Integrand size = 69, antiderivative size = 23 \[ \int \frac {\left (15 e^3-30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )+\left (30 x+\left (-15 e^3+30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{4 e^3 x^2-8 x^3} \, dx=\frac {15 \log (x) \log \left (\frac {5}{-e^3+2 x}\right )}{4 x} \]
Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\left (15 e^3-30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )+\left (30 x+\left (-15 e^3+30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{4 e^3 x^2-8 x^3} \, dx=\frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x} \]
Integrate[((15*E^3 - 30*x)*Log[-5/(E^3 - 2*x)] + (30*x + (-15*E^3 + 30*x)* Log[-5/(E^3 - 2*x)])*Log[x])/(4*E^3*x^2 - 8*x^3),x]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 1.57 (sec) , antiderivative size = 92, normalized size of antiderivative = 4.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.072, Rules used = {2026, 7239, 27, 2010, 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.
\(\displaystyle \int \frac {\left (15 e^3-30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )+\left (30 x+\left (30 x-15 e^3\right ) \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{4 e^3 x^2-8 x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (15 e^3-30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )+\left (30 x+\left (30 x-15 e^3\right ) \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{\left (4 e^3-8 x\right ) x^2}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {15 \left (\frac {2 x \log (x)}{e^3-2 x}-\log \left (-\frac {5}{e^3-2 x}\right ) (\log (x)-1)\right )}{4 x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {15}{4} \int \frac {\log \left (-\frac {5}{e^3-2 x}\right ) (1-\log (x))+\frac {2 x \log (x)}{e^3-2 x}}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {15}{4} \int \left (\frac {\log \left (-\frac {5}{e^3-2 x}\right )}{x^2}-\frac {\left (2 \log \left (-\frac {5}{e^3-2 x}\right ) x+2 x-e^3 \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{x^2 \left (2 x-e^3\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {15}{4} \left (\frac {2 \operatorname {PolyLog}\left (2,\frac {2 x}{e^3}\right )}{e^3}+\frac {2 \operatorname {PolyLog}\left (2,1-\frac {2 x}{e^3}\right )}{e^3}+\frac {2 \log (x) \log \left (e^3-2 x\right )}{e^3}-\frac {2 (3-\log (2)) \log \left (e^3-2 x\right )}{e^3}+\frac {\log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{x}-\frac {6 \log (x)}{e^3}\right )\) |
Int[((15*E^3 - 30*x)*Log[-5/(E^3 - 2*x)] + (30*x + (-15*E^3 + 30*x)*Log[-5 /(E^3 - 2*x)])*Log[x])/(4*E^3*x^2 - 8*x^3),x]
(15*((-2*(3 - Log[2])*Log[E^3 - 2*x])/E^3 - (6*Log[x])/E^3 + (Log[-5/(E^3 - 2*x)]*Log[x])/x + (2*Log[E^3 - 2*x]*Log[x])/E^3 + (2*PolyLog[2, (2*x)/E^ 3])/E^3 + (2*PolyLog[2, 1 - (2*x)/E^3])/E^3))/4
3.15.59.3.1 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_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.33 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83
method | result | size |
parallelrisch | \(\frac {15 \ln \left (x \right ) \ln \left (-\frac {5}{{\mathrm e}^{3}-2 x}\right )}{4 x}\) | \(19\) |
risch | \(-\frac {15 \ln \left (x \right ) \ln \left ({\mathrm e}^{3}-2 x \right )}{4 x}+\frac {15 \left (-2 i \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{3}-2 x}\right )^{2}+2 i \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{3}-2 x}\right )^{3}+2 i \pi +2 \ln \left (5\right )\right ) \ln \left (x \right )}{8 x}\) | \(68\) |
default | \(-\frac {15 \left (\ln \left (x \right ) \ln \left (-{\mathrm e}^{3}+2 x \right )-\ln \left (\frac {1}{-{\mathrm e}^{3}+2 x}\right )-\left (\ln \left (\frac {1}{-{\mathrm e}^{3}+2 x}\right )+\ln \left (-{\mathrm e}^{3}+2 x \right )\right ) \ln \left (x \right )+2 \,{\mathrm e}^{-3} \ln \left (x \right ) x -2 \,{\mathrm e}^{-3} x \ln \left (-{\mathrm e}^{3}+2 x \right )\right )}{4 x}+\frac {15 \ln \left (5\right ) \ln \left (x \right )}{4 x}+\frac {15 \ln \left (\frac {{\mathrm e}^{3}}{-{\mathrm e}^{3}+2 x}+1\right ) {\mathrm e}^{-3}}{2}-\frac {15 \ln \left (\frac {1}{-{\mathrm e}^{3}+2 x}\right )}{2 \left (-{\mathrm e}^{3}+2 x \right ) \left (\frac {{\mathrm e}^{3}}{-{\mathrm e}^{3}+2 x}+1\right )}\) | \(154\) |
parts | \(-\frac {15 \left (\ln \left (x \right ) \ln \left (-{\mathrm e}^{3}+2 x \right )-\ln \left (\frac {1}{-{\mathrm e}^{3}+2 x}\right )-\left (\ln \left (\frac {1}{-{\mathrm e}^{3}+2 x}\right )+\ln \left (-{\mathrm e}^{3}+2 x \right )\right ) \ln \left (x \right )+2 \,{\mathrm e}^{-3} \ln \left (x \right ) x -2 \,{\mathrm e}^{-3} x \ln \left (-{\mathrm e}^{3}+2 x \right )\right )}{4 x}-\frac {15 \ln \left (5\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )}{4}+\frac {15 \ln \left (\frac {5 \,{\mathrm e}^{3}}{-{\mathrm e}^{3}+2 x}+5\right ) {\mathrm e}^{-3}}{2}-\frac {75 \ln \left (\frac {5}{-{\mathrm e}^{3}+2 x}\right )}{2 \left (-{\mathrm e}^{3}+2 x \right ) \left (\frac {5 \,{\mathrm e}^{3}}{-{\mathrm e}^{3}+2 x}+5\right )}\) | \(166\) |
int((((-15*exp(3)+30*x)*ln(-5/(exp(3)-2*x))+30*x)*ln(x)+(15*exp(3)-30*x)*l n(-5/(exp(3)-2*x)))/(4*x^2*exp(3)-8*x^3),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {\left (15 e^3-30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )+\left (30 x+\left (-15 e^3+30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{4 e^3 x^2-8 x^3} \, dx=\frac {15 \, \log \left (x\right ) \log \left (\frac {5}{2 \, x - e^{3}}\right )}{4 \, x} \]
integrate((((-15*exp(3)+30*x)*log(-5/(exp(3)-2*x))+30*x)*log(x)+(15*exp(3) -30*x)*log(-5/(exp(3)-2*x)))/(4*x^2*exp(3)-8*x^3),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\left (15 e^3-30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )+\left (30 x+\left (-15 e^3+30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{4 e^3 x^2-8 x^3} \, dx=\frac {15 \log {\left (x \right )} \log {\left (- \frac {5}{- 2 x + e^{3}} \right )}}{4 x} \]
integrate((((-15*exp(3)+30*x)*ln(-5/(exp(3)-2*x))+30*x)*ln(x)+(15*exp(3)-3 0*x)*ln(-5/(exp(3)-2*x)))/(4*x**2*exp(3)-8*x**3),x)
Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {\left (15 e^3-30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )+\left (30 x+\left (-15 e^3+30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{4 e^3 x^2-8 x^3} \, dx=\frac {15 \, {\left (\log \left (5\right ) \log \left (x\right ) - \log \left (2 \, x - e^{3}\right ) \log \left (x\right )\right )}}{4 \, x} \]
integrate((((-15*exp(3)+30*x)*log(-5/(exp(3)-2*x))+30*x)*log(x)+(15*exp(3) -30*x)*log(-5/(exp(3)-2*x)))/(4*x^2*exp(3)-8*x^3),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (20) = 40\).
Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.00 \[ \int \frac {\left (15 e^3-30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )+\left (30 x+\left (-15 e^3+30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{4 e^3 x^2-8 x^3} \, dx=\frac {15 \, {\left (\pi ^{2} \mathrm {sgn}\left (2 \, x - e^{3}\right ) \mathrm {sgn}\left (x\right ) - \pi ^{2} \mathrm {sgn}\left (2 \, x - e^{3}\right ) + 3 \, \pi ^{2} \mathrm {sgn}\left (x\right ) - 3 \, \pi ^{2} + 4 \, \log \left (5\right ) \log \left ({\left | x \right |}\right ) - 4 \, \log \left ({\left | 2 \, x - e^{3} \right |}\right ) \log \left ({\left | x \right |}\right )\right )}}{16 \, x} \]
integrate((((-15*exp(3)+30*x)*log(-5/(exp(3)-2*x))+30*x)*log(x)+(15*exp(3) -30*x)*log(-5/(exp(3)-2*x)))/(4*x^2*exp(3)-8*x^3),x, algorithm=\
15/16*(pi^2*sgn(2*x - e^3)*sgn(x) - pi^2*sgn(2*x - e^3) + 3*pi^2*sgn(x) - 3*pi^2 + 4*log(5)*log(abs(x)) - 4*log(abs(2*x - e^3))*log(abs(x)))/x
Time = 11.45 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\left (15 e^3-30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )+\left (30 x+\left (-15 e^3+30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{4 e^3 x^2-8 x^3} \, dx=\frac {15\,\ln \left (x\right )\,\left (\ln \left (5\right )+\ln \left (\frac {1}{2\,x-{\mathrm {e}}^3}\right )\right )}{4\,x} \]