3.2.47 \(\int \frac {(a+b \tanh ^{-1}(\frac {c}{x}))^2}{x} \, dx\) [147]

Optimal. Leaf size=133 \[ -2 \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-\frac {c}{x}}\right )+b \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right ) \text {PolyLog}\left (2,1-\frac {2}{1-\frac {c}{x}}\right )-b \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right ) \text {PolyLog}\left (2,-1+\frac {2}{1-\frac {c}{x}}\right )-\frac {1}{2} b^2 \text {PolyLog}\left (3,1-\frac {2}{1-\frac {c}{x}}\right )+\frac {1}{2} b^2 \text {PolyLog}\left (3,-1+\frac {2}{1-\frac {c}{x}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 133, 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 = {6035, 6033, 6199, 6095, 6205, 6745} \begin {gather*} b \text {Li}_2\left (1-\frac {2}{1-\frac {c}{x}}\right ) \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )-b \text {Li}_2\left (\frac {2}{1-\frac {c}{x}}-1\right ) \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )-2 \tanh ^{-1}\left (1-\frac {2}{1-\frac {c}{x}}\right ) \left (a+b \coth ^{-1}\left (\frac {x}{c}\right )\right )^2-\frac {1}{2} b^2 \text {Li}_3\left (1-\frac {2}{1-\frac {c}{x}}\right )+\frac {1}{2} b^2 \text {Li}_3\left (\frac {2}{1-\frac {c}{x}}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c/x])^2/x,x]

[Out]

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

Rule 6033

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 6035

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 6095

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 6199

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 6205

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 6745

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

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Mathematica [A]
time = 0.07, size = 114, normalized size = 0.86 \begin {gather*} -2 \left (a+b \tanh ^{-1}\left (\frac {c}{x}\right )\right )^2 \tanh ^{-1}\left (\frac {c+x}{c-x}\right )+\frac {1}{2} b \left (2 \left (a+b \tanh ^{-1}\left (\frac {c}{x}\right )\right ) \text {PolyLog}\left (2,\frac {c+x}{c-x}\right )-2 \left (a+b \tanh ^{-1}\left (\frac {c}{x}\right )\right ) \text {PolyLog}\left (2,\frac {c+x}{-c+x}\right )+b \left (-\text {PolyLog}\left (3,\frac {c+x}{c-x}\right )+\text {PolyLog}\left (3,\frac {c+x}{-c+x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c/x])^2/x,x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 3.45, size = 780, normalized size = 5.86 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*ln(c/x)-b^2*ln(c/x)*arctanh(c/x)^2+b^2*arctanh(c/x)*polylog(2,-(1+c/x)^2/(1-c^2/x^2))-1/2*b^2*polylog(3,-
(1+c/x)^2/(1-c^2/x^2))+b^2*arctanh(c/x)^2*ln((1+c/x)^2/(1-c^2/x^2)-1)-b^2*arctanh(c/x)^2*ln(1+(1+c/x)/(1-c^2/x
^2)^(1/2))-2*b^2*arctanh(c/x)*polylog(2,-(1+c/x)/(1-c^2/x^2)^(1/2))+2*b^2*polylog(3,-(1+c/x)/(1-c^2/x^2)^(1/2)
)-b^2*arctanh(c/x)^2*ln(1-(1+c/x)/(1-c^2/x^2)^(1/2))-2*b^2*arctanh(c/x)*polylog(2,(1+c/x)/(1-c^2/x^2)^(1/2))+2
*b^2*polylog(3,(1+c/x)/(1-c^2/x^2)^(1/2))+1/2*I*b^2*Pi*csgn(I*((1+c/x)^2/(1-c^2/x^2)-1))*csgn(I*((1+c/x)^2/(1-
c^2/x^2)-1)/(1+(1+c/x)^2/(1-c^2/x^2)))^2*arctanh(c/x)^2-1/2*I*b^2*Pi*csgn(I*((1+c/x)^2/(1-c^2/x^2)-1)/(1+(1+c/
x)^2/(1-c^2/x^2)))^3*arctanh(c/x)^2-1/2*I*b^2*Pi*csgn(I*((1+c/x)^2/(1-c^2/x^2)-1))*csgn(I/(1+(1+c/x)^2/(1-c^2/
x^2)))*csgn(I*((1+c/x)^2/(1-c^2/x^2)-1)/(1+(1+c/x)^2/(1-c^2/x^2)))*arctanh(c/x)^2+1/2*I*b^2*Pi*csgn(I/(1+(1+c/
x)^2/(1-c^2/x^2)))*csgn(I*((1+c/x)^2/(1-c^2/x^2)-1)/(1+(1+c/x)^2/(1-c^2/x^2)))^2*arctanh(c/x)^2-2*a*b*ln(c/x)*
arctanh(c/x)+a*b*ln(c/x)*ln(1+c/x)+a*b*dilog(1+c/x)+a*b*dilog(c/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((a+b*arctanh(c/x))^2/x,x, algorithm="maxima")

[Out]

a^2*log(x) + integrate(1/4*b^2*(log(c/x + 1) - log(-c/x + 1))^2/x + a*b*(log(c/x + 1) - log(-c/x + 1))/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((a+b*arctanh(c/x))^2/x,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c/x))^2/x,x)

[Out]

int((a + b*atanh(c/x))^2/x, x)

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