3.5.43 \(\int \frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{x} \, dx\) [443]

Optimal. Leaf size=174 \[ 4 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)+\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2-2 \tanh ^{-1}(a x) \text {PolyLog}\left (2,-e^{\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \text {PolyLog}\left (2,e^{\tanh ^{-1}(a x)}\right )+2 i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )-2 i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )+2 \text {PolyLog}\left (3,-e^{\tanh ^{-1}(a x)}\right )-2 \text {PolyLog}\left (3,e^{\tanh ^{-1}(a x)}\right ) \]

[Out]

4*arctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)-2*arctanh((a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)^2-2*arc
tanh(a*x)*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))+2*arctanh(a*x)*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))+2*I*poly
log(2,-I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))-2*I*polylog(2,I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))+2*polylog(3,-(a*x+1)/(-a^
2*x^2+1)^(1/2))-2*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))+(-a^2*x^2+1)^(1/2)*arctanh(a*x)^2

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Rubi [A]
time = 0.24, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6161, 6167, 4267, 2611, 2320, 6724, 6141, 6097} \begin {gather*} \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+4 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)+2 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )-2 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )-2 \tanh ^{-1}(a x) \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+2 \text {Li}_3\left (-e^{\tanh ^{-1}(a x)}\right )-2 \text {Li}_3\left (e^{\tanh ^{-1}(a x)}\right )-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

4*ArcTan[Sqrt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x] + Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2 - 2*ArcTanh[E^ArcTanh[a*
x]]*ArcTanh[a*x]^2 - 2*ArcTanh[a*x]*PolyLog[2, -E^ArcTanh[a*x]] + 2*ArcTanh[a*x]*PolyLog[2, E^ArcTanh[a*x]] +
(2*I)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]] - (2*I)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]] + 2*P
olyLog[3, -E^ArcTanh[a*x]] - 2*PolyLog[3, E^ArcTanh[a*x]]

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 4267

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

Rule 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[-2*(a + b*ArcTanh[c*x])*(
ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]]/(c*Sqrt[d])), x] + (-Simp[I*b*(PolyLog[2, (-I)*(Sqrt[1 - c*x]/Sqrt[1 + c*x
])]/(c*Sqrt[d])), x] + Simp[I*b*(PolyLog[2, I*(Sqrt[1 - c*x]/Sqrt[1 + c*x])]/(c*Sqrt[d])), x]) /; FreeQ[{a, b,
c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0]

Rule 6141

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(d + e*x^2)^
(q + 1)*((a + b*ArcTanh[c*x])^p/(2*e*(q + 1))), x] + Dist[b*(p/(2*c*(q + 1))), Int[(d + e*x^2)^q*(a + b*ArcTan
h[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0] && NeQ[q, -1]

Rule 6161

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Dist
[d, Int[(f*x)^m*(d + e*x^2)^(q - 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[c^2*(d/f^2), Int[(f*x)^(m + 2)*(d +
e*x^2)^(q - 1)*(a + b*ArcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[q
, 0] && IGtQ[p, 0] && (RationalQ[m] || (EqQ[p, 1] && IntegerQ[q]))

Rule 6167

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Dist[1/Sqrt[d], Su
bst[Int[(a + b*x)^p*Csch[x], x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGt
Q[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 {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2}{x} \, dx &=-\left (a^2 \int \frac {x \tanh ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}} \, dx\right )+\int \frac {\tanh ^{-1}(a x)^2}{x \sqrt {1-a^2 x^2}} \, dx\\ &=\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-(2 a) \int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx+\text {Subst}\left (\int x^2 \text {csch}(x) \, dx,x,\tanh ^{-1}(a x)\right )\\ &=4 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)+\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2+2 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )-2 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )-2 \text {Subst}\left (\int x \log \left (1-e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )+2 \text {Subst}\left (\int x \log \left (1+e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )\\ &=4 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)+\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2-2 \tanh ^{-1}(a x) \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+2 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )-2 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )+2 \text {Subst}\left (\int \text {Li}_2\left (-e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )-2 \text {Subst}\left (\int \text {Li}_2\left (e^x\right ) \, dx,x,\tanh ^{-1}(a x)\right )\\ &=4 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)+\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2-2 \tanh ^{-1}(a x) \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+2 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )-2 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )+2 \text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )-2 \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{\tanh ^{-1}(a x)}\right )\\ &=4 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)+\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2-2 \tanh ^{-1}\left (e^{\tanh ^{-1}(a x)}\right ) \tanh ^{-1}(a x)^2-2 \tanh ^{-1}(a x) \text {Li}_2\left (-e^{\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \text {Li}_2\left (e^{\tanh ^{-1}(a x)}\right )+2 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )-2 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )+2 \text {Li}_3\left (-e^{\tanh ^{-1}(a x)}\right )-2 \text {Li}_3\left (e^{\tanh ^{-1}(a x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 203, normalized size = 1.17 \begin {gather*} \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)^2+\tanh ^{-1}(a x)^2 \log \left (1-e^{-\tanh ^{-1}(a x)}\right )+2 i \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-2 i \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x)^2 \log \left (1+e^{-\tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x) \text {PolyLog}\left (2,-e^{-\tanh ^{-1}(a x)}\right )+2 i \text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )-2 i \text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )-2 \tanh ^{-1}(a x) \text {PolyLog}\left (2,e^{-\tanh ^{-1}(a x)}\right )+2 \text {PolyLog}\left (3,-e^{-\tanh ^{-1}(a x)}\right )-2 \text {PolyLog}\left (3,e^{-\tanh ^{-1}(a x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2 + ArcTanh[a*x]^2*Log[1 - E^(-ArcTanh[a*x])] + (2*I)*ArcTanh[a*x]*Log[1 - I/E^
ArcTanh[a*x]] - (2*I)*ArcTanh[a*x]*Log[1 + I/E^ArcTanh[a*x]] - ArcTanh[a*x]^2*Log[1 + E^(-ArcTanh[a*x])] + 2*A
rcTanh[a*x]*PolyLog[2, -E^(-ArcTanh[a*x])] + (2*I)*PolyLog[2, (-I)/E^ArcTanh[a*x]] - (2*I)*PolyLog[2, I/E^ArcT
anh[a*x]] - 2*ArcTanh[a*x]*PolyLog[2, E^(-ArcTanh[a*x])] + 2*PolyLog[3, -E^(-ArcTanh[a*x])] - 2*PolyLog[3, E^(
-ArcTanh[a*x])]

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Maple [F]
time = 0.99, size = 0, normalized size = 0.00 \[\int \frac {\arctanh \left (a x \right )^{2} \sqrt {-a^{2} x^{2}+1}}{x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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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}}{x} \,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,x)

[Out]

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

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