Integrand size = 107, antiderivative size = 25 \[ \int \frac {275+128 x+15 x^2+e^{4 x^3} \left (5+300 x^2+60 x^3\right )+e^{2 x^3} \left (-90-20 x-1200 x^2-540 x^3-60 x^4\right )}{400+275 x+64 x^2+5 x^3+e^{4 x^3} (25+5 x)+e^{2 x^3} \left (-200-90 x-10 x^2\right )} \, dx=\log \left ((5+x) \left (-\frac {x}{5}+\left (4-e^{2 x^3}+x\right )^2\right )\right ) \]
Time = 15.80 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \[ \int \frac {275+128 x+15 x^2+e^{4 x^3} \left (5+300 x^2+60 x^3\right )+e^{2 x^3} \left (-90-20 x-1200 x^2-540 x^3-60 x^4\right )}{400+275 x+64 x^2+5 x^3+e^{4 x^3} (25+5 x)+e^{2 x^3} \left (-200-90 x-10 x^2\right )} \, dx=\log (5+x)+\log \left (80-40 e^{2 x^3}+5 e^{4 x^3}+39 x-10 e^{2 x^3} x+5 x^2\right ) \]
Integrate[(275 + 128*x + 15*x^2 + E^(4*x^3)*(5 + 300*x^2 + 60*x^3) + E^(2* x^3)*(-90 - 20*x - 1200*x^2 - 540*x^3 - 60*x^4))/(400 + 275*x + 64*x^2 + 5 *x^3 + E^(4*x^3)*(25 + 5*x) + E^(2*x^3)*(-200 - 90*x - 10*x^2)),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.
\(\displaystyle \int \frac {15 x^2+e^{4 x^3} \left (60 x^3+300 x^2+5\right )+e^{2 x^3} \left (-60 x^4-540 x^3-1200 x^2-20 x-90\right )+128 x+275}{5 x^3+e^{4 x^3} (5 x+25)+64 x^2+e^{2 x^3} \left (-10 x^2-90 x-200\right )+275 x+400} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {15 x^2+e^{4 x^3} \left (60 x^3+300 x^2+5\right )+e^{2 x^3} \left (-60 x^4-540 x^3-1200 x^2-20 x-90\right )+128 x+275}{(x+5) \left (-10 e^{2 x^3} x-40 e^{2 x^3}+5 e^{4 x^3}+5 x^2+39 x+80\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {12 x^3+60 x^2+1}{x+5}+\frac {-60 x^4+60 e^{2 x^3} x^3-468 x^3-10 e^{2 x^3}-960 x^2+240 e^{2 x^3} x^2+10 x+39}{-10 e^{2 x^3} x-40 e^{2 x^3}+5 e^{4 x^3}+5 x^2+39 x+80}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 39 \int \frac {1}{5 x^2-10 e^{2 x^3} x+39 x-40 e^{2 x^3}+5 e^{4 x^3}+80}dx-10 \int \frac {e^{2 x^3}}{5 x^2-10 e^{2 x^3} x+39 x-40 e^{2 x^3}+5 e^{4 x^3}+80}dx+10 \int \frac {x}{5 x^2-10 e^{2 x^3} x+39 x-40 e^{2 x^3}+5 e^{4 x^3}+80}dx-960 \int \frac {x^2}{5 x^2-10 e^{2 x^3} x+39 x-40 e^{2 x^3}+5 e^{4 x^3}+80}dx+240 \int \frac {e^{2 x^3} x^2}{5 x^2-10 e^{2 x^3} x+39 x-40 e^{2 x^3}+5 e^{4 x^3}+80}dx-468 \int \frac {x^3}{5 x^2-10 e^{2 x^3} x+39 x-40 e^{2 x^3}+5 e^{4 x^3}+80}dx+60 \int \frac {e^{2 x^3} x^3}{5 x^2-10 e^{2 x^3} x+39 x-40 e^{2 x^3}+5 e^{4 x^3}+80}dx-60 \int \frac {x^4}{5 x^2-10 e^{2 x^3} x+39 x-40 e^{2 x^3}+5 e^{4 x^3}+80}dx+4 x^3+\log (x+5)\) |
Int[(275 + 128*x + 15*x^2 + E^(4*x^3)*(5 + 300*x^2 + 60*x^3) + E^(2*x^3)*( -90 - 20*x - 1200*x^2 - 540*x^3 - 60*x^4))/(400 + 275*x + 64*x^2 + 5*x^3 + E^(4*x^3)*(25 + 5*x) + E^(2*x^3)*(-200 - 90*x - 10*x^2)),x]
3.24.77.3.1 Defintions of rubi rules used
Time = 0.14 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32
method | result | size |
risch | \(\ln \left (5+x \right )+\ln \left ({\mathrm e}^{4 x^{3}}+\left (-2 x -8\right ) {\mathrm e}^{2 x^{3}}+x^{2}+\frac {39 x}{5}+16\right )\) | \(33\) |
parallelrisch | \(\ln \left (5+x \right )+\ln \left ({\mathrm e}^{4 x^{3}}-2 \,{\mathrm e}^{2 x^{3}} x -8 \,{\mathrm e}^{2 x^{3}}+x^{2}+\frac {39 x}{5}+16\right )\) | \(38\) |
norman | \(\ln \left (5+x \right )+\ln \left (5 \,{\mathrm e}^{4 x^{3}}-10 \,{\mathrm e}^{2 x^{3}} x +5 x^{2}-40 \,{\mathrm e}^{2 x^{3}}+39 x +80\right )\) | \(42\) |
int(((60*x^3+300*x^2+5)*exp(x^3)^4+(-60*x^4-540*x^3-1200*x^2-20*x-90)*exp( x^3)^2+15*x^2+128*x+275)/((25+5*x)*exp(x^3)^4+(-10*x^2-90*x-200)*exp(x^3)^ 2+5*x^3+64*x^2+275*x+400),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {275+128 x+15 x^2+e^{4 x^3} \left (5+300 x^2+60 x^3\right )+e^{2 x^3} \left (-90-20 x-1200 x^2-540 x^3-60 x^4\right )}{400+275 x+64 x^2+5 x^3+e^{4 x^3} (25+5 x)+e^{2 x^3} \left (-200-90 x-10 x^2\right )} \, dx=\log \left (5 \, x^{2} - 10 \, {\left (x + 4\right )} e^{\left (2 \, x^{3}\right )} + 39 \, x + 5 \, e^{\left (4 \, x^{3}\right )} + 80\right ) + \log \left (x + 5\right ) \]
integrate(((60*x^3+300*x^2+5)*exp(x^3)^4+(-60*x^4-540*x^3-1200*x^2-20*x-90 )*exp(x^3)^2+15*x^2+128*x+275)/((25+5*x)*exp(x^3)^4+(-10*x^2-90*x-200)*exp (x^3)^2+5*x^3+64*x^2+275*x+400),x, algorithm=\
Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {275+128 x+15 x^2+e^{4 x^3} \left (5+300 x^2+60 x^3\right )+e^{2 x^3} \left (-90-20 x-1200 x^2-540 x^3-60 x^4\right )}{400+275 x+64 x^2+5 x^3+e^{4 x^3} (25+5 x)+e^{2 x^3} \left (-200-90 x-10 x^2\right )} \, dx=\log {\left (x + 5 \right )} + \log {\left (x^{2} + \frac {39 x}{5} + \left (- 2 x - 8\right ) e^{2 x^{3}} + e^{4 x^{3}} + 16 \right )} \]
integrate(((60*x**3+300*x**2+5)*exp(x**3)**4+(-60*x**4-540*x**3-1200*x**2- 20*x-90)*exp(x**3)**2+15*x**2+128*x+275)/((25+5*x)*exp(x**3)**4+(-10*x**2- 90*x-200)*exp(x**3)**2+5*x**3+64*x**2+275*x+400),x)
Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {275+128 x+15 x^2+e^{4 x^3} \left (5+300 x^2+60 x^3\right )+e^{2 x^3} \left (-90-20 x-1200 x^2-540 x^3-60 x^4\right )}{400+275 x+64 x^2+5 x^3+e^{4 x^3} (25+5 x)+e^{2 x^3} \left (-200-90 x-10 x^2\right )} \, dx=\log \left (x^{2} - 2 \, {\left (x + 4\right )} e^{\left (2 \, x^{3}\right )} + \frac {39}{5} \, x + e^{\left (4 \, x^{3}\right )} + 16\right ) + \log \left (x + 5\right ) \]
integrate(((60*x^3+300*x^2+5)*exp(x^3)^4+(-60*x^4-540*x^3-1200*x^2-20*x-90 )*exp(x^3)^2+15*x^2+128*x+275)/((25+5*x)*exp(x^3)^4+(-10*x^2-90*x-200)*exp (x^3)^2+5*x^3+64*x^2+275*x+400),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {275+128 x+15 x^2+e^{4 x^3} \left (5+300 x^2+60 x^3\right )+e^{2 x^3} \left (-90-20 x-1200 x^2-540 x^3-60 x^4\right )}{400+275 x+64 x^2+5 x^3+e^{4 x^3} (25+5 x)+e^{2 x^3} \left (-200-90 x-10 x^2\right )} \, dx=\log \left (5 \, x^{2} - 10 \, x e^{\left (2 \, x^{3}\right )} + 39 \, x + 5 \, e^{\left (4 \, x^{3}\right )} - 40 \, e^{\left (2 \, x^{3}\right )} + 80\right ) + \log \left (x + 5\right ) \]
integrate(((60*x^3+300*x^2+5)*exp(x^3)^4+(-60*x^4-540*x^3-1200*x^2-20*x-90 )*exp(x^3)^2+15*x^2+128*x+275)/((25+5*x)*exp(x^3)^4+(-10*x^2-90*x-200)*exp (x^3)^2+5*x^3+64*x^2+275*x+400),x, algorithm=\
Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {275+128 x+15 x^2+e^{4 x^3} \left (5+300 x^2+60 x^3\right )+e^{2 x^3} \left (-90-20 x-1200 x^2-540 x^3-60 x^4\right )}{400+275 x+64 x^2+5 x^3+e^{4 x^3} (25+5 x)+e^{2 x^3} \left (-200-90 x-10 x^2\right )} \, dx=\ln \left (x+5\right )+\ln \left (\frac {39\,x}{5}-8\,{\mathrm {e}}^{2\,x^3}+{\mathrm {e}}^{4\,x^3}-2\,x\,{\mathrm {e}}^{2\,x^3}+x^2+16\right ) \]