3.4.76 \(\int \frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx\) [376]

Optimal. Leaf size=103 \[ \frac {2 \text {ArcTan}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a}-\frac {2 i \tanh ^{-1}(a x) \text {PolyLog}\left (2,-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {2 i \tanh ^{-1}(a x) \text {PolyLog}\left (2,i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {2 i \text {PolyLog}\left (3,-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {2 i \text {PolyLog}\left (3,i e^{\tanh ^{-1}(a x)}\right )}{a} \]

[Out]

2*arctan((a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)^2/a-2*I*arctanh(a*x)*polylog(2,-I*(a*x+1)/(-a^2*x^2+1)^(1/2)
)/a+2*I*arctanh(a*x)*polylog(2,I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a+2*I*polylog(3,-I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a-
2*I*polylog(3,I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a

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Rubi [A]
time = 0.07, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {6099, 4265, 2611, 2320, 6724} \begin {gather*} \frac {2 \tanh ^{-1}(a x)^2 \text {ArcTan}\left (e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {2 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {2 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {2 i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {2 i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^2/Sqrt[1 - a^2*x^2],x]

[Out]

(2*ArcTan[E^ArcTanh[a*x]]*ArcTanh[a*x]^2)/a - ((2*I)*ArcTanh[a*x]*PolyLog[2, (-I)*E^ArcTanh[a*x]])/a + ((2*I)*
ArcTanh[a*x]*PolyLog[2, I*E^ArcTanh[a*x]])/a + ((2*I)*PolyLog[3, (-I)*E^ArcTanh[a*x]])/a - ((2*I)*PolyLog[3, I
*E^ArcTanh[a*x]])/a

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4265

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c +
d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^(I*k*Pi)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*
Log[1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 6099

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c*Sqrt[d]), Subs
t[Int[(a + b*x)^p*Sech[x], x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[
p, 0] && GtQ[d, 0]

Rule 6724

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

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx &=\frac {\text {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a}-\frac {(2 i) \text {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}+\frac {(2 i) \text {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a}-\frac {2 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {2 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {(2 i) \text {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}-\frac {(2 i) \text {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a}-\frac {2 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {2 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {(2 i) \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {(2 i) \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )}{a}\\ &=\frac {2 \tan ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2}{a}-\frac {2 i \tanh ^{-1}(a x) \text {Li}_2\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {2 i \tanh ^{-1}(a x) \text {Li}_2\left (i e^{\tanh ^{-1}(a x)}\right )}{a}+\frac {2 i \text {Li}_3\left (-i e^{\tanh ^{-1}(a x)}\right )}{a}-\frac {2 i \text {Li}_3\left (i e^{\tanh ^{-1}(a x)}\right )}{a}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 119, normalized size = 1.16 \begin {gather*} \frac {i \left (-\tanh ^{-1}(a x)^2 \left (\log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(a x)}\right )\right )-2 \tanh ^{-1}(a x) \left (\text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-\text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )\right )-2 \left (\text {PolyLog}\left (3,-i e^{-\tanh ^{-1}(a x)}\right )-\text {PolyLog}\left (3,i e^{-\tanh ^{-1}(a x)}\right )\right )\right )}{a} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[a*x]^2/Sqrt[1 - a^2*x^2],x]

[Out]

(I*(-(ArcTanh[a*x]^2*(Log[1 - I/E^ArcTanh[a*x]] - Log[1 + I/E^ArcTanh[a*x]])) - 2*ArcTanh[a*x]*(PolyLog[2, (-I
)/E^ArcTanh[a*x]] - PolyLog[2, I/E^ArcTanh[a*x]]) - 2*(PolyLog[3, (-I)/E^ArcTanh[a*x]] - PolyLog[3, I/E^ArcTan
h[a*x]])))/a

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Maple [F]
time = 1.33, size = 0, normalized size = 0.00 \[\int \frac {\arctanh \left (a x \right )^{2}}{\sqrt {-a^{2} x^{2}+1}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atanh(a*x)**2/sqrt(-(a*x - 1)*(a*x + 1)), 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(a*x)^2/(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {atanh}\left (a\,x\right )}^2}{\sqrt {1-a^2\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^2/(1 - a^2*x^2)^(1/2),x)

[Out]

int(atanh(a*x)^2/(1 - a^2*x^2)^(1/2), x)

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