Integrand size = 105, antiderivative size = 31 \[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=\log \left (\frac {3}{x+x^2-\left (e^{x^2+\frac {5+x}{x}}+5 x\right )^2}\right ) \] Output:
ln(3/(x-(5*x+exp(x^2+1/x*(5+x)))^2+x^2))
\[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=\int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx \] Input:
Integrate[(x^2 - 48*x^3 + E^((2*(5 + x + x^3))/x)*(10 - 4*x^3) + E^((5 + x + x^3)/x)*(50*x - 10*x^2 - 20*x^4))/(E^((2*(5 + x + x^3))/x)*x^2 - x^3 + 10*E^((5 + x + x^3)/x)*x^3 + 24*x^4),x]
Output:
Integrate[(x^2 - 48*x^3 + E^((2*(5 + x + x^3))/x)*(10 - 4*x^3) + E^((5 + x + x^3)/x)*(50*x - 10*x^2 - 20*x^4))/(E^((2*(5 + x + x^3))/x)*x^2 - x^3 + 10*E^((5 + x + x^3)/x)*x^3 + 24*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.
| \(\displaystyle \int \frac {-48 x^3+e^{\frac {2 \left (x^3+x+5\right )}{x}} \left (10-4 x^3\right )+x^2+e^{\frac {x^3+x+5}{x}} \left (-20 x^4-10 x^2+50 x\right )}{24 x^4+10 e^{\frac {x^3+x+5}{x}} x^3-x^3+e^{\frac {2 \left (x^3+x+5\right )}{x}} x^2} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {96 x^4-4 x^3-48 x^2-10 e^{x^2+\frac {5}{x}+1} x-50 e^{x^2+\frac {5}{x}+1}+20 e^{x^2+\frac {5}{x}+1} x^3-239 x+10}{x \left (24 x^2+10 e^{x^2+\frac {5}{x}+1} x+e^{2 x^2+\frac {10}{x}+2}-x\right )}-\frac {2 \left (2 x^3-5\right )}{x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -239 \int \frac {1}{24 x^2+10 e^{x^2+1+\frac {5}{x}} x-x+e^{2 x^2+2+\frac {10}{x}}}dx-10 \int \frac {e^{x^2+1+\frac {5}{x}}}{24 x^2+10 e^{x^2+1+\frac {5}{x}} x-x+e^{2 x^2+2+\frac {10}{x}}}dx+10 \int \frac {1}{x \left (24 x^2+10 e^{x^2+1+\frac {5}{x}} x-x+e^{2 x^2+2+\frac {10}{x}}\right )}dx-50 \int \frac {e^{x^2+1+\frac {5}{x}}}{x \left (24 x^2+10 e^{x^2+1+\frac {5}{x}} x-x+e^{2 x^2+2+\frac {10}{x}}\right )}dx-48 \int \frac {x}{24 x^2+10 e^{x^2+1+\frac {5}{x}} x-x+e^{2 x^2+2+\frac {10}{x}}}dx-4 \int \frac {x^2}{24 x^2+10 e^{x^2+1+\frac {5}{x}} x-x+e^{2 x^2+2+\frac {10}{x}}}dx+20 \int \frac {e^{x^2+1+\frac {5}{x}} x^2}{24 x^2+10 e^{x^2+1+\frac {5}{x}} x-x+e^{2 x^2+2+\frac {10}{x}}}dx+96 \int \frac {x^3}{24 x^2+10 e^{x^2+1+\frac {5}{x}} x-x+e^{2 x^2+2+\frac {10}{x}}}dx-2 x^2-\frac {10}{x}\) |
Input:
Int[(x^2 - 48*x^3 + E^((2*(5 + x + x^3))/x)*(10 - 4*x^3) + E^((5 + x + x^3 )/x)*(50*x - 10*x^2 - 20*x^4))/(E^((2*(5 + x + x^3))/x)*x^2 - x^3 + 10*E^( (5 + x + x^3)/x)*x^3 + 24*x^4),x]
Output:
$Aborted
Time = 0.36 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29
| method | result | size |
| norman | \(-\ln \left (10 x \,{\mathrm e}^{\frac {x^{3}+x +5}{x}}+24 x^{2}+{\mathrm e}^{\frac {2 x^{3}+2 x +10}{x}}-x \right )\) | \(40\) |
| parallelrisch | \(-\ln \left (x^{2}+\frac {5 x \,{\mathrm e}^{\frac {x^{3}+x +5}{x}}}{12}+\frac {{\mathrm e}^{\frac {2 x^{3}+2 x +10}{x}}}{24}-\frac {x}{24}\right )\) | \(40\) |
| risch | \(-2 x^{2}-\frac {10}{x}+\frac {2 x^{3}+2 x +10}{x}-\ln \left (10 x \,{\mathrm e}^{\frac {x^{3}+x +5}{x}}+24 x^{2}+{\mathrm e}^{\frac {2 x^{3}+2 x +10}{x}}-x \right )\) | \(61\) |
Input:
int(((-4*x^3+10)*exp((x^3+x+5)/x)^2+(-20*x^4-10*x^2+50*x)*exp((x^3+x+5)/x) -48*x^3+x^2)/(x^2*exp((x^3+x+5)/x)^2+10*x^3*exp((x^3+x+5)/x)+24*x^4-x^3),x ,method=_RETURNVERBOSE)
Output:
-ln(exp((x^3+x+5)/x)^2+10*x*exp((x^3+x+5)/x)+24*x^2-x)
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=-\log \left (24 \, x^{2} + 10 \, x e^{\left (\frac {x^{3} + x + 5}{x}\right )} - x + e^{\left (\frac {2 \, {\left (x^{3} + x + 5\right )}}{x}\right )}\right ) \] Input:
integrate(((-4*x^3+10)*exp((x^3+x+5)/x)^2+(-20*x^4-10*x^2+50*x)*exp((x^3+x +5)/x)-48*x^3+x^2)/(x^2*exp((x^3+x+5)/x)^2+10*x^3*exp((x^3+x+5)/x)+24*x^4- x^3),x, algorithm="fricas")
Output:
-log(24*x^2 + 10*x*e^((x^3 + x + 5)/x) - x + e^(2*(x^3 + x + 5)/x))
Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=- \log {\left (24 x^{2} + 10 x e^{\frac {x^{3} + x + 5}{x}} - x + e^{\frac {2 \left (x^{3} + x + 5\right )}{x}} \right )} \] Input:
integrate(((-4*x**3+10)*exp((x**3+x+5)/x)**2+(-20*x**4-10*x**2+50*x)*exp(( x**3+x+5)/x)-48*x**3+x**2)/(x**2*exp((x**3+x+5)/x)**2+10*x**3*exp((x**3+x+ 5)/x)+24*x**4-x**3),x)
Output:
-log(24*x**2 + 10*x*exp((x**3 + x + 5)/x) - x + exp(2*(x**3 + x + 5)/x))
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74 \[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=-\frac {10}{x} - \log \left ({\left (24 \, x^{2} + 10 \, x e^{\left (x^{2} + \frac {5}{x} + 1\right )} - x + e^{\left (2 \, x^{2} + \frac {10}{x} + 2\right )}\right )} e^{\left (-\frac {10}{x} - 2\right )}\right ) \] Input:
integrate(((-4*x^3+10)*exp((x^3+x+5)/x)^2+(-20*x^4-10*x^2+50*x)*exp((x^3+x +5)/x)-48*x^3+x^2)/(x^2*exp((x^3+x+5)/x)^2+10*x^3*exp((x^3+x+5)/x)+24*x^4- x^3),x, algorithm="maxima")
Output:
-10/x - log((24*x^2 + 10*x*e^(x^2 + 5/x + 1) - x + e^(2*x^2 + 10/x + 2))*e ^(-10/x - 2))
Time = 0.17 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=-\log \left (24 \, x^{2} + 10 \, x e^{\left (\frac {x^{3} + x + 5}{x}\right )} - x + e^{\left (\frac {2 \, {\left (x^{3} + x + 5\right )}}{x}\right )}\right ) \] Input:
integrate(((-4*x^3+10)*exp((x^3+x+5)/x)^2+(-20*x^4-10*x^2+50*x)*exp((x^3+x +5)/x)-48*x^3+x^2)/(x^2*exp((x^3+x+5)/x)^2+10*x^3*exp((x^3+x+5)/x)+24*x^4- x^3),x, algorithm="giac")
Output:
-log(24*x^2 + 10*x*e^((x^3 + x + 5)/x) - x + e^(2*(x^3 + x + 5)/x))
Timed out. \[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=\int -\frac {{\mathrm {e}}^{\frac {2\,\left (x^3+x+5\right )}{x}}\,\left (4\,x^3-10\right )+{\mathrm {e}}^{\frac {x^3+x+5}{x}}\,\left (20\,x^4+10\,x^2-50\,x\right )-x^2+48\,x^3}{10\,x^3\,{\mathrm {e}}^{\frac {x^3+x+5}{x}}+x^2\,{\mathrm {e}}^{\frac {2\,\left (x^3+x+5\right )}{x}}-x^3+24\,x^4} \,d x \] Input:
int(-(exp((2*(x + x^3 + 5))/x)*(4*x^3 - 10) + exp((x + x^3 + 5)/x)*(10*x^2 - 50*x + 20*x^4) - x^2 + 48*x^3)/(10*x^3*exp((x + x^3 + 5)/x) + x^2*exp(( 2*(x + x^3 + 5))/x) - x^3 + 24*x^4),x)
Output:
int(-(exp((2*(x + x^3 + 5))/x)*(4*x^3 - 10) + exp((x + x^3 + 5)/x)*(10*x^2 - 50*x + 20*x^4) - x^2 + 48*x^3)/(10*x^3*exp((x + x^3 + 5)/x) + x^2*exp(( 2*(x + x^3 + 5))/x) - x^3 + 24*x^4), x)
\[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=50 \left (\int \frac {e^{\frac {x^{3}+5}{x}}}{e^{\frac {2 x^{3}+10}{x}} e^{2} x +10 e^{\frac {x^{3}+5}{x}} e \,x^{2}+24 x^{3}-x^{2}}d x \right ) e -10 \left (\int \frac {e^{\frac {x^{3}+5}{x}}}{e^{\frac {2 x^{3}+10}{x}} e^{2}+10 e^{\frac {x^{3}+5}{x}} e x +24 x^{2}-x}d x \right ) e +10 \left (\int \frac {e^{\frac {2 x^{3}+10}{x}}}{e^{\frac {2 x^{3}+10}{x}} e^{2} x^{2}+10 e^{\frac {x^{3}+5}{x}} e \,x^{3}+24 x^{4}-x^{3}}d x \right ) e^{2}-20 \left (\int \frac {e^{\frac {x^{3}+5}{x}} x^{2}}{e^{\frac {2 x^{3}+10}{x}} e^{2}+10 e^{\frac {x^{3}+5}{x}} e x +24 x^{2}-x}d x \right ) e -4 \left (\int \frac {e^{\frac {2 x^{3}+10}{x}} x}{e^{\frac {2 x^{3}+10}{x}} e^{2}+10 e^{\frac {x^{3}+5}{x}} e x +24 x^{2}-x}d x \right ) e^{2}-48 \left (\int \frac {x}{e^{\frac {2 x^{3}+10}{x}} e^{2}+10 e^{\frac {x^{3}+5}{x}} e x +24 x^{2}-x}d x \right )+\int \frac {1}{e^{\frac {2 x^{3}+10}{x}} e^{2}+10 e^{\frac {x^{3}+5}{x}} e x +24 x^{2}-x}d x \] Input:
int(((-4*x^3+10)*exp((x^3+x+5)/x)^2+(-20*x^4-10*x^2+50*x)*exp((x^3+x+5)/x) -48*x^3+x^2)/(x^2*exp((x^3+x+5)/x)^2+10*x^3*exp((x^3+x+5)/x)+24*x^4-x^3),x )
Output:
50*int(e**((x**3 + 5)/x)/(e**((2*x**3 + 10)/x)*e**2*x + 10*e**((x**3 + 5)/ x)*e*x**2 + 24*x**3 - x**2),x)*e - 10*int(e**((x**3 + 5)/x)/(e**((2*x**3 + 10)/x)*e**2 + 10*e**((x**3 + 5)/x)*e*x + 24*x**2 - x),x)*e + 10*int(e**(( 2*x**3 + 10)/x)/(e**((2*x**3 + 10)/x)*e**2*x**2 + 10*e**((x**3 + 5)/x)*e*x **3 + 24*x**4 - x**3),x)*e**2 - 20*int((e**((x**3 + 5)/x)*x**2)/(e**((2*x* *3 + 10)/x)*e**2 + 10*e**((x**3 + 5)/x)*e*x + 24*x**2 - x),x)*e - 4*int((e **((2*x**3 + 10)/x)*x)/(e**((2*x**3 + 10)/x)*e**2 + 10*e**((x**3 + 5)/x)*e *x + 24*x**2 - x),x)*e**2 - 48*int(x/(e**((2*x**3 + 10)/x)*e**2 + 10*e**(( x**3 + 5)/x)*e*x + 24*x**2 - x),x) + int(1/(e**((2*x**3 + 10)/x)*e**2 + 10 *e**((x**3 + 5)/x)*e*x + 24*x**2 - x),x)