Integrand size = 39, antiderivative size = 49 \[ \int \frac {\log \left (\frac {2 x \left (d \sqrt {-\frac {e}{d}}+e x\right )}{d+e x^2}\right )}{d+e x^2} \, dx=-\frac {\sqrt {-\frac {e}{d}} \operatorname {PolyLog}\left (2,1-\frac {2 x \left (d \sqrt {-\frac {e}{d}}+e x\right )}{d+e x^2}\right )}{2 e} \] Output:
-1/2*(-e/d)^(1/2)*polylog(2,1-2*x*(d*(-e/d)^(1/2)+e*x)/(e*x^2+d))/e
Leaf count is larger than twice the leaf count of optimal. \(625\) vs. \(2(49)=98\).
Time = 0.54 (sec) , antiderivative size = 625, normalized size of antiderivative = 12.76 \[ \int \frac {\log \left (\frac {2 x \left (d \sqrt {-\frac {e}{d}}+e x\right )}{d+e x^2}\right )}{d+e x^2} \, dx=\frac {-2 \log \left (\frac {\sqrt {e} x}{\sqrt {-d}}\right ) \log \left (\sqrt {-d}-\sqrt {e} x\right )+\log ^2\left (\sqrt {-d}-\sqrt {e} x\right )+2 \log \left (\frac {d \sqrt {e} x}{(-d)^{3/2}}\right ) \log \left (\sqrt {-d}+\sqrt {e} x\right )-\log ^2\left (\sqrt {-d}+\sqrt {e} x\right )+2 \log \left (\sqrt {-d}-\sqrt {e} x\right ) \log \left (\frac {d-\sqrt {-d} \sqrt {e} x}{2 d}\right )-2 \log \left (\sqrt {-d}+\sqrt {e} x\right ) \log \left (\frac {d+\sqrt {-d} \sqrt {e} x}{2 d}\right )-2 \log \left (\sqrt {-d}-\sqrt {e} x\right ) \log \left (\frac {d \sqrt {-\frac {e}{d}}+e x}{\sqrt {-d} \sqrt {e}+d \sqrt {-\frac {e}{d}}}\right )+2 \log \left (\sqrt {-d}+\sqrt {e} x\right ) \log \left (\frac {e+d \left (-\frac {e}{d}\right )^{3/2} x}{e+\sqrt {-d} \sqrt {e} \sqrt {-\frac {e}{d}}}\right )+2 \log \left (\sqrt {-d}-\sqrt {e} x\right ) \log \left (\frac {2 x \left (d \sqrt {-\frac {e}{d}}+e x\right )}{d+e x^2}\right )-2 \log \left (\sqrt {-d}+\sqrt {e} x\right ) \log \left (\frac {2 x \left (d \sqrt {-\frac {e}{d}}+e x\right )}{d+e x^2}\right )+2 \operatorname {PolyLog}\left (2,\frac {\sqrt {-d}+\sqrt {e} x}{\sqrt {-d}+\frac {\sqrt {e}}{\sqrt {-\frac {e}{d}}}}\right )+2 \operatorname {PolyLog}\left (2,1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )-2 \operatorname {PolyLog}\left (2,\frac {d-\sqrt {-d} \sqrt {e} x}{2 d}\right )+2 \operatorname {PolyLog}\left (2,\frac {d+\sqrt {-d} \sqrt {e} x}{2 d}\right )-2 \operatorname {PolyLog}\left (2,1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )-2 \operatorname {PolyLog}\left (2,\frac {\sqrt {-d} \sqrt {e}-e x}{\sqrt {-d} \sqrt {e}+d \sqrt {-\frac {e}{d}}}\right )}{4 \sqrt {-d} \sqrt {e}} \] Input:
Integrate[Log[(2*x*(d*Sqrt[-(e/d)] + e*x))/(d + e*x^2)]/(d + e*x^2),x]
Output:
(-2*Log[(Sqrt[e]*x)/Sqrt[-d]]*Log[Sqrt[-d] - Sqrt[e]*x] + Log[Sqrt[-d] - S qrt[e]*x]^2 + 2*Log[(d*Sqrt[e]*x)/(-d)^(3/2)]*Log[Sqrt[-d] + Sqrt[e]*x] - Log[Sqrt[-d] + Sqrt[e]*x]^2 + 2*Log[Sqrt[-d] - Sqrt[e]*x]*Log[(d - Sqrt[-d ]*Sqrt[e]*x)/(2*d)] - 2*Log[Sqrt[-d] + Sqrt[e]*x]*Log[(d + Sqrt[-d]*Sqrt[e ]*x)/(2*d)] - 2*Log[Sqrt[-d] - Sqrt[e]*x]*Log[(d*Sqrt[-(e/d)] + e*x)/(Sqrt [-d]*Sqrt[e] + d*Sqrt[-(e/d)])] + 2*Log[Sqrt[-d] + Sqrt[e]*x]*Log[(e + d*( -(e/d))^(3/2)*x)/(e + Sqrt[-d]*Sqrt[e]*Sqrt[-(e/d)])] + 2*Log[Sqrt[-d] - S qrt[e]*x]*Log[(2*x*(d*Sqrt[-(e/d)] + e*x))/(d + e*x^2)] - 2*Log[Sqrt[-d] + Sqrt[e]*x]*Log[(2*x*(d*Sqrt[-(e/d)] + e*x))/(d + e*x^2)] + 2*PolyLog[2, ( Sqrt[-d] + Sqrt[e]*x)/(Sqrt[-d] + Sqrt[e]/Sqrt[-(e/d)])] + 2*PolyLog[2, 1 + (Sqrt[e]*x)/Sqrt[-d]] - 2*PolyLog[2, (d - Sqrt[-d]*Sqrt[e]*x)/(2*d)] + 2 *PolyLog[2, (d + Sqrt[-d]*Sqrt[e]*x)/(2*d)] - 2*PolyLog[2, 1 + (d*Sqrt[e]* x)/(-d)^(3/2)] - 2*PolyLog[2, (Sqrt[-d]*Sqrt[e] - e*x)/(Sqrt[-d]*Sqrt[e] + d*Sqrt[-(e/d)])])/(4*Sqrt[-d]*Sqrt[e])
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2897}
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.
\(\displaystyle \int \frac {\log \left (\frac {2 x \left (d \sqrt {-\frac {e}{d}}+e x\right )}{d+e x^2}\right )}{d+e x^2} \, dx\) |
\(\Big \downarrow \) 2897 |
\(\displaystyle -\frac {\sqrt {-\frac {e}{d}} \operatorname {PolyLog}\left (2,1-\frac {2 x \left (\sqrt {-\frac {e}{d}} d+e x\right )}{e x^2+d}\right )}{2 e}\) |
Input:
Int[Log[(2*x*(d*Sqrt[-(e/d)] + e*x))/(d + e*x^2)]/(d + e*x^2),x]
Output:
-1/2*(Sqrt[-(e/d)]*PolyLog[2, 1 - (2*x*(d*Sqrt[-(e/d)] + e*x))/(d + e*x^2) ])/e
Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/ D[u, x])]}, Simp[C*PolyLog[2, 1 - u], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponents[u, x][[2]], Expon[Pq, x]]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.82 (sec) , antiderivative size = 238, normalized size of antiderivative = 4.86
method | result | size |
risch | \(\frac {\ln \left (2\right ) \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{\sqrt {d e}}+\frac {\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (e \,\textit {\_Z}^{2}+d \right )}{\sum }\frac {2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x \left (d \sqrt {-\frac {e}{d}}+e x \right )}{e \,x^{2}+d}\right )-2 \operatorname {dilog}\left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )-2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )-2 \operatorname {dilog}\left (\frac {d \sqrt {-\frac {e}{d}}+e \underline {\hspace {1.25 ex}}\alpha +\left (x -\underline {\hspace {1.25 ex}}\alpha \right ) e}{e \underline {\hspace {1.25 ex}}\alpha +d \sqrt {-\frac {e}{d}}}\right )-2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {d \sqrt {-\frac {e}{d}}+e \underline {\hspace {1.25 ex}}\alpha +\left (x -\underline {\hspace {1.25 ex}}\alpha \right ) e}{e \underline {\hspace {1.25 ex}}\alpha +d \sqrt {-\frac {e}{d}}}\right )+e \left (\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{\underline {\hspace {1.25 ex}}\alpha e}+\frac {2 \underline {\hspace {1.25 ex}}\alpha \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right )}{d}+\frac {2 \underline {\hspace {1.25 ex}}\alpha \operatorname {dilog}\left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right )}{d}\right )}{\underline {\hspace {1.25 ex}}\alpha }}{4 e}\) | \(238\) |
Input:
int(ln(2*x*(d*(-e/d)^(1/2)+e*x)/(e*x^2+d))/(e*x^2+d),x,method=_RETURNVERBO SE)
Output:
ln(2)/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))+1/4/e*sum(1/_alpha*(2*ln(x-_alph a)*ln(x*(d*(-e/d)^(1/2)+e*x)/(e*x^2+d))-2*dilog(x/_alpha)-2*ln(x-_alpha)*l n(x/_alpha)-2*dilog((d*(-e/d)^(1/2)+e*_alpha+(x-_alpha)*e)/(e*_alpha+d*(-e /d)^(1/2)))-2*ln(x-_alpha)*ln((d*(-e/d)^(1/2)+e*_alpha+(x-_alpha)*e)/(e*_a lpha+d*(-e/d)^(1/2)))+e*(1/_alpha/e*ln(x-_alpha)^2+2*_alpha/d*ln(x-_alpha) *ln(1/2*(x+_alpha)/_alpha)+2*_alpha/d*dilog(1/2*(x+_alpha)/_alpha))),_alph a=RootOf(_Z^2*e+d))
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90 \[ \int \frac {\log \left (\frac {2 x \left (d \sqrt {-\frac {e}{d}}+e x\right )}{d+e x^2}\right )}{d+e x^2} \, dx=-\frac {\sqrt {-\frac {e}{d}} {\rm Li}_2\left (-\frac {2 \, {\left (e x^{2} + d x \sqrt {-\frac {e}{d}}\right )}}{e x^{2} + d} + 1\right )}{2 \, e} \] Input:
integrate(log(2*x*(d*(-e/d)^(1/2)+e*x)/(e*x^2+d))/(e*x^2+d),x, algorithm=" fricas")
Output:
-1/2*sqrt(-e/d)*dilog(-2*(e*x^2 + d*x*sqrt(-e/d))/(e*x^2 + d) + 1)/e
Exception generated. \[ \int \frac {\log \left (\frac {2 x \left (d \sqrt {-\frac {e}{d}}+e x\right )}{d+e x^2}\right )}{d+e x^2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(ln(2*x*(d*(-e/d)**(1/2)+e*x)/(e*x**2+d))/(e*x**2+d),x)
Output:
Exception raised: TypeError >> Invalid comparison of non-real zoo
Exception generated. \[ \int \frac {\log \left (\frac {2 x \left (d \sqrt {-\frac {e}{d}}+e x\right )}{d+e x^2}\right )}{d+e x^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(log(2*x*(d*(-e/d)^(1/2)+e*x)/(e*x^2+d))/(e*x^2+d),x, algorithm=" maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {\log \left (\frac {2 x \left (d \sqrt {-\frac {e}{d}}+e x\right )}{d+e x^2}\right )}{d+e x^2} \, dx=\int { \frac {\log \left (\frac {2 \, {\left (e x + d \sqrt {-\frac {e}{d}}\right )} x}{e x^{2} + d}\right )}{e x^{2} + d} \,d x } \] Input:
integrate(log(2*x*(d*(-e/d)^(1/2)+e*x)/(e*x^2+d))/(e*x^2+d),x, algorithm=" giac")
Output:
integrate(log(2*(e*x + d*sqrt(-e/d))*x/(e*x^2 + d))/(e*x^2 + d), x)
Timed out. \[ \int \frac {\log \left (\frac {2 x \left (d \sqrt {-\frac {e}{d}}+e x\right )}{d+e x^2}\right )}{d+e x^2} \, dx=\int \frac {\ln \left (\frac {2\,x\,\left (e\,x+d\,\sqrt {-\frac {e}{d}}\right )}{e\,x^2+d}\right )}{e\,x^2+d} \,d x \] Input:
int(log((2*x*(e*x + d*(-e/d)^(1/2)))/(d + e*x^2))/(d + e*x^2),x)
Output:
int(log((2*x*(e*x + d*(-e/d)^(1/2)))/(d + e*x^2))/(d + e*x^2), x)
\[ \int \frac {\log \left (\frac {2 x \left (d \sqrt {-\frac {e}{d}}+e x\right )}{d+e x^2}\right )}{d+e x^2} \, dx=\frac {\sqrt {e}\, \sqrt {d}\, i \left (-2 \left (\int \frac {\mathrm {log}\left (\frac {2 \sqrt {d}\, e \,x^{2}+2 \sqrt {e}\, d i x}{\sqrt {d}\, d +\sqrt {d}\, e \,x^{2}}\right )}{e \,x^{3}+d x}d x \right ) d +\mathrm {log}\left (\frac {2 \sqrt {d}\, e \,x^{2}+2 \sqrt {e}\, d i x}{\sqrt {d}\, d +\sqrt {d}\, e \,x^{2}}\right )^{2}\right )}{2 d e} \] Input:
int(log(2*x*(d*(-e/d)^(1/2)+e*x)/(e*x^2+d))/(e*x^2+d),x)
Output:
(sqrt(e)*sqrt(d)*i*( - 2*int(log((2*sqrt(d)*e*x**2 + 2*sqrt(e)*d*i*x)/(sqr t(d)*d + sqrt(d)*e*x**2))/(d*x + e*x**3),x)*d + log((2*sqrt(d)*e*x**2 + 2* sqrt(e)*d*i*x)/(sqrt(d)*d + sqrt(d)*e*x**2))**2))/(2*d*e)