Integrand size = 75, antiderivative size = 30 \[ \int \frac {184 x+e^{x+x^2} \left (-184 x-368 x^2\right )+e^{\frac {1}{16} \left (144-192 x+64 x^2+(-24+16 x) \log (x)+\log ^2(x)\right )} \left (-276-2024 x+1472 x^2+(23+184 x) \log (x)\right )}{32 x} \, dx=\frac {23}{4} \left (-e^{x (1+x)}+e^{\left (-3+2 x+\frac {\log (x)}{4}\right )^2}+x\right ) \] Output:
23/4*exp((2*x+1/4*ln(x)-3)^2)+23/4*x-23/4*exp(x*(1+x))
Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.73 \[ \int \frac {184 x+e^{x+x^2} \left (-184 x-368 x^2\right )+e^{\frac {1}{16} \left (144-192 x+64 x^2+(-24+16 x) \log (x)+\log ^2(x)\right )} \left (-276-2024 x+1472 x^2+(23+184 x) \log (x)\right )}{32 x} \, dx=\frac {23}{32} \left (-8 e^{x (1+x)}+8 x+8 e^{9-12 x+4 x^2+\frac {\log ^2(x)}{16}} x^{1+\frac {1}{2} (-5+2 x)}\right ) \] Input:
Integrate[(184*x + E^(x + x^2)*(-184*x - 368*x^2) + E^((144 - 192*x + 64*x ^2 + (-24 + 16*x)*Log[x] + Log[x]^2)/16)*(-276 - 2024*x + 1472*x^2 + (23 + 184*x)*Log[x]))/(32*x),x]
Output:
(23*(-8*E^(x*(1 + x)) + 8*x + 8*E^(9 - 12*x + 4*x^2 + Log[x]^2/16)*x^(1 + (-5 + 2*x)/2)))/32
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.
| \(\displaystyle \int \frac {\left (1472 x^2-2024 x+(184 x+23) \log (x)-276\right ) \exp \left (\frac {1}{16} \left (64 x^2-192 x+\log ^2(x)+(16 x-24) \log (x)+144\right )\right )+e^{x^2+x} \left (-368 x^2-184 x\right )+184 x}{32 x} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{32} \int \frac {23 \left (-e^{\frac {1}{16} \left (64 x^2-192 x+\log ^2(x)+144\right )} \left (-64 x^2+88 x-(8 x+1) \log (x)+12\right ) x^{\frac {1}{16} (16 x-24)}+8 x-8 e^{x^2+x} \left (2 x^2+x\right )\right )}{x}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {23}{32} \int \frac {-e^{\frac {1}{16} \left (64 x^2-192 x+\log ^2(x)+144\right )} \left (-64 x^2+88 x-(8 x+1) \log (x)+12\right ) x^{\frac {1}{2} (2 x-3)}+8 x-8 e^{x^2+x} \left (2 x^2+x\right )}{x}dx\) |
| \(\Big \downarrow \) 2010 |
| \(\displaystyle \frac {23}{32} \int \left (-12 e^{\frac {1}{16} \left (64 x^2-192 x+\log ^2(x)+144\right )} x^{x-\frac {5}{2}}+e^{\frac {1}{16} \left (64 x^2-192 x+\log ^2(x)+144\right )} \log (x) x^{x-\frac {5}{2}}-88 e^{\frac {1}{16} \left (64 x^2-192 x+\log ^2(x)+144\right )} x^{x-\frac {3}{2}}+8 e^{\frac {1}{16} \left (64 x^2-192 x+\log ^2(x)+144\right )} \log (x) x^{x-\frac {3}{2}}+64 e^{\frac {1}{16} \left (64 x^2-192 x+\log ^2(x)+144\right )} x^{x-\frac {1}{2}}-8 \left (2 e^{x^2+x} x+e^{x^2+x}-1\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {23}{32} \left (-12 \int e^{\frac {1}{16} \left (64 x^2-192 x+\log ^2(x)+144\right )} x^{x-\frac {5}{2}}dx-88 \int e^{\frac {1}{16} \left (64 x^2-192 x+\log ^2(x)+144\right )} x^{x-\frac {3}{2}}dx+64 \int e^{\frac {1}{16} \left (64 x^2-192 x+\log ^2(x)+144\right )} x^{x-\frac {1}{2}}dx+\int e^{\frac {1}{16} \left (64 x^2-192 x+\log ^2(x)+144\right )} x^{x-\frac {5}{2}} \log (x)dx+8 \int e^{\frac {1}{16} \left (64 x^2-192 x+\log ^2(x)+144\right )} x^{x-\frac {3}{2}} \log (x)dx-8 e^{x^2+x}+8 x\right )\) |
Input:
Int[(184*x + E^(x + x^2)*(-184*x - 368*x^2) + E^((144 - 192*x + 64*x^2 + ( -24 + 16*x)*Log[x] + Log[x]^2)/16)*(-276 - 2024*x + 1472*x^2 + (23 + 184*x )*Log[x]))/(32*x),x]
Output:
$Aborted
Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23
| method | result | size |
| risch | \(\frac {23 x}{4}-\frac {23 \,{\mathrm e}^{\left (1+x \right ) x}}{4}+\frac {23 x^{x -\frac {3}{2}} {\mathrm e}^{\frac {\ln \left (x \right )^{2}}{16}+9+4 x^{2}-12 x}}{4}\) | \(37\) |
| default | \(\frac {23 x}{4}-\frac {23 \,{\mathrm e}^{x^{2}+x}}{4}+\frac {23 \,{\mathrm e}^{\frac {\ln \left (x \right )^{2}}{16}+\frac {\left (16 x -24\right ) \ln \left (x \right )}{16}+4 x^{2}-12 x +9}}{4}\) | \(41\) |
| parallelrisch | \(\frac {23 x}{4}-\frac {23 \,{\mathrm e}^{x^{2}+x}}{4}+\frac {23 \,{\mathrm e}^{\frac {\ln \left (x \right )^{2}}{16}+\frac {\left (16 x -24\right ) \ln \left (x \right )}{16}+4 x^{2}-12 x +9}}{4}\) | \(41\) |
| parts | \(\frac {23 x}{4}-\frac {23 \,{\mathrm e}^{x^{2}+x}}{4}+\frac {23 \,{\mathrm e}^{\frac {\ln \left (x \right )^{2}}{16}+\frac {\left (16 x -24\right ) \ln \left (x \right )}{16}+4 x^{2}-12 x +9}}{4}\) | \(41\) |
Input:
int(1/32*(((184*x+23)*ln(x)+1472*x^2-2024*x-276)*exp(1/16*ln(x)^2+1/16*(16 *x-24)*ln(x)+4*x^2-12*x+9)+(-368*x^2-184*x)*exp(x^2+x)+184*x)/x,x,method=_ RETURNVERBOSE)
Output:
23/4*x-23/4*exp((1+x)*x)+23/4*x^(x-3/2)*exp(1/16*ln(x)^2+9+4*x^2-12*x)
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {184 x+e^{x+x^2} \left (-184 x-368 x^2\right )+e^{\frac {1}{16} \left (144-192 x+64 x^2+(-24+16 x) \log (x)+\log ^2(x)\right )} \left (-276-2024 x+1472 x^2+(23+184 x) \log (x)\right )}{32 x} \, dx=\frac {23}{4} \, x + \frac {23}{4} \, e^{\left (4 \, x^{2} + \frac {1}{2} \, {\left (2 \, x - 3\right )} \log \left (x\right ) + \frac {1}{16} \, \log \left (x\right )^{2} - 12 \, x + 9\right )} - \frac {23}{4} \, e^{\left (x^{2} + x\right )} \] Input:
integrate(1/32*(((184*x+23)*log(x)+1472*x^2-2024*x-276)*exp(1/16*log(x)^2+ 1/16*(16*x-24)*log(x)+4*x^2-12*x+9)+(-368*x^2-184*x)*exp(x^2+x)+184*x)/x,x , algorithm="fricas")
Output:
23/4*x + 23/4*e^(4*x^2 + 1/2*(2*x - 3)*log(x) + 1/16*log(x)^2 - 12*x + 9) - 23/4*e^(x^2 + x)
Time = 0.33 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.47 \[ \int \frac {184 x+e^{x+x^2} \left (-184 x-368 x^2\right )+e^{\frac {1}{16} \left (144-192 x+64 x^2+(-24+16 x) \log (x)+\log ^2(x)\right )} \left (-276-2024 x+1472 x^2+(23+184 x) \log (x)\right )}{32 x} \, dx=\frac {23 x}{4} - \frac {23 e^{x^{2} + x}}{4} + \frac {23 e^{4 x^{2} - 12 x + \left (x - \frac {3}{2}\right ) \log {\left (x \right )} + \frac {\log {\left (x \right )}^{2}}{16} + 9}}{4} \] Input:
integrate(1/32*(((184*x+23)*ln(x)+1472*x**2-2024*x-276)*exp(1/16*ln(x)**2+ 1/16*(16*x-24)*ln(x)+4*x**2-12*x+9)+(-368*x**2-184*x)*exp(x**2+x)+184*x)/x ,x)
Output:
23*x/4 - 23*exp(x**2 + x)/4 + 23*exp(4*x**2 - 12*x + (x - 3/2)*log(x) + lo g(x)**2/16 + 9)/4
\[ \int \frac {184 x+e^{x+x^2} \left (-184 x-368 x^2\right )+e^{\frac {1}{16} \left (144-192 x+64 x^2+(-24+16 x) \log (x)+\log ^2(x)\right )} \left (-276-2024 x+1472 x^2+(23+184 x) \log (x)\right )}{32 x} \, dx=\int { \frac {23 \, {\left ({\left (64 \, x^{2} + {\left (8 \, x + 1\right )} \log \left (x\right ) - 88 \, x - 12\right )} e^{\left (4 \, x^{2} + \frac {1}{2} \, {\left (2 \, x - 3\right )} \log \left (x\right ) + \frac {1}{16} \, \log \left (x\right )^{2} - 12 \, x + 9\right )} - 8 \, {\left (2 \, x^{2} + x\right )} e^{\left (x^{2} + x\right )} + 8 \, x\right )}}{32 \, x} \,d x } \] Input:
integrate(1/32*(((184*x+23)*log(x)+1472*x^2-2024*x-276)*exp(1/16*log(x)^2+ 1/16*(16*x-24)*log(x)+4*x^2-12*x+9)+(-368*x^2-184*x)*exp(x^2+x)+184*x)/x,x , algorithm="maxima")
Output:
23/8*I*sqrt(pi)*erf(I*x + 1/2*I)*e^(-1/4) + 23/8*(sqrt(pi)*(2*x + 1)*(erf( 1/2*sqrt(-(2*x + 1)^2)) - 1)/sqrt(-(2*x + 1)^2) - 2*e^(1/4*(2*x + 1)^2))*e ^(-1/4) + 23/4*x + 23/32*integrate((64*x^2*e^(4*x^2 + x*log(x) + 9) + 8*(e ^9*log(x) - 11*e^9)*x*e^(4*x^2 + x*log(x)) + (e^9*log(x) - 12*e^9)*e^(4*x^ 2 + x*log(x)))*e^(1/16*log(x)^2 - 12*x)/x^(5/2), x)
Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {184 x+e^{x+x^2} \left (-184 x-368 x^2\right )+e^{\frac {1}{16} \left (144-192 x+64 x^2+(-24+16 x) \log (x)+\log ^2(x)\right )} \left (-276-2024 x+1472 x^2+(23+184 x) \log (x)\right )}{32 x} \, dx=\frac {23}{4} \, x + \frac {23}{4} \, e^{\left (4 \, x^{2} + x \log \left (x\right ) + \frac {1}{16} \, \log \left (x\right )^{2} - 12 \, x - \frac {3}{2} \, \log \left (x\right ) + 9\right )} - \frac {23}{4} \, e^{\left (x^{2} + x\right )} \] Input:
integrate(1/32*(((184*x+23)*log(x)+1472*x^2-2024*x-276)*exp(1/16*log(x)^2+ 1/16*(16*x-24)*log(x)+4*x^2-12*x+9)+(-368*x^2-184*x)*exp(x^2+x)+184*x)/x,x , algorithm="giac")
Output:
23/4*x + 23/4*e^(4*x^2 + x*log(x) + 1/16*log(x)^2 - 12*x - 3/2*log(x) + 9) - 23/4*e^(x^2 + x)
Time = 3.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {184 x+e^{x+x^2} \left (-184 x-368 x^2\right )+e^{\frac {1}{16} \left (144-192 x+64 x^2+(-24+16 x) \log (x)+\log ^2(x)\right )} \left (-276-2024 x+1472 x^2+(23+184 x) \log (x)\right )}{32 x} \, dx=\frac {23\,x}{4}-\frac {23\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^x}{4}+\frac {23\,x^x\,{\mathrm {e}}^{\frac {{\ln \left (x\right )}^2}{16}}\,{\mathrm {e}}^{-12\,x}\,{\mathrm {e}}^9\,{\mathrm {e}}^{4\,x^2}}{4\,x^{3/2}} \] Input:
int(-((exp(log(x)^2/16 - 12*x + (log(x)*(16*x - 24))/16 + 4*x^2 + 9)*(2024 *x - log(x)*(184*x + 23) - 1472*x^2 + 276))/32 - (23*x)/4 + (exp(x + x^2)* (184*x + 368*x^2))/32)/x,x)
Output:
(23*x)/4 - (23*exp(x^2)*exp(x))/4 + (23*x^x*exp(log(x)^2/16)*exp(-12*x)*ex p(9)*exp(4*x^2))/(4*x^(3/2))
Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.97 \[ \int \frac {184 x+e^{x+x^2} \left (-184 x-368 x^2\right )+e^{\frac {1}{16} \left (144-192 x+64 x^2+(-24+16 x) \log (x)+\log ^2(x)\right )} \left (-276-2024 x+1472 x^2+(23+184 x) \log (x)\right )}{32 x} \, dx=\frac {\frac {23 x^{x +\frac {1}{2}} e^{\frac {\mathrm {log}\left (x \right )^{2}}{16}+4 x^{2}} e^{9}}{4}-\frac {23 e^{x^{2}+13 x} x^{2}}{4}+\frac {23 e^{12 x} x^{3}}{4}}{e^{12 x} x^{2}} \] Input:
int(1/32*(((184*x+23)*log(x)+1472*x^2-2024*x-276)*exp(1/16*log(x)^2+1/16*( 16*x-24)*log(x)+4*x^2-12*x+9)+(-368*x^2-184*x)*exp(x^2+x)+184*x)/x,x)
Output:
(23*(x**((2*x + 1)/2)*e**((log(x)**2 + 64*x**2)/16)*e**9 - e**(x**2 + 13*x )*x**2 + e**(12*x)*x**3))/(4*e**(12*x)*x**2)