3.456 \(\int \frac {(1-a^2 x^2)^{3/2} \tanh ^{-1}(a x)}{x^5} \, dx\)

Optimal. Leaf size=191 \[ \frac {3}{8} a^4 \text {Li}_2\left (-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {3}{8} a^4 \text {Li}_2\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {3}{4} a^4 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )+\frac {5 a^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 x^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{4 x^4}-\frac {a \sqrt {1-a^2 x^2}}{12 x^3}+\frac {11 a^3 \sqrt {1-a^2 x^2}}{24 x} \]

[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 = 0.57, 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 = {6014, 6010, 6026, 271, 264, 6018} \[ \frac {3}{8} a^4 \text {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\frac {3}{8} a^4 \text {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )+\frac {11 a^3 \sqrt {1-a^2 x^2}}{24 x}-\frac {a \sqrt {1-a^2 x^2}}{12 x^3}+\frac {5 a^2 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 x^2}-\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{4 x^4}-\frac {3}{4} a^4 \tanh ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Sqrt[1 - a^2*x^2])/(12*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 264

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 271

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 6010

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 6014

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 6018

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

Rule 6026

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 = 3.83, size = 282, normalized size = 1.48 \[ \frac {1}{192} a \left (72 a^3 \text {Li}_2\left (-e^{-\tanh ^{-1}(a x)}\right )-72 a^3 \text {Li}_2\left (e^{-\tanh ^{-1}(a x)}\right )-40 a^3 \tanh \left (\frac {1}{2} \tanh ^{-1}(a x)\right )+72 a^3 \tanh ^{-1}(a x) \log \left (1-e^{-\tanh ^{-1}(a x)}\right )-72 a^3 \tanh ^{-1}(a x) \log \left (e^{-\tanh ^{-1}(a x)}+1\right )+40 a^3 \coth \left (\frac {1}{2} \tanh ^{-1}(a x)\right )-3 a^3 \tanh ^{-1}(a x) \text {csch}^4\left (\frac {1}{2} \tanh ^{-1}(a x)\right )+18 a^3 \tanh ^{-1}(a x) \text {csch}^2\left (\frac {1}{2} \tanh ^{-1}(a x)\right )+3 a^3 \tanh ^{-1}(a x) \text {sech}^4\left (\frac {1}{2} \tanh ^{-1}(a x)\right )+18 a^3 \tanh ^{-1}(a x) \text {sech}^2\left (\frac {1}{2} \tanh ^{-1}(a x)\right )+\frac {16 a^2 \sqrt {1-a^2 x^2} \sinh ^4\left (\frac {1}{2} \tanh ^{-1}(a x)\right )}{x}-\frac {16 \sqrt {1-a^2 x^2} \sinh ^4\left (\frac {1}{2} \tanh ^{-1}(a x)\right )}{x^3}-\frac {a^4 x \text {csch}^4\left (\frac {1}{2} \tanh ^{-1}(a x)\right )}{\sqrt {1-a^2 x^2}}\right ) \]

Warning: Unable to verify antiderivative.

[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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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a^{2} x^{2} - 1\right )} \sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )}{x^{5}}, x\right ) \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.44, size = 164, normalized size = 0.86 \[ \frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (11 x^{3} a^{3}+15 a^{2} x^{2} \arctanh \left (a x \right )-2 a x -6 \arctanh \left (a x \right )\right )}{24 x^{4}}-\frac {3 a^{4} \arctanh \left (a x \right ) \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{8}-\frac {3 a^{4} \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{8}+\frac {3 a^{4} \arctanh \left (a x \right ) \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{8}+\frac {3 a^{4} \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{8} \]

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)

[Out]

1/24*(-(a*x-1)*(a*x+1))^(1/2)*(11*x^3*a^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 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \operatorname {artanh}\left (a x\right )}{x^{5}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{3/2}}{x^5} \,d x \]

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

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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