3.57 \(\int \frac{\tanh ^{-1}(a+b x)}{c+\frac{d}{x^2}} \, dx\)

Optimal. Leaf size=545 \[ \frac{\sqrt{d} \text{PolyLog}\left (2,\frac{\sqrt{-c} (-a-b x+1)}{a \left (-\sqrt{-c}\right )-b \sqrt{d}+\sqrt{-c}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \text{PolyLog}\left (2,\frac{\sqrt{-c} (-a-b x+1)}{(1-a) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{\sqrt{d} \text{PolyLog}\left (2,\frac{\sqrt{-c} (a+b x+1)}{(a+1) \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \text{PolyLog}\left (2,\frac{\sqrt{-c} (a+b x+1)}{(a+1) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{\sqrt{d} \log (-a-b x+1) \log \left (-\frac{b \left (\sqrt{d}-\sqrt{-c} x\right )}{(1-a) \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \log (-a-b x+1) \log \left (\frac{b \left (\sqrt{-c} x+\sqrt{d}\right )}{(1-a) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \log (a+b x+1) \log \left (\frac{b \left (\sqrt{d}-\sqrt{-c} x\right )}{(a+1) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{\sqrt{d} \log (a+b x+1) \log \left (-\frac{b \left (\sqrt{-c} x+\sqrt{d}\right )}{(a+1) \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{(-a-b x+1) \log (-a-b x+1)}{2 b c}+\frac{(a+b x+1) \log (a+b x+1)}{2 b c} \]

[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) + (Sqrt[d]*Log[1 - a - b*x
]*Log[-((b*(Sqrt[d] - Sqrt[-c]*x))/((1 - a)*Sqrt[-c] - b*Sqrt[d]))])/(4*(-c)^(3/2)) - (Sqrt[d]*Log[1 + a + b*x
]*Log[(b*(Sqrt[d] - Sqrt[-c]*x))/((1 + a)*Sqrt[-c] + b*Sqrt[d])])/(4*(-c)^(3/2)) + (Sqrt[d]*Log[1 + a + b*x]*L
og[-((b*(Sqrt[d] + Sqrt[-c]*x))/((1 + a)*Sqrt[-c] - b*Sqrt[d]))])/(4*(-c)^(3/2)) - (Sqrt[d]*Log[1 - a - b*x]*L
og[(b*(Sqrt[d] + Sqrt[-c]*x))/((1 - a)*Sqrt[-c] + b*Sqrt[d])])/(4*(-c)^(3/2)) + (Sqrt[d]*PolyLog[2, (Sqrt[-c]*
(1 - a - b*x))/(Sqrt[-c] - a*Sqrt[-c] - b*Sqrt[d])])/(4*(-c)^(3/2)) - (Sqrt[d]*PolyLog[2, (Sqrt[-c]*(1 - a - b
*x))/((1 - a)*Sqrt[-c] + b*Sqrt[d])])/(4*(-c)^(3/2)) + (Sqrt[d]*PolyLog[2, (Sqrt[-c]*(1 + a + b*x))/((1 + a)*S
qrt[-c] - b*Sqrt[d])])/(4*(-c)^(3/2)) - (Sqrt[d]*PolyLog[2, (Sqrt[-c]*(1 + a + b*x))/((1 + a)*Sqrt[-c] + b*Sqr
t[d])])/(4*(-c)^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.910714, antiderivative size = 545, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {6115, 2409, 2389, 2295, 2394, 2393, 2391} \[ \frac{\sqrt{d} \text{PolyLog}\left (2,\frac{\sqrt{-c} (-a-b x+1)}{a \left (-\sqrt{-c}\right )-b \sqrt{d}+\sqrt{-c}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \text{PolyLog}\left (2,\frac{\sqrt{-c} (-a-b x+1)}{(1-a) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{\sqrt{d} \text{PolyLog}\left (2,\frac{\sqrt{-c} (a+b x+1)}{(a+1) \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \text{PolyLog}\left (2,\frac{\sqrt{-c} (a+b x+1)}{(a+1) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{\sqrt{d} \log (-a-b x+1) \log \left (-\frac{b \left (\sqrt{d}-\sqrt{-c} x\right )}{(1-a) \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \log (-a-b x+1) \log \left (\frac{b \left (\sqrt{-c} x+\sqrt{d}\right )}{(1-a) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \log (a+b x+1) \log \left (\frac{b \left (\sqrt{d}-\sqrt{-c} x\right )}{(a+1) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{\sqrt{d} \log (a+b x+1) \log \left (-\frac{b \left (\sqrt{-c} x+\sqrt{d}\right )}{(a+1) \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{(-a-b x+1) \log (-a-b x+1)}{2 b c}+\frac{(a+b x+1) \log (a+b x+1)}{2 b c} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a + b*x]/(c + d/x^2),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) + (Sqrt[d]*Log[1 - a - b*x
]*Log[-((b*(Sqrt[d] - Sqrt[-c]*x))/((1 - a)*Sqrt[-c] - b*Sqrt[d]))])/(4*(-c)^(3/2)) - (Sqrt[d]*Log[1 + a + b*x
]*Log[(b*(Sqrt[d] - Sqrt[-c]*x))/((1 + a)*Sqrt[-c] + b*Sqrt[d])])/(4*(-c)^(3/2)) + (Sqrt[d]*Log[1 + a + b*x]*L
og[-((b*(Sqrt[d] + Sqrt[-c]*x))/((1 + a)*Sqrt[-c] - b*Sqrt[d]))])/(4*(-c)^(3/2)) - (Sqrt[d]*Log[1 - a - b*x]*L
og[(b*(Sqrt[d] + Sqrt[-c]*x))/((1 - a)*Sqrt[-c] + b*Sqrt[d])])/(4*(-c)^(3/2)) + (Sqrt[d]*PolyLog[2, (Sqrt[-c]*
(1 - a - b*x))/(Sqrt[-c] - a*Sqrt[-c] - b*Sqrt[d])])/(4*(-c)^(3/2)) - (Sqrt[d]*PolyLog[2, (Sqrt[-c]*(1 - a - b
*x))/((1 - a)*Sqrt[-c] + b*Sqrt[d])])/(4*(-c)^(3/2)) + (Sqrt[d]*PolyLog[2, (Sqrt[-c]*(1 + a + b*x))/((1 + a)*S
qrt[-c] - b*Sqrt[d])])/(4*(-c)^(3/2)) - (Sqrt[d]*PolyLog[2, (Sqrt[-c]*(1 + a + b*x))/((1 + a)*Sqrt[-c] + b*Sqr
t[d])])/(4*(-c)^(3/2))

Rule 6115

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]

Rule 2409

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 2389

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 2295

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

Rule 2394

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 2393

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 2391

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]

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(a+b x)}{c+\frac{d}{x^2}} \, dx &=-\left (\frac{1}{2} \int \frac{\log (1-a-b x)}{c+\frac{d}{x^2}} \, dx\right )+\frac{1}{2} \int \frac{\log (1+a+b x)}{c+\frac{d}{x^2}} \, dx\\ &=-\left (\frac{1}{2} \int \left (\frac{\log (1-a-b x)}{c}-\frac{d \log (1-a-b x)}{c \left (d+c x^2\right )}\right ) \, dx\right )+\frac{1}{2} \int \left (\frac{\log (1+a+b x)}{c}-\frac{d \log (1+a+b x)}{c \left (d+c x^2\right )}\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^2} \, dx}{2 c}-\frac{d \int \frac{\log (1+a+b x)}{d+c x^2} \, dx}{2 c}\\ &=\frac{\operatorname{Subst}(\int \log (x) \, dx,x,1-a-b x)}{2 b c}+\frac{\operatorname{Subst}(\int \log (x) \, dx,x,1+a+b x)}{2 b c}+\frac{d \int \left (\frac{\log (1-a-b x)}{2 \sqrt{d} \left (\sqrt{d}-\sqrt{-c} x\right )}+\frac{\log (1-a-b x)}{2 \sqrt{d} \left (\sqrt{d}+\sqrt{-c} x\right )}\right ) \, dx}{2 c}-\frac{d \int \left (\frac{\log (1+a+b x)}{2 \sqrt{d} \left (\sqrt{d}-\sqrt{-c} x\right )}+\frac{\log (1+a+b x)}{2 \sqrt{d} \left (\sqrt{d}+\sqrt{-c} x\right )}\right ) \, dx}{2 c}\\ &=\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{\sqrt{d} \int \frac{\log (1-a-b x)}{\sqrt{d}-\sqrt{-c} x} \, dx}{4 c}+\frac{\sqrt{d} \int \frac{\log (1-a-b x)}{\sqrt{d}+\sqrt{-c} x} \, dx}{4 c}-\frac{\sqrt{d} \int \frac{\log (1+a+b x)}{\sqrt{d}-\sqrt{-c} x} \, dx}{4 c}-\frac{\sqrt{d} \int \frac{\log (1+a+b x)}{\sqrt{d}+\sqrt{-c} x} \, dx}{4 c}\\ &=\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{\sqrt{d} \log (1-a-b x) \log \left (-\frac{b \left (\sqrt{d}-\sqrt{-c} x\right )}{(1-a) \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \log (1+a+b x) \log \left (\frac{b \left (\sqrt{d}-\sqrt{-c} x\right )}{(1+a) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{\sqrt{d} \log (1+a+b x) \log \left (-\frac{b \left (\sqrt{d}+\sqrt{-c} x\right )}{(1+a) \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \log (1-a-b x) \log \left (\frac{b \left (\sqrt{d}+\sqrt{-c} x\right )}{(1-a) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{\left (b \sqrt{d}\right ) \int \frac{\log \left (-\frac{b \left (\sqrt{d}-\sqrt{-c} x\right )}{(1-a) \sqrt{-c}-b \sqrt{d}}\right )}{1-a-b x} \, dx}{4 (-c)^{3/2}}+\frac{\left (b \sqrt{d}\right ) \int \frac{\log \left (\frac{b \left (\sqrt{d}-\sqrt{-c} x\right )}{(1+a) \sqrt{-c}+b \sqrt{d}}\right )}{1+a+b x} \, dx}{4 (-c)^{3/2}}-\frac{\left (b \sqrt{d}\right ) \int \frac{\log \left (-\frac{b \left (\sqrt{d}+\sqrt{-c} x\right )}{-(1-a) \sqrt{-c}-b \sqrt{d}}\right )}{1-a-b x} \, dx}{4 (-c)^{3/2}}-\frac{\left (b \sqrt{d}\right ) \int \frac{\log \left (\frac{b \left (\sqrt{d}+\sqrt{-c} x\right )}{-(1+a) \sqrt{-c}+b \sqrt{d}}\right )}{1+a+b x} \, dx}{4 (-c)^{3/2}}\\ &=\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{\sqrt{d} \log (1-a-b x) \log \left (-\frac{b \left (\sqrt{d}-\sqrt{-c} x\right )}{(1-a) \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \log (1+a+b x) \log \left (\frac{b \left (\sqrt{d}-\sqrt{-c} x\right )}{(1+a) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{\sqrt{d} \log (1+a+b x) \log \left (-\frac{b \left (\sqrt{d}+\sqrt{-c} x\right )}{(1+a) \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \log (1-a-b x) \log \left (\frac{b \left (\sqrt{d}+\sqrt{-c} x\right )}{(1-a) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{\sqrt{d} \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{\sqrt{-c} x}{-(1-a) \sqrt{-c}-b \sqrt{d}}\right )}{x} \, dx,x,1-a-b x\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{\sqrt{-c} x}{(1-a) \sqrt{-c}-b \sqrt{d}}\right )}{x} \, dx,x,1-a-b x\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{\sqrt{-c} x}{-(1+a) \sqrt{-c}+b \sqrt{d}}\right )}{x} \, dx,x,1+a+b x\right )}{4 (-c)^{3/2}}+\frac{\sqrt{d} \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{\sqrt{-c} x}{(1+a) \sqrt{-c}+b \sqrt{d}}\right )}{x} \, dx,x,1+a+b x\right )}{4 (-c)^{3/2}}\\ &=\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{\sqrt{d} \log (1-a-b x) \log \left (-\frac{b \left (\sqrt{d}-\sqrt{-c} x\right )}{(1-a) \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \log (1+a+b x) \log \left (\frac{b \left (\sqrt{d}-\sqrt{-c} x\right )}{(1+a) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{\sqrt{d} \log (1+a+b x) \log \left (-\frac{b \left (\sqrt{d}+\sqrt{-c} x\right )}{(1+a) \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \log (1-a-b x) \log \left (\frac{b \left (\sqrt{d}+\sqrt{-c} x\right )}{(1-a) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{\sqrt{d} \text{Li}_2\left (\frac{\sqrt{-c} (1-a-b x)}{\sqrt{-c}-a \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \text{Li}_2\left (\frac{\sqrt{-c} (1-a-b x)}{(1-a) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}+\frac{\sqrt{d} \text{Li}_2\left (\frac{\sqrt{-c} (1+a+b x)}{(1+a) \sqrt{-c}-b \sqrt{d}}\right )}{4 (-c)^{3/2}}-\frac{\sqrt{d} \text{Li}_2\left (\frac{\sqrt{-c} (1+a+b x)}{(1+a) \sqrt{-c}+b \sqrt{d}}\right )}{4 (-c)^{3/2}}\\ \end{align*}

Mathematica [C]  time = 24.5071, size = 1456, normalized size = 2.67 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[a + b*x]/(c + d/x^2),x]

[Out]

((a + b*x)*ArcTanh[a + b*x] - Log[1/Sqrt[1 - (a + b*x)^2]])/(b*c) + (Sqrt[d]*((2*I)*Sqrt[c]*ArcTan[((-1 + a)*S
qrt[c])/(b*Sqrt[d])]*ArcTan[(Sqrt[c]*x)/Sqrt[d]] - (2*I)*a^2*Sqrt[c]*ArcTan[((-1 + a)*Sqrt[c])/(b*Sqrt[d])]*Ar
cTan[(Sqrt[c]*x)/Sqrt[d]] - (2*I)*Sqrt[c]*ArcTan[((1 + a)*Sqrt[c])/(b*Sqrt[d])]*ArcTan[(Sqrt[c]*x)/Sqrt[d]] +
(2*I)*a^2*Sqrt[c]*ArcTan[((1 + a)*Sqrt[c])/(b*Sqrt[d])]*ArcTan[(Sqrt[c]*x)/Sqrt[d]] - 2*b*Sqrt[d]*ArcTan[(Sqrt
[c]*x)/Sqrt[d]]^2 + (b*Sqrt[d]*Sqrt[((-1 + a)^2*c + b^2*d)/(b^2*d)]*ArcTan[(Sqrt[c]*x)/Sqrt[d]]^2)/E^(I*ArcTan
[((-1 + a)*Sqrt[c])/(b*Sqrt[d])]) + (a*b*Sqrt[d]*Sqrt[((-1 + a)^2*c + b^2*d)/(b^2*d)]*ArcTan[(Sqrt[c]*x)/Sqrt[
d]]^2)/E^(I*ArcTan[((-1 + a)*Sqrt[c])/(b*Sqrt[d])]) + (b*Sqrt[d]*Sqrt[((1 + a)^2*c + b^2*d)/(b^2*d)]*ArcTan[(S
qrt[c]*x)/Sqrt[d]]^2)/E^(I*ArcTan[((1 + a)*Sqrt[c])/(b*Sqrt[d])]) - (a*b*Sqrt[d]*Sqrt[((1 + a)^2*c + b^2*d)/(b
^2*d)]*ArcTan[(Sqrt[c]*x)/Sqrt[d]]^2)/E^(I*ArcTan[((1 + a)*Sqrt[c])/(b*Sqrt[d])]) - 4*(-1 + a^2)*Sqrt[c]*ArcTa
n[(Sqrt[c]*x)/Sqrt[d]]*ArcTanh[a + b*x] + 2*Sqrt[c]*ArcTan[((-1 + a)*Sqrt[c])/(b*Sqrt[d])]*Log[1 - E^((-2*I)*(
ArcTan[((-1 + a)*Sqrt[c])/(b*Sqrt[d])] + ArcTan[(Sqrt[c]*x)/Sqrt[d]]))] - 2*a^2*Sqrt[c]*ArcTan[((-1 + a)*Sqrt[
c])/(b*Sqrt[d])]*Log[1 - E^((-2*I)*(ArcTan[((-1 + a)*Sqrt[c])/(b*Sqrt[d])] + ArcTan[(Sqrt[c]*x)/Sqrt[d]]))] +
2*Sqrt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[d]]*Log[1 - E^((-2*I)*(ArcTan[((-1 + a)*Sqrt[c])/(b*Sqrt[d])] + ArcTan[(Sqrt
[c]*x)/Sqrt[d]]))] - 2*a^2*Sqrt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[d]]*Log[1 - E^((-2*I)*(ArcTan[((-1 + a)*Sqrt[c])/(b
*Sqrt[d])] + ArcTan[(Sqrt[c]*x)/Sqrt[d]]))] - 2*Sqrt[c]*ArcTan[((1 + a)*Sqrt[c])/(b*Sqrt[d])]*Log[1 - E^((-2*I
)*(ArcTan[((1 + a)*Sqrt[c])/(b*Sqrt[d])] + ArcTan[(Sqrt[c]*x)/Sqrt[d]]))] + 2*a^2*Sqrt[c]*ArcTan[((1 + a)*Sqrt
[c])/(b*Sqrt[d])]*Log[1 - E^((-2*I)*(ArcTan[((1 + a)*Sqrt[c])/(b*Sqrt[d])] + ArcTan[(Sqrt[c]*x)/Sqrt[d]]))] -
2*Sqrt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[d]]*Log[1 - E^((-2*I)*(ArcTan[((1 + a)*Sqrt[c])/(b*Sqrt[d])] + ArcTan[(Sqrt[
c]*x)/Sqrt[d]]))] + 2*a^2*Sqrt[c]*ArcTan[(Sqrt[c]*x)/Sqrt[d]]*Log[1 - E^((-2*I)*(ArcTan[((1 + a)*Sqrt[c])/(b*S
qrt[d])] + ArcTan[(Sqrt[c]*x)/Sqrt[d]]))] - 2*Sqrt[c]*ArcTan[((-1 + a)*Sqrt[c])/(b*Sqrt[d])]*Log[-Sin[ArcTan[(
(-1 + a)*Sqrt[c])/(b*Sqrt[d])] + ArcTan[(Sqrt[c]*x)/Sqrt[d]]]] + 2*a^2*Sqrt[c]*ArcTan[((-1 + a)*Sqrt[c])/(b*Sq
rt[d])]*Log[-Sin[ArcTan[((-1 + a)*Sqrt[c])/(b*Sqrt[d])] + ArcTan[(Sqrt[c]*x)/Sqrt[d]]]] + 2*Sqrt[c]*ArcTan[((1
+ a)*Sqrt[c])/(b*Sqrt[d])]*Log[-Sin[ArcTan[((1 + a)*Sqrt[c])/(b*Sqrt[d])] + ArcTan[(Sqrt[c]*x)/Sqrt[d]]]] - 2
*a^2*Sqrt[c]*ArcTan[((1 + a)*Sqrt[c])/(b*Sqrt[d])]*Log[-Sin[ArcTan[((1 + a)*Sqrt[c])/(b*Sqrt[d])] + ArcTan[(Sq
rt[c]*x)/Sqrt[d]]]] - I*(-1 + a^2)*Sqrt[c]*PolyLog[2, E^((-2*I)*(ArcTan[((-1 + a)*Sqrt[c])/(b*Sqrt[d])] + ArcT
an[(Sqrt[c]*x)/Sqrt[d]]))] + I*(-1 + a^2)*Sqrt[c]*PolyLog[2, E^((-2*I)*(ArcTan[((1 + a)*Sqrt[c])/(b*Sqrt[d])]
+ ArcTan[(Sqrt[c]*x)/Sqrt[d]]))]))/(4*(-1 + a^2)*c^2)

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Maple [C]  time = 3.524, size = 20505, normalized size = 37.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(b*x+a)/(c+d/x^2),x)

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(b*x+a)/(c+d/x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{2} \operatorname{artanh}\left (b x + a\right )}{c x^{2} + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(b*x+a)/(c+d/x^2),x, algorithm="fricas")

[Out]

integral(x^2*arctanh(b*x + a)/(c*x^2 + d), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(b*x+a)/(c+d/x**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{artanh}\left (b x + a\right )}{c + \frac{d}{x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(b*x+a)/(c+d/x^2),x, algorithm="giac")

[Out]

integrate(arctanh(b*x + a)/(c + d/x^2), x)