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

Optimal. Leaf size=151 \[ \frac {3}{2} c^2 d \left (a+b \tanh ^{-1}(c x)\right )^2+2 b c^2 d \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {c d \left (a+b \tanh ^{-1}(c x)\right )^2}{x}-\frac {b c d \left (a+b \tanh ^{-1}(c x)\right )}{x}-b^2 c^2 d \text {Li}_2\left (\frac {2}{c x+1}-1\right )-\frac {1}{2} b^2 c^2 d \log \left (1-c^2 x^2\right )+b^2 c^2 d \log (x) \]

[Out]

-b*c*d*(a+b*arctanh(c*x))/x+3/2*c^2*d*(a+b*arctanh(c*x))^2-1/2*d*(a+b*arctanh(c*x))^2/x^2-c*d*(a+b*arctanh(c*x
))^2/x+b^2*c^2*d*ln(x)-1/2*b^2*c^2*d*ln(-c^2*x^2+1)+2*b*c^2*d*(a+b*arctanh(c*x))*ln(2-2/(c*x+1))-b^2*c^2*d*pol
ylog(2,-1+2/(c*x+1))

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Rubi [A]  time = 0.37, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {5940, 5916, 5982, 266, 36, 29, 31, 5948, 5988, 5932, 2447} \[ -b^2 c^2 d \text {PolyLog}\left (2,\frac {2}{c x+1}-1\right )+\frac {3}{2} c^2 d \left (a+b \tanh ^{-1}(c x)\right )^2+2 b c^2 d \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {c d \left (a+b \tanh ^{-1}(c x)\right )^2}{x}-\frac {b c d \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac {1}{2} b^2 c^2 d \log \left (1-c^2 x^2\right )+b^2 c^2 d \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

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

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5932

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

Rule 5940

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

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 5982

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

Rule 5988

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

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 206, normalized size = 1.36 \[ -\frac {d \left (2 a^2 c x+a^2-4 a b c^2 x^2 \log (c x)+a b c^2 x^2 \log (1-c x)-a b c^2 x^2 \log (c x+1)+2 a b c^2 x^2 \log \left (1-c^2 x^2\right )+2 b \tanh ^{-1}(c x) \left (2 a c x+a-2 b c^2 x^2 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+b c x\right )+2 a b c x+2 b^2 c^2 x^2 \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )-2 b^2 c^2 x^2 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )+b^2 \left (-3 c^2 x^2+2 c x+1\right ) \tanh ^{-1}(c x)^2\right )}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} c d x + a^{2} d + {\left (b^{2} c d x + b^{2} d\right )} \operatorname {artanh}\left (c x\right )^{2} + 2 \, {\left (a b c d x + a b d\right )} \operatorname {artanh}\left (c x\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c d x + d\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.08, size = 400, normalized size = 2.65 \[ -\frac {2 c d a b \arctanh \left (c x \right )}{x}-\frac {c^{2} d \,b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{2}+2 c^{2} d \,b^{2} \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {d \,b^{2} \arctanh \left (c x \right )^{2}}{2 x^{2}}-c^{2} d \,b^{2} \dilog \left (c x +1\right )+c^{2} d \,b^{2} \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )-\frac {c^{2} d \,b^{2} \ln \left (c x +1\right )}{2}+c^{2} d \,b^{2} \ln \left (c x \right )-\frac {c \,a^{2} d}{x}-c^{2} d \,b^{2} \dilog \left (c x \right )-\frac {3 c^{2} d \,b^{2} \ln \left (c x -1\right )^{2}}{8}+\frac {c^{2} d \,b^{2} \ln \left (c x +1\right )^{2}}{8}-\frac {c^{2} d \,b^{2} \ln \left (c x -1\right )}{2}-\frac {a^{2} d}{2 x^{2}}-\frac {c d \,b^{2} \arctanh \left (c x \right )}{x}-\frac {3 c^{2} d \,b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{2}-\frac {c d a b}{x}-c^{2} d \,b^{2} \ln \left (c x \right ) \ln \left (c x +1\right )-\frac {d a b \arctanh \left (c x \right )}{x^{2}}-\frac {c d \,b^{2} \arctanh \left (c x \right )^{2}}{x}+\frac {c^{2} d \,b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{4}+\frac {3 c^{2} d \,b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{4}-\frac {c^{2} d \,b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{4}-\frac {c^{2} d a b \ln \left (c x +1\right )}{2}-\frac {3 c^{2} d a b \ln \left (c x -1\right )}{2}+2 c^{2} d a b \ln \left (c x \right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*c*d*a*b*arctanh(c*x)/x-1/2*c^2*d*b^2*ln(c*x+1)+c^2*d*b^2*ln(c*x)-c^2*d*b^2*dilog(c*x)-1/2*d*b^2*arctanh(c*x
)^2/x^2-c*a^2*d/x-c^2*d*b^2*dilog(c*x+1)-3/8*c^2*d*b^2*ln(c*x-1)^2+c^2*d*b^2*dilog(1/2+1/2*c*x)+1/8*c^2*d*b^2*
ln(c*x+1)^2-1/2*c^2*d*b^2*ln(c*x-1)-d*a*b*arctanh(c*x)/x^2-1/2*a^2*d/x^2-c*d*a*b/x-c*d*b^2*arctanh(c*x)^2/x-c^
2*d*b^2*ln(c*x)*ln(c*x+1)-c*d*b^2*arctanh(c*x)/x+1/4*c^2*d*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-3/2*c^2*d*b^2*
arctanh(c*x)*ln(c*x-1)-1/2*c^2*d*b^2*arctanh(c*x)*ln(c*x+1)+3/4*c^2*d*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)-1/4*c^2*d*
b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+2*c^2*d*b^2*arctanh(c*x)*ln(c*x)-1/2*c^2*d*a*b*ln(c*x+1)-3/2*c^2*d*a*b*ln(c*x-1
)+2*c^2*d*a*b*ln(c*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -{\left (c {\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x}\right )} a b c d - \frac {1}{4} \, b^{2} c d {\left (\frac {\log \left (-c x + 1\right )^{2}}{x} + \int -\frac {{\left (c x - 1\right )} \log \left (c x + 1\right )^{2} + 2 \, {\left (c x - {\left (c x - 1\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{c x^{3} - x^{2}}\,{d x}\right )} + \frac {1}{2} \, {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{2}}\right )} a b d + \frac {1}{8} \, {\left ({\left (2 \, {\left (\log \left (c x - 1\right ) - 2\right )} \log \left (c x + 1\right ) - \log \left (c x + 1\right )^{2} - \log \left (c x - 1\right )^{2} - 4 \, \log \left (c x - 1\right ) + 8 \, \log \relax (x)\right )} c^{2} + 4 \, {\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c \operatorname {artanh}\left (c x\right )\right )} b^{2} d - \frac {a^{2} c d}{x} - \frac {b^{2} d \operatorname {artanh}\left (c x\right )^{2}}{2 \, x^{2}} - \frac {a^{2} d}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*a*b*c*d - 1/4*b^2*c*d*(log(-c*x + 1)^2/x + integrate(-((
c*x - 1)*log(c*x + 1)^2 + 2*(c*x - (c*x - 1)*log(c*x + 1))*log(-c*x + 1))/(c*x^3 - x^2), x)) + 1/2*((c*log(c*x
+ 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*a*b*d + 1/8*((2*(log(c*x - 1) - 2)*log(c*x + 1) - log(c*
x + 1)^2 - log(c*x - 1)^2 - 4*log(c*x - 1) + 8*log(x))*c^2 + 4*(c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c*arcta
nh(c*x))*b^2*d - a^2*c*d/x - 1/2*b^2*d*arctanh(c*x)^2/x^2 - 1/2*a^2*d/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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