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

Optimal. Leaf size=397 \[ -\frac {\text {Li}_2\left (-\frac {\sqrt {b} (1-x)}{\sqrt {-a}-\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}+\frac {\text {Li}_2\left (\frac {\sqrt {b} (1-x)}{\sqrt {-a}+\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}-\frac {\text {Li}_2\left (-\frac {\sqrt {b} (x+1)}{\sqrt {-a}-\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}+\frac {\text {Li}_2\left (\frac {\sqrt {b} (x+1)}{\sqrt {-a}+\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}-\frac {\log (1-x) \log \left (\frac {\sqrt {-a}-\sqrt {b} x}{\sqrt {-a}-\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}+\frac {\log (x+1) \log \left (\frac {\sqrt {-a}-\sqrt {b} x}{\sqrt {-a}+\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}-\frac {\log (x+1) \log \left (\frac {\sqrt {-a}+\sqrt {b} x}{\sqrt {-a}-\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}+\frac {\log (1-x) \log \left (\frac {\sqrt {-a}+\sqrt {b} x}{\sqrt {-a}+\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}} \]

[Out]

-1/4*ln(1-x)*ln(((-a)^(1/2)-x*b^(1/2))/((-a)^(1/2)-b^(1/2)))/(-a)^(1/2)/b^(1/2)+1/4*ln(1+x)*ln(((-a)^(1/2)-x*b
^(1/2))/((-a)^(1/2)+b^(1/2)))/(-a)^(1/2)/b^(1/2)-1/4*ln(1+x)*ln(((-a)^(1/2)+x*b^(1/2))/((-a)^(1/2)-b^(1/2)))/(
-a)^(1/2)/b^(1/2)+1/4*ln(1-x)*ln(((-a)^(1/2)+x*b^(1/2))/((-a)^(1/2)+b^(1/2)))/(-a)^(1/2)/b^(1/2)-1/4*polylog(2
,-(1-x)*b^(1/2)/((-a)^(1/2)-b^(1/2)))/(-a)^(1/2)/b^(1/2)-1/4*polylog(2,-(1+x)*b^(1/2)/((-a)^(1/2)-b^(1/2)))/(-
a)^(1/2)/b^(1/2)+1/4*polylog(2,(1-x)*b^(1/2)/((-a)^(1/2)+b^(1/2)))/(-a)^(1/2)/b^(1/2)+1/4*polylog(2,(1+x)*b^(1
/2)/((-a)^(1/2)+b^(1/2)))/(-a)^(1/2)/b^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 397, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5972, 2409, 2394, 2393, 2391} \[ -\frac {\text {PolyLog}\left (2,-\frac {\sqrt {b} (1-x)}{\sqrt {-a}-\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}+\frac {\text {PolyLog}\left (2,\frac {\sqrt {b} (1-x)}{\sqrt {-a}+\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}-\frac {\text {PolyLog}\left (2,-\frac {\sqrt {b} (x+1)}{\sqrt {-a}-\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}+\frac {\text {PolyLog}\left (2,\frac {\sqrt {b} (x+1)}{\sqrt {-a}+\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}-\frac {\log (1-x) \log \left (\frac {\sqrt {-a}-\sqrt {b} x}{\sqrt {-a}-\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}+\frac {\log (x+1) \log \left (\frac {\sqrt {-a}-\sqrt {b} x}{\sqrt {-a}+\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}-\frac {\log (x+1) \log \left (\frac {\sqrt {-a}+\sqrt {b} x}{\sqrt {-a}-\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}}+\frac {\log (1-x) \log \left (\frac {\sqrt {-a}+\sqrt {b} x}{\sqrt {-a}+\sqrt {b}}\right )}{4 \sqrt {-a} \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[x]/(a + b*x^2),x]

[Out]

-(Log[1 - x]*Log[(Sqrt[-a] - Sqrt[b]*x)/(Sqrt[-a] - Sqrt[b])])/(4*Sqrt[-a]*Sqrt[b]) + (Log[1 + x]*Log[(Sqrt[-a
] - Sqrt[b]*x)/(Sqrt[-a] + Sqrt[b])])/(4*Sqrt[-a]*Sqrt[b]) - (Log[1 + x]*Log[(Sqrt[-a] + Sqrt[b]*x)/(Sqrt[-a]
- Sqrt[b])])/(4*Sqrt[-a]*Sqrt[b]) + (Log[1 - x]*Log[(Sqrt[-a] + Sqrt[b]*x)/(Sqrt[-a] + Sqrt[b])])/(4*Sqrt[-a]*
Sqrt[b]) - PolyLog[2, -((Sqrt[b]*(1 - x))/(Sqrt[-a] - Sqrt[b]))]/(4*Sqrt[-a]*Sqrt[b]) + PolyLog[2, (Sqrt[b]*(1
- x))/(Sqrt[-a] + Sqrt[b])]/(4*Sqrt[-a]*Sqrt[b]) - PolyLog[2, -((Sqrt[b]*(1 + x))/(Sqrt[-a] - Sqrt[b]))]/(4*S
qrt[-a]*Sqrt[b]) + PolyLog[2, (Sqrt[b]*(1 + x))/(Sqrt[-a] + Sqrt[b])]/(4*Sqrt[-a]*Sqrt[b])

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]

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 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 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 5972

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

Rubi steps

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

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Mathematica [C]  time = 1.13, size = 485, normalized size = 1.22 \[ -\frac {i \left (\text {Li}_2\left (\frac {\left (-a+b+2 i \sqrt {a b}\right ) \left (i a+\sqrt {a b} x\right )}{(a+b) \left (\sqrt {a b} x-i a\right )}\right )-\text {Li}_2\left (\frac {\left (-a+b-2 i \sqrt {a b}\right ) \left (i a+\sqrt {a b} x\right )}{(a+b) \left (\sqrt {a b} x-i a\right )}\right )\right )-2 i \cos ^{-1}\left (\frac {b-a}{a+b}\right ) \tan ^{-1}\left (\frac {b x}{\sqrt {a b}}\right )+4 \tanh ^{-1}(x) \tan ^{-1}\left (\frac {a}{x \sqrt {a b}}\right )-\log \left (\frac {2 i a (x-1) \left (\sqrt {a b}+i b\right )}{(a+b) \left (a+i x \sqrt {a b}\right )}\right ) \left (2 \tan ^{-1}\left (\frac {b x}{\sqrt {a b}}\right )+\cos ^{-1}\left (\frac {b-a}{a+b}\right )\right )-\log \left (\frac {2 a (x+1) \left (b+i \sqrt {a b}\right )}{(a+b) \left (a+i x \sqrt {a b}\right )}\right ) \left (\cos ^{-1}\left (\frac {b-a}{a+b}\right )-2 \tan ^{-1}\left (\frac {b x}{\sqrt {a b}}\right )\right )+\left (2 \left (\tan ^{-1}\left (\frac {a}{x \sqrt {a b}}\right )+\tan ^{-1}\left (\frac {b x}{\sqrt {a b}}\right )\right )+\cos ^{-1}\left (\frac {b-a}{a+b}\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {a b} e^{-\tanh ^{-1}(x)}}{\sqrt {a+b} \sqrt {(a+b) \cosh \left (2 \tanh ^{-1}(x)\right )+a-b}}\right )+\left (\cos ^{-1}\left (\frac {b-a}{a+b}\right )-2 \left (\tan ^{-1}\left (\frac {a}{x \sqrt {a b}}\right )+\tan ^{-1}\left (\frac {b x}{\sqrt {a b}}\right )\right )\right ) \log \left (\frac {\sqrt {2} \sqrt {a b} e^{\tanh ^{-1}(x)}}{\sqrt {a+b} \sqrt {(a+b) \cosh \left (2 \tanh ^{-1}(x)\right )+a-b}}\right )}{4 \sqrt {a b}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[x]/(a + b*x^2),x]

[Out]

-1/4*((-2*I)*ArcCos[(-a + b)/(a + b)]*ArcTan[(b*x)/Sqrt[a*b]] + 4*ArcTan[a/(Sqrt[a*b]*x)]*ArcTanh[x] - (ArcCos
[(-a + b)/(a + b)] + 2*ArcTan[(b*x)/Sqrt[a*b]])*Log[((2*I)*a*(I*b + Sqrt[a*b])*(-1 + x))/((a + b)*(a + I*Sqrt[
a*b]*x))] - (ArcCos[(-a + b)/(a + b)] - 2*ArcTan[(b*x)/Sqrt[a*b]])*Log[(2*a*(b + I*Sqrt[a*b])*(1 + x))/((a + b
)*(a + I*Sqrt[a*b]*x))] + (ArcCos[(-a + b)/(a + b)] + 2*(ArcTan[a/(Sqrt[a*b]*x)] + ArcTan[(b*x)/Sqrt[a*b]]))*L
og[(Sqrt[2]*Sqrt[a*b])/(Sqrt[a + b]*E^ArcTanh[x]*Sqrt[a - b + (a + b)*Cosh[2*ArcTanh[x]]])] + (ArcCos[(-a + b)
/(a + b)] - 2*(ArcTan[a/(Sqrt[a*b]*x)] + ArcTan[(b*x)/Sqrt[a*b]]))*Log[(Sqrt[2]*Sqrt[a*b]*E^ArcTanh[x])/(Sqrt[
a + b]*Sqrt[a - b + (a + b)*Cosh[2*ArcTanh[x]]])] + I*(-PolyLog[2, ((-a + b - (2*I)*Sqrt[a*b])*(I*a + Sqrt[a*b
]*x))/((a + b)*((-I)*a + Sqrt[a*b]*x))] + PolyLog[2, ((-a + b + (2*I)*Sqrt[a*b])*(I*a + Sqrt[a*b]*x))/((a + b)
*((-I)*a + Sqrt[a*b]*x))]))/Sqrt[a*b]

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\relax (x)}{b x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctanh(x)/(b*x^2 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\relax (x)}{b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctanh(x)/(b*x^2 + a), x)

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maple [B]  time = 0.62, size = 606, normalized size = 1.53 \[ -\frac {\sqrt {-a b}\, \arctanh \relax (x ) \ln \left (1-\frac {\left (a +b \right ) \left (1+x \right )^{2}}{\left (-x^{2}+1\right ) \left (2 \sqrt {-a b}-a +b \right )}\right )}{2 a b}+\frac {\sqrt {-a b}\, \arctanh \relax (x )^{2}}{2 a b}-\frac {\sqrt {-a b}\, \polylog \left (2, \frac {\left (a +b \right ) \left (1+x \right )^{2}}{\left (-x^{2}+1\right ) \left (2 \sqrt {-a b}-a +b \right )}\right )}{4 a b}-\frac {\left (-2 \sqrt {-a b}+a -b \right ) \ln \left (1-\frac {\left (a +b \right ) \left (1+x \right )^{2}}{\left (-x^{2}+1\right ) \left (-2 \sqrt {-a b}-a +b \right )}\right ) \arctanh \relax (x )}{a^{2}+2 a b +b^{2}}+\frac {\left (2 a b +\sqrt {-a b}\, a -\sqrt {-a b}\, b \right ) \ln \left (1-\frac {\left (a +b \right ) \left (1+x \right )^{2}}{\left (-x^{2}+1\right ) \left (-2 \sqrt {-a b}-a +b \right )}\right ) \arctanh \relax (x )}{2 b \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (2 a b +\sqrt {-a b}\, a -\sqrt {-a b}\, b \right ) \ln \left (1-\frac {\left (a +b \right ) \left (1+x \right )^{2}}{\left (-x^{2}+1\right ) \left (-2 \sqrt {-a b}-a +b \right )}\right ) \arctanh \relax (x )}{2 a \left (a^{2}+2 a b +b^{2}\right )}-\frac {\arctanh \relax (x )^{2} \sqrt {-a b}}{a^{2}+2 a b +b^{2}}-\frac {\arctanh \relax (x )^{2} a \sqrt {-a b}}{2 b \left (a^{2}+2 a b +b^{2}\right )}-\frac {\arctanh \relax (x )^{2} b \sqrt {-a b}}{2 a \left (a^{2}+2 a b +b^{2}\right )}+\frac {\polylog \left (2, \frac {\left (a +b \right ) \left (1+x \right )^{2}}{\left (-x^{2}+1\right ) \left (-2 \sqrt {-a b}-a +b \right )}\right ) \sqrt {-a b}}{2 a^{2}+4 a b +2 b^{2}}+\frac {\polylog \left (2, \frac {\left (a +b \right ) \left (1+x \right )^{2}}{\left (-x^{2}+1\right ) \left (-2 \sqrt {-a b}-a +b \right )}\right ) a \sqrt {-a b}}{4 b \left (a^{2}+2 a b +b^{2}\right )}+\frac {\polylog \left (2, \frac {\left (a +b \right ) \left (1+x \right )^{2}}{\left (-x^{2}+1\right ) \left (-2 \sqrt {-a b}-a +b \right )}\right ) b \sqrt {-a b}}{4 a \left (a^{2}+2 a b +b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-a*b)^(1/2)/a/b*arctanh(x)*ln(1-(a+b)*(1+x)^2/(-x^2+1)/(2*(-a*b)^(1/2)-a+b))+1/2*(-a*b)^(1/2)/a/b*arctan
h(x)^2-1/4*(-a*b)^(1/2)/a/b*polylog(2,(a+b)*(1+x)^2/(-x^2+1)/(2*(-a*b)^(1/2)-a+b))-(-2*(-a*b)^(1/2)+a-b)/(a^2+
2*a*b+b^2)*ln(1-(a+b)*(1+x)^2/(-x^2+1)/(-2*(-a*b)^(1/2)-a+b))*arctanh(x)+1/2*(2*a*b+(-a*b)^(1/2)*a-(-a*b)^(1/2
)*b)/b/(a^2+2*a*b+b^2)*ln(1-(a+b)*(1+x)^2/(-x^2+1)/(-2*(-a*b)^(1/2)-a+b))*arctanh(x)-1/2*(2*a*b+(-a*b)^(1/2)*a
-(-a*b)^(1/2)*b)/a/(a^2+2*a*b+b^2)*ln(1-(a+b)*(1+x)^2/(-x^2+1)/(-2*(-a*b)^(1/2)-a+b))*arctanh(x)-1/(a^2+2*a*b+
b^2)*arctanh(x)^2*(-a*b)^(1/2)-1/2/b/(a^2+2*a*b+b^2)*arctanh(x)^2*a*(-a*b)^(1/2)-1/2/a/(a^2+2*a*b+b^2)*arctanh
(x)^2*b*(-a*b)^(1/2)+1/2/(a^2+2*a*b+b^2)*polylog(2,(a+b)*(1+x)^2/(-x^2+1)/(-2*(-a*b)^(1/2)-a+b))*(-a*b)^(1/2)+
1/4/b/(a^2+2*a*b+b^2)*polylog(2,(a+b)*(1+x)^2/(-x^2+1)/(-2*(-a*b)^(1/2)-a+b))*a*(-a*b)^(1/2)+1/4/a/(a^2+2*a*b+
b^2)*polylog(2,(a+b)*(1+x)^2/(-x^2+1)/(-2*(-a*b)^(1/2)-a+b))*b*(-a*b)^(1/2)

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maxima [C]  time = 0.52, size = 304, normalized size = 0.77 \[ \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right ) \operatorname {artanh}\relax (x)}{\sqrt {a b}} + \frac {{\left (\arctan \left (\frac {\sqrt {a} \sqrt {b} {\left (x + 1\right )}}{a + b}, \frac {b x + b}{a + b}\right ) - \arctan \left (\frac {\sqrt {a} \sqrt {b} {\left (x - 1\right )}}{a + b}, -\frac {b x - b}{a + b}\right )\right )} \log \left (b x^{2} + a\right ) - \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {b x^{2} + 2 \, b x + b}{a + b}\right ) + \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {b x^{2} - 2 \, b x + b}{a + b}\right ) - i \, {\rm Li}_2\left (-\frac {b x - \sqrt {a} \sqrt {b} {\left (i \, x + i\right )} - a}{a + 2 i \, \sqrt {a} \sqrt {b} - b}\right ) - i \, {\rm Li}_2\left (\frac {b x - \sqrt {a} \sqrt {b} {\left (i \, x - i\right )} + a}{a + 2 i \, \sqrt {a} \sqrt {b} - b}\right ) + i \, {\rm Li}_2\left (-\frac {b x + \sqrt {a} \sqrt {b} {\left (i \, x + i\right )} - a}{a - 2 i \, \sqrt {a} \sqrt {b} - b}\right ) + i \, {\rm Li}_2\left (\frac {b x + \sqrt {a} \sqrt {b} {\left (i \, x - i\right )} + a}{a - 2 i \, \sqrt {a} \sqrt {b} - b}\right )}{4 \, \sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(b*x/sqrt(a*b))*arctanh(x)/sqrt(a*b) + 1/4*((arctan2(sqrt(a)*sqrt(b)*(x + 1)/(a + b), (b*x + b)/(a + b))
- arctan2(sqrt(a)*sqrt(b)*(x - 1)/(a + b), -(b*x - b)/(a + b)))*log(b*x^2 + a) - arctan(sqrt(b)*x/sqrt(a))*lo
g((b*x^2 + 2*b*x + b)/(a + b)) + arctan(sqrt(b)*x/sqrt(a))*log((b*x^2 - 2*b*x + b)/(a + b)) - I*dilog(-(b*x -
sqrt(a)*sqrt(b)*(I*x + I) - a)/(a + 2*I*sqrt(a)*sqrt(b) - b)) - I*dilog((b*x - sqrt(a)*sqrt(b)*(I*x - I) + a)/
(a + 2*I*sqrt(a)*sqrt(b) - b)) + I*dilog(-(b*x + sqrt(a)*sqrt(b)*(I*x + I) - a)/(a - 2*I*sqrt(a)*sqrt(b) - b))
+ I*dilog((b*x + sqrt(a)*sqrt(b)*(I*x - I) + a)/(a - 2*I*sqrt(a)*sqrt(b) - b)))/sqrt(a*b)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {atanh}\relax (x)}{b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(x)/(a + b*x^2),x)

[Out]

int(atanh(x)/(a + b*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\relax (x )}}{a + b x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atanh(x)/(a + b*x**2), x)

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