Integrand size = 76, antiderivative size = 34 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=\frac {1}{2 \left (2-\frac {-e^5+x}{\frac {x}{4}+\log \left (7-\frac {1}{\log (2)}\right )}\right )} \] Output:
2/(8-4*(-exp(5)+x)/(ln(7-1/ln(2))+1/4*x))
Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=-\frac {e^5+4 \log \left (7-\frac {1}{\log (2)}\right )}{2 \left (-2 e^5+x-4 \log \left (7-\frac {1}{\log (2)}\right )\right )} \] Input:
Integrate[(E^5 + 4*Log[(-1 + 7*Log[2])/Log[2]])/(8*E^10 - 8*E^5*x + 2*x^2 + (32*E^5 - 16*x)*Log[(-1 + 7*Log[2])/Log[2]] + 32*Log[(-1 + 7*Log[2])/Log [2]]^2),x]
Output:
-1/2*(E^5 + 4*Log[7 - Log[2]^(-1)])/(-2*E^5 + x - 4*Log[7 - Log[2]^(-1)])
Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {27, 2080, 1077, 17}
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.
\(\displaystyle \int \frac {e^5+4 \log \left (\frac {7 \log (2)-1}{\log (2)}\right )}{2 x^2-8 e^5 x+\left (32 e^5-16 x\right ) \log \left (\frac {7 \log (2)-1}{\log (2)}\right )+8 e^{10}+32 \log ^2\left (\frac {7 \log (2)-1}{\log (2)}\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \left (e^5+4 \log \left (7-\frac {1}{\log (2)}\right )\right ) \int \frac {1}{2 x^2-8 e^5 x+8 \left (e^{10}+4 \log ^2\left (7-\frac {1}{\log (2)}\right )\right )+16 \left (2 e^5-x\right ) \log \left (7-\frac {1}{\log (2)}\right )}dx\) |
\(\Big \downarrow \) 2080 |
\(\displaystyle \left (e^5+4 \log \left (7-\frac {1}{\log (2)}\right )\right ) \int \frac {1}{2 x^2-8 \left (e^5+2 \log \left (7-\frac {1}{\log (2)}\right )\right ) x+8 \left (e^5+2 \log \left (7-\frac {1}{\log (2)}\right )\right )^2}dx\) |
\(\Big \downarrow \) 1077 |
\(\displaystyle 2 \left (e^5+4 \log \left (7-\frac {1}{\log (2)}\right )\right ) \int \frac {1}{\left (2 x-4 \left (e^5+2 \log \left (7-\frac {1}{\log (2)}\right )\right )\right )^2}dx\) |
\(\Big \downarrow \) 17 |
\(\displaystyle -\frac {e^5+4 \log \left (7-\frac {1}{\log (2)}\right )}{2 \left (x-2 \left (e^5+2 \log \left (7-\frac {1}{\log (2)}\right )\right )\right )}\) |
Input:
Int[(E^5 + 4*Log[(-1 + 7*Log[2])/Log[2]])/(8*E^10 - 8*E^5*x + 2*x^2 + (32* E^5 - 16*x)*Log[(-1 + 7*Log[2])/Log[2]] + 32*Log[(-1 + 7*Log[2])/Log[2]]^2 ),x]
Output:
-1/2*(E^5 + 4*Log[7 - Log[2]^(-1)])/(x - 2*(E^5 + 2*Log[7 - Log[2]^(-1)]))
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int [(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && Q uadraticQ[u, x] && !QuadraticMatchQ[u, x]
Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.29
method | result | size |
gosper | \(\frac {4 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+{\mathrm e}^{5}}{8 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+4 \,{\mathrm e}^{5}-2 x}\) | \(44\) |
parallelrisch | \(\frac {4 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+{\mathrm e}^{5}}{8 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+4 \,{\mathrm e}^{5}-2 x}\) | \(44\) |
norman | \(\frac {-2 \ln \left (\ln \left (2\right )\right )+2 \ln \left (7 \ln \left (2\right )-1\right )+\frac {{\mathrm e}^{5}}{2}}{4 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+2 \,{\mathrm e}^{5}-x}\) | \(45\) |
risch | \(-\frac {\ln \left (7 \ln \left (2\right )-1\right )}{2 \left (\ln \left (\ln \left (2\right )\right )-\ln \left (7 \ln \left (2\right )-1\right )-\frac {{\mathrm e}^{5}}{2}+\frac {x}{4}\right )}+\frac {\ln \left (\ln \left (2\right )\right )}{2 \ln \left (\ln \left (2\right )\right )-2 \ln \left (7 \ln \left (2\right )-1\right )-{\mathrm e}^{5}+\frac {x}{2}}-\frac {{\mathrm e}^{5}}{8 \left (\ln \left (\ln \left (2\right )\right )-\ln \left (7 \ln \left (2\right )-1\right )-\frac {{\mathrm e}^{5}}{2}+\frac {x}{4}\right )}\) | \(86\) |
orering | \(\frac {\left (-4 \ln \left (\ln \left (2\right )\right )+4 \ln \left (7 \ln \left (2\right )-1\right )+2 \,{\mathrm e}^{5}-x \right ) \left (4 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+{\mathrm e}^{5}\right )}{32 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )^{2}+\left (32 \,{\mathrm e}^{5}-16 x \right ) \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+8 \,{\mathrm e}^{10}-8 x \,{\mathrm e}^{5}+2 x^{2}}\) | \(97\) |
meijerg | \(-\frac {\ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right ) x}{\left (-4 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )-2 \,{\mathrm e}^{5}\right ) \left (2 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+{\mathrm e}^{5}\right ) \left (1-\frac {x}{2 \left (2 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+{\mathrm e}^{5}\right )}\right )}-\frac {{\mathrm e}^{5} x}{4 \left (-4 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )-2 \,{\mathrm e}^{5}\right ) \left (2 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+{\mathrm e}^{5}\right ) \left (1-\frac {x}{2 \left (2 \ln \left (\frac {7 \ln \left (2\right )-1}{\ln \left (2\right )}\right )+{\mathrm e}^{5}\right )}\right )}\) | \(154\) |
Input:
int((4*ln((7*ln(2)-1)/ln(2))+exp(5))/(32*ln((7*ln(2)-1)/ln(2))^2+(32*exp(5 )-16*x)*ln((7*ln(2)-1)/ln(2))+8*exp(5)^2-8*x*exp(5)+2*x^2),x,method=_RETUR NVERBOSE)
Output:
1/2*(4*ln((7*ln(2)-1)/ln(2))+exp(5))/(4*ln((7*ln(2)-1)/ln(2))+2*exp(5)-x)
Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=-\frac {e^{5} + 4 \, \log \left (\frac {7 \, \log \left (2\right ) - 1}{\log \left (2\right )}\right )}{2 \, {\left (x - 2 \, e^{5} - 4 \, \log \left (\frac {7 \, \log \left (2\right ) - 1}{\log \left (2\right )}\right )\right )}} \] Input:
integrate((4*log((7*log(2)-1)/log(2))+exp(5))/(32*log((7*log(2)-1)/log(2)) ^2+(32*exp(5)-16*x)*log((7*log(2)-1)/log(2))+8*exp(5)^2-8*x*exp(5)+2*x^2), x, algorithm="fricas")
Output:
-1/2*(e^5 + 4*log((7*log(2) - 1)/log(2)))/(x - 2*e^5 - 4*log((7*log(2) - 1 )/log(2)))
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=- \frac {- 4 \log {\left (\log {\left (2 \right )} \right )} + 4 \log {\left (-1 + 7 \log {\left (2 \right )} \right )} + e^{5}}{2 x - 4 e^{5} - 8 \log {\left (-1 + 7 \log {\left (2 \right )} \right )} + 8 \log {\left (\log {\left (2 \right )} \right )}} \] Input:
integrate((4*ln((7*ln(2)-1)/ln(2))+exp(5))/(32*ln((7*ln(2)-1)/ln(2))**2+(3 2*exp(5)-16*x)*ln((7*ln(2)-1)/ln(2))+8*exp(5)**2-8*x*exp(5)+2*x**2),x)
Output:
-(-4*log(log(2)) + 4*log(-1 + 7*log(2)) + exp(5))/(2*x - 4*exp(5) - 8*log( -1 + 7*log(2)) + 8*log(log(2)))
Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=-\frac {e^{5} + 4 \, \log \left (\frac {7 \, \log \left (2\right ) - 1}{\log \left (2\right )}\right )}{2 \, {\left (x - 2 \, e^{5} - 4 \, \log \left (\frac {7 \, \log \left (2\right ) - 1}{\log \left (2\right )}\right )\right )}} \] Input:
integrate((4*log((7*log(2)-1)/log(2))+exp(5))/(32*log((7*log(2)-1)/log(2)) ^2+(32*exp(5)-16*x)*log((7*log(2)-1)/log(2))+8*exp(5)^2-8*x*exp(5)+2*x^2), x, algorithm="maxima")
Output:
-1/2*(e^5 + 4*log((7*log(2) - 1)/log(2)))/(x - 2*e^5 - 4*log((7*log(2) - 1 )/log(2)))
Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=-\frac {e^{5} + 4 \, \log \left (\frac {7 \, \log \left (2\right ) - 1}{\log \left (2\right )}\right )}{2 \, {\left (x - 2 \, e^{5} - 4 \, \log \left (\frac {7 \, \log \left (2\right ) - 1}{\log \left (2\right )}\right )\right )}} \] Input:
integrate((4*log((7*log(2)-1)/log(2))+exp(5))/(32*log((7*log(2)-1)/log(2)) ^2+(32*exp(5)-16*x)*log((7*log(2)-1)/log(2))+8*exp(5)^2-8*x*exp(5)+2*x^2), x, algorithm="giac")
Output:
-1/2*(e^5 + 4*log((7*log(2) - 1)/log(2)))/(x - 2*e^5 - 4*log((7*log(2) - 1 )/log(2)))
Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=-\frac {2\,\ln \left (\ln \left (128\right )-1\right )+\frac {{\mathrm {e}}^5}{2}-2\,\ln \left (\ln \left (2\right )\right )}{x-4\,\ln \left (\ln \left (128\right )-1\right )-2\,{\mathrm {e}}^5+4\,\ln \left (\ln \left (2\right )\right )} \] Input:
int((exp(5) + 4*log((7*log(2) - 1)/log(2)))/(8*exp(10) - log((7*log(2) - 1 )/log(2))*(16*x - 32*exp(5)) + 32*log((7*log(2) - 1)/log(2))^2 - 8*x*exp(5 ) + 2*x^2),x)
Output:
-(2*log(log(128) - 1) + exp(5)/2 - 2*log(log(2)))/(x - 4*log(log(128) - 1) - 2*exp(5) + 4*log(log(2)))
Time = 0.15 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.41 \[ \int \frac {e^5+4 \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )}{8 e^{10}-8 e^5 x+2 x^2+\left (32 e^5-16 x\right ) \log \left (\frac {-1+7 \log (2)}{\log (2)}\right )+32 \log ^2\left (\frac {-1+7 \log (2)}{\log (2)}\right )} \, dx=\frac {x \left (4 \,\mathrm {log}\left (\frac {7 \,\mathrm {log}\left (2\right )-1}{\mathrm {log}\left (2\right )}\right )+e^{5}\right )}{32 \mathrm {log}\left (\frac {7 \,\mathrm {log}\left (2\right )-1}{\mathrm {log}\left (2\right )}\right )^{2}+32 \,\mathrm {log}\left (\frac {7 \,\mathrm {log}\left (2\right )-1}{\mathrm {log}\left (2\right )}\right ) e^{5}-8 \,\mathrm {log}\left (\frac {7 \,\mathrm {log}\left (2\right )-1}{\mathrm {log}\left (2\right )}\right ) x +8 e^{10}-4 e^{5} x} \] Input:
int((4*log((7*log(2)-1)/log(2))+exp(5))/(32*log((7*log(2)-1)/log(2))^2+(32 *exp(5)-16*x)*log((7*log(2)-1)/log(2))+8*exp(5)^2-8*x*exp(5)+2*x^2),x)
Output:
(x*(4*log((7*log(2) - 1)/log(2)) + e**5))/(4*(8*log((7*log(2) - 1)/log(2)) **2 + 8*log((7*log(2) - 1)/log(2))*e**5 - 2*log((7*log(2) - 1)/log(2))*x + 2*e**10 - e**5*x))