Integrand size = 152, antiderivative size = 35 \[ \int \frac {e^{3/x} (240-95 x)-2240 x^2+310 x^3-10 x^4+e^5 \left (1280 x-160 x^2+5 x^3\right )}{3 e^{6/x} x^3+588 x^5-84 x^6+3 x^7+e^{10} \left (768 x^3-96 x^4+3 x^5\right )+e^5 \left (-1344 x^4+180 x^5-6 x^6\right )+e^{3/x} \left (84 x^4-6 x^5+e^5 \left (-96 x^3+6 x^4\right )\right )} \, dx=\frac {5}{3 x \left (-e^5+\frac {e^{3/x}-2 x}{16-x}+x\right )} \]
Time = 0.82 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {e^{3/x} (240-95 x)-2240 x^2+310 x^3-10 x^4+e^5 \left (1280 x-160 x^2+5 x^3\right )}{3 e^{6/x} x^3+588 x^5-84 x^6+3 x^7+e^{10} \left (768 x^3-96 x^4+3 x^5\right )+e^5 \left (-1344 x^4+180 x^5-6 x^6\right )+e^{3/x} \left (84 x^4-6 x^5+e^5 \left (-96 x^3+6 x^4\right )\right )} \, dx=-\frac {5 (-16+x)}{3 x \left (e^{3/x}+e^5 (-16+x)-(-14+x) x\right )} \]
Integrate[(E^(3/x)*(240 - 95*x) - 2240*x^2 + 310*x^3 - 10*x^4 + E^5*(1280* x - 160*x^2 + 5*x^3))/(3*E^(6/x)*x^3 + 588*x^5 - 84*x^6 + 3*x^7 + E^10*(76 8*x^3 - 96*x^4 + 3*x^5) + E^5*(-1344*x^4 + 180*x^5 - 6*x^6) + E^(3/x)*(84* x^4 - 6*x^5 + E^5*(-96*x^3 + 6*x^4))),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.
\(\displaystyle \int \frac {-10 x^4+310 x^3-2240 x^2+e^5 \left (5 x^3-160 x^2+1280 x\right )+e^{3/x} (240-95 x)}{3 x^7-84 x^6+588 x^5+3 e^{6/x} x^3+e^5 \left (-6 x^6+180 x^5-1344 x^4\right )+e^{10} \left (3 x^5-96 x^4+768 x^3\right )+e^{3/x} \left (-6 x^5+84 x^4+e^5 \left (6 x^4-96 x^3\right )\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {5 \left (-2 x^2 \left (x^2-31 x+224\right )+e^5 x (x-16)^2-e^{3/x} (19 x-48)\right )}{3 x^3 \left (e^5 (x-16)+e^{3/x}-(x-14) x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{3} \int \frac {e^5 x (16-x)^2+e^{3/x} (48-19 x)-2 x^2 \left (x^2-31 x+224\right )}{x^3 \left (-e^5 (16-x)+e^{3/x}+(14-x) x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {5}{3} \int \left (\frac {19 x-48}{x^3 \left (x^2-14 \left (1+\frac {e^5}{14}\right ) x-e^{3/x}+16 e^5\right )}+\frac {(16-x) \left (2 x^3-\left (11+e^5\right ) x^2-3 \left (14+e^5\right ) x+48 e^5\right )}{x^3 \left (x^2-14 \left (1+\frac {e^5}{14}\right ) x-e^{3/x}+16 e^5\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5}{3} \left (\left (43+e^5\right ) \int \frac {1}{\left (x^2-14 \left (1+\frac {e^5}{14}\right ) x-e^{3/x}+16 e^5\right )^2}dx-96 \left (7+e^5\right ) \int \frac {1}{x^2 \left (x^2-14 \left (1+\frac {e^5}{14}\right ) x-e^{3/x}+16 e^5\right )^2}dx-\left (134+13 e^5\right ) \int \frac {1}{x \left (x^2-14 \left (1+\frac {e^5}{14}\right ) x-e^{3/x}+16 e^5\right )^2}dx-2 \int \frac {x}{\left (x^2-14 \left (1+\frac {e^5}{14}\right ) x-e^{3/x}+16 e^5\right )^2}dx+19 \int \frac {1}{x^2 \left (x^2-14 \left (1+\frac {e^5}{14}\right ) x-e^{3/x}+16 e^5\right )}dx+48 \int \frac {1}{x^3 \left (-x^2+14 \left (1+\frac {e^5}{14}\right ) x+e^{3/x}-16 e^5\right )}dx+768 e^5 \int \frac {1}{x^3 \left (x^2-14 \left (1+\frac {e^5}{14}\right ) x-e^{3/x}+16 e^5\right )^2}dx\right )\) |
Int[(E^(3/x)*(240 - 95*x) - 2240*x^2 + 310*x^3 - 10*x^4 + E^5*(1280*x - 16 0*x^2 + 5*x^3))/(3*E^(6/x)*x^3 + 588*x^5 - 84*x^6 + 3*x^7 + E^10*(768*x^3 - 96*x^4 + 3*x^5) + E^5*(-1344*x^4 + 180*x^5 - 6*x^6) + E^(3/x)*(84*x^4 - 6*x^5 + E^5*(-96*x^3 + 6*x^4))),x]
3.11.73.3.1 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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 5.50 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97
method | result | size |
risch | \(-\frac {5 \left (x -16\right )}{3 x \left (x \,{\mathrm e}^{5}-x^{2}-16 \,{\mathrm e}^{5}+{\mathrm e}^{\frac {3}{x}}+14 x \right )}\) | \(34\) |
parallelrisch | \(\frac {80-5 x}{3 x \left (x \,{\mathrm e}^{5}-x^{2}-16 \,{\mathrm e}^{5}+{\mathrm e}^{\frac {3}{x}}+14 x \right )}\) | \(36\) |
norman | \(\frac {-\frac {5}{3} x^{2}+\frac {80}{3} x}{x^{2} \left (x \,{\mathrm e}^{5}-x^{2}-16 \,{\mathrm e}^{5}+{\mathrm e}^{\frac {3}{x}}+14 x \right )}\) | \(39\) |
int(((-95*x+240)*exp(3/x)+(5*x^3-160*x^2+1280*x)*exp(5)-10*x^4+310*x^3-224 0*x^2)/(3*x^3*exp(3/x)^2+((6*x^4-96*x^3)*exp(5)-6*x^5+84*x^4)*exp(3/x)+(3* x^5-96*x^4+768*x^3)*exp(5)^2+(-6*x^6+180*x^5-1344*x^4)*exp(5)+3*x^7-84*x^6 +588*x^5),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {e^{3/x} (240-95 x)-2240 x^2+310 x^3-10 x^4+e^5 \left (1280 x-160 x^2+5 x^3\right )}{3 e^{6/x} x^3+588 x^5-84 x^6+3 x^7+e^{10} \left (768 x^3-96 x^4+3 x^5\right )+e^5 \left (-1344 x^4+180 x^5-6 x^6\right )+e^{3/x} \left (84 x^4-6 x^5+e^5 \left (-96 x^3+6 x^4\right )\right )} \, dx=\frac {5 \, {\left (x - 16\right )}}{3 \, {\left (x^{3} - 14 \, x^{2} - {\left (x^{2} - 16 \, x\right )} e^{5} - x e^{\frac {3}{x}}\right )}} \]
integrate(((-95*x+240)*exp(3/x)+(5*x^3-160*x^2+1280*x)*exp(5)-10*x^4+310*x ^3-2240*x^2)/(3*x^3*exp(3/x)^2+((6*x^4-96*x^3)*exp(5)-6*x^5+84*x^4)*exp(3/ x)+(3*x^5-96*x^4+768*x^3)*exp(5)^2+(-6*x^6+180*x^5-1344*x^4)*exp(5)+3*x^7- 84*x^6+588*x^5),x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {e^{3/x} (240-95 x)-2240 x^2+310 x^3-10 x^4+e^5 \left (1280 x-160 x^2+5 x^3\right )}{3 e^{6/x} x^3+588 x^5-84 x^6+3 x^7+e^{10} \left (768 x^3-96 x^4+3 x^5\right )+e^5 \left (-1344 x^4+180 x^5-6 x^6\right )+e^{3/x} \left (84 x^4-6 x^5+e^5 \left (-96 x^3+6 x^4\right )\right )} \, dx=\frac {80 - 5 x}{- 3 x^{3} + 42 x^{2} + 3 x^{2} e^{5} + 3 x e^{\frac {3}{x}} - 48 x e^{5}} \]
integrate(((-95*x+240)*exp(3/x)+(5*x**3-160*x**2+1280*x)*exp(5)-10*x**4+31 0*x**3-2240*x**2)/(3*x**3*exp(3/x)**2+((6*x**4-96*x**3)*exp(5)-6*x**5+84*x **4)*exp(3/x)+(3*x**5-96*x**4+768*x**3)*exp(5)**2+(-6*x**6+180*x**5-1344*x **4)*exp(5)+3*x**7-84*x**6+588*x**5),x)
Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int \frac {e^{3/x} (240-95 x)-2240 x^2+310 x^3-10 x^4+e^5 \left (1280 x-160 x^2+5 x^3\right )}{3 e^{6/x} x^3+588 x^5-84 x^6+3 x^7+e^{10} \left (768 x^3-96 x^4+3 x^5\right )+e^5 \left (-1344 x^4+180 x^5-6 x^6\right )+e^{3/x} \left (84 x^4-6 x^5+e^5 \left (-96 x^3+6 x^4\right )\right )} \, dx=\frac {5 \, {\left (x - 16\right )}}{3 \, {\left (x^{3} - x^{2} {\left (e^{5} + 14\right )} + 16 \, x e^{5} - x e^{\frac {3}{x}}\right )}} \]
integrate(((-95*x+240)*exp(3/x)+(5*x^3-160*x^2+1280*x)*exp(5)-10*x^4+310*x ^3-2240*x^2)/(3*x^3*exp(3/x)^2+((6*x^4-96*x^3)*exp(5)-6*x^5+84*x^4)*exp(3/ x)+(3*x^5-96*x^4+768*x^3)*exp(5)^2+(-6*x^6+180*x^5-1344*x^4)*exp(5)+3*x^7- 84*x^6+588*x^5),x, algorithm=\
Time = 0.49 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int \frac {e^{3/x} (240-95 x)-2240 x^2+310 x^3-10 x^4+e^5 \left (1280 x-160 x^2+5 x^3\right )}{3 e^{6/x} x^3+588 x^5-84 x^6+3 x^7+e^{10} \left (768 x^3-96 x^4+3 x^5\right )+e^5 \left (-1344 x^4+180 x^5-6 x^6\right )+e^{3/x} \left (84 x^4-6 x^5+e^5 \left (-96 x^3+6 x^4\right )\right )} \, dx=-\frac {5 \, {\left (\frac {1}{x^{2}} - \frac {16}{x^{3}}\right )}}{3 \, {\left (\frac {e^{5}}{x} + \frac {14}{x} - \frac {16 \, e^{5}}{x^{2}} + \frac {e^{\frac {3}{x}}}{x^{2}} - 1\right )}} \]
integrate(((-95*x+240)*exp(3/x)+(5*x^3-160*x^2+1280*x)*exp(5)-10*x^4+310*x ^3-2240*x^2)/(3*x^3*exp(3/x)^2+((6*x^4-96*x^3)*exp(5)-6*x^5+84*x^4)*exp(3/ x)+(3*x^5-96*x^4+768*x^3)*exp(5)^2+(-6*x^6+180*x^5-1344*x^4)*exp(5)+3*x^7- 84*x^6+588*x^5),x, algorithm=\
Time = 9.91 (sec) , antiderivative size = 106, normalized size of antiderivative = 3.03 \[ \int \frac {e^{3/x} (240-95 x)-2240 x^2+310 x^3-10 x^4+e^5 \left (1280 x-160 x^2+5 x^3\right )}{3 e^{6/x} x^3+588 x^5-84 x^6+3 x^7+e^{10} \left (768 x^3-96 x^4+3 x^5\right )+e^5 \left (-1344 x^4+180 x^5-6 x^6\right )+e^{3/x} \left (84 x^4-6 x^5+e^5 \left (-96 x^3+6 x^4\right )\right )} \, dx=\frac {\frac {10\,x^5}{3}+\left (-\frac {5\,{\mathrm {e}}^5}{3}-\frac {215}{3}\right )\,x^4+\left (\frac {65\,{\mathrm {e}}^5}{3}+\frac {670}{3}\right )\,x^3+\left (160\,{\mathrm {e}}^5+1120\right )\,x^2-1280\,{\mathrm {e}}^5\,x}{\left (14\,x-16\,{\mathrm {e}}^5+{\mathrm {e}}^{3/x}+x\,{\mathrm {e}}^5-x^2\right )\,\left (3\,x^3\,{\mathrm {e}}^5-48\,x^2\,{\mathrm {e}}^5+x^4\,{\mathrm {e}}^5+42\,x^3+11\,x^4-2\,x^5\right )} \]
int(-(exp(3/x)*(95*x - 240) - exp(5)*(1280*x - 160*x^2 + 5*x^3) + 2240*x^2 - 310*x^3 + 10*x^4)/(exp(10)*(768*x^3 - 96*x^4 + 3*x^5) - exp(3/x)*(exp(5 )*(96*x^3 - 6*x^4) - 84*x^4 + 6*x^5) - exp(5)*(1344*x^4 - 180*x^5 + 6*x^6) + 3*x^3*exp(6/x) + 588*x^5 - 84*x^6 + 3*x^7),x)