Integrand size = 19, antiderivative size = 32 \[ \int \frac {\text {arctanh}(a+b x)}{\frac {a d}{b}+d x} \, dx=-\frac {\operatorname {PolyLog}(2,-a-b x)}{2 d}+\frac {\operatorname {PolyLog}(2,a+b x)}{2 d} \] Output:
-1/2*polylog(2,-b*x-a)/d+1/2*polylog(2,b*x+a)/d
Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.62 \[ \int \frac {\text {arctanh}(a+b x)}{\frac {a d}{b}+d x} \, dx=b \left (-\frac {\operatorname {PolyLog}\left (2,-\frac {a d+b d x}{d}\right )}{2 b d}+\frac {\operatorname {PolyLog}\left (2,\frac {a d+b d x}{d}\right )}{2 b d}\right ) \] Input:
Integrate[ArcTanh[a + b*x]/((a*d)/b + d*x),x]
Output:
b*(-1/2*PolyLog[2, -((a*d + b*d*x)/d)]/(b*d) + PolyLog[2, (a*d + b*d*x)/d] /(2*b*d))
Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {6657, 27, 6446}
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.
\(\displaystyle \int \frac {\text {arctanh}(a+b x)}{\frac {a d}{b}+d x} \, dx\) |
\(\Big \downarrow \) 6657 |
\(\displaystyle \frac {\int \frac {b \text {arctanh}(a+b x)}{d (a+b x)}d(a+b x)}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\text {arctanh}(a+b x)}{a+b x}d(a+b x)}{d}\) |
\(\Big \downarrow \) 6446 |
\(\displaystyle \frac {\frac {1}{2} \operatorname {PolyLog}(2,a+b x)-\frac {1}{2} \operatorname {PolyLog}(2,-a-b x)}{d}\) |
Input:
Int[ArcTanh[a + b*x]/((a*d)/b + d*x),x]
Output:
(-1/2*PolyLog[2, -a - b*x] + PolyLog[2, a + b*x]/2)/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x ] + (-Simp[(b/2)*PolyLog[2, (-c)*x], x] + Simp[(b/2)*PolyLog[2, c*x], x]) / ; FreeQ[{a, b, c}, x]
Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( m_.), x_Symbol] :> Simp[1/d Subst[Int[(f*(x/d))^m*(a + b*ArcTanh[x])^p, x ], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f, 0] && IGtQ[p, 0]
Time = 0.63 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {\operatorname {dilog}\left (b x +a +1\right )}{2 d}+\frac {\operatorname {dilog}\left (-b x -a +1\right )}{2 d}\) | \(29\) |
parts | \(\frac {\ln \left (b x +a \right ) \operatorname {arctanh}\left (b x +a \right )}{d}-\frac {\frac {\operatorname {dilog}\left (b x +a +1\right )}{2}+\frac {\ln \left (b x +a \right ) \ln \left (b x +a +1\right )}{2}+\frac {\operatorname {dilog}\left (b x +a \right )}{2}}{d}\) | \(56\) |
derivativedivides | \(\frac {\frac {b \ln \left (b x +a \right ) \operatorname {arctanh}\left (b x +a \right )}{d}-\frac {b \left (\frac {\operatorname {dilog}\left (b x +a +1\right )}{2}+\frac {\ln \left (b x +a \right ) \ln \left (b x +a +1\right )}{2}+\frac {\operatorname {dilog}\left (b x +a \right )}{2}\right )}{d}}{b}\) | \(62\) |
default | \(\frac {\frac {b \ln \left (b x +a \right ) \operatorname {arctanh}\left (b x +a \right )}{d}-\frac {b \left (\frac {\operatorname {dilog}\left (b x +a +1\right )}{2}+\frac {\ln \left (b x +a \right ) \ln \left (b x +a +1\right )}{2}+\frac {\operatorname {dilog}\left (b x +a \right )}{2}\right )}{d}}{b}\) | \(62\) |
Input:
int(arctanh(b*x+a)/(a*d/b+d*x),x,method=_RETURNVERBOSE)
Output:
-1/2/d*dilog(b*x+a+1)+1/2/d*dilog(-b*x-a+1)
\[ \int \frac {\text {arctanh}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int { \frac {\operatorname {artanh}\left (b x + a\right )}{d x + \frac {a d}{b}} \,d x } \] Input:
integrate(arctanh(b*x+a)/(a*d/b+d*x),x, algorithm="fricas")
Output:
integral(b*arctanh(b*x + a)/(b*d*x + a*d), x)
\[ \int \frac {\text {arctanh}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {b \int \frac {\operatorname {atanh}{\left (a + b x \right )}}{a + b x}\, dx}{d} \] Input:
integrate(atanh(b*x+a)/(a*d/b+d*x),x)
Output:
b*Integral(atanh(a + b*x)/(a + b*x), x)/d
Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (26) = 52\).
Time = 0.03 (sec) , antiderivative size = 132, normalized size of antiderivative = 4.12 \[ \int \frac {\text {arctanh}(a+b x)}{\frac {a d}{b}+d x} \, dx=-\frac {1}{2} \, b {\left (\frac {\log \left (b x + a\right ) \log \left (b x + a - 1\right ) + {\rm Li}_2\left (-b x - a + 1\right )}{b d} - \frac {\log \left (b x + a + 1\right ) \log \left (-b x - a\right ) + {\rm Li}_2\left (b x + a + 1\right )}{b d}\right )} - \frac {b {\left (\frac {\log \left (b x + a + 1\right )}{b} - \frac {\log \left (b x + a - 1\right )}{b}\right )} \log \left (d x + \frac {a d}{b}\right )}{2 \, d} + \frac {\operatorname {artanh}\left (b x + a\right ) \log \left (d x + \frac {a d}{b}\right )}{d} \] Input:
integrate(arctanh(b*x+a)/(a*d/b+d*x),x, algorithm="maxima")
Output:
-1/2*b*((log(b*x + a)*log(b*x + a - 1) + dilog(-b*x - a + 1))/(b*d) - (log (b*x + a + 1)*log(-b*x - a) + dilog(b*x + a + 1))/(b*d)) - 1/2*b*(log(b*x + a + 1)/b - log(b*x + a - 1)/b)*log(d*x + a*d/b)/d + arctanh(b*x + a)*log (d*x + a*d/b)/d
\[ \int \frac {\text {arctanh}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int { \frac {\operatorname {artanh}\left (b x + a\right )}{d x + \frac {a d}{b}} \,d x } \] Input:
integrate(arctanh(b*x+a)/(a*d/b+d*x),x, algorithm="giac")
Output:
integrate(arctanh(b*x + a)/(d*x + a*d/b), x)
Timed out. \[ \int \frac {\text {arctanh}(a+b x)}{\frac {a d}{b}+d x} \, dx=\int \frac {\mathrm {atanh}\left (a+b\,x\right )}{d\,x+\frac {a\,d}{b}} \,d x \] Input:
int(atanh(a + b*x)/(d*x + (a*d)/b),x)
Output:
int(atanh(a + b*x)/(d*x + (a*d)/b), x)
\[ \int \frac {\text {arctanh}(a+b x)}{\frac {a d}{b}+d x} \, dx=\frac {\left (\int \frac {\mathit {atanh} \left (b x +a \right )}{b x +a}d x \right ) b}{d} \] Input:
int(atanh(b*x+a)/(a*d/b+d*x),x)
Output:
(int(atanh(a + b*x)/(a + b*x),x)*b)/d