Integrand size = 15, antiderivative size = 88 \[ \int \frac {\text {arctanh}(x)}{1-\sqrt {2} x} \, dx=-\frac {\text {arctanh}\left (\frac {1}{\sqrt {2}}\right ) \log \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\operatorname {PolyLog}\left (2,-\frac {\sqrt {2}-2 x}{2-\sqrt {2}}\right )}{2 \sqrt {2}}+\frac {\operatorname {PolyLog}\left (2,\frac {\sqrt {2}-2 x}{2+\sqrt {2}}\right )}{2 \sqrt {2}} \]
-1/2*arctanh(1/2*2^(1/2))*ln(1-x*2^(1/2))*2^(1/2)-1/4*polylog(2,(2*x-2^(1/ 2))/(2-2^(1/2)))*2^(1/2)+1/4*polylog(2,(-2*x+2^(1/2))/(2+2^(1/2)))*2^(1/2)
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 272, normalized size of antiderivative = 3.09 \[ \int \frac {\text {arctanh}(x)}{1-\sqrt {2} x} \, dx=\frac {\pi ^2-4 \text {arctanh}\left (\frac {1}{\sqrt {2}}\right )^2-4 i \pi \text {arctanh}(x)+8 \text {arctanh}\left (\frac {1}{\sqrt {2}}\right ) \text {arctanh}(x)-8 \text {arctanh}(x)^2+8 \text {arctanh}\left (\frac {1}{\sqrt {2}}\right ) \log \left (1-e^{2 \text {arctanh}\left (\frac {1}{\sqrt {2}}\right )-2 \text {arctanh}(x)}\right )-8 \text {arctanh}(x) \log \left (1-e^{2 \text {arctanh}\left (\frac {1}{\sqrt {2}}\right )-2 \text {arctanh}(x)}\right )+4 i \pi \log \left (1+e^{2 \text {arctanh}(x)}\right )+8 \text {arctanh}(x) \log \left (1+e^{2 \text {arctanh}(x)}\right )-4 i \pi \log \left (\frac {2}{\sqrt {1-x^2}}\right )-8 \text {arctanh}(x) \log \left (\frac {2}{\sqrt {1-x^2}}\right )-4 \text {arctanh}(x) \log \left (1-x^2\right )-8 \text {arctanh}(x) \log \left (-i \sinh \left (\text {arctanh}\left (\frac {1}{\sqrt {2}}\right )-\text {arctanh}(x)\right )\right )-8 \text {arctanh}\left (\frac {1}{\sqrt {2}}\right ) \log \left (-2 i \sinh \left (\text {arctanh}\left (\frac {1}{\sqrt {2}}\right )-\text {arctanh}(x)\right )\right )+8 \text {arctanh}(x) \log \left (-2 i \sinh \left (\text {arctanh}\left (\frac {1}{\sqrt {2}}\right )-\text {arctanh}(x)\right )\right )+4 \operatorname {PolyLog}\left (2,e^{2 \text {arctanh}\left (\frac {1}{\sqrt {2}}\right )-2 \text {arctanh}(x)}\right )+4 \operatorname {PolyLog}\left (2,-e^{2 \text {arctanh}(x)}\right )}{8 \sqrt {2}} \]
(Pi^2 - 4*ArcTanh[1/Sqrt[2]]^2 - (4*I)*Pi*ArcTanh[x] + 8*ArcTanh[1/Sqrt[2] ]*ArcTanh[x] - 8*ArcTanh[x]^2 + 8*ArcTanh[1/Sqrt[2]]*Log[1 - E^(2*ArcTanh[ 1/Sqrt[2]] - 2*ArcTanh[x])] - 8*ArcTanh[x]*Log[1 - E^(2*ArcTanh[1/Sqrt[2]] - 2*ArcTanh[x])] + (4*I)*Pi*Log[1 + E^(2*ArcTanh[x])] + 8*ArcTanh[x]*Log[ 1 + E^(2*ArcTanh[x])] - (4*I)*Pi*Log[2/Sqrt[1 - x^2]] - 8*ArcTanh[x]*Log[2 /Sqrt[1 - x^2]] - 4*ArcTanh[x]*Log[1 - x^2] - 8*ArcTanh[x]*Log[(-I)*Sinh[A rcTanh[1/Sqrt[2]] - ArcTanh[x]]] - 8*ArcTanh[1/Sqrt[2]]*Log[(-2*I)*Sinh[Ar cTanh[1/Sqrt[2]] - ArcTanh[x]]] + 8*ArcTanh[x]*Log[(-2*I)*Sinh[ArcTanh[1/S qrt[2]] - ArcTanh[x]]] + 4*PolyLog[2, E^(2*ArcTanh[1/Sqrt[2]] - 2*ArcTanh[ x])] + 4*PolyLog[2, -E^(2*ArcTanh[x])])/(8*Sqrt[2])
Time = 0.41 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.23, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6472, 2849, 2752, 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.
| \(\displaystyle \int \frac {\text {arctanh}(x)}{1-\sqrt {2} x} \, dx\) |
| \(\Big \downarrow \) 6472 |
| \(\displaystyle -\frac {\int \frac {\log \left (\frac {2}{x+1}\right )}{1-x^2}dx}{\sqrt {2}}+\frac {\int \frac {\log \left (-\frac {2 \left (1+\sqrt {2}\right ) \left (1-\sqrt {2} x\right )}{x+1}\right )}{1-x^2}dx}{\sqrt {2}}+\frac {\text {arctanh}(x) \log \left (\frac {2}{x+1}\right )}{\sqrt {2}}-\frac {\text {arctanh}(x) \log \left (-\frac {2 \left (1+\sqrt {2}\right ) \left (1-\sqrt {2} x\right )}{x+1}\right )}{\sqrt {2}}\) |
| \(\Big \downarrow \) 2849 |
| \(\displaystyle \frac {\int \frac {\log \left (-\frac {2 \left (1+\sqrt {2}\right ) \left (1-\sqrt {2} x\right )}{x+1}\right )}{1-x^2}dx}{\sqrt {2}}-\frac {\int \frac {\log \left (\frac {2}{x+1}\right )}{1-\frac {2}{x+1}}d\frac {1}{x+1}}{\sqrt {2}}+\frac {\text {arctanh}(x) \log \left (\frac {2}{x+1}\right )}{\sqrt {2}}-\frac {\text {arctanh}(x) \log \left (-\frac {2 \left (1+\sqrt {2}\right ) \left (1-\sqrt {2} x\right )}{x+1}\right )}{\sqrt {2}}\) |
| \(\Big \downarrow \) 2752 |
| \(\displaystyle \frac {\int \frac {\log \left (-\frac {2 \left (1+\sqrt {2}\right ) \left (1-\sqrt {2} x\right )}{x+1}\right )}{1-x^2}dx}{\sqrt {2}}+\frac {\text {arctanh}(x) \log \left (\frac {2}{x+1}\right )}{\sqrt {2}}-\frac {\text {arctanh}(x) \log \left (-\frac {2 \left (1+\sqrt {2}\right ) \left (1-\sqrt {2} x\right )}{x+1}\right )}{\sqrt {2}}-\frac {\operatorname {PolyLog}\left (2,1-\frac {2}{x+1}\right )}{2 \sqrt {2}}\) |
| \(\Big \downarrow \) 2897 |
| \(\displaystyle \frac {\text {arctanh}(x) \log \left (\frac {2}{x+1}\right )}{\sqrt {2}}-\frac {\text {arctanh}(x) \log \left (-\frac {2 \left (1+\sqrt {2}\right ) \left (1-\sqrt {2} x\right )}{x+1}\right )}{\sqrt {2}}-\frac {\operatorname {PolyLog}\left (2,1-\frac {2}{x+1}\right )}{2 \sqrt {2}}+\frac {\operatorname {PolyLog}\left (2,\frac {2 \left (1+\sqrt {2}\right ) \left (1-\sqrt {2} x\right )}{x+1}+1\right )}{2 \sqrt {2}}\) |
(ArcTanh[x]*Log[2/(1 + x)])/Sqrt[2] - (ArcTanh[x]*Log[(-2*(1 + Sqrt[2])*(1 - Sqrt[2]*x))/(1 + x)])/Sqrt[2] - PolyLog[2, 1 - 2/(1 + x)]/(2*Sqrt[2]) + PolyLog[2, 1 + (2*(1 + Sqrt[2])*(1 - Sqrt[2]*x))/(1 + x)]/(2*Sqrt[2])
3.1.22.3.1 Defintions of rubi rules used
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp [-e/g Subst[Int[Log[2*d*x]/(1 - 2*d*x), x], x, 1/(d + e*x)], x] /; FreeQ[ {c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]
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]]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> S imp[(-(a + b*ArcTanh[c*x]))*(Log[2/(1 + c*x)]/e), x] + (Simp[(a + b*ArcTanh [c*x])*(Log[2*c*((d + e*x)/((c*d + e)*(1 + c*x)))]/e), x] + Simp[b*(c/e) Int[Log[2/(1 + c*x)]/(1 - c^2*x^2), x], x] - Simp[b*(c/e) Int[Log[2*c*((d + e*x)/((c*d + e)*(1 + c*x)))]/(1 - c^2*x^2), x], x]) /; FreeQ[{a, b, c, d , e}, x] && NeQ[c^2*d^2 - e^2, 0]
Time = 1.47 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.44
| method | result | size |
| default | \(-\frac {\ln \left (x \sqrt {2}-1\right ) \sqrt {2}\, \operatorname {arctanh}\left (x \right )}{2}-\frac {\sqrt {2}\, \ln \left (x \sqrt {2}-1\right ) \ln \left (\frac {\sqrt {2}-x \sqrt {2}}{\sqrt {2}-1}\right )}{4}+\frac {\sqrt {2}\, \ln \left (x \sqrt {2}-1\right ) \ln \left (\frac {\sqrt {2}+x \sqrt {2}}{1+\sqrt {2}}\right )}{4}-\frac {\sqrt {2}\, \operatorname {dilog}\left (\frac {\sqrt {2}-x \sqrt {2}}{\sqrt {2}-1}\right )}{4}+\frac {\sqrt {2}\, \operatorname {dilog}\left (\frac {\sqrt {2}+x \sqrt {2}}{1+\sqrt {2}}\right )}{4}\) | \(127\) |
| parts | \(-\frac {\ln \left (x \sqrt {2}-1\right ) \sqrt {2}\, \operatorname {arctanh}\left (x \right )}{2}-\frac {\sqrt {2}\, \ln \left (x \sqrt {2}-1\right ) \ln \left (\frac {\sqrt {2}-x \sqrt {2}}{\sqrt {2}-1}\right )}{4}+\frac {\sqrt {2}\, \ln \left (x \sqrt {2}-1\right ) \ln \left (\frac {\sqrt {2}+x \sqrt {2}}{1+\sqrt {2}}\right )}{4}-\frac {\sqrt {2}\, \operatorname {dilog}\left (\frac {\sqrt {2}-x \sqrt {2}}{\sqrt {2}-1}\right )}{4}+\frac {\sqrt {2}\, \operatorname {dilog}\left (\frac {\sqrt {2}+x \sqrt {2}}{1+\sqrt {2}}\right )}{4}\) | \(127\) |
| risch | \(\frac {\sqrt {2}\, \ln \left (\frac {-2 x +\sqrt {2}}{2+\sqrt {2}}\right ) \ln \left (\frac {2+2 x}{2+\sqrt {2}}\right )}{4}-\frac {\sqrt {2}\, \ln \left (\frac {-2 x +\sqrt {2}}{2+\sqrt {2}}\right ) \ln \left (1+x \right )}{4}+\frac {\sqrt {2}\, \operatorname {dilog}\left (\frac {2+2 x}{2+\sqrt {2}}\right )}{4}-\frac {\sqrt {2}\, \ln \left (\frac {2 x -\sqrt {2}}{2-\sqrt {2}}\right ) \ln \left (\frac {2-2 x}{2-\sqrt {2}}\right )}{4}+\frac {\sqrt {2}\, \ln \left (\frac {2 x -\sqrt {2}}{2-\sqrt {2}}\right ) \ln \left (1-x \right )}{4}-\frac {\sqrt {2}\, \operatorname {dilog}\left (\frac {2-2 x}{2-\sqrt {2}}\right )}{4}\) | \(174\) |
-1/2*ln(x*2^(1/2)-1)*2^(1/2)*arctanh(x)-1/4*2^(1/2)*ln(x*2^(1/2)-1)*ln((2^ (1/2)-x*2^(1/2))/(2^(1/2)-1))+1/4*2^(1/2)*ln(x*2^(1/2)-1)*ln((2^(1/2)+x*2^ (1/2))/(1+2^(1/2)))-1/4*2^(1/2)*dilog((2^(1/2)-x*2^(1/2))/(2^(1/2)-1))+1/4 *2^(1/2)*dilog((2^(1/2)+x*2^(1/2))/(1+2^(1/2)))
\[ \int \frac {\text {arctanh}(x)}{1-\sqrt {2} x} \, dx=\int { -\frac {\operatorname {artanh}\left (x\right )}{\sqrt {2} x - 1} \,d x } \]
\[ \int \frac {\text {arctanh}(x)}{1-\sqrt {2} x} \, dx=- \int \frac {\operatorname {atanh}{\left (x \right )}}{\sqrt {2} x - 1}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (69) = 138\).
Time = 0.27 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.64 \[ \int \frac {\text {arctanh}(x)}{1-\sqrt {2} x} \, dx=\frac {1}{4} \, \sqrt {2} {\left (\log \left (x + 1\right ) - \log \left (x - 1\right )\right )} \log \left (\sqrt {2} x - 1\right ) - \frac {1}{2} \, \sqrt {2} \operatorname {artanh}\left (x\right ) \log \left (\sqrt {2} x - 1\right ) - \frac {1}{4} \, \sqrt {2} {\left (\log \left (x + 1\right ) \log \left (-\frac {\sqrt {2} x + \sqrt {2}}{\sqrt {2} + 1} + 1\right ) + {\rm Li}_2\left (\frac {\sqrt {2} x + \sqrt {2}}{\sqrt {2} + 1}\right )\right )} + \frac {1}{4} \, \sqrt {2} {\left (\log \left (x - 1\right ) \log \left (\frac {\sqrt {2} x - \sqrt {2}}{\sqrt {2} - 1} + 1\right ) + {\rm Li}_2\left (-\frac {\sqrt {2} x - \sqrt {2}}{\sqrt {2} - 1}\right )\right )} \]
1/4*sqrt(2)*(log(x + 1) - log(x - 1))*log(sqrt(2)*x - 1) - 1/2*sqrt(2)*arc tanh(x)*log(sqrt(2)*x - 1) - 1/4*sqrt(2)*(log(x + 1)*log(-(sqrt(2)*x + sqr t(2))/(sqrt(2) + 1) + 1) + dilog((sqrt(2)*x + sqrt(2))/(sqrt(2) + 1))) + 1 /4*sqrt(2)*(log(x - 1)*log((sqrt(2)*x - sqrt(2))/(sqrt(2) - 1) + 1) + dilo g(-(sqrt(2)*x - sqrt(2))/(sqrt(2) - 1)))
\[ \int \frac {\text {arctanh}(x)}{1-\sqrt {2} x} \, dx=\int { -\frac {\operatorname {artanh}\left (x\right )}{\sqrt {2} x - 1} \,d x } \]
Timed out. \[ \int \frac {\text {arctanh}(x)}{1-\sqrt {2} x} \, dx=-\int \frac {\mathrm {atanh}\left (x\right )}{\sqrt {2}\,x-1} \,d x \]