Integrand size = 17, antiderivative size = 67 \[ \int \frac {a+b \text {arctanh}(c x)}{1+2 c x} \, dx=\frac {\left (a-b \text {arctanh}\left (\frac {1}{2}\right )\right ) \log \left (-\frac {1+2 c x}{2 d}\right )}{2 c}-\frac {b \operatorname {PolyLog}(2,-1-2 c x)}{4 c}+\frac {b \operatorname {PolyLog}\left (2,\frac {1}{3} (1+2 c x)\right )}{4 c} \] Output:
1/2*(a-b*arctanh(1/2))*ln(-1/2*(2*c*x+1)/d)/c-1/4*b*polylog(2,-2*c*x-1)/c+ 1/4*b*polylog(2,2/3*c*x+1/3)/c
Result contains complex when optimal does not.
Time = 0.21 (sec) , antiderivative size = 240, normalized size of antiderivative = 3.58 \[ \int \frac {a+b \text {arctanh}(c x)}{1+2 c x} \, dx=\frac {a \log (1+2 c x)+b \text {arctanh}(c x) \left (\frac {1}{2} \log \left (1-c^2 x^2\right )+\log \left (i \sinh \left (\text {arctanh}\left (\frac {1}{2}\right )+\text {arctanh}(c x)\right )\right )\right )-\frac {1}{2} i b \left (-\frac {1}{4} i (\pi -2 i \text {arctanh}(c x))^2+i \left (\text {arctanh}\left (\frac {1}{2}\right )+\text {arctanh}(c x)\right )^2+(\pi -2 i \text {arctanh}(c x)) \log \left (1+e^{2 \text {arctanh}(c x)}\right )+2 i \left (\text {arctanh}\left (\frac {1}{2}\right )+\text {arctanh}(c x)\right ) \log \left (1-e^{-2 \left (\text {arctanh}\left (\frac {1}{2}\right )+\text {arctanh}(c x)\right )}\right )-(\pi -2 i \text {arctanh}(c x)) \log \left (\frac {2}{\sqrt {1-c^2 x^2}}\right )-2 i \left (\text {arctanh}\left (\frac {1}{2}\right )+\text {arctanh}(c x)\right ) \log \left (2 i \sinh \left (\text {arctanh}\left (\frac {1}{2}\right )+\text {arctanh}(c x)\right )\right )-i \operatorname {PolyLog}\left (2,-e^{2 \text {arctanh}(c x)}\right )-i \operatorname {PolyLog}\left (2,e^{-2 \left (\text {arctanh}\left (\frac {1}{2}\right )+\text {arctanh}(c x)\right )}\right )\right )}{2 c} \] Input:
Integrate[(a + b*ArcTanh[c*x])/(1 + 2*c*x),x]
Output:
(a*Log[1 + 2*c*x] + b*ArcTanh[c*x]*(Log[1 - c^2*x^2]/2 + Log[I*Sinh[ArcTan h[1/2] + ArcTanh[c*x]]]) - (I/2)*b*((-1/4*I)*(Pi - (2*I)*ArcTanh[c*x])^2 + I*(ArcTanh[1/2] + ArcTanh[c*x])^2 + (Pi - (2*I)*ArcTanh[c*x])*Log[1 + E^( 2*ArcTanh[c*x])] + (2*I)*(ArcTanh[1/2] + ArcTanh[c*x])*Log[1 - E^(-2*(ArcT anh[1/2] + ArcTanh[c*x]))] - (Pi - (2*I)*ArcTanh[c*x])*Log[2/Sqrt[1 - c^2* x^2]] - (2*I)*(ArcTanh[1/2] + ArcTanh[c*x])*Log[(2*I)*Sinh[ArcTanh[1/2] + ArcTanh[c*x]]] - I*PolyLog[2, -E^(2*ArcTanh[c*x])] - I*PolyLog[2, E^(-2*(A rcTanh[1/2] + ArcTanh[c*x]))]))/(2*c)
Time = 0.66 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.63, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, 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.
| \(\displaystyle \int \frac {a+b \text {arctanh}(c x)}{2 c x+1} \, dx\) |
| \(\Big \downarrow \) 6472 |
| \(\displaystyle \frac {1}{2} b \int \frac {\log \left (\frac {2}{c x+1}\right )}{1-c^2 x^2}dx-\frac {1}{2} b \int \frac {\log \left (\frac {2 (2 c x+1)}{3 (c x+1)}\right )}{1-c^2 x^2}dx-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{2 c}+\frac {\log \left (\frac {2 (2 c x+1)}{3 (c x+1)}\right ) (a+b \text {arctanh}(c x))}{2 c}\) |
| \(\Big \downarrow \) 2849 |
| \(\displaystyle -\frac {1}{2} b \int \frac {\log \left (\frac {2 (2 c x+1)}{3 (c x+1)}\right )}{1-c^2 x^2}dx+\frac {b \int \frac {\log \left (\frac {2}{c x+1}\right )}{1-\frac {2}{c x+1}}d\frac {1}{c x+1}}{2 c}-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{2 c}+\frac {\log \left (\frac {2 (2 c x+1)}{3 (c x+1)}\right ) (a+b \text {arctanh}(c x))}{2 c}\) |
| \(\Big \downarrow \) 2752 |
| \(\displaystyle -\frac {1}{2} b \int \frac {\log \left (\frac {2 (2 c x+1)}{3 (c x+1)}\right )}{1-c^2 x^2}dx-\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{2 c}+\frac {\log \left (\frac {2 (2 c x+1)}{3 (c x+1)}\right ) (a+b \text {arctanh}(c x))}{2 c}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{4 c}\) |
| \(\Big \downarrow \) 2897 |
| \(\displaystyle -\frac {\log \left (\frac {2}{c x+1}\right ) (a+b \text {arctanh}(c x))}{2 c}+\frac {\log \left (\frac {2 (2 c x+1)}{3 (c x+1)}\right ) (a+b \text {arctanh}(c x))}{2 c}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{4 c}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2 (2 c x+1)}{3 (c x+1)}\right )}{4 c}\) |
Input:
Int[(a + b*ArcTanh[c*x])/(1 + 2*c*x),x]
Output:
-1/2*((a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/c + ((a + b*ArcTanh[c*x])*Log [(2*(1 + 2*c*x))/(3*(1 + c*x))])/(2*c) + (b*PolyLog[2, 1 - 2/(1 + c*x)])/( 4*c) - (b*PolyLog[2, 1 - (2*(1 + 2*c*x))/(3*(1 + c*x))])/(4*c)
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 = 0.22 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.37
| method | result | size |
| derivativedivides | \(\frac {\frac {a \ln \left (2 c x +1\right )}{2}+b \left (\frac {\ln \left (2 c x +1\right ) \operatorname {arctanh}\left (c x \right )}{2}+\frac {\left (\ln \left (2 c x +1\right )-\ln \left (\frac {2 c x}{3}+\frac {1}{3}\right )\right ) \ln \left (\frac {2}{3}-\frac {2 c x}{3}\right )}{4}-\frac {\operatorname {dilog}\left (\frac {2 c x}{3}+\frac {1}{3}\right )}{4}-\frac {\operatorname {dilog}\left (2 c x +2\right )}{4}-\frac {\ln \left (2 c x +1\right ) \ln \left (2 c x +2\right )}{4}\right )}{c}\) | \(92\) |
| default | \(\frac {\frac {a \ln \left (2 c x +1\right )}{2}+b \left (\frac {\ln \left (2 c x +1\right ) \operatorname {arctanh}\left (c x \right )}{2}+\frac {\left (\ln \left (2 c x +1\right )-\ln \left (\frac {2 c x}{3}+\frac {1}{3}\right )\right ) \ln \left (\frac {2}{3}-\frac {2 c x}{3}\right )}{4}-\frac {\operatorname {dilog}\left (\frac {2 c x}{3}+\frac {1}{3}\right )}{4}-\frac {\operatorname {dilog}\left (2 c x +2\right )}{4}-\frac {\ln \left (2 c x +1\right ) \ln \left (2 c x +2\right )}{4}\right )}{c}\) | \(92\) |
| parts | \(\frac {a \ln \left (2 c x +1\right )}{2 c}+\frac {b \left (\frac {\ln \left (2 c x +1\right ) \operatorname {arctanh}\left (c x \right )}{2}+\frac {\left (\ln \left (2 c x +1\right )-\ln \left (\frac {2 c x}{3}+\frac {1}{3}\right )\right ) \ln \left (\frac {2}{3}-\frac {2 c x}{3}\right )}{4}-\frac {\operatorname {dilog}\left (\frac {2 c x}{3}+\frac {1}{3}\right )}{4}-\frac {\operatorname {dilog}\left (2 c x +2\right )}{4}-\frac {\ln \left (2 c x +1\right ) \ln \left (2 c x +2\right )}{4}\right )}{c}\) | \(94\) |
| risch | \(\frac {b \ln \left (\frac {2 c x}{3}+\frac {1}{3}\right ) \ln \left (\frac {2}{3}-\frac {2 c x}{3}\right )}{4 c}-\frac {b \ln \left (\frac {2 c x}{3}+\frac {1}{3}\right ) \ln \left (-c x +1\right )}{4 c}+\frac {b \operatorname {dilog}\left (\frac {2}{3}-\frac {2 c x}{3}\right )}{4 c}+\frac {a \ln \left (-2 c x -1\right )}{2 c}-\frac {b \ln \left (-2 c x -1\right ) \ln \left (2 c x +2\right )}{4 c}+\frac {b \ln \left (-2 c x -1\right ) \ln \left (c x +1\right )}{4 c}-\frac {b \operatorname {dilog}\left (2 c x +2\right )}{4 c}\) | \(120\) |
Input:
int((a+b*arctanh(c*x))/(2*c*x+1),x,method=_RETURNVERBOSE)
Output:
1/c*(1/2*a*ln(2*c*x+1)+b*(1/2*ln(2*c*x+1)*arctanh(c*x)+1/4*(ln(2*c*x+1)-ln (2/3*c*x+1/3))*ln(2/3-2/3*c*x)-1/4*dilog(2/3*c*x+1/3)-1/4*dilog(2*c*x+2)-1 /4*ln(2*c*x+1)*ln(2*c*x+2)))
\[ \int \frac {a+b \text {arctanh}(c x)}{1+2 c x} \, dx=\int { \frac {b \operatorname {artanh}\left (c x\right ) + a}{2 \, c x + 1} \,d x } \] Input:
integrate((a+b*arctanh(c*x))/(2*c*x+1),x, algorithm="fricas")
Output:
integral((b*arctanh(c*x) + a)/(2*c*x + 1), x)
\[ \int \frac {a+b \text {arctanh}(c x)}{1+2 c x} \, dx=\int \frac {a + b \operatorname {atanh}{\left (c x \right )}}{2 c x + 1}\, dx \] Input:
integrate((a+b*atanh(c*x))/(2*c*x+1),x)
Output:
Integral((a + b*atanh(c*x))/(2*c*x + 1), x)
\[ \int \frac {a+b \text {arctanh}(c x)}{1+2 c x} \, dx=\int { \frac {b \operatorname {artanh}\left (c x\right ) + a}{2 \, c x + 1} \,d x } \] Input:
integrate((a+b*arctanh(c*x))/(2*c*x+1),x, algorithm="maxima")
Output:
1/2*b*integrate((log(c*x + 1) - log(-c*x + 1))/(2*c*x + 1), x) + 1/2*a*log (2*c*x + 1)/c
\[ \int \frac {a+b \text {arctanh}(c x)}{1+2 c x} \, dx=\int { \frac {b \operatorname {artanh}\left (c x\right ) + a}{2 \, c x + 1} \,d x } \] Input:
integrate((a+b*arctanh(c*x))/(2*c*x+1),x, algorithm="giac")
Output:
integrate((b*arctanh(c*x) + a)/(2*c*x + 1), x)
Timed out. \[ \int \frac {a+b \text {arctanh}(c x)}{1+2 c x} \, dx=\int \frac {a+b\,\mathrm {atanh}\left (c\,x\right )}{2\,c\,x+1} \,d x \] Input:
int((a + b*atanh(c*x))/(2*c*x + 1),x)
Output:
int((a + b*atanh(c*x))/(2*c*x + 1), x)
\[ \int \frac {a+b \text {arctanh}(c x)}{1+2 c x} \, dx=\frac {2 \left (\int \frac {\mathit {atanh} \left (c x \right )}{2 c x +1}d x \right ) b c +\mathrm {log}\left (2 c x +1\right ) a}{2 c} \] Input:
int((a+b*atanh(c*x))/(2*c*x+1),x)
Output:
(2*int(atanh(c*x)/(2*c*x + 1),x)*b*c + log(2*c*x + 1)*a)/(2*c)