Integrand size = 96, antiderivative size = 25 \[ \int \frac {-32 x-4 e^x x-4 x^2+\left (16+12 x+2 x^2+e^x \left (2+3 x+x^2\right )\right ) \log \left (16+16 x+4 x^2\right )+e^{x^2} \left (4 x+2 x^2\right ) \log ^3\left (16+16 x+4 x^2\right )}{(2+x) \log ^3\left (16+16 x+4 x^2\right )} \, dx=2+e^{x^2}+\frac {x \left (8+e^x+x\right )}{\log ^2\left ((4+2 x)^2\right )} \] Output:
(x+exp(x)+8)*x/ln((4+2*x)^2)^2+exp(x^2)+2
Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-32 x-4 e^x x-4 x^2+\left (16+12 x+2 x^2+e^x \left (2+3 x+x^2\right )\right ) \log \left (16+16 x+4 x^2\right )+e^{x^2} \left (4 x+2 x^2\right ) \log ^3\left (16+16 x+4 x^2\right )}{(2+x) \log ^3\left (16+16 x+4 x^2\right )} \, dx=e^{x^2}+\frac {x \left (8+e^x+x\right )}{\log ^2\left (4 (2+x)^2\right )} \] Input:
Integrate[(-32*x - 4*E^x*x - 4*x^2 + (16 + 12*x + 2*x^2 + E^x*(2 + 3*x + x ^2))*Log[16 + 16*x + 4*x^2] + E^x^2*(4*x + 2*x^2)*Log[16 + 16*x + 4*x^2]^3 )/((2 + x)*Log[16 + 16*x + 4*x^2]^3),x]
Output:
E^x^2 + (x*(8 + E^x + x))/Log[4*(2 + x)^2]^2
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.
\(\displaystyle \int \frac {-4 x^2+e^{x^2} \left (2 x^2+4 x\right ) \log ^3\left (4 x^2+16 x+16\right )+\left (2 x^2+e^x \left (x^2+3 x+2\right )+12 x+16\right ) \log \left (4 x^2+16 x+16\right )-4 e^x x-32 x}{(x+2) \log ^3\left (4 x^2+16 x+16\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (2 e^{x^2} x-\frac {4 \left (x+e^x+8\right ) x}{(x+2) \log ^3\left (4 (x+2)^2\right )}+\frac {e^x (x+1)+2 (x+4)}{\log ^2\left (4 (x+2)^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \frac {e^x}{\log ^3\left (4 (x+2)^2\right )}dx+8 \int \frac {e^x}{(x+2) \log ^3\left (4 (x+2)^2\right )}dx-\int \frac {e^x}{\log ^2\left (4 (x+2)^2\right )}dx+\int \frac {e^x (x+2)}{\log ^2\left (4 (x+2)^2\right )}dx+e^{x^2}+\frac {x (x+2)}{\log ^2\left (4 (x+2)^2\right )}+\frac {6 (x+2)}{\log ^2\left (4 (x+2)^2\right )}-\frac {12}{\log ^2\left (4 (x+2)^2\right )}+\frac {x (x+2)}{\log \left (4 (x+2)^2\right )}-\frac {(x+4) (x+2)}{\log \left (4 (x+2)^2\right )}+\frac {4 (x+2)}{\log \left (4 (x+2)^2\right )}\) |
Input:
Int[(-32*x - 4*E^x*x - 4*x^2 + (16 + 12*x + 2*x^2 + E^x*(2 + 3*x + x^2))*L og[16 + 16*x + 4*x^2] + E^x^2*(4*x + 2*x^2)*Log[16 + 16*x + 4*x^2]^3)/((2 + x)*Log[16 + 16*x + 4*x^2]^3),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(48\) vs. \(2(23)=46\).
Time = 63.36 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96
method | result | size |
parallelrisch | \(-\frac {-4 \ln \left (4 x^{2}+16 x +16\right )^{2} {\mathrm e}^{x^{2}}-4 x^{2}-4 \,{\mathrm e}^{x} x -32 x}{4 \ln \left (4 x^{2}+16 x +16\right )^{2}}\) | \(49\) |
risch | \({\mathrm e}^{x^{2}}-\frac {4 x \left (x +{\mathrm e}^{x}+8\right )}{\left (\pi \operatorname {csgn}\left (i \left (2+x \right )\right )^{2} \operatorname {csgn}\left (i \left (2+x \right )^{2}\right )-2 \pi \,\operatorname {csgn}\left (i \left (2+x \right )\right ) \operatorname {csgn}\left (i \left (2+x \right )^{2}\right )^{2}+\pi \operatorname {csgn}\left (i \left (2+x \right )^{2}\right )^{3}+4 i \ln \left (2\right )+4 i \ln \left (2+x \right )\right )^{2}}\) | \(83\) |
Input:
int(((2*x^2+4*x)*exp(x^2)*ln(4*x^2+16*x+16)^3+((x^2+3*x+2)*exp(x)+2*x^2+12 *x+16)*ln(4*x^2+16*x+16)-4*exp(x)*x-4*x^2-32*x)/(2+x)/ln(4*x^2+16*x+16)^3, x,method=_RETURNVERBOSE)
Output:
-1/4*(-4*ln(4*x^2+16*x+16)^2*exp(x^2)-4*x^2-4*exp(x)*x-32*x)/ln(4*x^2+16*x +16)^2
Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72 \[ \int \frac {-32 x-4 e^x x-4 x^2+\left (16+12 x+2 x^2+e^x \left (2+3 x+x^2\right )\right ) \log \left (16+16 x+4 x^2\right )+e^{x^2} \left (4 x+2 x^2\right ) \log ^3\left (16+16 x+4 x^2\right )}{(2+x) \log ^3\left (16+16 x+4 x^2\right )} \, dx=\frac {e^{\left (x^{2}\right )} \log \left (4 \, x^{2} + 16 \, x + 16\right )^{2} + x^{2} + x e^{x} + 8 \, x}{\log \left (4 \, x^{2} + 16 \, x + 16\right )^{2}} \] Input:
integrate(((2*x^2+4*x)*exp(x^2)*log(4*x^2+16*x+16)^3+((x^2+3*x+2)*exp(x)+2 *x^2+12*x+16)*log(4*x^2+16*x+16)-4*exp(x)*x-4*x^2-32*x)/(2+x)/log(4*x^2+16 *x+16)^3,x, algorithm="fricas")
Output:
(e^(x^2)*log(4*x^2 + 16*x + 16)^2 + x^2 + x*e^x + 8*x)/log(4*x^2 + 16*x + 16)^2
Time = 0.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68 \[ \int \frac {-32 x-4 e^x x-4 x^2+\left (16+12 x+2 x^2+e^x \left (2+3 x+x^2\right )\right ) \log \left (16+16 x+4 x^2\right )+e^{x^2} \left (4 x+2 x^2\right ) \log ^3\left (16+16 x+4 x^2\right )}{(2+x) \log ^3\left (16+16 x+4 x^2\right )} \, dx=\frac {x e^{x}}{\log {\left (4 x^{2} + 16 x + 16 \right )}^{2}} + \frac {x^{2} + 8 x}{\log {\left (4 x^{2} + 16 x + 16 \right )}^{2}} + e^{x^{2}} \] Input:
integrate(((2*x**2+4*x)*exp(x**2)*ln(4*x**2+16*x+16)**3+((x**2+3*x+2)*exp( x)+2*x**2+12*x+16)*ln(4*x**2+16*x+16)-4*exp(x)*x-4*x**2-32*x)/(2+x)/ln(4*x **2+16*x+16)**3,x)
Output:
x*exp(x)/log(4*x**2 + 16*x + 16)**2 + (x**2 + 8*x)/log(4*x**2 + 16*x + 16) **2 + exp(x**2)
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (23) = 46\).
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.36 \[ \int \frac {-32 x-4 e^x x-4 x^2+\left (16+12 x+2 x^2+e^x \left (2+3 x+x^2\right )\right ) \log \left (16+16 x+4 x^2\right )+e^{x^2} \left (4 x+2 x^2\right ) \log ^3\left (16+16 x+4 x^2\right )}{(2+x) \log ^3\left (16+16 x+4 x^2\right )} \, dx=\frac {x^{2} + 4 \, {\left (\log \left (2\right )^{2} + 2 \, \log \left (2\right ) \log \left (x + 2\right ) + \log \left (x + 2\right )^{2}\right )} e^{\left (x^{2}\right )} + x e^{x} + 8 \, x}{4 \, {\left (\log \left (2\right )^{2} + 2 \, \log \left (2\right ) \log \left (x + 2\right ) + \log \left (x + 2\right )^{2}\right )}} \] Input:
integrate(((2*x^2+4*x)*exp(x^2)*log(4*x^2+16*x+16)^3+((x^2+3*x+2)*exp(x)+2 *x^2+12*x+16)*log(4*x^2+16*x+16)-4*exp(x)*x-4*x^2-32*x)/(2+x)/log(4*x^2+16 *x+16)^3,x, algorithm="maxima")
Output:
1/4*(x^2 + 4*(log(2)^2 + 2*log(2)*log(x + 2) + log(x + 2)^2)*e^(x^2) + x*e ^x + 8*x)/(log(2)^2 + 2*log(2)*log(x + 2) + log(x + 2)^2)
Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72 \[ \int \frac {-32 x-4 e^x x-4 x^2+\left (16+12 x+2 x^2+e^x \left (2+3 x+x^2\right )\right ) \log \left (16+16 x+4 x^2\right )+e^{x^2} \left (4 x+2 x^2\right ) \log ^3\left (16+16 x+4 x^2\right )}{(2+x) \log ^3\left (16+16 x+4 x^2\right )} \, dx=\frac {e^{\left (x^{2}\right )} \log \left (4 \, x^{2} + 16 \, x + 16\right )^{2} + x^{2} + x e^{x} + 8 \, x}{\log \left (4 \, x^{2} + 16 \, x + 16\right )^{2}} \] Input:
integrate(((2*x^2+4*x)*exp(x^2)*log(4*x^2+16*x+16)^3+((x^2+3*x+2)*exp(x)+2 *x^2+12*x+16)*log(4*x^2+16*x+16)-4*exp(x)*x-4*x^2-32*x)/(2+x)/log(4*x^2+16 *x+16)^3,x, algorithm="giac")
Output:
(e^(x^2)*log(4*x^2 + 16*x + 16)^2 + x^2 + x*e^x + 8*x)/log(4*x^2 + 16*x + 16)^2
Time = 3.93 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.20 \[ \int \frac {-32 x-4 e^x x-4 x^2+\left (16+12 x+2 x^2+e^x \left (2+3 x+x^2\right )\right ) \log \left (16+16 x+4 x^2\right )+e^{x^2} \left (4 x+2 x^2\right ) \log ^3\left (16+16 x+4 x^2\right )}{(2+x) \log ^3\left (16+16 x+4 x^2\right )} \, dx={\mathrm {e}}^{x^2}+\frac {8\,x}{{\ln \left (4\,x^2+16\,x+16\right )}^2}+\frac {x^2}{{\ln \left (4\,x^2+16\,x+16\right )}^2}+\frac {x\,{\mathrm {e}}^x}{{\ln \left (4\,x^2+16\,x+16\right )}^2} \] Input:
int(-(32*x + 4*x*exp(x) + 4*x^2 - log(16*x + 4*x^2 + 16)*(12*x + exp(x)*(3 *x + x^2 + 2) + 2*x^2 + 16) - exp(x^2)*log(16*x + 4*x^2 + 16)^3*(4*x + 2*x ^2))/(log(16*x + 4*x^2 + 16)^3*(x + 2)),x)
Output:
exp(x^2) + (8*x)/log(16*x + 4*x^2 + 16)^2 + x^2/log(16*x + 4*x^2 + 16)^2 + (x*exp(x))/log(16*x + 4*x^2 + 16)^2
Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \[ \int \frac {-32 x-4 e^x x-4 x^2+\left (16+12 x+2 x^2+e^x \left (2+3 x+x^2\right )\right ) \log \left (16+16 x+4 x^2\right )+e^{x^2} \left (4 x+2 x^2\right ) \log ^3\left (16+16 x+4 x^2\right )}{(2+x) \log ^3\left (16+16 x+4 x^2\right )} \, dx=\frac {e^{x^{2}} \mathrm {log}\left (4 x^{2}+16 x +16\right )^{2}+e^{x} x +x^{2}+8 x}{\mathrm {log}\left (4 x^{2}+16 x +16\right )^{2}} \] Input:
int(((2*x^2+4*x)*exp(x^2)*log(4*x^2+16*x+16)^3+((x^2+3*x+2)*exp(x)+2*x^2+1 2*x+16)*log(4*x^2+16*x+16)-4*exp(x)*x-4*x^2-32*x)/(2+x)/log(4*x^2+16*x+16) ^3,x)
Output:
(e**(x**2)*log(4*x**2 + 16*x + 16)**2 + e**x*x + x**2 + 8*x)/log(4*x**2 + 16*x + 16)**2