,x\right )+10\right )+2\,x\,{\mathrm {e}}^x\right )+x^2+x\,{\ln \left (4\,x\,{\mathrm {e}}^3\right )}^2}{x^2+5\,x} \,d x \] Input:
int((4*x + x*exp(2*x) + log(x + 5)*(5*x + exp(2*x)*(10*x + 2*x^2) + exp(x) *(2*x + 10) + x^2) + log(4*x*exp(3))*(log(x + 5)*(2*x + exp(x)*(10*x + 2*x ^2) + 10) + 2*x*exp(x)) + x^2 + x*log(4*x*exp(3))^2)/(5*x + x^2),x)
Output:
int((4*x + x*exp(2*x) + log(x + 5)*(5*x + exp(2*x)*(10*x + 2*x^2) + exp(x) *(2*x + 10) + x^2) + log(4*x*exp(3))*(log(x + 5)*(2*x + exp(x)*(10*x + 2*x ^2) + 10) + 2*x*exp(x)) + x^2 + x*log(4*x*exp(3))^2)/(5*x + x^2), x)
Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {4 x+e^{2 x} x+x^2+x \log ^2\left (4 e^3 x\right )+\left (5 x+x^2+e^x (10+2 x)+e^{2 x} \left (10 x+2 x^2\right )\right ) \log (5+x)+\log \left (4 e^3 x\right ) \left (2 e^x x+\left (10+2 x+e^x \left (10 x+2 x^2\right )\right ) \log (5+x)\right )}{5 x+x^2} \, dx=\mathrm {log}\left (x +5\right ) \left (e^{2 x}+2 e^{x} \mathrm {log}\left (4 e^{3} x \right )+\mathrm {log}\left (4 e^{3} x \right )^{2}+x +4\right ) \] Input:
int((x*log(4*x*exp(3))^2+(((2*x^2+10*x)*exp(x)+2*x+10)*log(5+x)+2*exp(x)*x )*log(4*x*exp(3))+((2*x^2+10*x)*exp(x)^2+(2*x+10)*exp(x)+x^2+5*x)*log(5+x) +x*exp(x)^2+x^2+4*x)/(x^2+5*x),x)
Output:
log(x + 5)*(e**(2*x) + 2*e**x*log(4*e**3*x) + log(4*e**3*x)**2 + x + 4)