3.73 \(\int \frac {(d+c d x) (a+b \tanh ^{-1}(c x))^2}{x^2} \, dx\)
Optimal. Leaf size=201 \[ -b c d \text {Li}_2\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+b c d \text {Li}_2\left (\frac {2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )+c d \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 c d \tanh ^{-1}\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2+2 b c d \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )+b^2 (-c) d \text {Li}_2\left (\frac {2}{c x+1}-1\right )+\frac {1}{2} b^2 c d \text {Li}_3\left (1-\frac {2}{1-c x}\right )-\frac {1}{2} b^2 c d \text {Li}_3\left (\frac {2}{1-c x}-1\right ) \]
[Out]
c*d*(a+b*arctanh(c*x))^2-d*(a+b*arctanh(c*x))^2/x-2*c*d*(a+b*arctanh(c*x))^2*arctanh(-1+2/(-c*x+1))+2*b*c*d*(a
+b*arctanh(c*x))*ln(2-2/(c*x+1))-b*c*d*(a+b*arctanh(c*x))*polylog(2,1-2/(-c*x+1))+b*c*d*(a+b*arctanh(c*x))*pol
ylog(2,-1+2/(-c*x+1))-b^2*c*d*polylog(2,-1+2/(c*x+1))+1/2*b^2*c*d*polylog(3,1-2/(-c*x+1))-1/2*b^2*c*d*polylog(
3,-1+2/(-c*x+1))
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Rubi [A] time = 0.48, antiderivative size = 201, normalized size of antiderivative =
1.00, number of steps used = 12, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\)
= 0.500, Rules used = {5940, 5916, 5988, 5932, 2447, 5914, 6052, 5948, 6058, 6610}
\[ -b c d \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+b c d \text {PolyLog}\left (2,\frac {2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )+b^2 (-c) d \text {PolyLog}\left (2,\frac {2}{c x+1}-1\right )+\frac {1}{2} b^2 c d \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )-\frac {1}{2} b^2 c d \text {PolyLog}\left (3,\frac {2}{1-c x}-1\right )+c d \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 c d \tanh ^{-1}\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2+2 b c d \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right ) \]
Antiderivative was successfully verified.
[In]
Int[((d + c*d*x)*(a + b*ArcTanh[c*x])^2)/x^2,x]
[Out]
c*d*(a + b*ArcTanh[c*x])^2 - (d*(a + b*ArcTanh[c*x])^2)/x + 2*c*d*(a + b*ArcTanh[c*x])^2*ArcTanh[1 - 2/(1 - c*
x)] + 2*b*c*d*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)] - b*c*d*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 - c*x
)] + b*c*d*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 - c*x)] - b^2*c*d*PolyLog[2, -1 + 2/(1 + c*x)] + (b^2*c*d
*PolyLog[3, 1 - 2/(1 - c*x)])/2 - (b^2*c*d*PolyLog[3, -1 + 2/(1 - c*x)])/2
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 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 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 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]
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 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 {(d+c d x) \left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx &=\int \left (\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{x^2}+\frac {c d \left (a+b \tanh ^{-1}(c x)\right )^2}{x}\right ) \, dx\\ &=d \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx+(c d) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx\\ &=-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 c d \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+(2 b c d) \int \frac {a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx-\left (4 b c^2 d\right ) \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\\ &=c d \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 c d \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+(2 b c d) \int \frac {a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx+\left (2 b c^2 d\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (2 b c^2 d\right ) \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\\ &=c d \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 c d \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+2 b c d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-b c d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )+b c d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )+\left (b^2 c^2 d\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (b^2 c^2 d\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (2 b^2 c^2 d\right ) \int \frac {\log \left (2-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=c d \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 c d \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+2 b c d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-b c d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )+b c d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )-b^2 c d \text {Li}_2\left (-1+\frac {2}{1+c x}\right )+\frac {1}{2} b^2 c d \text {Li}_3\left (1-\frac {2}{1-c x}\right )-\frac {1}{2} b^2 c d \text {Li}_3\left (-1+\frac {2}{1-c x}\right )\\ \end {align*}
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Mathematica [C] time = 0.52, size = 249, normalized size = 1.24 \[ -\frac {d \left (a^2 (-c) x \log (x)+a^2+a b \left (c x \left (\log \left (1-c^2 x^2\right )-2 \log (c x)\right )+2 \tanh ^{-1}(c x)\right )+a b c x (\text {Li}_2(-c x)-\text {Li}_2(c x))+b^2 \left (c x \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )+\tanh ^{-1}(c x) \left ((1-c x) \tanh ^{-1}(c x)-2 c x \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )\right )-b^2 c x \left (\tanh ^{-1}(c x) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+\tanh ^{-1}(c x) \text {Li}_2\left (e^{2 \tanh ^{-1}(c x)}\right )+\frac {1}{2} \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c x)}\right )-\frac {1}{2} \text {Li}_3\left (e^{2 \tanh ^{-1}(c x)}\right )-\frac {2}{3} \tanh ^{-1}(c x)^3-\tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+\tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+\frac {i \pi ^3}{24}\right )\right )}{x} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((d + c*d*x)*(a + b*ArcTanh[c*x])^2)/x^2,x]
[Out]
-((d*(a^2 - a^2*c*x*Log[x] + a*b*(2*ArcTanh[c*x] + c*x*(-2*Log[c*x] + Log[1 - c^2*x^2])) + b^2*(ArcTanh[c*x]*(
(1 - c*x)*ArcTanh[c*x] - 2*c*x*Log[1 - E^(-2*ArcTanh[c*x])]) + c*x*PolyLog[2, E^(-2*ArcTanh[c*x])]) + a*b*c*x*
(PolyLog[2, -(c*x)] - PolyLog[2, c*x]) - b^2*c*x*((I/24)*Pi^3 - (2*ArcTanh[c*x]^3)/3 - ArcTanh[c*x]^2*Log[1 +
E^(-2*ArcTanh[c*x])] + ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] + ArcTanh[c*x]*PolyLog[2, -E^(-2*ArcTanh[c*x
])] + ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] + PolyLog[3, -E^(-2*ArcTanh[c*x])]/2 - PolyLog[3, E^(2*ArcTa
nh[c*x])]/2)))/x)
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} c d x + a^{2} d + {\left (b^{2} c d x + b^{2} d\right )} \operatorname {artanh}\left (c x\right )^{2} + 2 \, {\left (a b c d x + a b d\right )} \operatorname {artanh}\left (c x\right )}{x^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*d*x+d)*(a+b*arctanh(c*x))^2/x^2,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^2, x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c d x + d\right )} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*d*x+d)*(a+b*arctanh(c*x))^2/x^2,x, algorithm="giac")
[Out]
integrate((c*d*x + d)*(b*arctanh(c*x) + a)^2/x^2, x)
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maple [C] time = 0.98, size = 3104, normalized size = 15.44 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c*d*x+d)*(a+b*arctanh(c*x))^2/x^2,x)
[Out]
-a^2*d/x-1/4*c*d*b^2*dilog((c*x+1)^2/(-c^2*x^2+1))+c*a^2*d*ln(c*x)-c*d*b^2*arctanh(c*x)^2-1/4*c*d*b^2*polylog(
2,-(c*x+1)^2/(-c^2*x^2+1))+1/2*c*d*b^2*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))+3/4*c*d*b^2*polylog(2,(c*x+1)^2/(-c^
2*x^2+1))-1/2*c*d*b^2*polylog(3,(c*x+1)^2/(-c^2*x^2+1))+1/4*c*d*b^2*dilog(1+(c*x+1)^2/(-c^2*x^2+1))-d*b^2*arct
anh(c*x)^2/x-1/8*I*c*d*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^
2/(-c^2*x^2+1)))^2*dilog(1+(c*x+1)^2/(-c^2*x^2+1))+1/8*I*c*d*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I*
((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))-1/8*I*c*d*b^2*Pi*c
sgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*dilog(1+(c*x
+1)^2/(-c^2*x^2+1))+1/8*I*c*d*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(
c*x+1)^2/(-c^2*x^2+1)))^2*dilog((c*x+1)^2/(-c^2*x^2+1))+1/8*I*c*d*b^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*cs
gn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+1/8*I*c*d*b^2
*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*polylog
(2,(c*x+1)^2/(-c^2*x^2+1))-1/4*I*c*d*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3*ar
ctanh(c*x)*ln(1-(c*x+1)^2/(-c^2*x^2+1))+1/8*I*c*d*b^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*((c*x+1)^2/
(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*polylog(2,(c*x+1)^2/(-c^2*x^2+1))+1/8*I*c*d*b^2*Pi*csgn(I/(1+(c*
x+1)^2/(-c^2*x^2+1)))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*dilog((c*x+1)^2/(-c^2*x^
2+1))-1/2*I*c*d*b^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2
*x^2+1)))^2*arctanh(c*x)^2-1/2*I*c*d*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-
1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*arctanh(c*x)^2-c*d*a*b*ln(c*x)*ln(c*x+1)+2*c*d*a*b*arctanh(c*x)*ln(c*x)-1/8*I
*c*d*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))
-1/8*I*c*d*b^2*dilog((c*x+1)^2/(-c^2*x^2+1))*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^
3+1/8*I*c*d*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3*dilog(1+(c*x+1)^2/(-c^2*x^2
+1))+1/2*I*c*d*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3*arctanh(c*x)^2-1/8*I*c*d
*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3*polylog(2,(c*x+1)^2/(-c^2*x^2+1))-2*d*
a*b*arctanh(c*x)/x-c*d*a*b*ln(c*x+1)-c*d*a*b*ln(c*x-1)+2*c*d*a*b*ln(c*x)-c*d*a*b*dilog(c*x)-c*d*a*b*dilog(c*x+
1)+c*d*b^2*arctanh(c*x)^2*ln(1-(c*x+1)^2/(-c^2*x^2+1))+c*d*b^2*arctanh(c*x)*polylog(2,(c*x+1)^2/(-c^2*x^2+1))+
3/2*c*d*b^2*arctanh(c*x)*ln(1-(c*x+1)^2/(-c^2*x^2+1))+c*d*b^2*arctanh(c*x)^2*ln(c*x)-c*d*b^2*arctanh(c*x)*poly
log(2,-(c*x+1)^2/(-c^2*x^2+1))-c*d*b^2*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)-1/4*I*c*d*b^2*Pi*csgn(I*((c
*x+1)^2/(-c^2*x^2+1)-1))*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c
^2*x^2+1)))*arctanh(c*x)*ln(1-(c*x+1)^2/(-c^2*x^2+1))+1/8*I*c*d*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn
(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))*dilog(1+(c*x+1)^2
/(-c^2*x^2+1))+1/2*I*c*d*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(
(c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))*arctanh(c*x)^2-1/8*I*c*d*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*
x^2+1)-1))*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))*po
lylog(2,(c*x+1)^2/(-c^2*x^2+1))-1/8*I*c*d*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I/(1+(c*x+1)^2/(-c^2*
x^2+1)))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))-1/8*
I*c*d*b^2*dilog((c*x+1)^2/(-c^2*x^2+1))*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)
))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))+1/4*I*c*d*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+
1)-1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*arctanh(c*x)*ln(1-(c*x+1)^2/(-c^2*x^2+1
))+1/4*I*c*d*b^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^
2+1)))^2*arctanh(c*x)*ln(1-(c*x+1)^2/(-c^2*x^2+1))
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} c d \log \relax (x) - {\left (c {\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x}\right )} a b d - \frac {b^{2} d \log \left (-c x + 1\right )^{2}}{4 \, x} - \frac {a^{2} d}{x} - \int -\frac {{\left (b^{2} c^{2} d x^{2} - b^{2} d\right )} \log \left (c x + 1\right )^{2} + 4 \, {\left (a b c^{2} d x^{2} - a b c d x\right )} \log \left (c x + 1\right ) - 2 \, {\left (2 \, a b c^{2} d x^{2} - {\left (2 \, a b c d + b^{2} c d\right )} x + {\left (b^{2} c^{2} d x^{2} - b^{2} d\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \, {\left (c x^{3} - x^{2}\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*d*x+d)*(a+b*arctanh(c*x))^2/x^2,x, algorithm="maxima")
[Out]
a^2*c*d*log(x) - (c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*a*b*d - 1/4*b^2*d*log(-c*x + 1)^2/x - a^
2*d/x - integrate(-1/4*((b^2*c^2*d*x^2 - b^2*d)*log(c*x + 1)^2 + 4*(a*b*c^2*d*x^2 - a*b*c*d*x)*log(c*x + 1) -
2*(2*a*b*c^2*d*x^2 - (2*a*b*c*d + b^2*c*d)*x + (b^2*c^2*d*x^2 - b^2*d)*log(c*x + 1))*log(-c*x + 1))/(c*x^3 - x
^2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2\,\left (d+c\,d\,x\right )}{x^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((a + b*atanh(c*x))^2*(d + c*d*x))/x^2,x)
[Out]
int(((a + b*atanh(c*x))^2*(d + c*d*x))/x^2, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ d \left (\int \frac {a^{2}}{x^{2}}\, dx + \int \frac {a^{2} c}{x}\, dx + \int \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{x^{2}}\, dx + \int \frac {2 a b \operatorname {atanh}{\left (c x \right )}}{x^{2}}\, dx + \int \frac {b^{2} c \operatorname {atanh}^{2}{\left (c x \right )}}{x}\, dx + \int \frac {2 a b c \operatorname {atanh}{\left (c x \right )}}{x}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*d*x+d)*(a+b*atanh(c*x))**2/x**2,x)
[Out]
d*(Integral(a**2/x**2, x) + Integral(a**2*c/x, x) + Integral(b**2*atanh(c*x)**2/x**2, x) + Integral(2*a*b*atan
h(c*x)/x**2, x) + Integral(b**2*c*atanh(c*x)**2/x, x) + Integral(2*a*b*c*atanh(c*x)/x, x))
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