Integrand size = 207, antiderivative size = 27 \[ \int \frac {\left (-2 e^x x+2 x^2\right ) \log ^2(x)+\left (\left (e^x (-10-2 x)+10 x+2 x^2\right ) \log ^2(x)+\left (10 x+12 x^2+2 x^3+e^{2 x} \left (10 x+2 x^2\right )+e^x \left (-10 x-12 x^2-2 x^3\right )\right ) \log ^3(x)\right ) \log (5+x)+\left (2 x \log (x)+\left (\left (10-8 x-2 x^2+e^x (10+2 x)\right ) \log (x)+\left (10 x+2 x^2+e^x \left (-10 x-2 x^2\right )\right ) \log ^2(x)\right ) \log (5+x)\right ) \log (x \log (5+x))+(-10-2 x) \log (5+x) \log ^2(x \log (5+x))}{\left (5 x+x^2\right ) \log ^3(x) \log (5+x)} \, dx=2+2 x+\left (e^x-x-\frac {\log (x \log (5+x))}{\log (x)}\right )^2 \]
Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93 \[ \int \frac {\left (-2 e^x x+2 x^2\right ) \log ^2(x)+\left (\left (e^x (-10-2 x)+10 x+2 x^2\right ) \log ^2(x)+\left (10 x+12 x^2+2 x^3+e^{2 x} \left (10 x+2 x^2\right )+e^x \left (-10 x-12 x^2-2 x^3\right )\right ) \log ^3(x)\right ) \log (5+x)+\left (2 x \log (x)+\left (\left (10-8 x-2 x^2+e^x (10+2 x)\right ) \log (x)+\left (10 x+2 x^2+e^x \left (-10 x-2 x^2\right )\right ) \log ^2(x)\right ) \log (5+x)\right ) \log (x \log (5+x))+(-10-2 x) \log (5+x) \log ^2(x \log (5+x))}{\left (5 x+x^2\right ) \log ^3(x) \log (5+x)} \, dx=e^{2 x}+2 x-2 e^x x+x^2-\frac {2 \left (e^x-x\right ) \log (x \log (5+x))}{\log (x)}+\frac {\log ^2(x \log (5+x))}{\log ^2(x)} \]
Integrate[((-2*E^x*x + 2*x^2)*Log[x]^2 + ((E^x*(-10 - 2*x) + 10*x + 2*x^2) *Log[x]^2 + (10*x + 12*x^2 + 2*x^3 + E^(2*x)*(10*x + 2*x^2) + E^x*(-10*x - 12*x^2 - 2*x^3))*Log[x]^3)*Log[5 + x] + (2*x*Log[x] + ((10 - 8*x - 2*x^2 + E^x*(10 + 2*x))*Log[x] + (10*x + 2*x^2 + E^x*(-10*x - 2*x^2))*Log[x]^2)* Log[5 + x])*Log[x*Log[5 + x]] + (-10 - 2*x)*Log[5 + x]*Log[x*Log[5 + x]]^2 )/((5*x + x^2)*Log[x]^3*Log[5 + x]),x]
E^(2*x) + 2*x - 2*E^x*x + x^2 - (2*(E^x - x)*Log[x*Log[5 + x]])/Log[x] + L og[x*Log[5 + x]]^2/Log[x]^2
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.
| \(\displaystyle \int \frac {\left (2 x^2-2 e^x x\right ) \log ^2(x)+\left (\left (\left (2 x^2+e^x \left (-2 x^2-10 x\right )+10 x\right ) \log ^2(x)+\left (-2 x^2-8 x+e^x (2 x+10)+10\right ) \log (x)\right ) \log (x+5)+2 x \log (x)\right ) \log (x \log (x+5))+\left (\left (2 x^2+10 x+e^x (-2 x-10)\right ) \log ^2(x)+\left (2 x^3+12 x^2+e^{2 x} \left (2 x^2+10 x\right )+e^x \left (-2 x^3-12 x^2-10 x\right )+10 x\right ) \log ^3(x)\right ) \log (x+5)+(-2 x-10) \log (x+5) \log ^2(x \log (x+5))}{\left (x^2+5 x\right ) \log ^3(x) \log (x+5)} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {\left (2 x^2-2 e^x x\right ) \log ^2(x)+\left (\left (\left (2 x^2+e^x \left (-2 x^2-10 x\right )+10 x\right ) \log ^2(x)+\left (-2 x^2-8 x+e^x (2 x+10)+10\right ) \log (x)\right ) \log (x+5)+2 x \log (x)\right ) \log (x \log (x+5))+\left (\left (2 x^2+10 x+e^x (-2 x-10)\right ) \log ^2(x)+\left (2 x^3+12 x^2+e^{2 x} \left (2 x^2+10 x\right )+e^x \left (-2 x^3-12 x^2-10 x\right )+10 x\right ) \log ^3(x)\right ) \log (x+5)+(-2 x-10) \log (x+5) \log ^2(x \log (x+5))}{x (x+5) \log ^3(x) \log (x+5)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 x^2}{x+5}-\frac {2 e^x \left (x^3 \log ^2(x) \log (x+5)+6 x^2 \log ^2(x) \log (x+5)+x^2 \log (x) \log (x+5) \log (x \log (x+5))+5 x \log ^2(x) \log (x+5)+x \log (x)+x \log (x) \log (x+5)+5 x \log (x) \log (x+5) \log (x \log (x+5))-x \log (x+5) \log (x \log (x+5))+5 \log (x) \log (x+5)-5 \log (x+5) \log (x \log (x+5))\right )}{(x+5) x \log ^2(x) \log (x+5)}+\frac {12 x}{x+5}+2 e^{2 x}+\frac {10}{x+5}-\frac {2 x \log (x \log (x+5))}{(x+5) \log ^2(x)}+\frac {2 \log (x \log (x+5))}{(x+5) \log ^2(x) \log (x+5)}-\frac {8 \log (x \log (x+5))}{(x+5) \log ^2(x)}+\frac {10 \log (x \log (x+5))}{(x+5) x \log ^2(x)}-\frac {2 \log ^2(x \log (x+5))}{x \log ^3(x)}+\frac {2 x \log (x \log (x+5))}{(x+5) \log (x)}+\frac {2 x}{(x+5) \log (x)}+\frac {2 x}{(x+5) \log (x) \log (x+5)}+\frac {10 \log (x \log (x+5))}{(x+5) \log (x)}+\frac {10}{(x+5) \log (x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 \int \frac {\operatorname {LogIntegral}(x)}{(x+5) \log (x+5)}dx-2 \int \frac {\log (x \log (x+5))}{\log ^2(x)}dx+2 \int \frac {\log (x \log (x+5))}{x \log ^2(x)}dx+2 \int \frac {e^x \log (x \log (x+5))}{x \log ^2(x)}dx+2 \int \frac {\log (x \log (x+5))}{(x+5) \log ^2(x) \log (x+5)}dx-2 \int \frac {\log ^2(x \log (x+5))}{x \log ^3(x)}dx-2 \int \frac {e^x}{x \log (x)}dx+10 \int \frac {1}{(x+5) \log (x)}dx+2 \int \frac {x}{(x+5) \log (x)}dx+2 \int \frac {1}{\log (x) \log (x+5)}dx-10 \int \frac {1}{(x+5) \log (x) \log (x+5)}dx-2 \int \frac {e^x}{(x+5) \log (x) \log (x+5)}dx-2 \int \frac {e^x \log (x \log (x+5))}{\log (x)}dx-2 \operatorname {LogIntegral}(x) \log (x)+2 \operatorname {LogIntegral}(x) \log (x \log (x+5))+x^2-2 e^x x+4 x+e^{2 x}\) |
Int[((-2*E^x*x + 2*x^2)*Log[x]^2 + ((E^x*(-10 - 2*x) + 10*x + 2*x^2)*Log[x ]^2 + (10*x + 12*x^2 + 2*x^3 + E^(2*x)*(10*x + 2*x^2) + E^x*(-10*x - 12*x^ 2 - 2*x^3))*Log[x]^3)*Log[5 + x] + (2*x*Log[x] + ((10 - 8*x - 2*x^2 + E^x* (10 + 2*x))*Log[x] + (10*x + 2*x^2 + E^x*(-10*x - 2*x^2))*Log[x]^2)*Log[5 + x])*Log[x*Log[5 + x]] + (-10 - 2*x)*Log[5 + x]*Log[x*Log[5 + x]]^2)/((5* x + x^2)*Log[x]^3*Log[5 + x]),x]
3.11.88.3.1 Defintions of rubi rules used
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. \(84\) vs. \(2(26)=52\).
Time = 77.58 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.15
| method | result | size |
| parallelrisch | \(-\frac {-50 x^{2} \ln \left (x \right )^{2}+100 x \,{\mathrm e}^{x} \ln \left (x \right )^{2}-50 \ln \left (x \right )^{2} {\mathrm e}^{2 x}-100 x \ln \left (x \right )^{2}-100 \ln \left (x \right ) \ln \left (x \ln \left (5+x \right )\right ) x +100 \ln \left (x \ln \left (5+x \right )\right ) \ln \left (x \right ) {\mathrm e}^{x}+250 \ln \left (x \right )^{2}-50 \ln \left (x \ln \left (5+x \right )\right )^{2}}{50 \ln \left (x \right )^{2}}\) | \(85\) |
| risch | \(\text {Expression too large to display}\) | \(696\) |
int(((-2*x-10)*ln(5+x)*ln(x*ln(5+x))^2+((((-2*x^2-10*x)*exp(x)+2*x^2+10*x) *ln(x)^2+((2*x+10)*exp(x)-2*x^2-8*x+10)*ln(x))*ln(5+x)+2*x*ln(x))*ln(x*ln( 5+x))+(((2*x^2+10*x)*exp(x)^2+(-2*x^3-12*x^2-10*x)*exp(x)+2*x^3+12*x^2+10* x)*ln(x)^3+((-2*x-10)*exp(x)+2*x^2+10*x)*ln(x)^2)*ln(5+x)+(-2*exp(x)*x+2*x ^2)*ln(x)^2)/(x^2+5*x)/ln(x)^3/ln(5+x),x,method=_RETURNVERBOSE)
-1/50*(-50*x^2*ln(x)^2+100*x*exp(x)*ln(x)^2-50*exp(x)^2*ln(x)^2-100*x*ln(x )^2-100*ln(x)*ln(x*ln(5+x))*x+100*ln(x*ln(5+x))*ln(x)*exp(x)+250*ln(x)^2-5 0*ln(x*ln(5+x))^2)/ln(x)^2
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (25) = 50\).
Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {\left (-2 e^x x+2 x^2\right ) \log ^2(x)+\left (\left (e^x (-10-2 x)+10 x+2 x^2\right ) \log ^2(x)+\left (10 x+12 x^2+2 x^3+e^{2 x} \left (10 x+2 x^2\right )+e^x \left (-10 x-12 x^2-2 x^3\right )\right ) \log ^3(x)\right ) \log (5+x)+\left (2 x \log (x)+\left (\left (10-8 x-2 x^2+e^x (10+2 x)\right ) \log (x)+\left (10 x+2 x^2+e^x \left (-10 x-2 x^2\right )\right ) \log ^2(x)\right ) \log (5+x)\right ) \log (x \log (5+x))+(-10-2 x) \log (5+x) \log ^2(x \log (5+x))}{\left (5 x+x^2\right ) \log ^3(x) \log (5+x)} \, dx=\frac {2 \, {\left (x - e^{x}\right )} \log \left (x \log \left (x + 5\right )\right ) \log \left (x\right ) + {\left (x^{2} - 2 \, x e^{x} + 2 \, x + e^{\left (2 \, x\right )}\right )} \log \left (x\right )^{2} + \log \left (x \log \left (x + 5\right )\right )^{2}}{\log \left (x\right )^{2}} \]
integrate(((-2*x-10)*log(5+x)*log(x*log(5+x))^2+((((-2*x^2-10*x)*exp(x)+2* x^2+10*x)*log(x)^2+((2*x+10)*exp(x)-2*x^2-8*x+10)*log(x))*log(5+x)+2*x*log (x))*log(x*log(5+x))+(((2*x^2+10*x)*exp(x)^2+(-2*x^3-12*x^2-10*x)*exp(x)+2 *x^3+12*x^2+10*x)*log(x)^3+((-2*x-10)*exp(x)+2*x^2+10*x)*log(x)^2)*log(5+x )+(-2*exp(x)*x+2*x^2)*log(x)^2)/(x^2+5*x)/log(x)^3/log(5+x),x, algorithm=\
(2*(x - e^x)*log(x*log(x + 5))*log(x) + (x^2 - 2*x*e^x + 2*x + e^(2*x))*lo g(x)^2 + log(x*log(x + 5))^2)/log(x)^2
Exception generated. \[ \int \frac {\left (-2 e^x x+2 x^2\right ) \log ^2(x)+\left (\left (e^x (-10-2 x)+10 x+2 x^2\right ) \log ^2(x)+\left (10 x+12 x^2+2 x^3+e^{2 x} \left (10 x+2 x^2\right )+e^x \left (-10 x-12 x^2-2 x^3\right )\right ) \log ^3(x)\right ) \log (5+x)+\left (2 x \log (x)+\left (\left (10-8 x-2 x^2+e^x (10+2 x)\right ) \log (x)+\left (10 x+2 x^2+e^x \left (-10 x-2 x^2\right )\right ) \log ^2(x)\right ) \log (5+x)\right ) \log (x \log (5+x))+(-10-2 x) \log (5+x) \log ^2(x \log (5+x))}{\left (5 x+x^2\right ) \log ^3(x) \log (5+x)} \, dx=\text {Exception raised: TypeError} \]
integrate(((-2*x-10)*ln(5+x)*ln(x*ln(5+x))**2+((((-2*x**2-10*x)*exp(x)+2*x **2+10*x)*ln(x)**2+((2*x+10)*exp(x)-2*x**2-8*x+10)*ln(x))*ln(5+x)+2*x*ln(x ))*ln(x*ln(5+x))+(((2*x**2+10*x)*exp(x)**2+(-2*x**3-12*x**2-10*x)*exp(x)+2 *x**3+12*x**2+10*x)*ln(x)**3+((-2*x-10)*exp(x)+2*x**2+10*x)*ln(x)**2)*ln(5 +x)+(-2*exp(x)*x+2*x**2)*ln(x)**2)/(x**2+5*x)/ln(x)**3/ln(5+x),x)
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (25) = 50\).
Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.59 \[ \int \frac {\left (-2 e^x x+2 x^2\right ) \log ^2(x)+\left (\left (e^x (-10-2 x)+10 x+2 x^2\right ) \log ^2(x)+\left (10 x+12 x^2+2 x^3+e^{2 x} \left (10 x+2 x^2\right )+e^x \left (-10 x-12 x^2-2 x^3\right )\right ) \log ^3(x)\right ) \log (5+x)+\left (2 x \log (x)+\left (\left (10-8 x-2 x^2+e^x (10+2 x)\right ) \log (x)+\left (10 x+2 x^2+e^x \left (-10 x-2 x^2\right )\right ) \log ^2(x)\right ) \log (5+x)\right ) \log (x \log (5+x))+(-10-2 x) \log (5+x) \log ^2(x \log (5+x))}{\left (5 x+x^2\right ) \log ^3(x) \log (5+x)} \, dx=-\frac {2 \, {\left (x + 1\right )} e^{x} \log \left (x\right )^{2} - {\left (x^{2} + 4 \, x\right )} \log \left (x\right )^{2} - e^{\left (2 \, x\right )} \log \left (x\right )^{2} - 2 \, {\left ({\left (x + 1\right )} \log \left (x\right ) - e^{x} \log \left (x\right )\right )} \log \left (\log \left (x + 5\right )\right ) - \log \left (\log \left (x + 5\right )\right )^{2}}{\log \left (x\right )^{2}} \]
integrate(((-2*x-10)*log(5+x)*log(x*log(5+x))^2+((((-2*x^2-10*x)*exp(x)+2* x^2+10*x)*log(x)^2+((2*x+10)*exp(x)-2*x^2-8*x+10)*log(x))*log(5+x)+2*x*log (x))*log(x*log(5+x))+(((2*x^2+10*x)*exp(x)^2+(-2*x^3-12*x^2-10*x)*exp(x)+2 *x^3+12*x^2+10*x)*log(x)^3+((-2*x-10)*exp(x)+2*x^2+10*x)*log(x)^2)*log(5+x )+(-2*exp(x)*x+2*x^2)*log(x)^2)/(x^2+5*x)/log(x)^3/log(5+x),x, algorithm=\
-(2*(x + 1)*e^x*log(x)^2 - (x^2 + 4*x)*log(x)^2 - e^(2*x)*log(x)^2 - 2*((x + 1)*log(x) - e^x*log(x))*log(log(x + 5)) - log(log(x + 5))^2)/log(x)^2
Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (25) = 50\).
Time = 0.60 (sec) , antiderivative size = 84, normalized size of antiderivative = 3.11 \[ \int \frac {\left (-2 e^x x+2 x^2\right ) \log ^2(x)+\left (\left (e^x (-10-2 x)+10 x+2 x^2\right ) \log ^2(x)+\left (10 x+12 x^2+2 x^3+e^{2 x} \left (10 x+2 x^2\right )+e^x \left (-10 x-12 x^2-2 x^3\right )\right ) \log ^3(x)\right ) \log (5+x)+\left (2 x \log (x)+\left (\left (10-8 x-2 x^2+e^x (10+2 x)\right ) \log (x)+\left (10 x+2 x^2+e^x \left (-10 x-2 x^2\right )\right ) \log ^2(x)\right ) \log (5+x)\right ) \log (x \log (5+x))+(-10-2 x) \log (5+x) \log ^2(x \log (5+x))}{\left (5 x+x^2\right ) \log ^3(x) \log (5+x)} \, dx=\frac {x^{2} \log \left (x\right )^{2} - 2 \, x e^{x} \log \left (x\right )^{2} + 4 \, x \log \left (x\right )^{2} + e^{\left (2 \, x\right )} \log \left (x\right )^{2} - 2 \, e^{x} \log \left (x\right )^{2} + 2 \, x \log \left (x\right ) \log \left (\log \left (x + 5\right )\right ) - 2 \, e^{x} \log \left (x\right ) \log \left (\log \left (x + 5\right )\right ) + 2 \, \log \left (x\right ) \log \left (\log \left (x + 5\right )\right ) + \log \left (\log \left (x + 5\right )\right )^{2}}{\log \left (x\right )^{2}} \]
integrate(((-2*x-10)*log(5+x)*log(x*log(5+x))^2+((((-2*x^2-10*x)*exp(x)+2* x^2+10*x)*log(x)^2+((2*x+10)*exp(x)-2*x^2-8*x+10)*log(x))*log(5+x)+2*x*log (x))*log(x*log(5+x))+(((2*x^2+10*x)*exp(x)^2+(-2*x^3-12*x^2-10*x)*exp(x)+2 *x^3+12*x^2+10*x)*log(x)^3+((-2*x-10)*exp(x)+2*x^2+10*x)*log(x)^2)*log(5+x )+(-2*exp(x)*x+2*x^2)*log(x)^2)/(x^2+5*x)/log(x)^3/log(5+x),x, algorithm=\
(x^2*log(x)^2 - 2*x*e^x*log(x)^2 + 4*x*log(x)^2 + e^(2*x)*log(x)^2 - 2*e^x *log(x)^2 + 2*x*log(x)*log(log(x + 5)) - 2*e^x*log(x)*log(log(x + 5)) + 2* log(x)*log(log(x + 5)) + log(log(x + 5))^2)/log(x)^2
Time = 10.46 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.81 \[ \int \frac {\left (-2 e^x x+2 x^2\right ) \log ^2(x)+\left (\left (e^x (-10-2 x)+10 x+2 x^2\right ) \log ^2(x)+\left (10 x+12 x^2+2 x^3+e^{2 x} \left (10 x+2 x^2\right )+e^x \left (-10 x-12 x^2-2 x^3\right )\right ) \log ^3(x)\right ) \log (5+x)+\left (2 x \log (x)+\left (\left (10-8 x-2 x^2+e^x (10+2 x)\right ) \log (x)+\left (10 x+2 x^2+e^x \left (-10 x-2 x^2\right )\right ) \log ^2(x)\right ) \log (5+x)\right ) \log (x \log (5+x))+(-10-2 x) \log (5+x) \log ^2(x \log (5+x))}{\left (5 x+x^2\right ) \log ^3(x) \log (5+x)} \, dx=2\,x+{\mathrm {e}}^{2\,x}+\frac {{\ln \left (x\,\ln \left (x+5\right )\right )}^2}{{\ln \left (x\right )}^2}-2\,x\,{\mathrm {e}}^x+x^2+\frac {2\,\ln \left (x\,\ln \left (x+5\right )\right )\,\left (x-{\mathrm {e}}^x\right )}{\ln \left (x\right )} \]
int((log(x + 5)*(log(x)^3*(10*x + exp(2*x)*(10*x + 2*x^2) + 12*x^2 + 2*x^3 - exp(x)*(10*x + 12*x^2 + 2*x^3)) + log(x)^2*(10*x - exp(x)*(2*x + 10) + 2*x^2)) + log(x*log(x + 5))*(2*x*log(x) - log(x + 5)*(log(x)*(8*x - exp(x) *(2*x + 10) + 2*x^2 - 10) - log(x)^2*(10*x - exp(x)*(10*x + 2*x^2) + 2*x^2 ))) - log(x)^2*(2*x*exp(x) - 2*x^2) - log(x + 5)*log(x*log(x + 5))^2*(2*x + 10))/(log(x + 5)*log(x)^3*(5*x + x^2)),x)