Integrand size = 77, antiderivative size = 31 \[ \int \frac {24 x+24 x^2+e^8 \left (x+x^2\right )+e^4 \left (10 x+10 x^2\right )+e^{\frac {5 x^2-\log (1+x)}{x}} \left (6 x^2+5 x^3+(1+x) \log (1+x)\right )}{x+x^2} \, dx=-x+\left (e^{5 x-\frac {\log (1+x)}{x}}+\left (5+e^4\right )^2\right ) x+\log (2) \] Output:
x*(exp(5*x-ln(1+x)/x)+(5+exp(4))^2)+ln(2)-x
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {24 x+24 x^2+e^8 \left (x+x^2\right )+e^4 \left (10 x+10 x^2\right )+e^{\frac {5 x^2-\log (1+x)}{x}} \left (6 x^2+5 x^3+(1+x) \log (1+x)\right )}{x+x^2} \, dx=\left (24+10 e^4+e^8\right ) x+e^{5 x} x (1+x)^{-1/x} \] Input:
Integrate[(24*x + 24*x^2 + E^8*(x + x^2) + E^4*(10*x + 10*x^2) + E^((5*x^2 - Log[1 + x])/x)*(6*x^2 + 5*x^3 + (1 + x)*Log[1 + x]))/(x + x^2),x]
Output:
(24 + 10*E^4 + E^8)*x + (E^(5*x)*x)/(1 + x)^x^(-1)
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.
\(\displaystyle \int \frac {24 x^2+e^8 \left (x^2+x\right )+e^4 \left (10 x^2+10 x\right )+e^{\frac {5 x^2-\log (x+1)}{x}} \left (5 x^3+6 x^2+(x+1) \log (x+1)\right )+24 x}{x^2+x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {24 x^2+e^8 \left (x^2+x\right )+e^4 \left (10 x^2+10 x\right )+e^{\frac {5 x^2-\log (x+1)}{x}} \left (5 x^3+6 x^2+(x+1) \log (x+1)\right )+24 x}{x (x+1)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {24 x^2+\left (1+\frac {10}{e^4}\right ) e^8 \left (x^2+x\right )+e^{\frac {5 x^2-\log (x+1)}{x}} \left (5 x^3+6 x^2+(x+1) \log (x+1)\right )+24 x}{x (x+1)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{5 x} (x+1)^{-\frac {1}{x}-1} \left (5 x^3+6 x^2+x \log (x+1)+\log (x+1)\right )}{x}+24 \left (1+\frac {1}{24} e^4 \left (10+e^4\right )\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \int e^{5 x} x^2 (x+1)^{-1-\frac {1}{x}}dx+6 \int e^{5 x} x (x+1)^{-1-\frac {1}{x}}dx-\int \frac {\int \frac {e^{5 x} (x+1)^{-1/x}}{x}dx}{x+1}dx+\log (x+1) \int \frac {e^{5 x} (x+1)^{-1/x}}{x}dx+\left (24+10 e^4+e^8\right ) x\) |
Input:
Int[(24*x + 24*x^2 + E^8*(x + x^2) + E^4*(10*x + 10*x^2) + E^((5*x^2 - Log [1 + x])/x)*(6*x^2 + 5*x^3 + (1 + x)*Log[1 + x]))/(x + x^2),x]
Output:
$Aborted
Time = 0.82 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94
method | result | size |
risch | \(x \left (1+x \right )^{-\frac {1}{x}} {\mathrm e}^{5 x}+10 x \,{\mathrm e}^{4}+x \,{\mathrm e}^{8}+24 x\) | \(29\) |
norman | \(x \,{\mathrm e}^{\frac {-\ln \left (1+x \right )+5 x^{2}}{x}}+\left ({\mathrm e}^{8}+10 \,{\mathrm e}^{4}+24\right ) x\) | \(33\) |
default | \(24 x +x \,{\mathrm e}^{8}+x \,{\mathrm e}^{\frac {-\ln \left (1+x \right )+5 x^{2}}{x}}+10 x \,{\mathrm e}^{4}\) | \(35\) |
parts | \(24 x +x \,{\mathrm e}^{8}+x \,{\mathrm e}^{\frac {-\ln \left (1+x \right )+5 x^{2}}{x}}+10 x \,{\mathrm e}^{4}\) | \(35\) |
parallelrisch | \(x \,{\mathrm e}^{8}-2 \,{\mathrm e}^{8}+10 x \,{\mathrm e}^{4}+{\mathrm e}^{-\frac {-5 x^{2}+\ln \left (1+x \right )}{x}} x -20 \,{\mathrm e}^{4}+24 x -48\) | \(45\) |
Input:
int((((1+x)*ln(1+x)+5*x^3+6*x^2)*exp((-ln(1+x)+5*x^2)/x)+(x^2+x)*exp(4)^2+ (10*x^2+10*x)*exp(4)+24*x^2+24*x)/(x^2+x),x,method=_RETURNVERBOSE)
Output:
x*(1+x)^(-1/x)*exp(5*x)+10*x*exp(4)+x*exp(8)+24*x
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {24 x+24 x^2+e^8 \left (x+x^2\right )+e^4 \left (10 x+10 x^2\right )+e^{\frac {5 x^2-\log (1+x)}{x}} \left (6 x^2+5 x^3+(1+x) \log (1+x)\right )}{x+x^2} \, dx=x e^{8} + 10 \, x e^{4} + x e^{\left (\frac {5 \, x^{2} - \log \left (x + 1\right )}{x}\right )} + 24 \, x \] Input:
integrate((((1+x)*log(1+x)+5*x^3+6*x^2)*exp((-log(1+x)+5*x^2)/x)+(x^2+x)*e xp(4)^2+(10*x^2+10*x)*exp(4)+24*x^2+24*x)/(x^2+x),x, algorithm="fricas")
Output:
x*e^8 + 10*x*e^4 + x*e^((5*x^2 - log(x + 1))/x) + 24*x
Time = 4.71 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {24 x+24 x^2+e^8 \left (x+x^2\right )+e^4 \left (10 x+10 x^2\right )+e^{\frac {5 x^2-\log (1+x)}{x}} \left (6 x^2+5 x^3+(1+x) \log (1+x)\right )}{x+x^2} \, dx=x e^{\frac {5 x^{2} - \log {\left (x + 1 \right )}}{x}} + x \left (24 + 10 e^{4} + e^{8}\right ) \] Input:
integrate((((1+x)*ln(1+x)+5*x**3+6*x**2)*exp((-ln(1+x)+5*x**2)/x)+(x**2+x) *exp(4)**2+(10*x**2+10*x)*exp(4)+24*x**2+24*x)/(x**2+x),x)
Output:
x*exp((5*x**2 - log(x + 1))/x) + x*(24 + 10*exp(4) + exp(8))
Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.87 \[ \int \frac {24 x+24 x^2+e^8 \left (x+x^2\right )+e^4 \left (10 x+10 x^2\right )+e^{\frac {5 x^2-\log (1+x)}{x}} \left (6 x^2+5 x^3+(1+x) \log (1+x)\right )}{x+x^2} \, dx={\left (x - \log \left (x + 1\right )\right )} e^{8} + 10 \, {\left (x - \log \left (x + 1\right )\right )} e^{4} + x e^{\left (5 \, x - \frac {\log \left (x + 1\right )}{x}\right )} + e^{8} \log \left (x + 1\right ) + 10 \, e^{4} \log \left (x + 1\right ) + 24 \, x \] Input:
integrate((((1+x)*log(1+x)+5*x^3+6*x^2)*exp((-log(1+x)+5*x^2)/x)+(x^2+x)*e xp(4)^2+(10*x^2+10*x)*exp(4)+24*x^2+24*x)/(x^2+x),x, algorithm="maxima")
Output:
(x - log(x + 1))*e^8 + 10*(x - log(x + 1))*e^4 + x*e^(5*x - log(x + 1)/x) + e^8*log(x + 1) + 10*e^4*log(x + 1) + 24*x
Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {24 x+24 x^2+e^8 \left (x+x^2\right )+e^4 \left (10 x+10 x^2\right )+e^{\frac {5 x^2-\log (1+x)}{x}} \left (6 x^2+5 x^3+(1+x) \log (1+x)\right )}{x+x^2} \, dx=x e^{8} + 10 \, x e^{4} + x e^{\left (\frac {5 \, x^{2} - \log \left (x + 1\right )}{x}\right )} + 24 \, x \] Input:
integrate((((1+x)*log(1+x)+5*x^3+6*x^2)*exp((-log(1+x)+5*x^2)/x)+(x^2+x)*e xp(4)^2+(10*x^2+10*x)*exp(4)+24*x^2+24*x)/(x^2+x),x, algorithm="giac")
Output:
x*e^8 + 10*x*e^4 + x*e^((5*x^2 - log(x + 1))/x) + 24*x
Time = 1.71 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {24 x+24 x^2+e^8 \left (x+x^2\right )+e^4 \left (10 x+10 x^2\right )+e^{\frac {5 x^2-\log (1+x)}{x}} \left (6 x^2+5 x^3+(1+x) \log (1+x)\right )}{x+x^2} \, dx=24\,x+10\,x\,{\mathrm {e}}^4+x\,{\mathrm {e}}^8+\frac {x\,{\mathrm {e}}^{5\,x}}{{\left (x+1\right )}^{1/x}} \] Input:
int((24*x + exp(-(log(x + 1) - 5*x^2)/x)*(log(x + 1)*(x + 1) + 6*x^2 + 5*x ^3) + exp(4)*(10*x + 10*x^2) + exp(8)*(x + x^2) + 24*x^2)/(x + x^2),x)
Output:
24*x + 10*x*exp(4) + x*exp(8) + (x*exp(5*x))/(x + 1)^(1/x)
Time = 0.23 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.97 \[ \int \frac {24 x+24 x^2+e^8 \left (x+x^2\right )+e^4 \left (10 x+10 x^2\right )+e^{\frac {5 x^2-\log (1+x)}{x}} \left (6 x^2+5 x^3+(1+x) \log (1+x)\right )}{x+x^2} \, dx=\frac {x \left (e^{\frac {\mathrm {log}\left (x +1\right )}{x}} e^{8}+10 e^{\frac {\mathrm {log}\left (x +1\right )}{x}} e^{4}+24 e^{\frac {\mathrm {log}\left (x +1\right )}{x}}+e^{5 x}\right )}{e^{\frac {\mathrm {log}\left (x +1\right )}{x}}} \] Input:
int((((1+x)*log(1+x)+5*x^3+6*x^2)*exp((-log(1+x)+5*x^2)/x)+(x^2+x)*exp(4)^ 2+(10*x^2+10*x)*exp(4)+24*x^2+24*x)/(x^2+x),x)
Output:
(x*(e**(log(x + 1)/x)*e**8 + 10*e**(log(x + 1)/x)*e**4 + 24*e**(log(x + 1) /x) + e**(5*x)))/e**(log(x + 1)/x)