Integrand size = 78, antiderivative size = 27 \[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=\frac {5 \left (-2-e^{\left (\frac {3+2 x}{x}\right )^{5 x}}+x\right )}{2 x} \]
Time = 0.74 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=-\frac {5 \left (2+e^{\left (2+\frac {3}{x}\right )^{5 x}}\right )}{2 x} \]
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]
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 |
\(\displaystyle \frac {5}{2} \int \frac {e^{\left (2+\frac {3}{x}\right )^{5 x}}}{x^2}dx+\frac {25}{2} \int \frac {e^{\left (2+\frac {3}{x}\right )^{5 x}} \left (2+\frac {3}{x}\right )^{5 x}}{x}dx-25 \int \frac {e^{\left (2+\frac {3}{x}\right )^{5 x}} \left (2+\frac {3}{x}\right )^{5 x}}{2 x+3}dx-\frac {25}{2} \int \frac {\int \frac {e^{\left (2+\frac {3}{x}\right )^{5 x}} \left (2+\frac {3}{x}\right )^{5 x}}{x}dx}{x}dx+25 \int \frac {\int \frac {e^{\left (2+\frac {3}{x}\right )^{5 x}} \left (2+\frac {3}{x}\right )^{5 x}}{x}dx}{2 x+3}dx-\frac {25}{2} \log \left (\frac {3}{x}+2\right ) \int \frac {e^{\left (2+\frac {3}{x}\right )^{5 x}} \left (2+\frac {3}{x}\right )^{5 x}}{x}dx-\frac {5}{x}\) |
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]
3.9.15.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 0.73 (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 | \(\frac {-360+120 x -180 \,{\mathrm e}^{{\mathrm e}^{5 x \ln \left (\frac {3+2 x}{x}\right )}}}{72 x}\) | \(28\) |
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 )
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=-\frac {5 \, {\left (e^{\left (\left (\frac {2 \, x + 3}{x}\right )^{5 \, x}\right )} + 2\right )}}{2 \, x} \]
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=\
Time = 0.39 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=- \frac {5 e^{e^{5 x \log {\left (\frac {2 x + 3}{x} \right )}}}}{2 x} - \frac {5}{x} \]
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)
Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=-\frac {5 \, e^{\left (e^{\left (5 \, x \log \left (2 \, x + 3\right ) - 5 \, x \log \left (x\right )\right )}\right )}}{2 \, x} - \frac {5}{x} \]
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=\
\[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=\int { -\frac {5 \, {\left ({\left (5 \, {\left ({\left (2 \, x^{2} + 3 \, x\right )} \log \left (\frac {2 \, x + 3}{x}\right ) - 3 \, x\right )} \left (\frac {2 \, x + 3}{x}\right )^{5 \, x} - 2 \, x - 3\right )} e^{\left (\left (\frac {2 \, x + 3}{x}\right )^{5 \, x}\right )} - 4 \, x - 6\right )}}{2 \, {\left (2 \, x^{3} + 3 \, x^{2}\right )}} \,d x } \]
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=\
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 = 13.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {30+20 x+e^{\left (\frac {3+2 x}{x}\right )^{5 x}} \left (15+10 x+\left (\frac {3+2 x}{x}\right )^{5 x} \left (75 x+\left (-75 x-50 x^2\right ) \log \left (\frac {3+2 x}{x}\right )\right )\right )}{6 x^2+4 x^3} \, dx=-\frac {5\,\left ({\mathrm {e}}^{{\left (\frac {3}{x}+2\right )}^{5\,x}}+2\right )}{2\,x} \]