3.1.56 \(\int \frac {\tanh ^{-1}(a+b x)}{c+\frac {d}{x}} \, dx\) [56]

Optimal. Leaf size=186 \[ \frac {(1-a-b x) \log (1-a-b x)}{2 b c}+\frac {(1+a+b x) \log (1+a+b x)}{2 b c}-\frac {d \log (1+a+b x) \log \left (-\frac {b (d+c x)}{c+a c-b d}\right )}{2 c^2}+\frac {d \log (1-a-b x) \log \left (\frac {b (d+c x)}{c-a c+b d}\right )}{2 c^2}+\frac {d \text {PolyLog}\left (2,\frac {c (1-a-b x)}{c-a c+b d}\right )}{2 c^2}-\frac {d \text {PolyLog}\left (2,\frac {c (1+a+b x)}{c+a c-b d}\right )}{2 c^2} \]

[Out]

1/2*(-b*x-a+1)*ln(-b*x-a+1)/b/c+1/2*(b*x+a+1)*ln(b*x+a+1)/b/c-1/2*d*ln(b*x+a+1)*ln(-b*(c*x+d)/(a*c-b*d+c))/c^2
+1/2*d*ln(-b*x-a+1)*ln(b*(c*x+d)/(-a*c+b*d+c))/c^2+1/2*d*polylog(2,c*(-b*x-a+1)/(-a*c+b*d+c))/c^2-1/2*d*polylo
g(2,c*(b*x+a+1)/(a*c-b*d+c))/c^2

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Rubi [A]
time = 0.17, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6250, 2456, 2436, 2332, 2441, 2440, 2438} \begin {gather*} \frac {d \text {Li}_2\left (\frac {c (-a-b x+1)}{-a c+c+b d}\right )}{2 c^2}-\frac {d \text {Li}_2\left (\frac {c (a+b x+1)}{a c+c-b d}\right )}{2 c^2}+\frac {d \log (-a-b x+1) \log \left (\frac {b (c x+d)}{-a c+b d+c}\right )}{2 c^2}-\frac {d \log (a+b x+1) \log \left (-\frac {b (c x+d)}{a c-b d+c}\right )}{2 c^2}+\frac {(-a-b x+1) \log (-a-b x+1)}{2 b c}+\frac {(a+b x+1) \log (a+b x+1)}{2 b c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a + b*x]/(c + d/x),x]

[Out]

((1 - a - b*x)*Log[1 - a - b*x])/(2*b*c) + ((1 + a + b*x)*Log[1 + a + b*x])/(2*b*c) - (d*Log[1 + a + b*x]*Log[
-((b*(d + c*x))/(c + a*c - b*d))])/(2*c^2) + (d*Log[1 - a - b*x]*Log[(b*(d + c*x))/(c - a*c + b*d)])/(2*c^2) +
(d*PolyLog[2, (c*(1 - a - b*x))/(c - a*c + b*d)])/(2*c^2) - (d*PolyLog[2, (c*(1 + a + b*x))/(c + a*c - b*d)])
/(2*c^2)

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + c*e*(x/g)])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2456

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_)^(r_))^(q_.), x_Symbol] :> In
t[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x]
&& IGtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))

Rule 6250

Int[ArcTanh[(c_) + (d_.)*(x_)]/((e_) + (f_.)*(x_)^(n_.)), x_Symbol] :> Dist[1/2, Int[Log[1 + c + d*x]/(e + f*x
^n), x], x] - Dist[1/2, Int[Log[1 - c - d*x]/(e + f*x^n), x], x] /; FreeQ[{c, d, e, f}, x] && RationalQ[n]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(a+b x)}{c+\frac {d}{x}} \, dx &=-\left (\frac {1}{2} \int \frac {\log (1-a-b x)}{c+\frac {d}{x}} \, dx\right )+\frac {1}{2} \int \frac {\log (1+a+b x)}{c+\frac {d}{x}} \, dx\\ &=-\left (\frac {1}{2} \int \left (\frac {\log (1-a-b x)}{c}-\frac {d \log (1-a-b x)}{c (d+c x)}\right ) \, dx\right )+\frac {1}{2} \int \left (\frac {\log (1+a+b x)}{c}-\frac {d \log (1+a+b x)}{c (d+c x)}\right ) \, dx\\ &=-\frac {\int \log (1-a-b x) \, dx}{2 c}+\frac {\int \log (1+a+b x) \, dx}{2 c}+\frac {d \int \frac {\log (1-a-b x)}{d+c x} \, dx}{2 c}-\frac {d \int \frac {\log (1+a+b x)}{d+c x} \, dx}{2 c}\\ &=-\frac {d \log (1+a+b x) \log \left (-\frac {b (d+c x)}{c+a c-b d}\right )}{2 c^2}+\frac {d \log (1-a-b x) \log \left (\frac {b (d+c x)}{c-a c+b d}\right )}{2 c^2}+\frac {\text {Subst}(\int \log (x) \, dx,x,1-a-b x)}{2 b c}+\frac {\text {Subst}(\int \log (x) \, dx,x,1+a+b x)}{2 b c}+\frac {(b d) \int \frac {\log \left (-\frac {b (d+c x)}{-(1-a) c-b d}\right )}{1-a-b x} \, dx}{2 c^2}+\frac {(b d) \int \frac {\log \left (\frac {b (d+c x)}{-(1+a) c+b d}\right )}{1+a+b x} \, dx}{2 c^2}\\ &=\frac {(1-a-b x) \log (1-a-b x)}{2 b c}+\frac {(1+a+b x) \log (1+a+b x)}{2 b c}-\frac {d \log (1+a+b x) \log \left (-\frac {b (d+c x)}{c+a c-b d}\right )}{2 c^2}+\frac {d \log (1-a-b x) \log \left (\frac {b (d+c x)}{c-a c+b d}\right )}{2 c^2}-\frac {d \text {Subst}\left (\int \frac {\log \left (1+\frac {c x}{-(1-a) c-b d}\right )}{x} \, dx,x,1-a-b x\right )}{2 c^2}+\frac {d \text {Subst}\left (\int \frac {\log \left (1+\frac {c x}{-(1+a) c+b d}\right )}{x} \, dx,x,1+a+b x\right )}{2 c^2}\\ &=\frac {(1-a-b x) \log (1-a-b x)}{2 b c}+\frac {(1+a+b x) \log (1+a+b x)}{2 b c}-\frac {d \log (1+a+b x) \log \left (-\frac {b (d+c x)}{c+a c-b d}\right )}{2 c^2}+\frac {d \log (1-a-b x) \log \left (\frac {b (d+c x)}{c-a c+b d}\right )}{2 c^2}+\frac {d \text {Li}_2\left (\frac {c (1-a-b x)}{c-a c+b d}\right )}{2 c^2}-\frac {d \text {Li}_2\left (\frac {c (1+a+b x)}{c+a c-b d}\right )}{2 c^2}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 3.16, size = 759, normalized size = 4.08 \begin {gather*} \frac {-2 a^2 c^2 \tanh ^{-1}(a+b x)+2 a b c d \tanh ^{-1}(a+b x)+i a b c d \pi \tanh ^{-1}(a+b x)-i b^2 d^2 \pi \tanh ^{-1}(a+b x)-2 a b c^2 x \tanh ^{-1}(a+b x)+2 b^2 c d x \tanh ^{-1}(a+b x)-2 a b c d \tanh ^{-1}\left (a-\frac {b d}{c}\right ) \tanh ^{-1}(a+b x)+2 b^2 d^2 \tanh ^{-1}\left (a-\frac {b d}{c}\right ) \tanh ^{-1}(a+b x)-b c d \tanh ^{-1}(a+b x)^2-a b c d \tanh ^{-1}(a+b x)^2+b^2 d^2 \tanh ^{-1}(a+b x)^2+b c d \sqrt {1-a^2+\frac {2 a b d}{c}-\frac {b^2 d^2}{c^2}} e^{\tanh ^{-1}\left (a-\frac {b d}{c}\right )} \tanh ^{-1}(a+b x)^2-2 a b c d \tanh ^{-1}\left (a-\frac {b d}{c}\right ) \log \left (1-e^{2 \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )}\right )+2 b^2 d^2 \tanh ^{-1}\left (a-\frac {b d}{c}\right ) \log \left (1-e^{2 \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )}\right )+2 a b c d \tanh ^{-1}(a+b x) \log \left (1-e^{2 \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )}\right )-2 b^2 d^2 \tanh ^{-1}(a+b x) \log \left (1-e^{2 \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )}\right )-2 a b c d \tanh ^{-1}(a+b x) \log \left (1+e^{-2 \tanh ^{-1}(a+b x)}\right )+2 b^2 d^2 \tanh ^{-1}(a+b x) \log \left (1+e^{-2 \tanh ^{-1}(a+b x)}\right )-i a b c d \pi \log \left (1+e^{2 \tanh ^{-1}(a+b x)}\right )+i b^2 d^2 \pi \log \left (1+e^{2 \tanh ^{-1}(a+b x)}\right )+2 a c^2 \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )-2 b c d \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )+i a b c d \pi \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )-i b^2 d^2 \pi \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )+2 a b c d \tanh ^{-1}\left (a-\frac {b d}{c}\right ) \log \left (-i \sinh \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )-2 b^2 d^2 \tanh ^{-1}\left (a-\frac {b d}{c}\right ) \log \left (-i \sinh \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )\right )+b d (-a c+b d) \text {PolyLog}\left (2,e^{2 \left (\tanh ^{-1}\left (a-\frac {b d}{c}\right )-\tanh ^{-1}(a+b x)\right )}\right )+b d (a c-b d) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a+b x)}\right )}{2 b c^2 (-a c+b d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[a + b*x]/(c + d/x),x]

[Out]

(-2*a^2*c^2*ArcTanh[a + b*x] + 2*a*b*c*d*ArcTanh[a + b*x] + I*a*b*c*d*Pi*ArcTanh[a + b*x] - I*b^2*d^2*Pi*ArcTa
nh[a + b*x] - 2*a*b*c^2*x*ArcTanh[a + b*x] + 2*b^2*c*d*x*ArcTanh[a + b*x] - 2*a*b*c*d*ArcTanh[a - (b*d)/c]*Arc
Tanh[a + b*x] + 2*b^2*d^2*ArcTanh[a - (b*d)/c]*ArcTanh[a + b*x] - b*c*d*ArcTanh[a + b*x]^2 - a*b*c*d*ArcTanh[a
+ b*x]^2 + b^2*d^2*ArcTanh[a + b*x]^2 + b*c*d*Sqrt[1 - a^2 + (2*a*b*d)/c - (b^2*d^2)/c^2]*E^ArcTanh[a - (b*d)
/c]*ArcTanh[a + b*x]^2 - 2*a*b*c*d*ArcTanh[a - (b*d)/c]*Log[1 - E^(2*(ArcTanh[a - (b*d)/c] - ArcTanh[a + b*x])
)] + 2*b^2*d^2*ArcTanh[a - (b*d)/c]*Log[1 - E^(2*(ArcTanh[a - (b*d)/c] - ArcTanh[a + b*x]))] + 2*a*b*c*d*ArcTa
nh[a + b*x]*Log[1 - E^(2*(ArcTanh[a - (b*d)/c] - ArcTanh[a + b*x]))] - 2*b^2*d^2*ArcTanh[a + b*x]*Log[1 - E^(2
*(ArcTanh[a - (b*d)/c] - ArcTanh[a + b*x]))] - 2*a*b*c*d*ArcTanh[a + b*x]*Log[1 + E^(-2*ArcTanh[a + b*x])] + 2
*b^2*d^2*ArcTanh[a + b*x]*Log[1 + E^(-2*ArcTanh[a + b*x])] - I*a*b*c*d*Pi*Log[1 + E^(2*ArcTanh[a + b*x])] + I*
b^2*d^2*Pi*Log[1 + E^(2*ArcTanh[a + b*x])] + 2*a*c^2*Log[1/Sqrt[1 - (a + b*x)^2]] - 2*b*c*d*Log[1/Sqrt[1 - (a
+ b*x)^2]] + I*a*b*c*d*Pi*Log[1/Sqrt[1 - (a + b*x)^2]] - I*b^2*d^2*Pi*Log[1/Sqrt[1 - (a + b*x)^2]] + 2*a*b*c*d
*ArcTanh[a - (b*d)/c]*Log[(-I)*Sinh[ArcTanh[a - (b*d)/c] - ArcTanh[a + b*x]]] - 2*b^2*d^2*ArcTanh[a - (b*d)/c]
*Log[(-I)*Sinh[ArcTanh[a - (b*d)/c] - ArcTanh[a + b*x]]] + b*d*(-(a*c) + b*d)*PolyLog[2, E^(2*(ArcTanh[a - (b*
d)/c] - ArcTanh[a + b*x]))] + b*d*(a*c - b*d)*PolyLog[2, -E^(-2*ArcTanh[a + b*x])])/(2*b*c^2*(-(a*c) + b*d))

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Maple [A]
time = 8.19, size = 305, normalized size = 1.64

method	result	size
risch	\(-\frac {\ln \left (-b x -a +1\right ) x}{2 c}-\frac {\ln \left (-b x -a +1\right ) a}{2 b c}+\frac {\ln \left (-b x -a +1\right )}{2 b c}-\frac {1}{b c}+\frac {d \dilog \left (\frac {c \left (-b x -a +1\right )+a c -d b -c}{a c -d b -c}\right )}{2 c^{2}}+\frac {d \ln \left (-b x -a +1\right ) \ln \left (\frac {c \left (-b x -a +1\right )+a c -d b -c}{a c -d b -c}\right )}{2 c^{2}}+\frac {\ln \left (b x +a +1\right ) x}{2 c}+\frac {\ln \left (b x +a +1\right ) a}{2 b c}+\frac {\ln \left (b x +a +1\right )}{2 b c}-\frac {d \dilog \left (\frac {c \left (b x +a +1\right )-a c +d b -c}{-a c +d b -c}\right )}{2 c^{2}}-\frac {d \ln \left (b x +a +1\right ) \ln \left (\frac {c \left (b x +a +1\right )-a c +d b -c}{-a c +d b -c}\right )}{2 c^{2}}\)	\(290\)
derivativedivides	\(\frac {\frac {\arctanh \left (b x +a \right ) \left (b x +a \right )}{c}-\frac {\arctanh \left (b x +a \right ) d b \ln \left (a c -d b -c \left (b x +a \right )\right )}{c^{2}}+\frac {-\frac {b d \ln \left (a c -d b -c \left (b x +a \right )\right ) \ln \left (\frac {-c \left (b x +a \right )+c}{-a c +d b +c}\right )}{2 c}-\frac {b d \dilog \left (\frac {-c \left (b x +a \right )+c}{-a c +d b +c}\right )}{2 c}+\frac {b d \ln \left (a c -d b -c \left (b x +a \right )\right ) \ln \left (\frac {-c \left (b x +a \right )-c}{-a c +d b -c}\right )}{2 c}+\frac {b d \dilog \left (\frac {-c \left (b x +a \right )-c}{-a c +d b -c}\right )}{2 c}+\frac {\ln \left (a^{2} c^{2}-2 b a d c +d^{2} b^{2}-2 a c \left (a c -d b -c \left (b x +a \right )\right )+2 b d \left (a c -d b -c \left (b x +a \right )\right )-c^{2}+\left (a c -d b -c \left (b x +a \right )\right )^{2}\right )}{2}}{c}}{b}\)	\(305\)
default	\(\frac {\frac {\arctanh \left (b x +a \right ) \left (b x +a \right )}{c}-\frac {\arctanh \left (b x +a \right ) d b \ln \left (a c -d b -c \left (b x +a \right )\right )}{c^{2}}+\frac {-\frac {b d \ln \left (a c -d b -c \left (b x +a \right )\right ) \ln \left (\frac {-c \left (b x +a \right )+c}{-a c +d b +c}\right )}{2 c}-\frac {b d \dilog \left (\frac {-c \left (b x +a \right )+c}{-a c +d b +c}\right )}{2 c}+\frac {b d \ln \left (a c -d b -c \left (b x +a \right )\right ) \ln \left (\frac {-c \left (b x +a \right )-c}{-a c +d b -c}\right )}{2 c}+\frac {b d \dilog \left (\frac {-c \left (b x +a \right )-c}{-a c +d b -c}\right )}{2 c}+\frac {\ln \left (a^{2} c^{2}-2 b a d c +d^{2} b^{2}-2 a c \left (a c -d b -c \left (b x +a \right )\right )+2 b d \left (a c -d b -c \left (b x +a \right )\right )-c^{2}+\left (a c -d b -c \left (b x +a \right )\right )^{2}\right )}{2}}{c}}{b}\)	\(305\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(b*x+a)/(c+d/x),x,method=_RETURNVERBOSE)

[Out]

1/b*(arctanh(b*x+a)/c*(b*x+a)-arctanh(b*x+a)*d*b/c^2*ln(a*c-d*b-c*(b*x+a))+1/c*(-1/2*b*d/c*ln(a*c-d*b-c*(b*x+a
))*ln((-c*(b*x+a)+c)/(-a*c+b*d+c))-1/2*b*d/c*dilog((-c*(b*x+a)+c)/(-a*c+b*d+c))+1/2*b*d/c*ln(a*c-d*b-c*(b*x+a)
)*ln((-c*(b*x+a)-c)/(-a*c+b*d-c))+1/2*b*d/c*dilog((-c*(b*x+a)-c)/(-a*c+b*d-c))+1/2*ln(a^2*c^2-2*b*a*d*c+d^2*b^
2-2*a*c*(a*c-d*b-c*(b*x+a))+2*b*d*(a*c-d*b-c*(b*x+a))-c^2+(a*c-d*b-c*(b*x+a))^2)))

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Maxima [A]
time = 0.27, size = 192, normalized size = 1.03 \begin {gather*} \frac {1}{2} \, b {\left (\frac {{\left (\log \left (c x + d\right ) \log \left (\frac {b c x + b d}{a c - b d + c} + 1\right ) + {\rm Li}_2\left (-\frac {b c x + b d}{a c - b d + c}\right )\right )} d}{b c^{2}} - \frac {{\left (\log \left (c x + d\right ) \log \left (\frac {b c x + b d}{a c - b d - c} + 1\right ) + {\rm Li}_2\left (-\frac {b c x + b d}{a c - b d - c}\right )\right )} d}{b c^{2}} + \frac {{\left (a + 1\right )} \log \left (b x + a + 1\right )}{b^{2} c} - \frac {{\left (a - 1\right )} \log \left (b x + a - 1\right )}{b^{2} c}\right )} + {\left (\frac {x}{c} - \frac {d \log \left (c x + d\right )}{c^{2}}\right )} \operatorname {artanh}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(b*x+a)/(c+d/x),x, algorithm="maxima")

[Out]

1/2*b*((log(c*x + d)*log((b*c*x + b*d)/(a*c - b*d + c) + 1) + dilog(-(b*c*x + b*d)/(a*c - b*d + c)))*d/(b*c^2)
- (log(c*x + d)*log((b*c*x + b*d)/(a*c - b*d - c) + 1) + dilog(-(b*c*x + b*d)/(a*c - b*d - c)))*d/(b*c^2) + (
a + 1)*log(b*x + a + 1)/(b^2*c) - (a - 1)*log(b*x + a - 1)/(b^2*c)) + (x/c - d*log(c*x + d)/c^2)*arctanh(b*x +
a)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(b*x+a)/(c+d/x),x, algorithm="fricas")

[Out]

integral(x*arctanh(b*x + a)/(c*x + d), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \operatorname {atanh}{\left (a + b x \right )}}{c x + d}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(b*x+a)/(c+d/x),x)

[Out]

Integral(x*atanh(a + b*x)/(c*x + d), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(b*x+a)/(c+d/x),x, algorithm="giac")

[Out]

integrate(arctanh(b*x + a)/(c + d/x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {atanh}\left (a+b\,x\right )}{c+\frac {d}{x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a + b*x)/(c + d/x),x)

[Out]

int(atanh(a + b*x)/(c + d/x), x)

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