Integrand size = 110, antiderivative size = 23 \[ \int \frac {-10 x^2-11 x^3+120 x^6+120 x^7+\left (3 x^2+3 x^3-40 x^6-40 x^7\right ) \log \left (x+x^2\right )}{e^{-20+10 x^4} (9+9 x)+e^{-20+10 x^4} (-6-6 x) \log \left (x+x^2\right )+e^{-20+10 x^4} (1+x) \log ^2\left (x+x^2\right )} \, dx=\frac {e^{-10 \left (-2+x^4\right )} x^3}{-3+\log (x (1+x))} \] Output:
x^3/(ln(x*(1+x))-3)/exp(5*x^4-10)^2
Time = 0.41 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-10 x^2-11 x^3+120 x^6+120 x^7+\left (3 x^2+3 x^3-40 x^6-40 x^7\right ) \log \left (x+x^2\right )}{e^{-20+10 x^4} (9+9 x)+e^{-20+10 x^4} (-6-6 x) \log \left (x+x^2\right )+e^{-20+10 x^4} (1+x) \log ^2\left (x+x^2\right )} \, dx=-\frac {e^{20-10 x^4} x^3}{3-\log (x (1+x))} \] Input:
Integrate[(-10*x^2 - 11*x^3 + 120*x^6 + 120*x^7 + (3*x^2 + 3*x^3 - 40*x^6 - 40*x^7)*Log[x + x^2])/(E^(-20 + 10*x^4)*(9 + 9*x) + E^(-20 + 10*x^4)*(-6 - 6*x)*Log[x + x^2] + E^(-20 + 10*x^4)*(1 + x)*Log[x + x^2]^2),x]
Output:
-((E^(20 - 10*x^4)*x^3)/(3 - Log[x*(1 + x)]))
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.
\(\displaystyle \int \frac {120 x^7+120 x^6-11 x^3-10 x^2+\left (-40 x^7-40 x^6+3 x^3+3 x^2\right ) \log \left (x^2+x\right )}{e^{10 x^4-20} (9 x+9)+e^{10 x^4-20} (x+1) \log ^2\left (x^2+x\right )+e^{10 x^4-20} (-6 x-6) \log \left (x^2+x\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{20-10 x^4} \left (120 x^7+120 x^6-11 x^3-10 x^2+\left (-40 x^7-40 x^6+3 x^3+3 x^2\right ) \log \left (x^2+x\right )\right )}{(x+1) (3-\log (x (x+1)))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {120 e^{20-10 x^4} x^7}{(x+1) (\log (x (x+1))-3)^2}+\frac {120 e^{20-10 x^4} x^6}{(x+1) (\log (x (x+1))-3)^2}-\frac {11 e^{20-10 x^4} x^3}{(x+1) (\log (x (x+1))-3)^2}-\frac {e^{20-10 x^4} \left (40 x^4-3\right ) x^2 \log (x (x+1))}{(\log (x (x+1))-3)^2}-\frac {10 e^{20-10 x^4} x^2}{(x+1) (\log (x (x+1))-3)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^{20-10 x^4}}{(\log (x (x+1))-3)^2}dx+\int \frac {e^{20-10 x^4} x}{(\log (x (x+1))-3)^2}dx+\int \frac {e^{20-10 x^4}}{(x+1) (\log (x (x+1))-3)^2}dx-40 \int \frac {e^{20-10 x^4} x^6}{\log (x (x+1))-3}dx-2 \int \frac {e^{20-10 x^4} x^2}{(\log (x (x+1))-3)^2}dx+3 \int \frac {e^{20-10 x^4} x^2}{\log (x (x+1))-3}dx\) |
Input:
Int[(-10*x^2 - 11*x^3 + 120*x^6 + 120*x^7 + (3*x^2 + 3*x^3 - 40*x^6 - 40*x ^7)*Log[x + x^2])/(E^(-20 + 10*x^4)*(9 + 9*x) + E^(-20 + 10*x^4)*(-6 - 6*x )*Log[x + x^2] + E^(-20 + 10*x^4)*(1 + x)*Log[x + x^2]^2),x]
Output:
$Aborted
Time = 0.91 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09
method | result | size |
parallelrisch | \(\frac {x^{3} {\mathrm e}^{-10 x^{4}+20}}{\ln \left (x^{2}+x \right )-3}\) | \(25\) |
risch | \(\frac {2 i x^{3} {\mathrm e}^{-10 x^{4}+20}}{\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (1+x \right )\right ) \operatorname {csgn}\left (i x \left (1+x \right )\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \left (1+x \right )\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (1+x \right )\right ) \operatorname {csgn}\left (i x \left (1+x \right )\right )^{2}+\pi \operatorname {csgn}\left (i x \left (1+x \right )\right )^{3}+2 i \ln \left (x \right )+2 i \ln \left (1+x \right )-6 i}\) | \(104\) |
Input:
int(((-40*x^7-40*x^6+3*x^3+3*x^2)*ln(x^2+x)+120*x^7+120*x^6-11*x^3-10*x^2) /((1+x)*exp(5*x^4-10)^2*ln(x^2+x)^2+(-6*x-6)*exp(5*x^4-10)^2*ln(x^2+x)+(9* x+9)*exp(5*x^4-10)^2),x,method=_RETURNVERBOSE)
Output:
x^3/(ln(x^2+x)-3)/exp(5*x^4-10)^2
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {-10 x^2-11 x^3+120 x^6+120 x^7+\left (3 x^2+3 x^3-40 x^6-40 x^7\right ) \log \left (x+x^2\right )}{e^{-20+10 x^4} (9+9 x)+e^{-20+10 x^4} (-6-6 x) \log \left (x+x^2\right )+e^{-20+10 x^4} (1+x) \log ^2\left (x+x^2\right )} \, dx=\frac {x^{3}}{e^{\left (10 \, x^{4} - 20\right )} \log \left (x^{2} + x\right ) - 3 \, e^{\left (10 \, x^{4} - 20\right )}} \] Input:
integrate(((-40*x^7-40*x^6+3*x^3+3*x^2)*log(x^2+x)+120*x^7+120*x^6-11*x^3- 10*x^2)/((1+x)*exp(5*x^4-10)^2*log(x^2+x)^2+(-6*x-6)*exp(5*x^4-10)^2*log(x ^2+x)+(9*x+9)*exp(5*x^4-10)^2),x, algorithm="fricas")
Output:
x^3/(e^(10*x^4 - 20)*log(x^2 + x) - 3*e^(10*x^4 - 20))
Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-10 x^2-11 x^3+120 x^6+120 x^7+\left (3 x^2+3 x^3-40 x^6-40 x^7\right ) \log \left (x+x^2\right )}{e^{-20+10 x^4} (9+9 x)+e^{-20+10 x^4} (-6-6 x) \log \left (x+x^2\right )+e^{-20+10 x^4} (1+x) \log ^2\left (x+x^2\right )} \, dx=\frac {x^{3} e^{20 - 10 x^{4}}}{\log {\left (x^{2} + x \right )} - 3} \] Input:
integrate(((-40*x**7-40*x**6+3*x**3+3*x**2)*ln(x**2+x)+120*x**7+120*x**6-1 1*x**3-10*x**2)/((1+x)*exp(5*x**4-10)**2*ln(x**2+x)**2+(-6*x-6)*exp(5*x**4 -10)**2*ln(x**2+x)+(9*x+9)*exp(5*x**4-10)**2),x)
Output:
x**3*exp(20 - 10*x**4)/(log(x**2 + x) - 3)
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-10 x^2-11 x^3+120 x^6+120 x^7+\left (3 x^2+3 x^3-40 x^6-40 x^7\right ) \log \left (x+x^2\right )}{e^{-20+10 x^4} (9+9 x)+e^{-20+10 x^4} (-6-6 x) \log \left (x+x^2\right )+e^{-20+10 x^4} (1+x) \log ^2\left (x+x^2\right )} \, dx=\frac {x^{3} e^{\left (-10 \, x^{4} + 20\right )}}{\log \left (x + 1\right ) + \log \left (x\right ) - 3} \] Input:
integrate(((-40*x^7-40*x^6+3*x^3+3*x^2)*log(x^2+x)+120*x^7+120*x^6-11*x^3- 10*x^2)/((1+x)*exp(5*x^4-10)^2*log(x^2+x)^2+(-6*x-6)*exp(5*x^4-10)^2*log(x ^2+x)+(9*x+9)*exp(5*x^4-10)^2),x, algorithm="maxima")
Output:
x^3*e^(-10*x^4 + 20)/(log(x + 1) + log(x) - 3)
Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-10 x^2-11 x^3+120 x^6+120 x^7+\left (3 x^2+3 x^3-40 x^6-40 x^7\right ) \log \left (x+x^2\right )}{e^{-20+10 x^4} (9+9 x)+e^{-20+10 x^4} (-6-6 x) \log \left (x+x^2\right )+e^{-20+10 x^4} (1+x) \log ^2\left (x+x^2\right )} \, dx=\frac {x^{3} e^{\left (-10 \, x^{4} + 20\right )}}{\log \left (x^{2} + x\right ) - 3} \] Input:
integrate(((-40*x^7-40*x^6+3*x^3+3*x^2)*log(x^2+x)+120*x^7+120*x^6-11*x^3- 10*x^2)/((1+x)*exp(5*x^4-10)^2*log(x^2+x)^2+(-6*x-6)*exp(5*x^4-10)^2*log(x ^2+x)+(9*x+9)*exp(5*x^4-10)^2),x, algorithm="giac")
Output:
x^3*e^(-10*x^4 + 20)/(log(x^2 + x) - 3)
Time = 0.79 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-10 x^2-11 x^3+120 x^6+120 x^7+\left (3 x^2+3 x^3-40 x^6-40 x^7\right ) \log \left (x+x^2\right )}{e^{-20+10 x^4} (9+9 x)+e^{-20+10 x^4} (-6-6 x) \log \left (x+x^2\right )+e^{-20+10 x^4} (1+x) \log ^2\left (x+x^2\right )} \, dx=\frac {x^3\,{\mathrm {e}}^{20}\,{\mathrm {e}}^{-10\,x^4}}{\ln \left (x^2+x\right )-3} \] Input:
int((log(x + x^2)*(3*x^2 + 3*x^3 - 40*x^6 - 40*x^7) - 10*x^2 - 11*x^3 + 12 0*x^6 + 120*x^7)/(exp(10*x^4 - 20)*(9*x + 9) - log(x + x^2)*exp(10*x^4 - 2 0)*(6*x + 6) + log(x + x^2)^2*exp(10*x^4 - 20)*(x + 1)),x)
Output:
(x^3*exp(20)*exp(-10*x^4))/(log(x + x^2) - 3)
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-10 x^2-11 x^3+120 x^6+120 x^7+\left (3 x^2+3 x^3-40 x^6-40 x^7\right ) \log \left (x+x^2\right )}{e^{-20+10 x^4} (9+9 x)+e^{-20+10 x^4} (-6-6 x) \log \left (x+x^2\right )+e^{-20+10 x^4} (1+x) \log ^2\left (x+x^2\right )} \, dx=\frac {e^{20} x^{3}}{e^{10 x^{4}} \left (\mathrm {log}\left (x^{2}+x \right )-3\right )} \] Input:
int(((-40*x^7-40*x^6+3*x^3+3*x^2)*log(x^2+x)+120*x^7+120*x^6-11*x^3-10*x^2 )/((1+x)*exp(5*x^4-10)^2*log(x^2+x)^2+(-6*x-6)*exp(5*x^4-10)^2*log(x^2+x)+ (9*x+9)*exp(5*x^4-10)^2),x)
Output:
(e**20*x**3)/(e**(10*x**4)*(log(x**2 + x) - 3))