Integrand size = 85, antiderivative size = 28 \[ \int \frac {-24 x-8 e x-80 x^2+e^{\frac {4+x}{x}} \left (12+23 x+10 x^2+e (4+x)\right )}{192 x+64 e x+e^{\frac {4+x}{x}} (-48 x-16 e x)+e^{\frac {2 (4+x)}{x}} (3 x+e x)} \, dx=3+\frac {x+\frac {5 x^2}{3+e}}{-8+e^{\frac {4+x}{x}}} \]
Time = 0.42 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {-24 x-8 e x-80 x^2+e^{\frac {4+x}{x}} \left (12+23 x+10 x^2+e (4+x)\right )}{192 x+64 e x+e^{\frac {4+x}{x}} (-48 x-16 e x)+e^{\frac {2 (4+x)}{x}} (3 x+e x)} \, dx=\frac {x (3+e+5 x)}{(3+e) \left (-8+e^{1+\frac {4}{x}}\right )} \]
Integrate[(-24*x - 8*E*x - 80*x^2 + E^((4 + x)/x)*(12 + 23*x + 10*x^2 + E* (4 + x)))/(192*x + 64*E*x + E^((4 + x)/x)*(-48*x - 16*E*x) + E^((2*(4 + x) )/x)*(3*x + E*x)),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.
\(\displaystyle \int \frac {-80 x^2+e^{\frac {x+4}{x}} \left (10 x^2+23 x+e (x+4)+12\right )-8 e x-24 x}{64 e x+192 x+e^{\frac {x+4}{x}} (-16 e x-48 x)+e^{\frac {2 (x+4)}{x}} (e x+3 x)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-80 x^2+e^{\frac {x+4}{x}} \left (10 x^2+23 x+e (x+4)+12\right )-8 e x-24 x}{(192+64 e) x+e^{\frac {x+4}{x}} (-16 e x-48 x)+e^{\frac {2 (x+4)}{x}} (e x+3 x)}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-80 x^2+e^{\frac {x+4}{x}} \left (10 x^2+23 x+e (x+4)+12\right )+(-24-8 e) x}{(192+64 e) x+e^{\frac {x+4}{x}} (-16 e x-48 x)+e^{\frac {2 (x+4)}{x}} (e x+3 x)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-80 x^2+e^{\frac {x+4}{x}} \left (10 x^2+23 x+e (x+4)+12\right )+(-24-8 e) x}{(3+e) \left (8-e^{\frac {4}{x}+1}\right )^2 x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {80 x^2+8 (3+e) x-e^{\frac {x+4}{x}} \left (10 x^2+23 x+e (x+4)+12\right )}{\left (8-e^{1+\frac {4}{x}}\right )^2 x}dx}{3+e}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {80 x^2+8 (3+e) x-e^{\frac {x+4}{x}} \left (10 x^2+23 x+e (x+4)+12\right )}{\left (8-e^{1+\frac {4}{x}}\right )^2 x}dx}{3+e}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (\frac {10 x^2+(23+e) x+4 (3+e)}{\left (8-e^{1+\frac {4}{x}}\right ) x}-\frac {32 (5 x+e+3)}{\left (-8+e^{1+\frac {4}{x}}\right )^2 x}\right )dx}{3+e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {32 (3+e) \text {Subst}\left (\int \frac {1}{\left (-8+e^{4 x+1}\right )^2 x}dx,x,\frac {1}{x}\right )+4 (3+e) \text {Subst}\left (\int \frac {1}{\left (-8+e^{4 x+1}\right ) x}dx,x,\frac {1}{x}\right )-160 \int \frac {1}{\left (-8+e^{1+\frac {4}{x}}\right )^2}dx-(23+e) \int \frac {1}{-8+e^{1+\frac {4}{x}}}dx-10 \int \frac {x}{-8+e^{1+\frac {4}{x}}}dx}{3+e}\) |
Int[(-24*x - 8*E*x - 80*x^2 + E^((4 + x)/x)*(12 + 23*x + 10*x^2 + E*(4 + x )))/(192*x + 64*E*x + E^((4 + x)/x)*(-48*x - 16*E*x) + E^((2*(4 + x))/x)*( 3*x + E*x)),x]
3.28.93.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 3.83 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96
method | result | size |
norman | \(\frac {\frac {5 x^{2}}{3+{\mathrm e}}+x}{{\mathrm e}^{\frac {4+x}{x}}-8}\) | \(27\) |
risch | \(\frac {x \left ({\mathrm e}+5 x +3\right )}{\left (3+{\mathrm e}\right ) \left ({\mathrm e}^{\frac {4+x}{x}}-8\right )}\) | \(28\) |
parallelrisch | \(\frac {x \,{\mathrm e}+5 x^{2}+3 x}{\left (3+{\mathrm e}\right ) \left ({\mathrm e}^{\frac {4+x}{x}}-8\right )}\) | \(33\) |
int((((4+x)*exp(1)+10*x^2+23*x+12)*exp((4+x)/x)-8*x*exp(1)-80*x^2-24*x)/(( x*exp(1)+3*x)*exp((4+x)/x)^2+(-16*x*exp(1)-48*x)*exp((4+x)/x)+64*x*exp(1)+ 192*x),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {-24 x-8 e x-80 x^2+e^{\frac {4+x}{x}} \left (12+23 x+10 x^2+e (4+x)\right )}{192 x+64 e x+e^{\frac {4+x}{x}} (-48 x-16 e x)+e^{\frac {2 (4+x)}{x}} (3 x+e x)} \, dx=\frac {5 \, x^{2} + x e + 3 \, x}{{\left (e + 3\right )} e^{\left (\frac {x + 4}{x}\right )} - 8 \, e - 24} \]
integrate((((4+x)*exp(1)+10*x^2+23*x+12)*exp((4+x)/x)-8*x*exp(1)-80*x^2-24 *x)/((x*exp(1)+3*x)*exp((4+x)/x)^2+(-16*x*exp(1)-48*x)*exp((4+x)/x)+64*x*e xp(1)+192*x),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {-24 x-8 e x-80 x^2+e^{\frac {4+x}{x}} \left (12+23 x+10 x^2+e (4+x)\right )}{192 x+64 e x+e^{\frac {4+x}{x}} (-48 x-16 e x)+e^{\frac {2 (4+x)}{x}} (3 x+e x)} \, dx=\frac {5 x^{2} + e x + 3 x}{\left (e + 3\right ) e^{\frac {x + 4}{x}} - 24 - 8 e} \]
integrate((((4+x)*exp(1)+10*x**2+23*x+12)*exp((4+x)/x)-8*x*exp(1)-80*x**2- 24*x)/((x*exp(1)+3*x)*exp((4+x)/x)**2+(-16*x*exp(1)-48*x)*exp((4+x)/x)+64* x*exp(1)+192*x),x)
Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {-24 x-8 e x-80 x^2+e^{\frac {4+x}{x}} \left (12+23 x+10 x^2+e (4+x)\right )}{192 x+64 e x+e^{\frac {4+x}{x}} (-48 x-16 e x)+e^{\frac {2 (4+x)}{x}} (3 x+e x)} \, dx=\frac {5 \, x^{2} + x {\left (e + 3\right )}}{{\left (e^{2} + 3 \, e\right )} e^{\frac {4}{x}} - 8 \, e - 24} \]
integrate((((4+x)*exp(1)+10*x^2+23*x+12)*exp((4+x)/x)-8*x*exp(1)-80*x^2-24 *x)/((x*exp(1)+3*x)*exp((4+x)/x)^2+(-16*x*exp(1)-48*x)*exp((4+x)/x)+64*x*e xp(1)+192*x),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.00 \[ \int \frac {-24 x-8 e x-80 x^2+e^{\frac {4+x}{x}} \left (12+23 x+10 x^2+e (4+x)\right )}{192 x+64 e x+e^{\frac {4+x}{x}} (-48 x-16 e x)+e^{\frac {2 (4+x)}{x}} (3 x+e x)} \, dx=-\frac {\frac {e}{x} + \frac {3}{x} + 5}{\frac {8 \, e}{x^{2}} - \frac {e^{\left (\frac {4}{x} + 2\right )}}{x^{2}} - \frac {3 \, e^{\left (\frac {4}{x} + 1\right )}}{x^{2}} + \frac {24}{x^{2}}} \]
integrate((((4+x)*exp(1)+10*x^2+23*x+12)*exp((4+x)/x)-8*x*exp(1)-80*x^2-24 *x)/((x*exp(1)+3*x)*exp((4+x)/x)^2+(-16*x*exp(1)-48*x)*exp((4+x)/x)+64*x*e xp(1)+192*x),x, algorithm=\
Time = 9.38 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-24 x-8 e x-80 x^2+e^{\frac {4+x}{x}} \left (12+23 x+10 x^2+e (4+x)\right )}{192 x+64 e x+e^{\frac {4+x}{x}} (-48 x-16 e x)+e^{\frac {2 (4+x)}{x}} (3 x+e x)} \, dx=\frac {x\,\left (5\,x+\mathrm {e}+3\right )}{\left ({\mathrm {e}}^{\frac {4}{x}+1}-8\right )\,\left (\mathrm {e}+3\right )} \]