Integrand size = 78, antiderivative size = 27 \[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=\frac {5 \left (-2-e^{\left (\frac {3+2 x}{x}\right )^{5 x}}+x\right )}{2 x} \] Output:
5/2*(x-2-exp(exp(5*x*ln((3+2*x)/x))))/x
Time = 0.47 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=-\frac {5 \left (2+e^{\left (2+\frac {3}{x}\right )^{5 x}}\right )}{2 x} \] Input:
Integrate[(30 + 20*x + E^((3 + 2*x)/x)^(5*x)*(15 + 10*x + ((3 + 2*x)/x)^(5 *x)*(75*x + (-75*x - 50*x^2)*Log[(3 + 2*x)/x])))/(6*x^2 + 4*x^3),x]
Output:
(-5*(2 + E^(2 + 3/x)^(5*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.
| \(\displaystyle \int \frac {e^{\left (\frac {2 x+3}{x}\right )^{5 x}} \left (\left (\frac {2 x+3}{x}\right )^{5 x} \left (\left (-50 x^2-75 x\right ) \log \left (\frac {2 x+3}{x}\right )+75 x\right )+10 x+15\right )+20 x+30}{4 x^3+6 x^2} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {e^{\left (\frac {2 x+3}{x}\right )^{5 x}} \left (\left (\frac {2 x+3}{x}\right )^{5 x} \left (\left (-50 x^2-75 x\right ) \log \left (\frac {2 x+3}{x}\right )+75 x\right )+10 x+15\right )+20 x+30}{x^2 (4 x+6)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {5 \left (e^{\left (\frac {3}{x}+2\right )^{5 x}}+2\right )}{2 x^2}-\frac {25 e^{\left (\frac {3}{x}+2\right )^{5 x}} \left (\frac {3}{x}+2\right )^{5 x} \left (2 x \log \left (\frac {3}{x}+2\right )+3 \log \left (\frac {3}{x}+2\right )-3\right )}{2 x (2 x+3)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {5}{2} \int \frac {e^{\left (2+\frac {3}{x}\right )^{5 x}}}{x^2}dx+\frac {25}{2} \int \frac {e^{\left (2+\frac {3}{x}\right )^{5 x}} \left (2+\frac {3}{x}\right )^{5 x}}{x}dx-25 \int \frac {e^{\left (2+\frac {3}{x}\right )^{5 x}} \left (2+\frac {3}{x}\right )^{5 x}}{2 x+3}dx-\frac {25}{2} \int \frac {\int \frac {e^{\left (2+\frac {3}{x}\right )^{5 x}} \left (2+\frac {3}{x}\right )^{5 x}}{x}dx}{x}dx+25 \int \frac {\int \frac {e^{\left (2+\frac {3}{x}\right )^{5 x}} \left (2+\frac {3}{x}\right )^{5 x}}{x}dx}{2 x+3}dx-\frac {25}{2} \log \left (\frac {3}{x}+2\right ) \int \frac {e^{\left (2+\frac {3}{x}\right )^{5 x}} \left (2+\frac {3}{x}\right )^{5 x}}{x}dx-\frac {5}{x}\) |
Input:
Int[(30 + 20*x + E^((3 + 2*x)/x)^(5*x)*(15 + 10*x + ((3 + 2*x)/x)^(5*x)*(7 5*x + (-75*x - 50*x^2)*Log[(3 + 2*x)/x])))/(6*x^2 + 4*x^3),x]
Output:
$Aborted
Time = 1.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96
| method | result | size |
| risch | \(-\frac {5}{x}-\frac {5 \,{\mathrm e}^{\left (\frac {3+2 x}{x}\right )^{5 x}}}{2 x}\) | \(26\) |
| parallelrisch | \(\frac {-360+120 x -180 \,{\mathrm e}^{{\mathrm e}^{5 x \ln \left (\frac {3+2 x}{x}\right )}}}{72 x}\) | \(28\) |
Input:
int(((((-50*x^2-75*x)*ln((3+2*x)/x)+75*x)*exp(5*x*ln((3+2*x)/x))+10*x+15)* exp(exp(5*x*ln((3+2*x)/x)))+20*x+30)/(4*x^3+6*x^2),x,method=_RETURNVERBOSE )
Output:
-5/x-5/2/x*exp(((3+2*x)/x)^(5*x))
Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=-\frac {5 \, {\left (e^{\left (\left (\frac {2 \, x + 3}{x}\right )^{5 \, x}\right )} + 2\right )}}{2 \, x} \] Input:
integrate(((((-50*x^2-75*x)*log((3+2*x)/x)+75*x)*exp(5*x*log((3+2*x)/x))+1 0*x+15)*exp(exp(5*x*log((3+2*x)/x)))+20*x+30)/(4*x^3+6*x^2),x, algorithm=" fricas")
Output:
-5/2*(e^(((2*x + 3)/x)^(5*x)) + 2)/x
Time = 0.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=- \frac {5 e^{e^{5 x \log {\left (\frac {2 x + 3}{x} \right )}}}}{2 x} - \frac {5}{x} \] Input:
integrate(((((-50*x**2-75*x)*ln((3+2*x)/x)+75*x)*exp(5*x*ln((3+2*x)/x))+10 *x+15)*exp(exp(5*x*ln((3+2*x)/x)))+20*x+30)/(4*x**3+6*x**2),x)
Output:
-5*exp(exp(5*x*log((2*x + 3)/x)))/(2*x) - 5/x
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=-\frac {5 \, e^{\left (e^{\left (5 \, x \log \left (2 \, x + 3\right ) - 5 \, x \log \left (x\right )\right )}\right )}}{2 \, x} - \frac {5}{x} \] Input:
integrate(((((-50*x^2-75*x)*log((3+2*x)/x)+75*x)*exp(5*x*log((3+2*x)/x))+1 0*x+15)*exp(exp(5*x*log((3+2*x)/x)))+20*x+30)/(4*x^3+6*x^2),x, algorithm=" maxima")
Output:
-5/2*e^(e^(5*x*log(2*x + 3) - 5*x*log(x)))/x - 5/x
\[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=\int { -\frac {5 \, {\left ({\left (5 \, {\left ({\left (2 \, x^{2} + 3 \, x\right )} \log \left (\frac {2 \, x + 3}{x}\right ) - 3 \, x\right )} \left (\frac {2 \, x + 3}{x}\right )^{5 \, x} - 2 \, x - 3\right )} e^{\left (\left (\frac {2 \, x + 3}{x}\right )^{5 \, x}\right )} - 4 \, x - 6\right )}}{2 \, {\left (2 \, x^{3} + 3 \, x^{2}\right )}} \,d x } \] Input:
integrate(((((-50*x^2-75*x)*log((3+2*x)/x)+75*x)*exp(5*x*log((3+2*x)/x))+1 0*x+15)*exp(exp(5*x*log((3+2*x)/x)))+20*x+30)/(4*x^3+6*x^2),x, algorithm=" giac")
Output:
integrate(-5/2*((5*((2*x^2 + 3*x)*log((2*x + 3)/x) - 3*x)*((2*x + 3)/x)^(5 *x) - 2*x - 3)*e^(((2*x + 3)/x)^(5*x)) - 4*x - 6)/(2*x^3 + 3*x^2), x)
Time = 3.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=-\frac {5\,\left ({\mathrm {e}}^{{\left (\frac {3}{x}+2\right )}^{5\,x}}+2\right )}{2\,x} \] Input:
int((20*x + exp(exp(5*x*log((2*x + 3)/x)))*(10*x + exp(5*x*log((2*x + 3)/x ))*(75*x - log((2*x + 3)/x)*(75*x + 50*x^2)) + 15) + 30)/(6*x^2 + 4*x^3),x )
Output:
-(5*(exp((3/x + 2)^(5*x)) + 2))/(2*x)
\[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=\frac {\frac {15 \left (\int \frac {e^{\frac {\left (3+2 x \right )^{5 x}}{x^{5 x}}}}{2 x^{3}+3 x^{2}}d x \right ) x}{2}+5 \left (\int \frac {e^{\frac {\left (3+2 x \right )^{5 x}}{x^{5 x}}}}{2 x^{2}+3 x}d x \right ) x -\frac {75 \left (\int \frac {e^{\frac {\left (3+2 x \right )^{5 x}}{x^{5 x}}} \left (3+2 x \right )^{5 x} \mathrm {log}\left (\frac {3+2 x}{x}\right )}{2 x^{5 x} x^{2}+3 x^{5 x} x}d x \right ) x}{2}-25 \left (\int \frac {e^{\frac {\left (3+2 x \right )^{5 x}}{x^{5 x}}} \left (3+2 x \right )^{5 x} \mathrm {log}\left (\frac {3+2 x}{x}\right )}{2 x^{5 x} x +3 x^{5 x}}d x \right ) x +\frac {75 \left (\int \frac {e^{\frac {\left (3+2 x \right )^{5 x}}{x^{5 x}}} \left (3+2 x \right )^{5 x}}{2 x^{5 x} x^{2}+3 x^{5 x} x}d x \right ) x}{2}-5}{x} \] Input:
int(((((-50*x^2-75*x)*log((3+2*x)/x)+75*x)*exp(5*x*log((3+2*x)/x))+10*x+15 )*exp(exp(5*x*log((3+2*x)/x)))+20*x+30)/(4*x^3+6*x^2),x)
Output:
(5*(3*int(e**((2*x + 3)**(5*x)/x**(5*x))/(2*x**3 + 3*x**2),x)*x + 2*int(e* *((2*x + 3)**(5*x)/x**(5*x))/(2*x**2 + 3*x),x)*x - 15*int((e**((2*x + 3)** (5*x)/x**(5*x))*(2*x + 3)**(5*x)*log((2*x + 3)/x))/(2*x**(5*x)*x**2 + 3*x* *(5*x)*x),x)*x - 10*int((e**((2*x + 3)**(5*x)/x**(5*x))*(2*x + 3)**(5*x)*l og((2*x + 3)/x))/(2*x**(5*x)*x + 3*x**(5*x)),x)*x + 15*int((e**((2*x + 3)* *(5*x)/x**(5*x))*(2*x + 3)**(5*x))/(2*x**(5*x)*x**2 + 3*x**(5*x)*x),x)*x - 2))/(2*x)