Integrand size = 95, antiderivative size = 25 \[ \int \frac {48 \log (2)+e^x (16-16 x) \log ^2(2)+4 x^2 \log ^2(2)}{144-192 x+64 x^2+\left (-24 x^2+16 x^3\right ) \log (2)+16 e^{2 x} \log ^2(2)+x^4 \log ^2(2)+e^x \left ((96-64 x) \log (2)-8 x^2 \log ^2(2)\right )} \, dx=\frac {x}{e^x-\frac {x^2}{4}+\frac {3-2 x}{\log (2)}} \]
Time = 5.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {48 \log (2)+e^x (16-16 x) \log ^2(2)+4 x^2 \log ^2(2)}{144-192 x+64 x^2+\left (-24 x^2+16 x^3\right ) \log (2)+16 e^{2 x} \log ^2(2)+x^4 \log ^2(2)+e^x \left ((96-64 x) \log (2)-8 x^2 \log ^2(2)\right )} \, dx=\frac {4 x \log (2)}{12-8 x+4 e^x \log (2)-x^2 \log (2)} \]
Integrate[(48*Log[2] + E^x*(16 - 16*x)*Log[2]^2 + 4*x^2*Log[2]^2)/(144 - 1 92*x + 64*x^2 + (-24*x^2 + 16*x^3)*Log[2] + 16*E^(2*x)*Log[2]^2 + x^4*Log[ 2]^2 + E^x*((96 - 64*x)*Log[2] - 8*x^2*Log[2]^2)),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.
\(\displaystyle \int \frac {4 x^2 \log ^2(2)+e^x (16-16 x) \log ^2(2)+48 \log (2)}{x^4 \log ^2(2)+64 x^2+e^x \left ((96-64 x) \log (2)-8 x^2 \log ^2(2)\right )+\left (16 x^3-24 x^2\right ) \log (2)-192 x+16 e^{2 x} \log ^2(2)+144} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 \log (2) \left (x^2 \log (2)-4 e^x (x-1) \log (2)+12\right )}{\left (x^2 \log (2)+8 x-4 \left (e^x \log (2)+3\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \log (2) \int \frac {\log (2) x^2+4 e^x (1-x) \log (2)+12}{\left (\log (2) x^2+8 x-4 \left (3+e^x \log (2)\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 4 \log (2) \int \left (\frac {x-1}{\log (2) x^2+8 x-4 e^x \log (2)-12}+\frac {x \left (-\log (2) x^2-2 (4-\log (2)) x+20\right )}{\left (-\log (2) x^2-8 x+4 e^x \log (2)+12\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \log (2) \left (20 \int \frac {x}{\left (\log (2) x^2+8 x-4 e^x \log (2)-12\right )^2}dx-2 (4-\log (2)) \int \frac {x^2}{\left (\log (2) x^2+8 x-4 e^x \log (2)-12\right )^2}dx+\int \frac {x}{\log (2) x^2+8 x-4 e^x \log (2)-12}dx-\int \frac {1}{\log (2) x^2+8 x-e^x \log (16)-12}dx-\log (2) \int \frac {x^3}{\left (\log (2) x^2+8 x-4 e^x \log (2)-12\right )^2}dx\right )\) |
Int[(48*Log[2] + E^x*(16 - 16*x)*Log[2]^2 + 4*x^2*Log[2]^2)/(144 - 192*x + 64*x^2 + (-24*x^2 + 16*x^3)*Log[2] + 16*E^(2*x)*Log[2]^2 + x^4*Log[2]^2 + E^x*((96 - 64*x)*Log[2] - 8*x^2*Log[2]^2)),x]
3.5.56.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {4 x \ln \left (2\right )}{x^{2} \ln \left (2\right )-4 \,{\mathrm e}^{x} \ln \left (2\right )+8 x -12}\) | \(25\) |
norman | \(\frac {-2 \ln \left (2\right )^{2} {\mathrm e}^{x}+\frac {x^{2} \ln \left (2\right )^{2}}{2}-6 \ln \left (2\right )}{x^{2} \ln \left (2\right )-4 \,{\mathrm e}^{x} \ln \left (2\right )+8 x -12}\) | \(43\) |
parallelrisch | \(\frac {4 x^{2} \ln \left (2\right )^{2}-16 \ln \left (2\right )^{2} {\mathrm e}^{x}-48 \ln \left (2\right )}{8 x^{2} \ln \left (2\right )-32 \,{\mathrm e}^{x} \ln \left (2\right )+64 x -96}\) | \(44\) |
int(((-16*x+16)*ln(2)^2*exp(x)+4*x^2*ln(2)^2+48*ln(2))/(16*ln(2)^2*exp(x)^ 2+(-8*x^2*ln(2)^2+(-64*x+96)*ln(2))*exp(x)+x^4*ln(2)^2+(16*x^3-24*x^2)*ln( 2)+64*x^2-192*x+144),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {48 \log (2)+e^x (16-16 x) \log ^2(2)+4 x^2 \log ^2(2)}{144-192 x+64 x^2+\left (-24 x^2+16 x^3\right ) \log (2)+16 e^{2 x} \log ^2(2)+x^4 \log ^2(2)+e^x \left ((96-64 x) \log (2)-8 x^2 \log ^2(2)\right )} \, dx=-\frac {4 \, x \log \left (2\right )}{x^{2} \log \left (2\right ) - 4 \, e^{x} \log \left (2\right ) + 8 \, x - 12} \]
integrate(((-16*x+16)*log(2)^2*exp(x)+4*x^2*log(2)^2+48*log(2))/(16*log(2) ^2*exp(x)^2+(-8*x^2*log(2)^2+(-64*x+96)*log(2))*exp(x)+x^4*log(2)^2+(16*x^ 3-24*x^2)*log(2)+64*x^2-192*x+144),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {48 \log (2)+e^x (16-16 x) \log ^2(2)+4 x^2 \log ^2(2)}{144-192 x+64 x^2+\left (-24 x^2+16 x^3\right ) \log (2)+16 e^{2 x} \log ^2(2)+x^4 \log ^2(2)+e^x \left ((96-64 x) \log (2)-8 x^2 \log ^2(2)\right )} \, dx=\frac {4 x \log {\left (2 \right )}}{- x^{2} \log {\left (2 \right )} - 8 x + 4 e^{x} \log {\left (2 \right )} + 12} \]
integrate(((-16*x+16)*ln(2)**2*exp(x)+4*x**2*ln(2)**2+48*ln(2))/(16*ln(2)* *2*exp(x)**2+(-8*x**2*ln(2)**2+(-64*x+96)*ln(2))*exp(x)+x**4*ln(2)**2+(16* x**3-24*x**2)*ln(2)+64*x**2-192*x+144),x)
Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {48 \log (2)+e^x (16-16 x) \log ^2(2)+4 x^2 \log ^2(2)}{144-192 x+64 x^2+\left (-24 x^2+16 x^3\right ) \log (2)+16 e^{2 x} \log ^2(2)+x^4 \log ^2(2)+e^x \left ((96-64 x) \log (2)-8 x^2 \log ^2(2)\right )} \, dx=-\frac {4 \, x \log \left (2\right )}{x^{2} \log \left (2\right ) - 4 \, e^{x} \log \left (2\right ) + 8 \, x - 12} \]
integrate(((-16*x+16)*log(2)^2*exp(x)+4*x^2*log(2)^2+48*log(2))/(16*log(2) ^2*exp(x)^2+(-8*x^2*log(2)^2+(-64*x+96)*log(2))*exp(x)+x^4*log(2)^2+(16*x^ 3-24*x^2)*log(2)+64*x^2-192*x+144),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {48 \log (2)+e^x (16-16 x) \log ^2(2)+4 x^2 \log ^2(2)}{144-192 x+64 x^2+\left (-24 x^2+16 x^3\right ) \log (2)+16 e^{2 x} \log ^2(2)+x^4 \log ^2(2)+e^x \left ((96-64 x) \log (2)-8 x^2 \log ^2(2)\right )} \, dx=-\frac {4 \, x \log \left (2\right )}{x^{2} \log \left (2\right ) - 4 \, e^{x} \log \left (2\right ) + 8 \, x - 12} \]
integrate(((-16*x+16)*log(2)^2*exp(x)+4*x^2*log(2)^2+48*log(2))/(16*log(2) ^2*exp(x)^2+(-8*x^2*log(2)^2+(-64*x+96)*log(2))*exp(x)+x^4*log(2)^2+(16*x^ 3-24*x^2)*log(2)+64*x^2-192*x+144),x, algorithm=\
Time = 10.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {48 \log (2)+e^x (16-16 x) \log ^2(2)+4 x^2 \log ^2(2)}{144-192 x+64 x^2+\left (-24 x^2+16 x^3\right ) \log (2)+16 e^{2 x} \log ^2(2)+x^4 \log ^2(2)+e^x \left ((96-64 x) \log (2)-8 x^2 \log ^2(2)\right )} \, dx=-\frac {4\,x\,\ln \left (2\right )}{8\,x+x^2\,\ln \left (2\right )-4\,{\mathrm {e}}^x\,\ln \left (2\right )-12} \]