Integrand size = 150, antiderivative size = 25 \begin {dmath*} \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=\left (e^{\frac {x^2 (x+\log (4))}{5+e^4+\log (x)}}-x\right ) x \end {dmath*}
\begin {dmath*} \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=\int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx \end {dmath*}
Integrate[(-50*x - 20*E^4*x - 2*E^8*x + (-20*x - 4*E^4*x)*Log[x] - 2*x*Log [x]^2 + E^((x^3 + x^2*Log[4])/(5 + E^4 + Log[x]))*(25 + E^8 + 14*x^3 + E^4 *(10 + 3*x^3) + (9*x^2 + 2*E^4*x^2)*Log[4] + (10 + 2*E^4 + 3*x^3 + 2*x^2*L og[4])*Log[x] + Log[x]^2))/(25 + 10*E^4 + E^8 + (10 + 2*E^4)*Log[x] + Log[ x]^2),x]
Integrate[(-50*x - 20*E^4*x - 2*E^8*x + (-20*x - 4*E^4*x)*Log[x] - 2*x*Log [x]^2 + E^((x^3 + x^2*Log[4])/(5 + E^4 + Log[x]))*(25 + E^8 + 14*x^3 + E^4 *(10 + 3*x^3) + (9*x^2 + 2*E^4*x^2)*Log[4] + (10 + 2*E^4 + 3*x^3 + 2*x^2*L og[4])*Log[x] + Log[x]^2))/(25 + 10*E^4 + E^8 + (10 + 2*E^4)*Log[x] + Log[ x]^2), x]
Leaf count is larger than twice the leaf count of optimal. \(99\) vs. \(2(25)=50\).
Time = 2.25 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {6, 6, 7292, 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.
| \(\displaystyle \int \frac {e^{\frac {x^3+x^2 \log (4)}{\log (x)+e^4+5}} \left (14 x^3+e^4 \left (3 x^3+10\right )+\left (2 e^4 x^2+9 x^2\right ) \log (4)+\left (3 x^3+2 x^2 \log (4)+2 e^4+10\right ) \log (x)+\log ^2(x)+e^8+25\right )-2 e^8 x-20 e^4 x-50 x-2 x \log ^2(x)+\left (-4 e^4 x-20 x\right ) \log (x)}{\log ^2(x)+\left (10+2 e^4\right ) \log (x)+e^8+10 e^4+25} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^{\frac {x^3+x^2 \log (4)}{\log (x)+e^4+5}} \left (14 x^3+e^4 \left (3 x^3+10\right )+\left (2 e^4 x^2+9 x^2\right ) \log (4)+\left (3 x^3+2 x^2 \log (4)+2 e^4+10\right ) \log (x)+\log ^2(x)+e^8+25\right )+\left (-50-20 e^4\right ) x-2 e^8 x-2 x \log ^2(x)+\left (-4 e^4 x-20 x\right ) \log (x)}{\log ^2(x)+\left (10+2 e^4\right ) \log (x)+e^8+10 e^4+25}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^{\frac {x^3+x^2 \log (4)}{\log (x)+e^4+5}} \left (14 x^3+e^4 \left (3 x^3+10\right )+\left (2 e^4 x^2+9 x^2\right ) \log (4)+\left (3 x^3+2 x^2 \log (4)+2 e^4+10\right ) \log (x)+\log ^2(x)+e^8+25\right )+\left (-50-20 e^4-2 e^8\right ) x-2 x \log ^2(x)+\left (-4 e^4 x-20 x\right ) \log (x)}{\log ^2(x)+\left (10+2 e^4\right ) \log (x)+e^8+10 e^4+25}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{\frac {x^3+x^2 \log (4)}{\log (x)+e^4+5}} \left (14 x^3+e^4 \left (3 x^3+10\right )+\left (2 e^4 x^2+9 x^2\right ) \log (4)+\left (3 x^3+2 x^2 \log (4)+2 e^4+10\right ) \log (x)+\log ^2(x)+e^8+25\right )+\left (-50-20 e^4-2 e^8\right ) x-2 x \log ^2(x)+\left (-4 e^4 x-20 x\right ) \log (x)}{\left (\log (x)+5 \left (1+\frac {e^4}{5}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {4^{\frac {x^2}{\log (x)+5 \left (1+\frac {e^4}{5}\right )}} e^{\frac {x^3}{\log (x)+5 \left (1+\frac {e^4}{5}\right )}} \left (14 \left (1+\frac {3 e^4}{14}\right ) x^3+3 x^3 \log (x)+x^2 \log (16) \log (x)+9 x^2 \log (4) \left (1+\frac {e^4 \log (16)}{9 \log (4)}\right )+\log ^2(x)+10 \left (1+\frac {e^4}{5}\right ) \log (x)+25 \left (1+\frac {1}{25} e^4 \left (10+e^4\right )\right )\right )}{\left (\log (x)+5 \left (1+\frac {e^4}{5}\right )\right )^2}-2 x\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -x^2-\frac {4^{\frac {x^2}{\log (x)+e^4+5}} e^{\frac {x^3}{\log (x)+e^4+5}} \left (x^2 \log (16) \log (x)+\left (9+2 e^4\right ) x^2 \log (4)\right )}{\log (4) \left (\log (x)+e^4+5\right )^2 \left (\frac {x}{\left (\log (x)+e^4+5\right )^2}-\frac {2 x}{\log (x)+e^4+5}\right )}\) |
Int[(-50*x - 20*E^4*x - 2*E^8*x + (-20*x - 4*E^4*x)*Log[x] - 2*x*Log[x]^2 + E^((x^3 + x^2*Log[4])/(5 + E^4 + Log[x]))*(25 + E^8 + 14*x^3 + E^4*(10 + 3*x^3) + (9*x^2 + 2*E^4*x^2)*Log[4] + (10 + 2*E^4 + 3*x^3 + 2*x^2*Log[4]) *Log[x] + Log[x]^2))/(25 + 10*E^4 + E^8 + (10 + 2*E^4)*Log[x] + Log[x]^2), x]
-x^2 - (4^(x^2/(5 + E^4 + Log[x]))*E^(x^3/(5 + E^4 + Log[x]))*((9 + 2*E^4) *x^2*Log[4] + x^2*Log[16]*Log[x]))/(Log[4]*(5 + E^4 + Log[x])^2*(x/(5 + E^ 4 + Log[x])^2 - (2*x)/(5 + E^4 + Log[x])))
3.8.49.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Time = 4.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12
| method | result | size |
| risch | \(-x^{2}+{\mathrm e}^{\frac {\left (x +2 \ln \left (2\right )\right ) x^{2}}{\ln \left (x \right )+5+{\mathrm e}^{4}}} x\) | \(28\) |
| parallelrisch | \(-x^{2}+{\mathrm e}^{\frac {\left (x +2 \ln \left (2\right )\right ) x^{2}}{\ln \left (x \right )+5+{\mathrm e}^{4}}} x\) | \(28\) |
int(((ln(x)^2+(4*x^2*ln(2)+2*exp(4)+3*x^3+10)*ln(x)+2*(2*x^2*exp(4)+9*x^2) *ln(2)+exp(4)^2+(3*x^3+10)*exp(4)+14*x^3+25)*exp((2*x^2*ln(2)+x^3)/(ln(x)+ 5+exp(4)))-2*x*ln(x)^2+(-4*x*exp(4)-20*x)*ln(x)-2*x*exp(4)^2-20*x*exp(4)-5 0*x)/(ln(x)^2+(2*exp(4)+10)*ln(x)+exp(4)^2+10*exp(4)+25),x,method=_RETURNV ERBOSE)
Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \begin {dmath*} \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=-x^{2} + x e^{\left (\frac {x^{3} + 2 \, x^{2} \log \left (2\right )}{e^{4} + \log \left (x\right ) + 5}\right )} \end {dmath*}
integrate(((log(x)^2+(4*x^2*log(2)+2*exp(4)+3*x^3+10)*log(x)+2*(2*x^2*exp( 4)+9*x^2)*log(2)+exp(4)^2+(3*x^3+10)*exp(4)+14*x^3+25)*exp((2*x^2*log(2)+x ^3)/(log(x)+5+exp(4)))-2*x*log(x)^2+(-4*x*exp(4)-20*x)*log(x)-2*x*exp(4)^2 -20*x*exp(4)-50*x)/(log(x)^2+(2*exp(4)+10)*log(x)+exp(4)^2+10*exp(4)+25),x , algorithm=\
Exception generated. \begin {dmath*} \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=\text {Exception raised: TypeError} \end {dmath*}
integrate(((ln(x)**2+(4*x**2*ln(2)+2*exp(4)+3*x**3+10)*ln(x)+2*(2*x**2*exp (4)+9*x**2)*ln(2)+exp(4)**2+(3*x**3+10)*exp(4)+14*x**3+25)*exp((2*x**2*ln( 2)+x**3)/(ln(x)+5+exp(4)))-2*x*ln(x)**2+(-4*x*exp(4)-20*x)*ln(x)-2*x*exp(4 )**2-20*x*exp(4)-50*x)/(ln(x)**2+(2*exp(4)+10)*ln(x)+exp(4)**2+10*exp(4)+2 5),x)
Exception generated. \begin {dmath*} \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=\text {Exception raised: RuntimeError} \end {dmath*}
integrate(((log(x)^2+(4*x^2*log(2)+2*exp(4)+3*x^3+10)*log(x)+2*(2*x^2*exp( 4)+9*x^2)*log(2)+exp(4)^2+(3*x^3+10)*exp(4)+14*x^3+25)*exp((2*x^2*log(2)+x ^3)/(log(x)+5+exp(4)))-2*x*log(x)^2+(-4*x*exp(4)-20*x)*log(x)-2*x*exp(4)^2 -20*x*exp(4)-50*x)/(log(x)^2+(2*exp(4)+10)*log(x)+exp(4)^2+10*exp(4)+25),x , algorithm=\
Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 0which is not of the expected type LIST
Time = 0.95 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \begin {dmath*} \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=-x^{2} + x e^{\left (\frac {x^{3} + 2 \, x^{2} \log \left (2\right )}{e^{4} + \log \left (x\right ) + 5}\right )} \end {dmath*}
integrate(((log(x)^2+(4*x^2*log(2)+2*exp(4)+3*x^3+10)*log(x)+2*(2*x^2*exp( 4)+9*x^2)*log(2)+exp(4)^2+(3*x^3+10)*exp(4)+14*x^3+25)*exp((2*x^2*log(2)+x ^3)/(log(x)+5+exp(4)))-2*x*log(x)^2+(-4*x*exp(4)-20*x)*log(x)-2*x*exp(4)^2 -20*x*exp(4)-50*x)/(log(x)^2+(2*exp(4)+10)*log(x)+exp(4)^2+10*exp(4)+25),x , algorithm=\
Timed out. \begin {dmath*} \int \frac {-50 x-20 e^4 x-2 e^8 x+\left (-20 x-4 e^4 x\right ) \log (x)-2 x \log ^2(x)+e^{\frac {x^3+x^2 \log (4)}{5+e^4+\log (x)}} \left (25+e^8+14 x^3+e^4 \left (10+3 x^3\right )+\left (9 x^2+2 e^4 x^2\right ) \log (4)+\left (10+2 e^4+3 x^3+2 x^2 \log (4)\right ) \log (x)+\log ^2(x)\right )}{25+10 e^4+e^8+\left (10+2 e^4\right ) \log (x)+\log ^2(x)} \, dx=-\int \frac {50\,x+2\,x\,{\ln \left (x\right )}^2+20\,x\,{\mathrm {e}}^4+2\,x\,{\mathrm {e}}^8+\ln \left (x\right )\,\left (20\,x+4\,x\,{\mathrm {e}}^4\right )-{\mathrm {e}}^{\frac {x^3+2\,\ln \left (2\right )\,x^2}{{\mathrm {e}}^4+\ln \left (x\right )+5}}\,\left ({\mathrm {e}}^8+\ln \left (x\right )\,\left (3\,x^3+4\,\ln \left (2\right )\,x^2+2\,{\mathrm {e}}^4+10\right )+{\ln \left (x\right )}^2+{\mathrm {e}}^4\,\left (3\,x^3+10\right )+14\,x^3+2\,\ln \left (2\right )\,\left (2\,x^2\,{\mathrm {e}}^4+9\,x^2\right )+25\right )}{{\ln \left (x\right )}^2+\left (2\,{\mathrm {e}}^4+10\right )\,\ln \left (x\right )+10\,{\mathrm {e}}^4+{\mathrm {e}}^8+25} \,d x \end {dmath*}
int(-(50*x + 2*x*log(x)^2 + 20*x*exp(4) + 2*x*exp(8) + log(x)*(20*x + 4*x* exp(4)) - exp((2*x^2*log(2) + x^3)/(exp(4) + log(x) + 5))*(exp(8) + log(x) *(2*exp(4) + 4*x^2*log(2) + 3*x^3 + 10) + log(x)^2 + exp(4)*(3*x^3 + 10) + 14*x^3 + 2*log(2)*(2*x^2*exp(4) + 9*x^2) + 25))/(10*exp(4) + exp(8) + log (x)^2 + log(x)*(2*exp(4) + 10) + 25),x)
-int((50*x + 2*x*log(x)^2 + 20*x*exp(4) + 2*x*exp(8) + log(x)*(20*x + 4*x* exp(4)) - exp((2*x^2*log(2) + x^3)/(exp(4) + log(x) + 5))*(exp(8) + log(x) *(2*exp(4) + 4*x^2*log(2) + 3*x^3 + 10) + log(x)^2 + exp(4)*(3*x^3 + 10) + 14*x^3 + 2*log(2)*(2*x^2*exp(4) + 9*x^2) + 25))/(10*exp(4) + exp(8) + log (x)^2 + log(x)*(2*exp(4) + 10) + 25), x)