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

Optimal. Leaf size=89 \[ \frac{b \text{PolyLog}\left (2,-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}{2 c}-\frac{b \text{PolyLog}\left (2,\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}{2 c}-\frac{a \log \left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}{c} \]

[Out]

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

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Rubi [A]  time = 0.0506488, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {206, 6681, 5912} \[ \frac{b \text{PolyLog}\left (2,-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}{2 c}-\frac{b \text{PolyLog}\left (2,\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}{2 c}-\frac{a \log \left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])/(1 - c^2*x^2),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 6681

Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.)*(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^
2), x_Symbol] :> Dist[(2*e*g)/(C*(e*f - d*g)), Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x
]], x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0] && IGtQ[n, 0]

Rule 5912

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b*PolyLog[2, -(c*x)])/2
, x] + Simp[(b*PolyLog[2, c*x])/2, x]) /; FreeQ[{a, b, c}, x]

Rubi steps

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

Mathematica [A]  time = 0.263249, size = 43, normalized size = 0.48 \[ \frac{b \left (\text{PolyLog}\left (2,-e^{-\tanh ^{-1}(c x)}\right )-\text{PolyLog}\left (2,e^{-\tanh ^{-1}(c x)}\right )\right )}{2 c}+\frac{a \tanh ^{-1}(c x)}{c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])/(1 - c^2*x^2),x]

[Out]

(a*ArcTanh[c*x])/c + (b*(PolyLog[2, -E^(-ArcTanh[c*x])] - PolyLog[2, E^(-ArcTanh[c*x])]))/(2*c)

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Maple [A]  time = 0.514, size = 118, normalized size = 1.3 \begin{align*} -{\frac{a\ln \left ( cx-1 \right ) }{2\,c}}+{\frac{a\ln \left ( cx+1 \right ) }{2\,c}}-{\frac{b}{c}{\it dilog} \left ({ \left ( -{\frac{-cx+1}{cx+1}}+1 \right ) \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}+1 \right ) ^{-2}} \right ) }+{\frac{b}{4\,c}{\it dilog} \left ({ \left ( -{\frac{-cx+1}{cx+1}}+1 \right ) ^{2} \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}+1 \right ) ^{-4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))/(-c^2*x^2+1),x)

[Out]

-1/2*a/c*ln(c*x-1)+1/2*a/c*ln(c*x+1)-b/c*dilog(1/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)^2*(-(-c*x+1)/(c*x+1)+1))+1/4
*b/c*dilog(1/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)^4*(-(-c*x+1)/(c*x+1)+1)^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, b{\left (\frac{{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )} \log \left (\sqrt{c x + 1} + \sqrt{-c x + 1}\right ) -{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )} \log \left (\sqrt{c x + 1} - \sqrt{-c x + 1}\right )}{c} - 2 \, \int -\frac{\sqrt{c x + 1}{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{2 \,{\left ({\left (c^{2} x^{2} - 1\right )} \sqrt{c x + 1} +{\left (c^{2} x^{2} - 1\right )} \sqrt{-c x + 1}\right )}}\,{d x} - 2 \, \int \frac{\sqrt{c x + 1}{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{2 \,{\left ({\left (c^{2} x^{2} - 1\right )} \sqrt{c x + 1} -{\left (c^{2} x^{2} - 1\right )} \sqrt{-c x + 1}\right )}}\,{d x}\right )} + \frac{1}{2} \, a{\left (\frac{\log \left (c x + 1\right )}{c} - \frac{\log \left (c x - 1\right )}{c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))/(-c^2*x^2+1),x, algorithm="maxima")

[Out]

1/4*b*(((log(c*x + 1) - log(-c*x + 1))*log(sqrt(c*x + 1) + sqrt(-c*x + 1)) - (log(c*x + 1) - log(-c*x + 1))*lo
g(sqrt(c*x + 1) - sqrt(-c*x + 1)))/c - 2*integrate(-1/2*sqrt(c*x + 1)*(log(c*x + 1) - log(-c*x + 1))/((c^2*x^2
- 1)*sqrt(c*x + 1) + (c^2*x^2 - 1)*sqrt(-c*x + 1)), x) - 2*integrate(1/2*sqrt(c*x + 1)*(log(c*x + 1) - log(-c
*x + 1))/((c^2*x^2 - 1)*sqrt(c*x + 1) - (c^2*x^2 - 1)*sqrt(-c*x + 1)), x)) + 1/2*a*(log(c*x + 1)/c - log(c*x -
1)/c)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b \operatorname{artanh}\left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right ) + a}{c^{2} x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))/(-c^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-(b*arctanh(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)/(c^2*x^2 - 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{a}{c^{2} x^{2} - 1}\, dx - \int \frac{b \operatorname{atanh}{\left (\frac{\sqrt{- c x + 1}}{\sqrt{c x + 1}} \right )}}{c^{2} x^{2} - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh((-c*x+1)**(1/2)/(c*x+1)**(1/2)))/(-c**2*x**2+1),x)

[Out]

-Integral(a/(c**2*x**2 - 1), x) - Integral(b*atanh(sqrt(-c*x + 1)/sqrt(c*x + 1))/(c**2*x**2 - 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{b \operatorname{artanh}\left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right ) + a}{c^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))/(-c^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-(b*arctanh(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)/(c^2*x^2 - 1), x)