Integrand size = 99, antiderivative size = 26 \[ \int \frac {e^{3 x} (2-6 x) \log (5)+e^x (4-4 x) \log (5)-8 e^{2 x} x^2 \log (5)}{e^{6 x}+16 e^x x+16 e^{3 x} x+4 e^{5 x} x+16 x^2+e^{4 x} \left (4+4 x^2\right )+e^{2 x} \left (4+16 x^2\right )} \, dx=\frac {\log (5)}{\left (2+e^{2 x}\right ) \left (1+\frac {e^x}{2 x}\right )} \]
Time = 1.98 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {e^{3 x} (2-6 x) \log (5)+e^x (4-4 x) \log (5)-8 e^{2 x} x^2 \log (5)}{e^{6 x}+16 e^x x+16 e^{3 x} x+4 e^{5 x} x+16 x^2+e^{4 x} \left (4+4 x^2\right )+e^{2 x} \left (4+16 x^2\right )} \, dx=\frac {2 x \log (5)}{\left (2+e^{2 x}\right ) \left (e^x+2 x\right )} \]
Integrate[(E^(3*x)*(2 - 6*x)*Log[5] + E^x*(4 - 4*x)*Log[5] - 8*E^(2*x)*x^2 *Log[5])/(E^(6*x) + 16*E^x*x + 16*E^(3*x)*x + 4*E^(5*x)*x + 16*x^2 + E^(4* x)*(4 + 4*x^2) + E^(2*x)*(4 + 16*x^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.
\(\displaystyle \int \frac {-8 e^{2 x} x^2 \log (5)+e^{3 x} (2-6 x) \log (5)+e^x (4-4 x) \log (5)}{16 x^2+e^{4 x} \left (4 x^2+4\right )+e^{2 x} \left (16 x^2+4\right )+16 e^x x+16 e^{3 x} x+4 e^{5 x} x+e^{6 x}} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 e^x \left (-4 e^x x^2-2 (x-1)-e^{2 x} (3 x-1)\right ) \log (5)}{\left (e^{2 x}+2\right )^2 \left (2 x+e^x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \log (5) \int \frac {e^x \left (-4 e^x x^2+e^{2 x} (1-3 x)+2 (1-x)\right )}{\left (2+e^{2 x}\right )^2 \left (2 x+e^x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \log (5) \int \left (-\frac {e^x (x-1)}{2 \left (2 x+e^x\right )^2 \left (2 x^2+1\right )}-\frac {e^x \left (-2 x^3-2 x^2+2 e^x x-x+1\right )}{2 \left (2+e^{2 x}\right ) \left (2 x^2+1\right )^2}-\frac {2 e^x x \left (e^x x+1\right )}{\left (2+e^{2 x}\right )^2 \left (2 x^2+1\right )}+\frac {e^x x}{\left (2 x+e^x\right ) \left (2 x^2+1\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \log (5) \left (-\int \frac {e^x}{\left (2+e^{2 x}\right ) \left (2 x^2+1\right )^2}dx-\int \frac {e^{2 x} x}{\left (2+e^{2 x}\right ) \left (2 x^2+1\right )^2}dx+\int \frac {e^x x}{\left (2 x+e^x\right ) \left (2 x^2+1\right )^2}dx+\frac {i \int \frac {1}{\left (2+e^{2 x}\right ) \left (i-\sqrt {2} x\right )^2}dx}{2 \sqrt {2}}+\frac {\int \frac {1}{\left (i \sqrt {2}-e^x\right ) \left (i-\sqrt {2} x\right )}dx}{8 \sqrt {2}}-\frac {1}{8} i \int \frac {1}{\left (i \sqrt {2}-e^x\right ) \left (i-\sqrt {2} x\right )}dx-\frac {\int \frac {1}{\left (i \sqrt {2}+e^x\right ) \left (i-\sqrt {2} x\right )}dx}{8 \sqrt {2}}+\frac {1}{8} i \int \frac {1}{\left (i \sqrt {2}+e^x\right ) \left (i-\sqrt {2} x\right )}dx+\frac {\int \frac {e^x}{\left (2+e^{2 x}\right )^2 \left (i-\sqrt {2} x\right )}dx}{\sqrt {2}}+\frac {\int \frac {e^x}{\left (2 x+e^x\right )^2 \left (i-\sqrt {2} x\right )}dx}{4 \sqrt {2}}+\frac {1}{4} i \int \frac {e^x}{\left (2 x+e^x\right )^2 \left (i-\sqrt {2} x\right )}dx-\frac {i \int \frac {1}{\left (2+e^{2 x}\right ) \left (\sqrt {2} x+i\right )^2}dx}{2 \sqrt {2}}-\frac {\int \frac {1}{\left (i \sqrt {2}-e^x\right ) \left (\sqrt {2} x+i\right )}dx}{8 \sqrt {2}}-\frac {1}{8} i \int \frac {1}{\left (i \sqrt {2}-e^x\right ) \left (\sqrt {2} x+i\right )}dx+\frac {\int \frac {1}{\left (i \sqrt {2}+e^x\right ) \left (\sqrt {2} x+i\right )}dx}{8 \sqrt {2}}+\frac {1}{8} i \int \frac {1}{\left (i \sqrt {2}+e^x\right ) \left (\sqrt {2} x+i\right )}dx-\frac {\int \frac {e^x}{\left (2+e^{2 x}\right )^2 \left (\sqrt {2} x+i\right )}dx}{\sqrt {2}}-\frac {\int \frac {e^x}{\left (2 x+e^x\right )^2 \left (\sqrt {2} x+i\right )}dx}{4 \sqrt {2}}+\frac {1}{4} i \int \frac {e^x}{\left (2 x+e^x\right )^2 \left (\sqrt {2} x+i\right )}dx+\frac {1}{2 \left (e^{2 x}+2\right )}-\frac {i}{4 \left (e^{2 x}+2\right ) \left (-\sqrt {2} x+i\right )}-\frac {i}{4 \left (e^{2 x}+2\right ) \left (\sqrt {2} x+i\right )}\right )\) |
Int[(E^(3*x)*(2 - 6*x)*Log[5] + E^x*(4 - 4*x)*Log[5] - 8*E^(2*x)*x^2*Log[5 ])/(E^(6*x) + 16*E^x*x + 16*E^(3*x)*x + 4*E^(5*x)*x + 16*x^2 + E^(4*x)*(4 + 4*x^2) + E^(2*x)*(4 + 16*x^2)),x]
3.17.58.3.1 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 = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85
method | result | size |
norman | \(\frac {2 x \ln \left (5\right )}{\left (2+{\mathrm e}^{2 x}\right ) \left ({\mathrm e}^{x}+2 x \right )}\) | \(22\) |
parallelrisch | \(\frac {2 x \ln \left (5\right )}{\left (2+{\mathrm e}^{2 x}\right ) \left ({\mathrm e}^{x}+2 x \right )}\) | \(22\) |
risch | \(\frac {2 \ln \left (5\right ) x}{{\mathrm e}^{3 x}+2 x \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x}+4 x}\) | \(27\) |
int(((-6*x+2)*ln(5)*exp(x)^3-8*x^2*ln(5)*exp(x)^2+(4-4*x)*ln(5)*exp(x))/(e xp(x)^6+4*x*exp(x)^5+(4*x^2+4)*exp(x)^4+16*x*exp(x)^3+(16*x^2+4)*exp(x)^2+ 16*exp(x)*x+16*x^2),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3 x} (2-6 x) \log (5)+e^x (4-4 x) \log (5)-8 e^{2 x} x^2 \log (5)}{e^{6 x}+16 e^x x+16 e^{3 x} x+4 e^{5 x} x+16 x^2+e^{4 x} \left (4+4 x^2\right )+e^{2 x} \left (4+16 x^2\right )} \, dx=\frac {2 \, x \log \left (5\right )}{2 \, x e^{\left (2 \, x\right )} + 4 \, x + e^{\left (3 \, x\right )} + 2 \, e^{x}} \]
integrate(((-6*x+2)*log(5)*exp(x)^3-8*x^2*log(5)*exp(x)^2+(4-4*x)*log(5)*e xp(x))/(exp(x)^6+4*x*exp(x)^5+(4*x^2+4)*exp(x)^4+16*x*exp(x)^3+(16*x^2+4)* exp(x)^2+16*exp(x)*x+16*x^2),x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^{3 x} (2-6 x) \log (5)+e^x (4-4 x) \log (5)-8 e^{2 x} x^2 \log (5)}{e^{6 x}+16 e^x x+16 e^{3 x} x+4 e^{5 x} x+16 x^2+e^{4 x} \left (4+4 x^2\right )+e^{2 x} \left (4+16 x^2\right )} \, dx=\frac {2 x \log {\left (5 \right )}}{2 x e^{2 x} + 4 x + e^{3 x} + 2 e^{x}} \]
integrate(((-6*x+2)*ln(5)*exp(x)**3-8*x**2*ln(5)*exp(x)**2+(4-4*x)*ln(5)*e xp(x))/(exp(x)**6+4*x*exp(x)**5+(4*x**2+4)*exp(x)**4+16*x*exp(x)**3+(16*x* *2+4)*exp(x)**2+16*exp(x)*x+16*x**2),x)
Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3 x} (2-6 x) \log (5)+e^x (4-4 x) \log (5)-8 e^{2 x} x^2 \log (5)}{e^{6 x}+16 e^x x+16 e^{3 x} x+4 e^{5 x} x+16 x^2+e^{4 x} \left (4+4 x^2\right )+e^{2 x} \left (4+16 x^2\right )} \, dx=\frac {2 \, x \log \left (5\right )}{2 \, x e^{\left (2 \, x\right )} + 4 \, x + e^{\left (3 \, x\right )} + 2 \, e^{x}} \]
integrate(((-6*x+2)*log(5)*exp(x)^3-8*x^2*log(5)*exp(x)^2+(4-4*x)*log(5)*e xp(x))/(exp(x)^6+4*x*exp(x)^5+(4*x^2+4)*exp(x)^4+16*x*exp(x)^3+(16*x^2+4)* exp(x)^2+16*exp(x)*x+16*x^2),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (22) = 44\).
Time = 0.50 (sec) , antiderivative size = 182, normalized size of antiderivative = 7.00 \[ \int \frac {e^{3 x} (2-6 x) \log (5)+e^x (4-4 x) \log (5)-8 e^{2 x} x^2 \log (5)}{e^{6 x}+16 e^x x+16 e^{3 x} x+4 e^{5 x} x+16 x^2+e^{4 x} \left (4+4 x^2\right )+e^{2 x} \left (4+16 x^2\right )} \, dx=\frac {32 \, x^{5} \log \left (5\right ) + 4 \, x^{4} e^{x} \log \left (5\right ) + 2 \, x^{3} e^{\left (2 \, x\right )} \log \left (5\right ) - 4 \, x^{3} e^{x} \log \left (5\right ) + 40 \, x^{3} \log \left (5\right ) - 2 \, x^{2} e^{\left (2 \, x\right )} \log \left (5\right ) + 2 \, x^{2} e^{x} \log \left (5\right ) - 8 \, x^{2} \log \left (5\right ) - 3 \, x e^{\left (2 \, x\right )} \log \left (5\right ) - 2 \, x e^{x} \log \left (5\right ) + 4 \, x \log \left (5\right ) + e^{\left (2 \, x\right )} \log \left (5\right )}{2 \, {\left (8 \, x^{5} e^{\left (2 \, x\right )} + 16 \, x^{5} + 4 \, x^{4} e^{\left (3 \, x\right )} + 8 \, x^{4} e^{x} + 8 \, x^{3} e^{\left (2 \, x\right )} + 16 \, x^{3} + 4 \, x^{2} e^{\left (3 \, x\right )} + 8 \, x^{2} e^{x} + 2 \, x e^{\left (2 \, x\right )} + 4 \, x + e^{\left (3 \, x\right )} + 2 \, e^{x}\right )}} \]
integrate(((-6*x+2)*log(5)*exp(x)^3-8*x^2*log(5)*exp(x)^2+(4-4*x)*log(5)*e xp(x))/(exp(x)^6+4*x*exp(x)^5+(4*x^2+4)*exp(x)^4+16*x*exp(x)^3+(16*x^2+4)* exp(x)^2+16*exp(x)*x+16*x^2),x, algorithm=\
1/2*(32*x^5*log(5) + 4*x^4*e^x*log(5) + 2*x^3*e^(2*x)*log(5) - 4*x^3*e^x*l og(5) + 40*x^3*log(5) - 2*x^2*e^(2*x)*log(5) + 2*x^2*e^x*log(5) - 8*x^2*lo g(5) - 3*x*e^(2*x)*log(5) - 2*x*e^x*log(5) + 4*x*log(5) + e^(2*x)*log(5))/ (8*x^5*e^(2*x) + 16*x^5 + 4*x^4*e^(3*x) + 8*x^4*e^x + 8*x^3*e^(2*x) + 16*x ^3 + 4*x^2*e^(3*x) + 8*x^2*e^x + 2*x*e^(2*x) + 4*x + e^(3*x) + 2*e^x)
Time = 10.44 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3 x} (2-6 x) \log (5)+e^x (4-4 x) \log (5)-8 e^{2 x} x^2 \log (5)}{e^{6 x}+16 e^x x+16 e^{3 x} x+4 e^{5 x} x+16 x^2+e^{4 x} \left (4+4 x^2\right )+e^{2 x} \left (4+16 x^2\right )} \, dx=\frac {2\,x\,\ln \left (5\right )}{4\,x+{\mathrm {e}}^{3\,x}+2\,{\mathrm {e}}^x+2\,x\,{\mathrm {e}}^{2\,x}} \]