3.5.56 \(\int \frac {(1-a^2 x^2)^{3/2} \tanh ^{-1}(a x)}{x^5} \, dx\) [456]

Optimal. Leaf size=191 \[ -\frac {a \sqrt {1-a^2 x^2}}{12 x^3}+\frac {11 a^3 \sqrt {1-a^2 x^2}}{24 x}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{4 x^4}+\frac {5 a^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 x^2}-\frac {3}{4} a^4 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )+\frac {3}{8} a^4 \text {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\frac {3}{8} a^4 \text {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \]

[Out]

-3/4*a^4*arctanh(a*x)*arctanh((-a*x+1)^(1/2)/(a*x+1)^(1/2))+3/8*a^4*polylog(2,-(-a*x+1)^(1/2)/(a*x+1)^(1/2))-3
/8*a^4*polylog(2,(-a*x+1)^(1/2)/(a*x+1)^(1/2))-1/12*a*(-a^2*x^2+1)^(1/2)/x^3+11/24*a^3*(-a^2*x^2+1)^(1/2)/x-1/
4*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/x^4+5/8*a^2*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/x^2

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Rubi [A]
time = 0.39, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6161, 6157, 6173, 277, 270, 6165} \begin {gather*} \frac {3}{8} a^4 \text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {3}{8} a^4 \text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {3}{4} a^4 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )+\frac {5 a^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 x^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{4 x^4}-\frac {a \sqrt {1-a^2 x^2}}{12 x^3}+\frac {11 a^3 \sqrt {1-a^2 x^2}}{24 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/x^5,x]

[Out]

-1/12*(a*Sqrt[1 - a^2*x^2])/x^3 + (11*a^3*Sqrt[1 - a^2*x^2])/(24*x) - (Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(4*x^4)
+ (5*a^2*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(8*x^2) - (3*a^4*ArcTanh[a*x]*ArcTanh[Sqrt[1 - a*x]/Sqrt[1 + a*x]])/
4 + (3*a^4*PolyLog[2, -(Sqrt[1 - a*x]/Sqrt[1 + a*x])])/8 - (3*a^4*PolyLog[2, Sqrt[1 - a*x]/Sqrt[1 + a*x]])/8

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 6157

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(f*x)^(
m + 1)*Sqrt[d + e*x^2]*((a + b*ArcTanh[c*x])/(f*(m + 2))), x] + (Dist[d/(m + 2), Int[(f*x)^m*((a + b*ArcTanh[c
*x])/Sqrt[d + e*x^2]), x], x] - Dist[b*c*(d/(f*(m + 2))), Int[(f*x)^(m + 1)/Sqrt[d + e*x^2], x], x]) /; FreeQ[
{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && NeQ[m, -2]

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 6165

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

Rule 6173

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

Rubi steps

\begin {align*} \int \frac {\left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x)}{x^5} \, dx &=-\left (a^2 \int \frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{x^3} \, dx\right )+\int \frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{x^5} \, dx\\ &=-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{3 x^4}+\frac {a^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{x^2}-\frac {1}{3} \int \frac {\tanh ^{-1}(a x)}{x^5 \sqrt {1-a^2 x^2}} \, dx+\frac {1}{3} a \int \frac {1}{x^4 \sqrt {1-a^2 x^2}} \, dx+a^2 \int \frac {\tanh ^{-1}(a x)}{x^3 \sqrt {1-a^2 x^2}} \, dx-a^3 \int \frac {1}{x^2 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {a \sqrt {1-a^2 x^2}}{9 x^3}+\frac {a^3 \sqrt {1-a^2 x^2}}{x}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{4 x^4}+\frac {a^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{2 x^2}-\frac {1}{12} a \int \frac {1}{x^4 \sqrt {1-a^2 x^2}} \, dx-\frac {1}{4} a^2 \int \frac {\tanh ^{-1}(a x)}{x^3 \sqrt {1-a^2 x^2}} \, dx+\frac {1}{9} \left (2 a^3\right ) \int \frac {1}{x^2 \sqrt {1-a^2 x^2}} \, dx+\frac {1}{2} a^3 \int \frac {1}{x^2 \sqrt {1-a^2 x^2}} \, dx+\frac {1}{2} a^4 \int \frac {\tanh ^{-1}(a x)}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {a \sqrt {1-a^2 x^2}}{12 x^3}+\frac {5 a^3 \sqrt {1-a^2 x^2}}{18 x}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{4 x^4}+\frac {5 a^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 x^2}-a^4 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )+\frac {1}{2} a^4 \text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\frac {1}{2} a^4 \text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\frac {1}{18} a^3 \int \frac {1}{x^2 \sqrt {1-a^2 x^2}} \, dx-\frac {1}{8} a^3 \int \frac {1}{x^2 \sqrt {1-a^2 x^2}} \, dx-\frac {1}{8} a^4 \int \frac {\tanh ^{-1}(a x)}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {a \sqrt {1-a^2 x^2}}{12 x^3}+\frac {11 a^3 \sqrt {1-a^2 x^2}}{24 x}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{4 x^4}+\frac {5 a^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 x^2}-\frac {3}{4} a^4 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )+\frac {3}{8} a^4 \text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\frac {3}{8} a^4 \text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )\\ \end {align*}

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Mathematica [A]
time = 2.90, size = 282, normalized size = 1.48 \begin {gather*} \frac {1}{192} a \left (40 a^3 \coth \left (\frac {1}{2} \tanh ^{-1}(a x)\right )+18 a^3 \tanh ^{-1}(a x) \text {csch}^2\left (\frac {1}{2} \tanh ^{-1}(a x)\right )-\frac {a^4 x \text {csch}^4\left (\frac {1}{2} \tanh ^{-1}(a x)\right )}{\sqrt {1-a^2 x^2}}-3 a^3 \tanh ^{-1}(a x) \text {csch}^4\left (\frac {1}{2} \tanh ^{-1}(a x)\right )+72 a^3 \tanh ^{-1}(a x) \log \left (1-e^{-\tanh ^{-1}(a x)}\right )-72 a^3 \tanh ^{-1}(a x) \log \left (1+e^{-\tanh ^{-1}(a x)}\right )+72 a^3 \text {PolyLog}\left (2,-e^{-\tanh ^{-1}(a x)}\right )-72 a^3 \text {PolyLog}\left (2,e^{-\tanh ^{-1}(a x)}\right )+18 a^3 \tanh ^{-1}(a x) \text {sech}^2\left (\frac {1}{2} \tanh ^{-1}(a x)\right )+3 a^3 \tanh ^{-1}(a x) \text {sech}^4\left (\frac {1}{2} \tanh ^{-1}(a x)\right )-\frac {16 \sqrt {1-a^2 x^2} \sinh ^4\left (\frac {1}{2} \tanh ^{-1}(a x)\right )}{x^3}+\frac {16 a^2 \sqrt {1-a^2 x^2} \sinh ^4\left (\frac {1}{2} \tanh ^{-1}(a x)\right )}{x}-40 a^3 \tanh \left (\frac {1}{2} \tanh ^{-1}(a x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 - a^2*x^2)^(3/2)*ArcTanh[a*x])/x^5,x]

[Out]

(a*(40*a^3*Coth[ArcTanh[a*x]/2] + 18*a^3*ArcTanh[a*x]*Csch[ArcTanh[a*x]/2]^2 - (a^4*x*Csch[ArcTanh[a*x]/2]^4)/
Sqrt[1 - a^2*x^2] - 3*a^3*ArcTanh[a*x]*Csch[ArcTanh[a*x]/2]^4 + 72*a^3*ArcTanh[a*x]*Log[1 - E^(-ArcTanh[a*x])]
- 72*a^3*ArcTanh[a*x]*Log[1 + E^(-ArcTanh[a*x])] + 72*a^3*PolyLog[2, -E^(-ArcTanh[a*x])] - 72*a^3*PolyLog[2,
E^(-ArcTanh[a*x])] + 18*a^3*ArcTanh[a*x]*Sech[ArcTanh[a*x]/2]^2 + 3*a^3*ArcTanh[a*x]*Sech[ArcTanh[a*x]/2]^4 -
(16*Sqrt[1 - a^2*x^2]*Sinh[ArcTanh[a*x]/2]^4)/x^3 + (16*a^2*Sqrt[1 - a^2*x^2]*Sinh[ArcTanh[a*x]/2]^4)/x - 40*a
^3*Tanh[ArcTanh[a*x]/2]))/192

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Maple [A]
time = 1.41, size = 164, normalized size = 0.86

method	result	size
default	\(\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (11 a^{3} x^{3}+15 a^{2} x^{2} \arctanh \left (a x \right )-2 a x -6 \arctanh \left (a x \right )\right )}{24 x^{4}}+\frac {3 a^{4} \arctanh \left (a x \right ) \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{8}+\frac {3 a^{4} \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{8}-\frac {3 a^{4} \arctanh \left (a x \right ) \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{8}-\frac {3 a^{4} \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{8}\)	\(164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*x^2+1)^(3/2)*arctanh(a*x)/x^5,x,method=_RETURNVERBOSE)

[Out]

1/24*(-(a*x-1)*(a*x+1))^(1/2)*(11*a^3*x^3+15*a^2*x^2*arctanh(a*x)-2*a*x-6*arctanh(a*x))/x^4+3/8*a^4*arctanh(a*
x)*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))+3/8*a^4*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))-3/8*a^4*arctanh(a*x)*ln(1+(a
*x+1)/(-a^2*x^2+1)^(1/2))-3/8*a^4*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))

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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((-a^2*x^2+1)^(3/2)*arctanh(a*x)/x^5,x, algorithm="maxima")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*arctanh(a*x)/x^5, 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((-a^2*x^2+1)^(3/2)*arctanh(a*x)/x^5,x, algorithm="fricas")

[Out]

integral(-(a^2*x^2 - 1)*sqrt(-a^2*x^2 + 1)*arctanh(a*x)/x^5, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \operatorname {atanh}{\left (a x \right )}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*x**2+1)**(3/2)*atanh(a*x)/x**5,x)

[Out]

Integral((-(a*x - 1)*(a*x + 1))**(3/2)*atanh(a*x)/x**5, 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((-a^2*x^2+1)^(3/2)*arctanh(a*x)/x^5,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 )\,{\left (1-a^2\,x^2\right )}^{3/2}}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atanh(a*x)*(1 - a^2*x^2)^(3/2))/x^5,x)

[Out]

int((atanh(a*x)*(1 - a^2*x^2)^(3/2))/x^5, x)

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