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

Optimal. Leaf size=397 \[ -\frac{\text{PolyLog}\left (2,-\frac{\sqrt{b} (1-x)}{\sqrt{-a}-\sqrt{b}}\right )}{4 \sqrt{-a} \sqrt{b}}+\frac{\text{PolyLog}\left (2,\frac{\sqrt{b} (1-x)}{\sqrt{-a}+\sqrt{b}}\right )}{4 \sqrt{-a} \sqrt{b}}-\frac{\text{PolyLog}\left (2,-\frac{\sqrt{b} (x+1)}{\sqrt{-a}-\sqrt{b}}\right )}{4 \sqrt{-a} \sqrt{b}}+\frac{\text{PolyLog}\left (2,\frac{\sqrt{b} (x+1)}{\sqrt{-a}+\sqrt{b}}\right )}{4 \sqrt{-a} \sqrt{b}}-\frac{\log (1-x) \log \left (\frac{\sqrt{-a}-\sqrt{b} x}{\sqrt{-a}-\sqrt{b}}\right )}{4 \sqrt{-a} \sqrt{b}}+\frac{\log (x+1) \log \left (\frac{\sqrt{-a}-\sqrt{b} x}{\sqrt{-a}+\sqrt{b}}\right )}{4 \sqrt{-a} \sqrt{b}}-\frac{\log (x+1) \log \left (\frac{\sqrt{-a}+\sqrt{b} x}{\sqrt{-a}-\sqrt{b}}\right )}{4 \sqrt{-a} \sqrt{b}}+\frac{\log (1-x) \log \left (\frac{\sqrt{-a}+\sqrt{b} x}{\sqrt{-a}+\sqrt{b}}\right )}{4 \sqrt{-a} \sqrt{b}} \]

[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])

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Rubi [A]  time = 0.372074, antiderivative size = 397, normalized size of antiderivative = 1., 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 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]

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 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}(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*}

Mathematica [C]  time = 1.0085, size = 485, normalized size = 1.22 \[ -\frac{i \left (\text{PolyLog}\left (2,\frac{\left (2 i \sqrt{a b}-a+b\right ) \left (x \sqrt{a b}+i a\right )}{(a+b) \left (x \sqrt{a b}-i a\right )}\right )-\text{PolyLog}\left (2,\frac{\left (-2 i \sqrt{a b}-a+b\right ) \left (x \sqrt{a b}+i a\right )}{(a+b) \left (x \sqrt{a b}-i a\right )}\right )\right )-2 i \cos ^{-1}\left (\frac{b-a}{a+b}\right ) \tan ^{-1}\left (\frac{b x}{\sqrt{a b}}\right )+4 \tanh ^{-1}(x) \tan ^{-1}\left (\frac{a}{x \sqrt{a b}}\right )-\log \left (\frac{2 i a (x-1) \left (\sqrt{a b}+i b\right )}{(a+b) \left (a+i x \sqrt{a b}\right )}\right ) \left (2 \tan ^{-1}\left (\frac{b x}{\sqrt{a b}}\right )+\cos ^{-1}\left (\frac{b-a}{a+b}\right )\right )-\log \left (\frac{2 a (x+1) \left (b+i \sqrt{a b}\right )}{(a+b) \left (a+i x \sqrt{a b}\right )}\right ) \left (\cos ^{-1}\left (\frac{b-a}{a+b}\right )-2 \tan ^{-1}\left (\frac{b x}{\sqrt{a b}}\right )\right )+\left (2 \left (\tan ^{-1}\left (\frac{a}{x \sqrt{a b}}\right )+\tan ^{-1}\left (\frac{b x}{\sqrt{a b}}\right )\right )+\cos ^{-1}\left (\frac{b-a}{a+b}\right )\right ) \log \left (\frac{\sqrt{2} \sqrt{a b} e^{-\tanh ^{-1}(x)}}{\sqrt{a+b} \sqrt{(a+b) \cosh \left (2 \tanh ^{-1}(x)\right )+a-b}}\right )+\left (\cos ^{-1}\left (\frac{b-a}{a+b}\right )-2 \left (\tan ^{-1}\left (\frac{a}{x \sqrt{a b}}\right )+\tan ^{-1}\left (\frac{b x}{\sqrt{a b}}\right )\right )\right ) \log \left (\frac{\sqrt{2} \sqrt{a b} e^{\tanh ^{-1}(x)}}{\sqrt{a+b} \sqrt{(a+b) \cosh \left (2 \tanh ^{-1}(x)\right )+a-b}}\right )}{4 \sqrt{a b}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[x]/(a + b*x^2),x]

[Out]

-((-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]]))*Log[(
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))]))/(4*Sqrt[a*b])

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Maple [B]  time = 0.186, size = 606, normalized size = 1.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-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)-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*arctanh(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))

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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(x)/(b*x^2+a),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{\operatorname{artanh}\left (x\right )}{b x^{2} + a}, x\right ) \end{align*}

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

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

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)