3.174 \(\int \frac{(a+b \tanh ^{-1}(\frac{c}{x^2}))^2}{x^3} \, dx\)

Optimal. Leaf size=99 \[ \frac{b^2 \text{PolyLog}\left (2,1-\frac{2}{1-\frac{c}{x^2}}\right )}{2 c}-\frac{\left (a+b \coth ^{-1}\left (\frac{x^2}{c}\right )\right )^2}{2 c}-\frac{\left (a+b \coth ^{-1}\left (\frac{x^2}{c}\right )\right )^2}{2 x^2}+\frac{b \log \left (\frac{2}{1-\frac{c}{x^2}}\right ) \left (a+b \coth ^{-1}\left (\frac{x^2}{c}\right )\right )}{c} \]

[Out]

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

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Rubi [B]  time = 0.528957, antiderivative size = 207, normalized size of antiderivative = 2.09, number of steps used = 28, number of rules used = 12, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {6099, 2454, 2389, 2296, 2295, 6715, 2430, 43, 2416, 2394, 2393, 2391} \[ \frac{b^2 \text{PolyLog}\left (2,\frac{1}{2} \left (1-\frac{c}{x^2}\right )\right )}{4 c}-\frac{b^2 \text{PolyLog}\left (2,\frac{1}{2} \left (\frac{c}{x^2}+1\right )\right )}{4 c}-\frac{b \log \left (\frac{1}{2} \left (\frac{c}{x^2}+1\right )\right ) \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right )}{4 c}-\frac{b \log \left (\frac{c}{x^2}+1\right ) \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right )}{4 x^2}+\frac{\left (1-\frac{c}{x^2}\right ) \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right )^2}{8 c}-\frac{b^2 \left (\frac{c}{x^2}+1\right ) \log ^2\left (\frac{c}{x^2}+1\right )}{8 c}-\frac{b^2 \log \left (\frac{1}{2} \left (1-\frac{c}{x^2}\right )\right ) \log \left (\frac{c}{x^2}+1\right )}{4 c} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*ArcTanh[c/x^2])^2/x^3,x]

[Out]

((1 - c/x^2)*(2*a - b*Log[1 - c/x^2])^2)/(8*c) - (b*(2*a - b*Log[1 - c/x^2])*Log[(1 + c/x^2)/2])/(4*c) - (b^2*
Log[(1 - c/x^2)/2]*Log[1 + c/x^2])/(4*c) - (b*(2*a - b*Log[1 - c/x^2])*Log[1 + c/x^2])/(4*x^2) - (b^2*(1 + c/x
^2)*Log[1 + c/x^2]^2)/(8*c) + (b^2*PolyLog[2, (1 - c/x^2)/2])/(4*c) - (b^2*PolyLog[2, (1 + c/x^2)/2])/(4*c)

Rule 6099

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d*x)^
m*(a + (b*Log[1 + c*x^n])/2 - (b*Log[1 - c*x^n])/2)^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IGtQ[p, 0] &&
IntegerQ[m] && IntegerQ[n]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 2430

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.)), x_Symbol] :> Simp[x*(a + b*Log[c*(d + e*x)^n])^p*(f + g*Log[h*(i + j*x)^m]), x] + (-Dist[g*j*m, Int[(x
*(a + b*Log[c*(d + e*x)^n])^p)/(i + j*x), x], x] - Dist[b*e*n*p, Int[(x*(a + b*Log[c*(d + e*x)^n])^(p - 1)*(f
+ g*Log[h*(i + j*x)^m]))/(d + e*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, i, j, m, n}, x] && IGtQ[p, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )\right )^2}{x^3} \, dx &=\int \left (\frac{\left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right )^2}{4 x^3}-\frac{b \left (-2 a+b \log \left (1-\frac{c}{x^2}\right )\right ) \log \left (1+\frac{c}{x^2}\right )}{2 x^3}+\frac{b^2 \log ^2\left (1+\frac{c}{x^2}\right )}{4 x^3}\right ) \, dx\\ &=\frac{1}{4} \int \frac{\left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right )^2}{x^3} \, dx-\frac{1}{2} b \int \frac{\left (-2 a+b \log \left (1-\frac{c}{x^2}\right )\right ) \log \left (1+\frac{c}{x^2}\right )}{x^3} \, dx+\frac{1}{4} b^2 \int \frac{\log ^2\left (1+\frac{c}{x^2}\right )}{x^3} \, dx\\ &=-\left (\frac{1}{8} \operatorname{Subst}\left (\int (2 a-b \log (1-c x))^2 \, dx,x,\frac{1}{x^2}\right )\right )+\frac{1}{4} b \operatorname{Subst}\left (\int (-2 a+b \log (1-c x)) \log (1+c x) \, dx,x,\frac{1}{x^2}\right )-\frac{1}{8} b^2 \operatorname{Subst}\left (\int \log ^2(1+c x) \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{b \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right ) \log \left (1+\frac{c}{x^2}\right )}{4 x^2}+\frac{\operatorname{Subst}\left (\int (2 a-b \log (x))^2 \, dx,x,1-\frac{c}{x^2}\right )}{8 c}-\frac{b^2 \operatorname{Subst}\left (\int \log ^2(x) \, dx,x,1+\frac{c}{x^2}\right )}{8 c}-\frac{1}{4} (b c) \operatorname{Subst}\left (\int \frac{x (-2 a+b \log (1-c x))}{1+c x} \, dx,x,\frac{1}{x^2}\right )+\frac{1}{4} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{x \log (1+c x)}{1-c x} \, dx,x,\frac{1}{x^2}\right )\\ &=\frac{\left (1-\frac{c}{x^2}\right ) \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right )^2}{8 c}-\frac{b \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right ) \log \left (1+\frac{c}{x^2}\right )}{4 x^2}-\frac{b^2 \left (1+\frac{c}{x^2}\right ) \log ^2\left (1+\frac{c}{x^2}\right )}{8 c}+\frac{b \operatorname{Subst}\left (\int (2 a-b \log (x)) \, dx,x,1-\frac{c}{x^2}\right )}{4 c}+\frac{b^2 \operatorname{Subst}\left (\int \log (x) \, dx,x,1+\frac{c}{x^2}\right )}{4 c}-\frac{1}{4} (b c) \operatorname{Subst}\left (\int \left (\frac{-2 a+b \log (1-c x)}{c}-\frac{-2 a+b \log (1-c x)}{c (1+c x)}\right ) \, dx,x,\frac{1}{x^2}\right )+\frac{1}{4} \left (b^2 c\right ) \operatorname{Subst}\left (\int \left (-\frac{\log (1+c x)}{c}-\frac{\log (1+c x)}{c (-1+c x)}\right ) \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{a b}{2 x^2}-\frac{b^2}{4 x^2}+\frac{\left (1-\frac{c}{x^2}\right ) \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right )^2}{8 c}+\frac{b^2 \left (1+\frac{c}{x^2}\right ) \log \left (1+\frac{c}{x^2}\right )}{4 c}-\frac{b \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right ) \log \left (1+\frac{c}{x^2}\right )}{4 x^2}-\frac{b^2 \left (1+\frac{c}{x^2}\right ) \log ^2\left (1+\frac{c}{x^2}\right )}{8 c}-\frac{1}{4} b \operatorname{Subst}\left (\int (-2 a+b \log (1-c x)) \, dx,x,\frac{1}{x^2}\right )+\frac{1}{4} b \operatorname{Subst}\left (\int \frac{-2 a+b \log (1-c x)}{1+c x} \, dx,x,\frac{1}{x^2}\right )-\frac{1}{4} b^2 \operatorname{Subst}\left (\int \log (1+c x) \, dx,x,\frac{1}{x^2}\right )-\frac{1}{4} b^2 \operatorname{Subst}\left (\int \frac{\log (1+c x)}{-1+c x} \, dx,x,\frac{1}{x^2}\right )-\frac{b^2 \operatorname{Subst}\left (\int \log (x) \, dx,x,1-\frac{c}{x^2}\right )}{4 c}\\ &=-\frac{b^2}{2 x^2}-\frac{b^2 \left (1-\frac{c}{x^2}\right ) \log \left (1-\frac{c}{x^2}\right )}{4 c}+\frac{\left (1-\frac{c}{x^2}\right ) \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right )^2}{8 c}-\frac{b \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right ) \log \left (\frac{1}{2} \left (1+\frac{c}{x^2}\right )\right )}{4 c}+\frac{b^2 \left (1+\frac{c}{x^2}\right ) \log \left (1+\frac{c}{x^2}\right )}{4 c}-\frac{b^2 \log \left (\frac{1}{2} \left (1-\frac{c}{x^2}\right )\right ) \log \left (1+\frac{c}{x^2}\right )}{4 c}-\frac{b \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right ) \log \left (1+\frac{c}{x^2}\right )}{4 x^2}-\frac{b^2 \left (1+\frac{c}{x^2}\right ) \log ^2\left (1+\frac{c}{x^2}\right )}{8 c}+\frac{1}{4} b^2 \operatorname{Subst}\left (\int \frac{\log \left (\frac{1}{2} (1-c x)\right )}{1+c x} \, dx,x,\frac{1}{x^2}\right )-\frac{1}{4} b^2 \operatorname{Subst}\left (\int \log (1-c x) \, dx,x,\frac{1}{x^2}\right )+\frac{1}{4} b^2 \operatorname{Subst}\left (\int \frac{\log \left (\frac{1}{2} (1+c x)\right )}{1-c x} \, dx,x,\frac{1}{x^2}\right )-\frac{b^2 \operatorname{Subst}\left (\int \log (x) \, dx,x,1+\frac{c}{x^2}\right )}{4 c}\\ &=-\frac{b^2}{4 x^2}-\frac{b^2 \left (1-\frac{c}{x^2}\right ) \log \left (1-\frac{c}{x^2}\right )}{4 c}+\frac{\left (1-\frac{c}{x^2}\right ) \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right )^2}{8 c}-\frac{b \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right ) \log \left (\frac{1}{2} \left (1+\frac{c}{x^2}\right )\right )}{4 c}-\frac{b^2 \log \left (\frac{1}{2} \left (1-\frac{c}{x^2}\right )\right ) \log \left (1+\frac{c}{x^2}\right )}{4 c}-\frac{b \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right ) \log \left (1+\frac{c}{x^2}\right )}{4 x^2}-\frac{b^2 \left (1+\frac{c}{x^2}\right ) \log ^2\left (1+\frac{c}{x^2}\right )}{8 c}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1-\frac{c}{x^2}\right )}{4 c}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1+\frac{c}{x^2}\right )}{4 c}+\frac{b^2 \operatorname{Subst}\left (\int \log (x) \, dx,x,1-\frac{c}{x^2}\right )}{4 c}\\ &=\frac{\left (1-\frac{c}{x^2}\right ) \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right )^2}{8 c}-\frac{b \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right ) \log \left (\frac{1}{2} \left (1+\frac{c}{x^2}\right )\right )}{4 c}-\frac{b^2 \log \left (\frac{1}{2} \left (1-\frac{c}{x^2}\right )\right ) \log \left (1+\frac{c}{x^2}\right )}{4 c}-\frac{b \left (2 a-b \log \left (1-\frac{c}{x^2}\right )\right ) \log \left (1+\frac{c}{x^2}\right )}{4 x^2}-\frac{b^2 \left (1+\frac{c}{x^2}\right ) \log ^2\left (1+\frac{c}{x^2}\right )}{8 c}+\frac{b^2 \text{Li}_2\left (\frac{1}{2} \left (1-\frac{c}{x^2}\right )\right )}{4 c}-\frac{b^2 \text{Li}_2\left (\frac{1}{2} \left (1+\frac{c}{x^2}\right )\right )}{4 c}\\ \end{align*}

Mathematica [A]  time = 0.0899924, size = 114, normalized size = 1.15 \[ -\frac{b^2 \left (\text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}\left (\frac{c}{x^2}\right )}\right )+\tanh ^{-1}\left (\frac{c}{x^2}\right ) \left (\frac{c \tanh ^{-1}\left (\frac{c}{x^2}\right )}{x^2}-\tanh ^{-1}\left (\frac{c}{x^2}\right )-2 \log \left (e^{-2 \tanh ^{-1}\left (\frac{c}{x^2}\right )}+1\right )\right )\right )}{2 c}-\frac{a^2}{2 x^2}-\frac{a b \left (\frac{c \tanh ^{-1}\left (\frac{c}{x^2}\right )}{x^2}-\log \left (\frac{1}{\sqrt{1-\frac{c^2}{x^4}}}\right )\right )}{c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 144, normalized size = 1.5 \begin{align*} -{\frac{{a}^{2}}{2\,{x}^{2}}}-{\frac{{b}^{2}}{2\,{x}^{2}} \left ({\it Artanh} \left ({\frac{c}{{x}^{2}}} \right ) \right ) ^{2}}-{\frac{{b}^{2}}{2\,c} \left ({\it Artanh} \left ({\frac{c}{{x}^{2}}} \right ) \right ) ^{2}}+{\frac{{b}^{2}}{c}{\it Artanh} \left ({\frac{c}{{x}^{2}}} \right ) \ln \left ({ \left ( 1+{\frac{c}{{x}^{2}}} \right ) ^{2} \left ( 1-{\frac{{c}^{2}}{{x}^{4}}} \right ) ^{-1}}+1 \right ) }+{\frac{{b}^{2}}{2\,c}{\it polylog} \left ( 2,-{ \left ( 1+{\frac{c}{{x}^{2}}} \right ) ^{2} \left ( 1-{\frac{{c}^{2}}{{x}^{4}}} \right ) ^{-1}} \right ) }-{\frac{ab}{{x}^{2}}{\it Artanh} \left ({\frac{c}{{x}^{2}}} \right ) }-{\frac{ab}{2\,c}\ln \left ( 1-{\frac{{c}^{2}}{{x}^{4}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c/x^2))^2/x^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{8} \,{\left (8 \, c^{3} \int \frac{\log \left (x\right )^{2}}{c x^{7} - c^{3} x^{3}}\,{d x} + c^{2}{\left (\frac{\log \left (x^{2} + c\right )}{c^{3}} + \frac{\log \left (x^{2} - c\right )}{c^{3}} - \frac{4 \, \log \left (x\right )}{c^{3}}\right )} - 8 \, c^{2} \int \frac{x^{2} \log \left (x^{2} + c\right )}{c x^{7} - c^{3} x^{3}}\,{d x} + 8 \, c^{2} \int \frac{x^{2} \log \left (x\right )}{c x^{7} - c^{3} x^{3}}\,{d x} + 2 \, c{\left (\frac{\log \left (x^{2} - c\right )}{c^{2}} - \frac{\log \left (x^{2}\right )}{c^{2}} + \frac{1}{c x^{2}}\right )} \log \left (-\frac{c}{x^{2}} + 1\right ) - c{\left (\frac{\log \left (x^{2} + c\right )}{c^{2}} - \frac{\log \left (x^{2} - c\right )}{c^{2}}\right )} - 8 \, c \int \frac{x^{4} \log \left (x\right )^{2}}{c x^{7} - c^{3} x^{3}}\,{d x} - 4 \, c \int \frac{x^{4} \log \left (x^{2} + c\right )}{c x^{7} - c^{3} x^{3}}\,{d x} + 16 \, c \int \frac{x^{4} \log \left (x\right )}{c x^{7} - c^{3} x^{3}}\,{d x} - \frac{\log \left (-\frac{c}{x^{2}} + 1\right )^{2}}{x^{2}} - \frac{x^{2} \log \left (x^{2} - c\right )^{2} + 4 \, x^{2} \log \left (x\right )^{2} - 4 \, x^{2} \log \left (x\right ) - 2 \,{\left (2 \, x^{2} \log \left (x\right ) - x^{2}\right )} \log \left (x^{2} - c\right ) + 2 \, c}{c x^{2}} - \frac{c \log \left (x^{2} + c\right )^{2} - 2 \,{\left ({\left (x^{2} + c\right )} \log \left (x^{2} + c\right ) - 2 \,{\left (x^{2} + c\right )} \log \left (x\right ) - c\right )} \log \left (x^{2} - c\right )}{c x^{2}} - 4 \, \int \frac{x^{6} \log \left (x^{2} + c\right )}{c x^{7} - c^{3} x^{3}}\,{d x} + 8 \, \int \frac{x^{6} \log \left (x\right )}{c x^{7} - c^{3} x^{3}}\,{d x}\right )} b^{2} - \frac{a b{\left (\frac{2 \, c \operatorname{artanh}\left (\frac{c}{x^{2}}\right )}{x^{2}} + \log \left (-\frac{c^{2}}{x^{4}} + 1\right )\right )}}{2 \, c} - \frac{a^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c/x^2))^2/x^3,x, algorithm="maxima")

[Out]

1/8*(8*c^3*integrate(log(x)^2/(c*x^7 - c^3*x^3), x) + c^2*(log(x^2 + c)/c^3 + log(x^2 - c)/c^3 - 4*log(x)/c^3)
- 8*c^2*integrate(x^2*log(x^2 + c)/(c*x^7 - c^3*x^3), x) + 8*c^2*integrate(x^2*log(x)/(c*x^7 - c^3*x^3), x) +
2*c*(log(x^2 - c)/c^2 - log(x^2)/c^2 + 1/(c*x^2))*log(-c/x^2 + 1) - c*(log(x^2 + c)/c^2 - log(x^2 - c)/c^2) -
8*c*integrate(x^4*log(x)^2/(c*x^7 - c^3*x^3), x) - 4*c*integrate(x^4*log(x^2 + c)/(c*x^7 - c^3*x^3), x) + 16*
c*integrate(x^4*log(x)/(c*x^7 - c^3*x^3), x) - log(-c/x^2 + 1)^2/x^2 - (x^2*log(x^2 - c)^2 + 4*x^2*log(x)^2 -
4*x^2*log(x) - 2*(2*x^2*log(x) - x^2)*log(x^2 - c) + 2*c)/(c*x^2) - (c*log(x^2 + c)^2 - 2*((x^2 + c)*log(x^2 +
c) - 2*(x^2 + c)*log(x) - c)*log(x^2 - c))/(c*x^2) - 4*integrate(x^6*log(x^2 + c)/(c*x^7 - c^3*x^3), x) + 8*i
ntegrate(x^6*log(x)/(c*x^7 - c^3*x^3), x))*b^2 - 1/2*a*b*(2*c*arctanh(c/x^2)/x^2 + log(-c^2/x^4 + 1))/c - 1/2*
a^2/x^2

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c/x^2))^2/x^3,x, algorithm="fricas")

[Out]

integral((b^2*arctanh(c/x^2)^2 + 2*a*b*arctanh(c/x^2) + a^2)/x^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atanh}{\left (\frac{c}{x^{2}} \right )}\right )^{2}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c/x**2))**2/x**3,x)

[Out]

Integral((a + b*atanh(c/x**2))**2/x**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (\frac{c}{x^{2}}\right ) + a\right )}^{2}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c/x^2))^2/x^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c/x^2) + a)^2/x^3, x)