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

Optimal. Leaf size=168 \[ -\frac {b \text {Li}_2\left (1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e}+\frac {b \text {Li}_2\left (\frac {2}{-c-d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{-c-d x+1}\right )}{2 d e}-\frac {b^2 \text {Li}_3\left (\frac {2}{-c-d x+1}-1\right )}{2 d e} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6107, 12, 5914, 6052, 5948, 6058, 6610} \[ -\frac {b \text {PolyLog}\left (2,1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e}+\frac {b \text {PolyLog}\left (2,\frac {2}{-c-d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{-c-d x+1}\right )}{2 d e}-\frac {b^2 \text {PolyLog}\left (3,\frac {2}{-c-d x+1}-1\right )}{2 d e}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 6107

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((f*x)/d)^m*(a + b*ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f,
0] && 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}(c+d x)\right )^2}{c e+d e x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c-d x}\right )}{d e}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {\tanh ^{-1}\left (1-\frac {2}{1-x}\right ) \left (a+b \tanh ^{-1}(x)\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c-d x}\right )}{d e}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right ) \log \left (2-\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right ) \log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c-d x}\right )}{d e}-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{d e}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c-d x}\right )}{d e}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (1-\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}-\frac {b^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-1+\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c-d x}\right )}{d e}-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (1-\frac {2}{1-c-d x}\right )}{d e}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c-d x}\right )}{d e}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1-c-d x}\right )}{2 d e}-\frac {b^2 \text {Li}_3\left (-1+\frac {2}{1-c-d x}\right )}{2 d e}\\ \end {align*}

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Mathematica [C]  time = 0.39, size = 424, normalized size = 2.52 \[ \frac {a^2 \log (c+d x)-\frac {1}{4} a b \left (4 \text {Li}_2\left (e^{-2 \tanh ^{-1}(c+d x)}\right )+4 \text {Li}_2\left (-e^{2 \tanh ^{-1}(c+d x)}\right )-4 i \pi \log \left (\frac {2}{\sqrt {1-(c+d x)^2}}\right )-8 \tanh ^{-1}(c+d x)^2-4 i \pi \tanh ^{-1}(c+d x)-8 \tanh ^{-1}(c+d x) \log \left (1-e^{-2 \tanh ^{-1}(c+d x)}\right )+8 \tanh ^{-1}(c+d x) \log \left (e^{2 \tanh ^{-1}(c+d x)}+1\right )-8 \log \left (\frac {2}{\sqrt {1-(c+d x)^2}}\right ) \tanh ^{-1}(c+d x)+8 \log \left (\frac {2 i (c+d x)}{\sqrt {1-(c+d x)^2}}\right ) \tanh ^{-1}(c+d x)+4 i \pi \log \left (e^{2 \tanh ^{-1}(c+d x)}+1\right )+\pi ^2\right )+2 a b \left (-\log \left (\frac {1}{\sqrt {1-(c+d x)^2}}\right )+\log \left (\frac {i (c+d x)}{\sqrt {1-(c+d x)^2}}\right )\right ) \tanh ^{-1}(c+d x)+b^2 \left (\tanh ^{-1}(c+d x) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )+\tanh ^{-1}(c+d x) \text {Li}_2\left (e^{2 \tanh ^{-1}(c+d x)}\right )+\frac {1}{2} \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c+d x)}\right )-\frac {1}{2} \text {Li}_3\left (e^{2 \tanh ^{-1}(c+d x)}\right )-\frac {2}{3} \tanh ^{-1}(c+d x)^3-\tanh ^{-1}(c+d x)^2 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )+\tanh ^{-1}(c+d x)^2 \log \left (1-e^{2 \tanh ^{-1}(c+d x)}\right )+\frac {i \pi ^3}{24}\right )}{d e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*Log[c + d*x] + 2*a*b*ArcTanh[c + d*x]*(-Log[1/Sqrt[1 - (c + d*x)^2]] + Log[(I*(c + d*x))/Sqrt[1 - (c + d*
x)^2]]) - (a*b*(Pi^2 - (4*I)*Pi*ArcTanh[c + d*x] - 8*ArcTanh[c + d*x]^2 - 8*ArcTanh[c + d*x]*Log[1 - E^(-2*Arc
Tanh[c + d*x])] + (4*I)*Pi*Log[1 + E^(2*ArcTanh[c + d*x])] + 8*ArcTanh[c + d*x]*Log[1 + E^(2*ArcTanh[c + d*x])
] - (4*I)*Pi*Log[2/Sqrt[1 - (c + d*x)^2]] - 8*ArcTanh[c + d*x]*Log[2/Sqrt[1 - (c + d*x)^2]] + 8*ArcTanh[c + d*
x]*Log[((2*I)*(c + d*x))/Sqrt[1 - (c + d*x)^2]] + 4*PolyLog[2, E^(-2*ArcTanh[c + d*x])] + 4*PolyLog[2, -E^(2*A
rcTanh[c + d*x])]))/4 + b^2*((I/24)*Pi^3 - (2*ArcTanh[c + d*x]^3)/3 - ArcTanh[c + d*x]^2*Log[1 + E^(-2*ArcTanh
[c + d*x])] + ArcTanh[c + d*x]^2*Log[1 - E^(2*ArcTanh[c + d*x])] + ArcTanh[c + d*x]*PolyLog[2, -E^(-2*ArcTanh[
c + d*x])] + ArcTanh[c + d*x]*PolyLog[2, E^(2*ArcTanh[c + d*x])] + PolyLog[3, -E^(-2*ArcTanh[c + d*x])]/2 - Po
lyLog[3, E^(2*ArcTanh[c + d*x])]/2))/(d*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.22, size = 893, normalized size = 5.32 \[ \frac {a^{2} \ln \left (d x +c \right )}{d e}+\frac {b^{2} \ln \left (d x +c \right ) \arctanh \left (d x +c \right )^{2}}{d e}-\frac {b^{2} \arctanh \left (d x +c \right ) \polylog \left (2, -\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}\right )}{d e}+\frac {b^{2} \polylog \left (3, -\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}\right )}{2 d e}-\frac {b^{2} \arctanh \left (d x +c \right )^{2} \ln \left (\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}-1\right )}{d e}+\frac {b^{2} \arctanh \left (d x +c \right )^{2} \ln \left (1-\frac {d x +c +1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{d e}+\frac {2 b^{2} \arctanh \left (d x +c \right ) \polylog \left (2, \frac {d x +c +1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{d e}-\frac {2 b^{2} \polylog \left (3, \frac {d x +c +1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{d e}+\frac {b^{2} \arctanh \left (d x +c \right )^{2} \ln \left (1+\frac {d x +c +1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{d e}+\frac {2 b^{2} \arctanh \left (d x +c \right ) \polylog \left (2, -\frac {d x +c +1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{d e}-\frac {2 b^{2} \polylog \left (3, -\frac {d x +c +1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{d e}-\frac {i b^{2} \pi \,\mathrm {csgn}\left (\frac {i}{1+\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}-1\right )}{1+\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}}\right )^{2} \arctanh \left (d x +c \right )^{2}}{2 d e}+\frac {i b^{2} \pi \mathrm {csgn}\left (\frac {i \left (\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}-1\right )}{1+\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}}\right )^{3} \arctanh \left (d x +c \right )^{2}}{2 d e}-\frac {i b^{2} \pi \,\mathrm {csgn}\left (i \left (\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}-1\right )}{1+\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}}\right )^{2} \arctanh \left (d x +c \right )^{2}}{2 d e}+\frac {i b^{2} \pi \,\mathrm {csgn}\left (i \left (\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}-1\right )\right ) \mathrm {csgn}\left (\frac {i}{1+\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}-1\right )}{1+\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}}\right ) \arctanh \left (d x +c \right )^{2}}{2 d e}+\frac {2 a b \ln \left (d x +c \right ) \arctanh \left (d x +c \right )}{d e}-\frac {a b \dilog \left (d x +c \right )}{d e}-\frac {a b \dilog \left (d x +c +1\right )}{d e}-\frac {a b \ln \left (d x +c \right ) \ln \left (d x +c +1\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a^2/e*ln(d*x+c)+1/d*b^2/e*ln(d*x+c)*arctanh(d*x+c)^2-1/d*b^2/e*arctanh(d*x+c)*polylog(2,-(d*x+c+1)^2/(1-(d
*x+c)^2))+1/2/d*b^2/e*polylog(3,-(d*x+c+1)^2/(1-(d*x+c)^2))-1/d*b^2/e*arctanh(d*x+c)^2*ln((d*x+c+1)^2/(1-(d*x+
c)^2)-1)+1/d*b^2/e*arctanh(d*x+c)^2*ln(1-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+2/d*b^2/e*arctanh(d*x+c)*polylog(2,(d*
x+c+1)/(1-(d*x+c)^2)^(1/2))-2/d*b^2/e*polylog(3,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+1/d*b^2/e*arctanh(d*x+c)^2*ln(1
+(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+2/d*b^2/e*arctanh(d*x+c)*polylog(2,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-2/d*b^2/e*p
olylog(3,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+1/2*I/d*b^2/e*Pi*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1))*csgn(I/(1+(d*x
+c+1)^2/(1-(d*x+c)^2)))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*arctanh(d*x+c)^2-1
/2*I/d*b^2/e*Pi*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(
d*x+c)^2)))^2*arctanh(d*x+c)^2+1/2*I/d*b^2/e*Pi*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)
^2)))^3*arctanh(d*x+c)^2-1/2*I/d*b^2/e*Pi*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)
^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^2*arctanh(d*x+c)^2+2/d*a*b/e*ln(d*x+c)*arctanh(d*x+c)-1/d*a*b/e*dilog(d*
x+c)-1/d*a*b/e*dilog(d*x+c+1)-1/d*a*b/e*ln(d*x+c)*ln(d*x+c+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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