Integrand size = 91, antiderivative size = 28 \[ \int \frac {80+200 x^3+125 x^6+e^{\frac {2 \left (-4-5 x^3+5 x \log (x)\right )}{4+5 x^3}} \left (-144+40 x-160 x^3+50 x^4+25 x^6+\left (-160+40 x+400 x^3-100 x^4\right ) \log (x)\right )}{16+40 x^3+25 x^6} \, dx=\left (5+e^{-2+\frac {2 \log (x)}{x \left (\frac {4}{5 x^2}+x\right )}}\right ) (-4+x) \]
Time = 5.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {80+200 x^3+125 x^6+e^{\frac {2 \left (-4-5 x^3+5 x \log (x)\right )}{4+5 x^3}} \left (-144+40 x-160 x^3+50 x^4+25 x^6+\left (-160+40 x+400 x^3-100 x^4\right ) \log (x)\right )}{16+40 x^3+25 x^6} \, dx=5 x+\frac {(-4+x) x^{\frac {10 x}{4+5 x^3}}}{e^2} \]
Integrate[(80 + 200*x^3 + 125*x^6 + E^((2*(-4 - 5*x^3 + 5*x*Log[x]))/(4 + 5*x^3))*(-144 + 40*x - 160*x^3 + 50*x^4 + 25*x^6 + (-160 + 40*x + 400*x^3 - 100*x^4)*Log[x]))/(16 + 40*x^3 + 25*x^6),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.
\(\displaystyle \int \frac {125 x^6+200 x^3+e^{\frac {2 \left (-5 x^3+5 x \log (x)-4\right )}{5 x^3+4}} \left (25 x^6+50 x^4-160 x^3+\left (-100 x^4+400 x^3+40 x-160\right ) \log (x)+40 x-144\right )+80}{25 x^6+40 x^3+16} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle 25 \int \frac {-\frac {\left (-25 x^6-50 x^4+160 x^3-40 x+20 \left (5 x^4-20 x^3-2 x+8\right ) \log (x)+144\right ) x^{\frac {10 x}{5 x^3+4}}}{e^2}+125 x^6+200 x^3+80}{25 \left (5 x^3+4\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {125 x^6+200 x^3-\frac {x^{\frac {10 x}{5 x^3+4}} \left (-25 x^6-50 x^4+160 x^3+20 \left (5 x^4-20 x^3-2 x+8\right ) \log (x)-40 x+144\right )}{e^2}+80}{\left (5 x^3+4\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^{\frac {10 x}{5 x^3+4}} \left (25 x^6+50 x^4-100 x^4 \log (x)-160 x^3+400 x^3 \log (x)+40 x+40 x \log (x)-160 \log (x)-144\right )}{e^2 \left (5 x^3+4\right )^2}+5\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\frac {x^{\frac {10 x}{5 x^3+4}} \left (25 x^6+50 x^4-100 x^4 \log (x)-160 x^3+400 x^3 \log (x)+40 x+40 x \log (x)-160 \log (x)-144\right )}{e^2 \left (5 x^3+4\right )^2}+5\right )dx\) |
Int[(80 + 200*x^3 + 125*x^6 + E^((2*(-4 - 5*x^3 + 5*x*Log[x]))/(4 + 5*x^3) )*(-144 + 40*x - 160*x^3 + 50*x^4 + 25*x^6 + (-160 + 40*x + 400*x^3 - 100* x^4)*Log[x]))/(16 + 40*x^3 + 25*x^6),x]
3.28.98.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 4.50 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96
method | result | size |
risch | \(5 x +\left (x -4\right ) x^{\frac {10 x}{5 x^{3}+4}} {\mathrm e}^{-2}\) | \(27\) |
parallelrisch | \({\mathrm e}^{\frac {10 x \ln \left (x \right )-10 x^{3}-8}{5 x^{3}+4}} x -4 \,{\mathrm e}^{\frac {10 x \ln \left (x \right )-10 x^{3}-8}{5 x^{3}+4}}+5 x\) | \(59\) |
int((((-100*x^4+400*x^3+40*x-160)*ln(x)+25*x^6+50*x^4-160*x^3+40*x-144)*ex p((5*x*ln(x)-5*x^3-4)/(5*x^3+4))^2+125*x^6+200*x^3+80)/(25*x^6+40*x^3+16), x,method=_RETURNVERBOSE)
Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {80+200 x^3+125 x^6+e^{\frac {2 \left (-4-5 x^3+5 x \log (x)\right )}{4+5 x^3}} \left (-144+40 x-160 x^3+50 x^4+25 x^6+\left (-160+40 x+400 x^3-100 x^4\right ) \log (x)\right )}{16+40 x^3+25 x^6} \, dx={\left (x - 4\right )} e^{\left (-\frac {2 \, {\left (5 \, x^{3} - 5 \, x \log \left (x\right ) + 4\right )}}{5 \, x^{3} + 4}\right )} + 5 \, x \]
integrate((((-100*x^4+400*x^3+40*x-160)*log(x)+25*x^6+50*x^4-160*x^3+40*x- 144)*exp((5*x*log(x)-5*x^3-4)/(5*x^3+4))^2+125*x^6+200*x^3+80)/(25*x^6+40* x^3+16),x, algorithm=\
Time = 62.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {80+200 x^3+125 x^6+e^{\frac {2 \left (-4-5 x^3+5 x \log (x)\right )}{4+5 x^3}} \left (-144+40 x-160 x^3+50 x^4+25 x^6+\left (-160+40 x+400 x^3-100 x^4\right ) \log (x)\right )}{16+40 x^3+25 x^6} \, dx=5 x + \left (x - 4\right ) e^{\frac {2 \left (- 5 x^{3} + 5 x \log {\left (x \right )} - 4\right )}{5 x^{3} + 4}} \]
integrate((((-100*x**4+400*x**3+40*x-160)*ln(x)+25*x**6+50*x**4-160*x**3+4 0*x-144)*exp((5*x*ln(x)-5*x**3-4)/(5*x**3+4))**2+125*x**6+200*x**3+80)/(25 *x**6+40*x**3+16),x)
Time = 0.37 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {80+200 x^3+125 x^6+e^{\frac {2 \left (-4-5 x^3+5 x \log (x)\right )}{4+5 x^3}} \left (-144+40 x-160 x^3+50 x^4+25 x^6+\left (-160+40 x+400 x^3-100 x^4\right ) \log (x)\right )}{16+40 x^3+25 x^6} \, dx={\left (x - 4\right )} e^{\left (\frac {10 \, x \log \left (x\right )}{5 \, x^{3} + 4} - 2\right )} + 5 \, x \]
integrate((((-100*x^4+400*x^3+40*x-160)*log(x)+25*x^6+50*x^4-160*x^3+40*x- 144)*exp((5*x*log(x)-5*x^3-4)/(5*x^3+4))^2+125*x^6+200*x^3+80)/(25*x^6+40* x^3+16),x, algorithm=\
\[ \int \frac {80+200 x^3+125 x^6+e^{\frac {2 \left (-4-5 x^3+5 x \log (x)\right )}{4+5 x^3}} \left (-144+40 x-160 x^3+50 x^4+25 x^6+\left (-160+40 x+400 x^3-100 x^4\right ) \log (x)\right )}{16+40 x^3+25 x^6} \, dx=\int { \frac {125 \, x^{6} + 200 \, x^{3} + {\left (25 \, x^{6} + 50 \, x^{4} - 160 \, x^{3} - 20 \, {\left (5 \, x^{4} - 20 \, x^{3} - 2 \, x + 8\right )} \log \left (x\right ) + 40 \, x - 144\right )} e^{\left (-\frac {2 \, {\left (5 \, x^{3} - 5 \, x \log \left (x\right ) + 4\right )}}{5 \, x^{3} + 4}\right )} + 80}{25 \, x^{6} + 40 \, x^{3} + 16} \,d x } \]
integrate((((-100*x^4+400*x^3+40*x-160)*log(x)+25*x^6+50*x^4-160*x^3+40*x- 144)*exp((5*x*log(x)-5*x^3-4)/(5*x^3+4))^2+125*x^6+200*x^3+80)/(25*x^6+40* x^3+16),x, algorithm=\
integrate((125*x^6 + 200*x^3 + (25*x^6 + 50*x^4 - 160*x^3 - 20*(5*x^4 - 20 *x^3 - 2*x + 8)*log(x) + 40*x - 144)*e^(-2*(5*x^3 - 5*x*log(x) + 4)/(5*x^3 + 4)) + 80)/(25*x^6 + 40*x^3 + 16), x)
Time = 14.41 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {80+200 x^3+125 x^6+e^{\frac {2 \left (-4-5 x^3+5 x \log (x)\right )}{4+5 x^3}} \left (-144+40 x-160 x^3+50 x^4+25 x^6+\left (-160+40 x+400 x^3-100 x^4\right ) \log (x)\right )}{16+40 x^3+25 x^6} \, dx=5\,x+x^{\frac {10\,x}{5\,x^3+4}}\,{\mathrm {e}}^{-\frac {10\,x^3}{5\,x^3+4}-\frac {8}{5\,x^3+4}}\,\left (x-4\right ) \]