Integrand size = 59, antiderivative size = 27 \[ \int \frac {4-2 \log (5)+e^{2 x} (4+4 x+(1-2 x) \log (5))}{216+288 x+96 x^2+\left (-144 x-96 x^2\right ) \log (5)+24 x^2 \log ^2(5)} \, dx=\frac {2-e^{2 x}}{24 x \left (-2-\frac {3}{x}+\log (5)\right )} \]
Time = 0.71 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {4-2 \log (5)+e^{2 x} (4+4 x+(1-2 x) \log (5))}{216+288 x+96 x^2+\left (-144 x-96 x^2\right ) \log (5)+24 x^2 \log ^2(5)} \, dx=\frac {2-e^{2 x}}{24 (-3+x (-2+\log (5)))} \]
Integrate[(4 - 2*Log[5] + E^(2*x)*(4 + 4*x + (1 - 2*x)*Log[5]))/(216 + 288 *x + 96*x^2 + (-144*x - 96*x^2)*Log[5] + 24*x^2*Log[5]^2),x]
Time = 0.71 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.102, Rules used = {6, 7292, 7277, 27, 7293, 2009}
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.
\(\displaystyle \int \frac {e^{2 x} (4 x+(1-2 x) \log (5)+4)+4-2 \log (5)}{96 x^2+24 x^2 \log ^2(5)+\left (-96 x^2-144 x\right ) \log (5)+288 x+216} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{2 x} (4 x+(1-2 x) \log (5)+4)+4-2 \log (5)}{x^2 \left (96+24 \log ^2(5)\right )+\left (-96 x^2-144 x\right ) \log (5)+288 x+216}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 x} (4 x+(1-2 x) \log (5)+4)+4 \left (1-\frac {\log (5)}{2}\right )}{24 x^2 (2-\log (5))^2+144 x (2-\log (5))+216}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 96 (2-\log (5))^2 \int \frac {e^{2 x} (\log (5) (1-2 x)+4 x+4)+2 (2-\log (5))}{2304 ((2-\log (5)) x+3)^2 (2-\log (5))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{24} \int \frac {e^{2 x} (\log (5) (1-2 x)+4 x+4)+2 (2-\log (5))}{((2-\log (5)) x+3)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{24} \int \left (\frac {e^{2 x} (2 (2-\log (5)) x+\log (5)+4)}{((2-\log (5)) x+3)^2}+\frac {2 (2-\log (5))}{((2-\log (5)) x+3)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{24} \left (\frac {e^{2 x}}{x (2-\log (5))+3}-\frac {2}{x (2-\log (5))+3}\right )\) |
Int[(4 - 2*Log[5] + E^(2*x)*(4 + 4*x + (1 - 2*x)*Log[5]))/(216 + 288*x + 9 6*x^2 + (-144*x - 96*x^2)*Log[5] + 24*x^2*Log[5]^2),x]
3.3.67.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
norman | \(\frac {\left (-\frac {1}{18}+\frac {\ln \left (5\right )}{36}\right ) x -\frac {{\mathrm e}^{2 x}}{24}}{x \ln \left (5\right )-2 x -3}\) | \(28\) |
parallelrisch | \(-\frac {-2 x \ln \left (5\right )+4 x +3 \,{\mathrm e}^{2 x}}{72 \left (x \ln \left (5\right )-2 x -3\right )}\) | \(29\) |
risch | \(\frac {1}{12 x \ln \left (5\right )-24 x -36}-\frac {{\mathrm e}^{2 x}}{24 \left (x \ln \left (5\right )-2 x -3\right )}\) | \(32\) |
parts | \(-\frac {\frac {1}{6}-\frac {\ln \left (5\right )}{12}}{\left (2-\ln \left (5\right )\right ) \left (-x \ln \left (5\right )+2 x +3\right )}+\frac {\ln \left (5\right ) {\mathrm e}^{2 x}}{2 \left (\ln \left (5\right )-2\right )^{3} \left (2 x -\frac {6}{\ln \left (5\right )-2}\right )}+\frac {\ln \left (5\right ) {\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )}{2 \left (\ln \left (5\right )-2\right )^{3}}-\frac {{\mathrm e}^{2 x}}{3 \left (\ln \left (5\right )-2\right )^{2} \left (2 x -\frac {6}{\ln \left (5\right )-2}\right )}-\frac {{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )}{2 \left (\ln \left (5\right )-2\right )^{2}}-\frac {{\mathrm e}^{2 x}}{\left (\ln \left (5\right )-2\right )^{3} \left (2 x -\frac {6}{\ln \left (5\right )-2}\right )}-\frac {{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )}{\left (\ln \left (5\right )-2\right )^{3}}-\frac {\ln \left (5\right ) {\mathrm e}^{2 x}}{12 \left (\ln \left (5\right )-2\right )^{2} \left (2 x -\frac {6}{\ln \left (5\right )-2}\right )}\) | \(233\) |
derivativedivides | \(\frac {\ln \left (5\right ) \left (-\frac {{\mathrm e}^{2 x}}{2 x -\frac {6}{\ln \left (5\right )-2}}-{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )\right )}{12 \left (\ln \left (5\right )-2\right )^{2}}-\frac {1}{3 \left (2-\ln \left (5\right )\right ) \left (-2 x \ln \left (5\right )+4 x +6\right )}+\frac {-\frac {{\mathrm e}^{2 x}}{2 x -\frac {6}{\ln \left (5\right )-2}}-{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )}{3 \left (\ln \left (5\right )-2\right )^{2}}+\frac {\ln \left (5\right )}{6 \left (2-\ln \left (5\right )\right ) \left (-2 x \ln \left (5\right )+4 x +6\right )}+\frac {-\frac {{\mathrm e}^{2 x}}{2 x -\frac {6}{\ln \left (5\right )-2}}-{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )}{\left (\ln \left (5\right )-2\right )^{3}}-\frac {{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )}{6 \left (\ln \left (5\right )-2\right )^{2}}-\frac {\ln \left (5\right ) \left (\frac {-\frac {6 \,{\mathrm e}^{2 x}}{2 x -\frac {6}{\ln \left (5\right )-2}}-6 \,{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )}{\left (\ln \left (5\right )-2\right )^{3}}-\frac {{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )}{\left (\ln \left (5\right )-2\right )^{2}}\right )}{12}\) | \(332\) |
default | \(\frac {\ln \left (5\right ) \left (-\frac {{\mathrm e}^{2 x}}{2 x -\frac {6}{\ln \left (5\right )-2}}-{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )\right )}{12 \left (\ln \left (5\right )-2\right )^{2}}-\frac {1}{3 \left (2-\ln \left (5\right )\right ) \left (-2 x \ln \left (5\right )+4 x +6\right )}+\frac {-\frac {{\mathrm e}^{2 x}}{2 x -\frac {6}{\ln \left (5\right )-2}}-{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )}{3 \left (\ln \left (5\right )-2\right )^{2}}+\frac {\ln \left (5\right )}{6 \left (2-\ln \left (5\right )\right ) \left (-2 x \ln \left (5\right )+4 x +6\right )}+\frac {-\frac {{\mathrm e}^{2 x}}{2 x -\frac {6}{\ln \left (5\right )-2}}-{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )}{\left (\ln \left (5\right )-2\right )^{3}}-\frac {{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )}{6 \left (\ln \left (5\right )-2\right )^{2}}-\frac {\ln \left (5\right ) \left (\frac {-\frac {6 \,{\mathrm e}^{2 x}}{2 x -\frac {6}{\ln \left (5\right )-2}}-6 \,{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )}{\left (\ln \left (5\right )-2\right )^{3}}-\frac {{\mathrm e}^{\frac {6}{\ln \left (5\right )-2}} \operatorname {Ei}_{1}\left (-2 x +\frac {6}{\ln \left (5\right )-2}\right )}{\left (\ln \left (5\right )-2\right )^{2}}\right )}{12}\) | \(332\) |
int((((1-2*x)*ln(5)+4*x+4)*exp(2*x)-2*ln(5)+4)/(24*x^2*ln(5)^2+(-96*x^2-14 4*x)*ln(5)+96*x^2+288*x+216),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {4-2 \log (5)+e^{2 x} (4+4 x+(1-2 x) \log (5))}{216+288 x+96 x^2+\left (-144 x-96 x^2\right ) \log (5)+24 x^2 \log ^2(5)} \, dx=-\frac {e^{\left (2 \, x\right )} - 2}{24 \, {\left (x \log \left (5\right ) - 2 \, x - 3\right )}} \]
integrate((((1-2*x)*log(5)+4*x+4)*exp(2*x)-2*log(5)+4)/(24*x^2*log(5)^2+(- 96*x^2-144*x)*log(5)+96*x^2+288*x+216),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).
Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {4-2 \log (5)+e^{2 x} (4+4 x+(1-2 x) \log (5))}{216+288 x+96 x^2+\left (-144 x-96 x^2\right ) \log (5)+24 x^2 \log ^2(5)} \, dx=- \frac {2 - \log {\left (5 \right )}}{x \left (- 48 \log {\left (5 \right )} + 12 \log {\left (5 \right )}^{2} + 48\right ) - 36 \log {\left (5 \right )} + 72} - \frac {e^{2 x}}{- 48 x + 24 x \log {\left (5 \right )} - 72} \]
integrate((((1-2*x)*ln(5)+4*x+4)*exp(2*x)-2*ln(5)+4)/(24*x**2*ln(5)**2+(-9 6*x**2-144*x)*ln(5)+96*x**2+288*x+216),x)
-(2 - log(5))/(x*(-48*log(5) + 12*log(5)**2 + 48) - 36*log(5) + 72) - exp( 2*x)/(-48*x + 24*x*log(5) - 72)
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (24) = 48\).
Time = 0.32 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.33 \[ \int \frac {4-2 \log (5)+e^{2 x} (4+4 x+(1-2 x) \log (5))}{216+288 x+96 x^2+\left (-144 x-96 x^2\right ) \log (5)+24 x^2 \log ^2(5)} \, dx=-\frac {e^{\left (2 \, x\right )}}{24 \, {\left (x {\left (\log \left (5\right ) - 2\right )} - 3\right )}} + \frac {\log \left (5\right )}{12 \, {\left ({\left (\log \left (5\right )^{2} - 4 \, \log \left (5\right ) + 4\right )} x - 3 \, \log \left (5\right ) + 6\right )}} - \frac {1}{6 \, {\left ({\left (\log \left (5\right )^{2} - 4 \, \log \left (5\right ) + 4\right )} x - 3 \, \log \left (5\right ) + 6\right )}} \]
integrate((((1-2*x)*log(5)+4*x+4)*exp(2*x)-2*log(5)+4)/(24*x^2*log(5)^2+(- 96*x^2-144*x)*log(5)+96*x^2+288*x+216),x, algorithm=\
-1/24*e^(2*x)/(x*(log(5) - 2) - 3) + 1/12*log(5)/((log(5)^2 - 4*log(5) + 4 )*x - 3*log(5) + 6) - 1/6/((log(5)^2 - 4*log(5) + 4)*x - 3*log(5) + 6)
Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {4-2 \log (5)+e^{2 x} (4+4 x+(1-2 x) \log (5))}{216+288 x+96 x^2+\left (-144 x-96 x^2\right ) \log (5)+24 x^2 \log ^2(5)} \, dx=-\frac {e^{\left (2 \, x\right )} - 2}{24 \, {\left (x \log \left (5\right ) - 2 \, x - 3\right )}} \]
integrate((((1-2*x)*log(5)+4*x+4)*exp(2*x)-2*log(5)+4)/(24*x^2*log(5)^2+(- 96*x^2-144*x)*log(5)+96*x^2+288*x+216),x, algorithm=\
Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {4-2 \log (5)+e^{2 x} (4+4 x+(1-2 x) \log (5))}{216+288 x+96 x^2+\left (-144 x-96 x^2\right ) \log (5)+24 x^2 \log ^2(5)} \, dx=-\frac {\frac {{\mathrm {e}}^{2\,x}}{24}-\frac {1}{12}}{x\,\left (\ln \left (5\right )-2\right )-3} \]