Integrand size = 66, antiderivative size = 27 \[ \int \frac {4 x^3+2 x^4+e^x (-100+100 x) \log \left (\frac {2+x}{x}\right )+e^x \left (-100+50 x-25 x^3\right ) \log ^2\left (\frac {2+x}{x}\right )}{4 x^3+2 x^4} \, dx=x+\frac {25 e^x (1-x) \log ^2\left (1+\frac {2}{x}\right )}{2 x^2} \] Output:
x+25/2*(1-x)*ln(2/x+1)^2/x^2*exp(x)
\[ \int \frac {4 x^3+2 x^4+e^x (-100+100 x) \log \left (\frac {2+x}{x}\right )+e^x \left (-100+50 x-25 x^3\right ) \log ^2\left (\frac {2+x}{x}\right )}{4 x^3+2 x^4} \, dx=\int \frac {4 x^3+2 x^4+e^x (-100+100 x) \log \left (\frac {2+x}{x}\right )+e^x \left (-100+50 x-25 x^3\right ) \log ^2\left (\frac {2+x}{x}\right )}{4 x^3+2 x^4} \, dx \] Input:
Integrate[(4*x^3 + 2*x^4 + E^x*(-100 + 100*x)*Log[(2 + x)/x] + E^x*(-100 + 50*x - 25*x^3)*Log[(2 + x)/x]^2)/(4*x^3 + 2*x^4),x]
Output:
Integrate[(4*x^3 + 2*x^4 + E^x*(-100 + 100*x)*Log[(2 + x)/x] + E^x*(-100 + 50*x - 25*x^3)*Log[(2 + x)/x]^2)/(4*x^3 + 2*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.
\(\displaystyle \int \frac {2 x^4+4 x^3+e^x \left (-25 x^3+50 x-100\right ) \log ^2\left (\frac {x+2}{x}\right )+e^x (100 x-100) \log \left (\frac {x+2}{x}\right )}{2 x^4+4 x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {2 x^4+4 x^3+e^x \left (-25 x^3+50 x-100\right ) \log ^2\left (\frac {x+2}{x}\right )+e^x (100 x-100) \log \left (\frac {x+2}{x}\right )}{x^3 (2 x+4)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {25 e^x \log \left (\frac {2}{x}+1\right ) \left (x^3 \left (-\log \left (\frac {x+2}{x}\right )\right )+4 x+2 x \log \left (\frac {x+2}{x}\right )-4 \log \left (\frac {x+2}{x}\right )-4\right )}{2 x^3 (x+2)}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {25}{4} \int \frac {\operatorname {ExpIntegralEi}(x)}{x+2}dx+\frac {75 \int \frac {\operatorname {ExpIntegralEi}(x+2)}{x}dx}{4 e^2}-\frac {75 \int \frac {\operatorname {ExpIntegralEi}(x+2)}{x+2}dx}{4 e^2}-25 \int \frac {e^x \log ^2\left (1+\frac {2}{x}\right )}{x^3}dx+25 \int \frac {e^x \log ^2\left (1+\frac {2}{x}\right )}{x^2}dx-\frac {25}{2} \int \frac {e^x \log ^2\left (1+\frac {2}{x}\right )}{x}dx+\frac {25}{4} x \, _3F_3(1,1,1;2,2,2;x)+\frac {25}{4} \log (x) (\operatorname {ExpIntegralE}(1,-x)+\operatorname {ExpIntegralEi}(x))-\frac {75 \operatorname {ExpIntegralEi}(x)}{8}-\frac {125 \operatorname {ExpIntegralEi}(x+2)}{8 e^2}+\frac {25}{4} \operatorname {ExpIntegralEi}(x) \log \left (\frac {2}{x}+1\right )+\frac {75 \operatorname {ExpIntegralEi}(x+2) \log \left (\frac {2}{x}+1\right )}{4 e^2}-\frac {25 e^x}{4 x^2}+\frac {25 e^x \log \left (\frac {2}{x}+1\right )}{2 x^2}+x+\frac {25 e^x}{x}+\frac {25}{8} \log ^2(-x)-\frac {25 e^x \log \left (\frac {2}{x}+1\right )}{x}+\frac {25}{4} \gamma \log (x)\) |
Input:
Int[(4*x^3 + 2*x^4 + E^x*(-100 + 100*x)*Log[(2 + x)/x] + E^x*(-100 + 50*x - 25*x^3)*Log[(2 + x)/x]^2)/(4*x^3 + 2*x^4),x]
Output:
$Aborted
Time = 1.58 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70
method | result | size |
parallelrisch | \(\frac {-100 \,{\mathrm e}^{x} \ln \left (\frac {2+x}{x}\right )^{2} x +8 x^{3}+100 \,{\mathrm e}^{x} \ln \left (\frac {2+x}{x}\right )^{2}-8 x^{2}}{8 x^{2}}\) | \(46\) |
risch | \(\text {Expression too large to display}\) | \(1010\) |
orering | \(\text {Expression too large to display}\) | \(3737\) |
Input:
int(((-25*x^3+50*x-100)*exp(x)*ln((2+x)/x)^2+(100*x-100)*exp(x)*ln((2+x)/x )+2*x^4+4*x^3)/(2*x^4+4*x^3),x,method=_RETURNVERBOSE)
Output:
1/8*(-100*exp(x)*ln((2+x)/x)^2*x+8*x^3+100*exp(x)*ln((2+x)/x)^2-8*x^2)/x^2
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {4 x^3+2 x^4+e^x (-100+100 x) \log \left (\frac {2+x}{x}\right )+e^x \left (-100+50 x-25 x^3\right ) \log ^2\left (\frac {2+x}{x}\right )}{4 x^3+2 x^4} \, dx=-\frac {25 \, {\left (x - 1\right )} e^{x} \log \left (\frac {x + 2}{x}\right )^{2} - 2 \, x^{3}}{2 \, x^{2}} \] Input:
integrate(((-25*x^3+50*x-100)*exp(x)*log((2+x)/x)^2+(100*x-100)*exp(x)*log ((2+x)/x)+2*x^4+4*x^3)/(2*x^4+4*x^3),x, algorithm="fricas")
Output:
-1/2*(25*(x - 1)*e^x*log((x + 2)/x)^2 - 2*x^3)/x^2
Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {4 x^3+2 x^4+e^x (-100+100 x) \log \left (\frac {2+x}{x}\right )+e^x \left (-100+50 x-25 x^3\right ) \log ^2\left (\frac {2+x}{x}\right )}{4 x^3+2 x^4} \, dx=x + \frac {\left (- 25 x \log {\left (\frac {x + 2}{x} \right )}^{2} + 25 \log {\left (\frac {x + 2}{x} \right )}^{2}\right ) e^{x}}{2 x^{2}} \] Input:
integrate(((-25*x**3+50*x-100)*exp(x)*ln((2+x)/x)**2+(100*x-100)*exp(x)*ln ((2+x)/x)+2*x**4+4*x**3)/(2*x**4+4*x**3),x)
Output:
x + (-25*x*log((x + 2)/x)**2 + 25*log((x + 2)/x)**2)*exp(x)/(2*x**2)
Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {4 x^3+2 x^4+e^x (-100+100 x) \log \left (\frac {2+x}{x}\right )+e^x \left (-100+50 x-25 x^3\right ) \log ^2\left (\frac {2+x}{x}\right )}{4 x^3+2 x^4} \, dx=x - \frac {25 \, {\left ({\left (x - 1\right )} e^{x} \log \left (x + 2\right )^{2} - 2 \, {\left (x - 1\right )} e^{x} \log \left (x + 2\right ) \log \left (x\right ) + {\left (x - 1\right )} e^{x} \log \left (x\right )^{2}\right )}}{2 \, x^{2}} \] Input:
integrate(((-25*x^3+50*x-100)*exp(x)*log((2+x)/x)^2+(100*x-100)*exp(x)*log ((2+x)/x)+2*x^4+4*x^3)/(2*x^4+4*x^3),x, algorithm="maxima")
Output:
x - 25/2*((x - 1)*e^x*log(x + 2)^2 - 2*(x - 1)*e^x*log(x + 2)*log(x) + (x - 1)*e^x*log(x)^2)/x^2
Time = 0.14 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {4 x^3+2 x^4+e^x (-100+100 x) \log \left (\frac {2+x}{x}\right )+e^x \left (-100+50 x-25 x^3\right ) \log ^2\left (\frac {2+x}{x}\right )}{4 x^3+2 x^4} \, dx=-\frac {25 \, x e^{x} \log \left (\frac {x + 2}{x}\right )^{2} - 2 \, x^{3} - 25 \, e^{x} \log \left (\frac {x + 2}{x}\right )^{2}}{2 \, x^{2}} \] Input:
integrate(((-25*x^3+50*x-100)*exp(x)*log((2+x)/x)^2+(100*x-100)*exp(x)*log ((2+x)/x)+2*x^4+4*x^3)/(2*x^4+4*x^3),x, algorithm="giac")
Output:
-1/2*(25*x*e^x*log((x + 2)/x)^2 - 2*x^3 - 25*e^x*log((x + 2)/x)^2)/x^2
Timed out. \[ \int \frac {4 x^3+2 x^4+e^x (-100+100 x) \log \left (\frac {2+x}{x}\right )+e^x \left (-100+50 x-25 x^3\right ) \log ^2\left (\frac {2+x}{x}\right )}{4 x^3+2 x^4} \, dx=\int \frac {4\,x^3+2\,x^4-{\mathrm {e}}^x\,{\ln \left (\frac {x+2}{x}\right )}^2\,\left (25\,x^3-50\,x+100\right )+{\mathrm {e}}^x\,\ln \left (\frac {x+2}{x}\right )\,\left (100\,x-100\right )}{2\,x^4+4\,x^3} \,d x \] Input:
int((4*x^3 + 2*x^4 - exp(x)*log((x + 2)/x)^2*(25*x^3 - 50*x + 100) + exp(x )*log((x + 2)/x)*(100*x - 100))/(4*x^3 + 2*x^4),x)
Output:
int((4*x^3 + 2*x^4 - exp(x)*log((x + 2)/x)^2*(25*x^3 - 50*x + 100) + exp(x )*log((x + 2)/x)*(100*x - 100))/(4*x^3 + 2*x^4), x)
Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {4 x^3+2 x^4+e^x (-100+100 x) \log \left (\frac {2+x}{x}\right )+e^x \left (-100+50 x-25 x^3\right ) \log ^2\left (\frac {2+x}{x}\right )}{4 x^3+2 x^4} \, dx=\frac {-25 e^{x} \mathrm {log}\left (\frac {x +2}{x}\right )^{2} x +25 e^{x} \mathrm {log}\left (\frac {x +2}{x}\right )^{2}+2 x^{3}}{2 x^{2}} \] Input:
int(((-25*x^3+50*x-100)*exp(x)*log((2+x)/x)^2+(100*x-100)*exp(x)*log((2+x) /x)+2*x^4+4*x^3)/(2*x^4+4*x^3),x)
Output:
( - 25*e**x*log((x + 2)/x)**2*x + 25*e**x*log((x + 2)/x)**2 + 2*x**3)/(2*x **2)