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

Optimal. Leaf size=144 \[ \frac {1}{2} b \text {Li}_2\left (1-\frac {2}{1-\frac {c}{x^2}}\right ) \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )-\frac {1}{2} b \text {Li}_2\left (\frac {2}{1-\frac {c}{x^2}}-1\right ) \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )-\tanh ^{-1}\left (1-\frac {2}{1-\frac {c}{x^2}}\right ) \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2-\frac {1}{4} b^2 \text {Li}_3\left (1-\frac {2}{1-\frac {c}{x^2}}\right )+\frac {1}{4} b^2 \text {Li}_3\left (\frac {2}{1-\frac {c}{x^2}}-1\right ) \]

[Out]

(a+b*arccoth(x^2/c))^2*arctanh(-1+2/(1-c/x^2))+1/2*b*(a+b*arccoth(x^2/c))*polylog(2,1-2/(1-c/x^2))-1/2*b*(a+b*
arccoth(x^2/c))*polylog(2,-1+2/(1-c/x^2))-1/4*b^2*polylog(3,1-2/(1-c/x^2))+1/4*b^2*polylog(3,-1+2/(1-c/x^2))

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Rubi [A]  time = 0.32, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6095, 5914, 6052, 5948, 6058, 6610} \[ \frac {1}{2} b \text {PolyLog}\left (2,1-\frac {2}{1-\frac {c}{x^2}}\right ) \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )-\frac {1}{2} b \text {PolyLog}\left (2,\frac {2}{1-\frac {c}{x^2}}-1\right ) \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )-\frac {1}{4} b^2 \text {PolyLog}\left (3,1-\frac {2}{1-\frac {c}{x^2}}\right )+\frac {1}{4} b^2 \text {PolyLog}\left (3,\frac {2}{1-\frac {c}{x^2}}-1\right )-\tanh ^{-1}\left (1-\frac {2}{1-\frac {c}{x^2}}\right ) \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c/x^2])^2/x,x]

[Out]

-((a + b*ArcCoth[x^2/c])^2*ArcTanh[1 - 2/(1 - c/x^2)]) + (b*(a + b*ArcCoth[x^2/c])*PolyLog[2, 1 - 2/(1 - c/x^2
)])/2 - (b*(a + b*ArcCoth[x^2/c])*PolyLog[2, -1 + 2/(1 - c/x^2)])/2 - (b^2*PolyLog[3, 1 - 2/(1 - c/x^2)])/4 +
(b^2*PolyLog[3, -1 + 2/(1 - c/x^2)])/4

Rule 5914

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTanh[c*x])^p*ArcTanh[1 - 2/(1
- c*x)], x] - Dist[2*b*c*p, Int[((a + b*ArcTanh[c*x])^(p - 1)*ArcTanh[1 - 2/(1 - c*x)])/(1 - c^2*x^2), x], x]
/; FreeQ[{a, b, c}, x] && IGtQ[p, 1]

Rule 5948

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

Rule 6052

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

Rule 6058

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcT
anh[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p)/2, Int[((a + b*ArcTanh[c*x])^(p - 1)*PolyLog[2, 1 - u]
)/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 -
2/(1 - c*x))^2, 0]

Rule 6095

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTanh[c*x])
^p/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )^2}{x} \, dx &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx,x,\frac {1}{x^2}\right )\right )\\ &=-\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-\frac {c}{x^2}}\right )+(2 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\frac {1}{x^2}\right )\\ &=-\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-\frac {c}{x^2}}\right )-(b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\frac {1}{x^2}\right )+(b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\frac {1}{x^2}\right )\\ &=-\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-\frac {c}{x^2}}\right )+\frac {1}{2} b \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right ) \text {Li}_2\left (1-\frac {2}{1-\frac {c}{x^2}}\right )-\frac {1}{2} b \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right ) \text {Li}_2\left (-1+\frac {2}{1-\frac {c}{x^2}}\right )-\frac {1}{2} \left (b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\frac {1}{x^2}\right )+\frac {1}{2} \left (b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\frac {1}{x^2}\right )\\ &=-\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-\frac {c}{x^2}}\right )+\frac {1}{2} b \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right ) \text {Li}_2\left (1-\frac {2}{1-\frac {c}{x^2}}\right )-\frac {1}{2} b \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right ) \text {Li}_2\left (-1+\frac {2}{1-\frac {c}{x^2}}\right )-\frac {1}{4} b^2 \text {Li}_3\left (1-\frac {2}{1-\frac {c}{x^2}}\right )+\frac {1}{4} b^2 \text {Li}_3\left (-1+\frac {2}{1-\frac {c}{x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 183, normalized size = 1.27 \[ \frac {1}{2} \left (4 b c \left (\frac {1}{2} \left (\frac {\text {Li}_2\left (\frac {-\frac {c}{x^2}-1}{\frac {c}{x^2}-1}\right ) \left (-a-b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )}{2 c}+\frac {b \text {Li}_3\left (\frac {-\frac {c}{x^2}-1}{\frac {c}{x^2}-1}\right )}{4 c}\right )+\frac {1}{2} \left (-\frac {\text {Li}_2\left (\frac {\frac {c}{x^2}+1}{\frac {c}{x^2}-1}\right ) \left (-a-b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )}{2 c}-\frac {b \text {Li}_3\left (\frac {\frac {c}{x^2}+1}{\frac {c}{x^2}-1}\right )}{4 c}\right )\right )-2 \tanh ^{-1}\left (1-\frac {2}{1-\frac {c}{x^2}}\right ) \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )^2\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*ArcTanh[1 - 2/(1 - c/x^2)]*(a + b*ArcTanh[c/x^2])^2 + 4*b*c*((((-a - b*ArcTanh[c/x^2])*PolyLog[2, (-1 - c/
x^2)/(-1 + c/x^2)])/(2*c) + (b*PolyLog[3, (-1 - c/x^2)/(-1 + c/x^2)])/(4*c))/2 + (-1/2*((-a - b*ArcTanh[c/x^2]
)*PolyLog[2, (1 + c/x^2)/(-1 + c/x^2)])/c - (b*PolyLog[3, (1 + c/x^2)/(-1 + c/x^2)])/(4*c))/2))/2

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {artanh}\left (\frac {c}{x^{2}}\right )^{2} + 2 \, a b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a^{2}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + a\right )}^{2}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.71, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arctanh \left (\frac {c}{x^{2}}\right )\right )^{2}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \log \relax (x) + \int \frac {b^{2} {\left (\log \left (\frac {c}{x^{2}} + 1\right ) - \log \left (-\frac {c}{x^{2}} + 1\right )\right )}^{2}}{4 \, x} + \frac {a b {\left (\log \left (\frac {c}{x^{2}} + 1\right ) - \log \left (-\frac {c}{x^{2}} + 1\right )\right )}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c/x^2))^2/x,x)

[Out]

int((a + b*atanh(c/x^2))^2/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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