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

Optimal. Leaf size=313 \[ -b c^2 d^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+b c^2 d^2 \text {Li}_2\left (\frac {2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac {5}{2} c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2+2 c^2 d^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2+4 b c^2 d^2 \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}-\frac {b c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}-2 b^2 c^2 d^2 \text {Li}_2\left (\frac {2}{c x+1}-1\right )+\frac {1}{2} b^2 c^2 d^2 \text {Li}_3\left (1-\frac {2}{1-c x}\right )-\frac {1}{2} b^2 c^2 d^2 \text {Li}_3\left (\frac {2}{1-c x}-1\right )-\frac {1}{2} b^2 c^2 d^2 \log \left (1-c^2 x^2\right )+b^2 c^2 d^2 \log (x) \]

[Out]

-b*c*d^2*(a+b*arctanh(c*x))/x+5/2*c^2*d^2*(a+b*arctanh(c*x))^2-1/2*d^2*(a+b*arctanh(c*x))^2/x^2-2*c*d^2*(a+b*a
rctanh(c*x))^2/x-2*c^2*d^2*(a+b*arctanh(c*x))^2*arctanh(-1+2/(-c*x+1))+b^2*c^2*d^2*ln(x)-1/2*b^2*c^2*d^2*ln(-c
^2*x^2+1)+4*b*c^2*d^2*(a+b*arctanh(c*x))*ln(2-2/(c*x+1))-b*c^2*d^2*(a+b*arctanh(c*x))*polylog(2,1-2/(-c*x+1))+
b*c^2*d^2*(a+b*arctanh(c*x))*polylog(2,-1+2/(-c*x+1))-2*b^2*c^2*d^2*polylog(2,-1+2/(c*x+1))+1/2*b^2*c^2*d^2*po
lylog(3,1-2/(-c*x+1))-1/2*b^2*c^2*d^2*polylog(3,-1+2/(-c*x+1))

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Rubi [A]  time = 0.67, antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 15, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.682, Rules used = {5940, 5916, 5982, 266, 36, 29, 31, 5948, 5988, 5932, 2447, 5914, 6052, 6058, 6610} \[ -b c^2 d^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+b c^2 d^2 \text {PolyLog}\left (2,\frac {2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )-2 b^2 c^2 d^2 \text {PolyLog}\left (2,\frac {2}{c x+1}-1\right )+\frac {1}{2} b^2 c^2 d^2 \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )-\frac {1}{2} b^2 c^2 d^2 \text {PolyLog}\left (3,\frac {2}{1-c x}-1\right )+\frac {5}{2} c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2+2 c^2 d^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2+4 b c^2 d^2 \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}-\frac {b c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac {1}{2} b^2 c^2 d^2 \log \left (1-c^2 x^2\right )+b^2 c^2 d^2 \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*d^2*(a + b*ArcTanh[c*x]))/x) + (5*c^2*d^2*(a + b*ArcTanh[c*x])^2)/2 - (d^2*(a + b*ArcTanh[c*x])^2)/(2*x
^2) - (2*c*d^2*(a + b*ArcTanh[c*x])^2)/x + 2*c^2*d^2*(a + b*ArcTanh[c*x])^2*ArcTanh[1 - 2/(1 - c*x)] + b^2*c^2
*d^2*Log[x] - (b^2*c^2*d^2*Log[1 - c^2*x^2])/2 + 4*b*c^2*d^2*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)] - b*c^2
*d^2*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 - c*x)] + b*c^2*d^2*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 -
c*x)] - 2*b^2*c^2*d^2*PolyLog[2, -1 + 2/(1 + c*x)] + (b^2*c^2*d^2*PolyLog[3, 1 - 2/(1 - c*x)])/2 - (b^2*c^2*d^
2*PolyLog[3, -1 + 2/(1 - c*x)])/2

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 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 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]

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)^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{x^3} \, dx &=\int \left (\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{x^3}+\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{x^2}+\frac {c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}\right ) \, dx\\ &=d^2 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^3} \, dx+\left (2 c d^2\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx+\left (c^2 d^2\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx\\ &=-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+\left (b c d^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^2 \left (1-c^2 x^2\right )} \, dx+\left (4 b c^2 d^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx-\left (4 b c^3 d^2\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\\ &=2 c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+\left (b c d^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx+\left (4 b c^2 d^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx+\left (b c^3 d^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx+\left (2 b c^3 d^2\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^3 d^2\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\\ &=-\frac {b c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {5}{2} c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+4 b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )+b c^2 d^2 \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^2\right ) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx+\left (b^2 c^3 d^2\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (b^2 c^3 d^2\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (4 b^2 c^3 d^2\right ) \int \frac {\log \left (2-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac {b c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {5}{2} c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+4 b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )+b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )-2 b^2 c^2 d^2 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )+\frac {1}{2} b^2 c^2 d^2 \text {Li}_3\left (1-\frac {2}{1-c x}\right )-\frac {1}{2} b^2 c^2 d^2 \text {Li}_3\left (-1+\frac {2}{1-c x}\right )+\frac {1}{2} \left (b^2 c^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {b c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {5}{2} c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+4 b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )+b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )-2 b^2 c^2 d^2 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )+\frac {1}{2} b^2 c^2 d^2 \text {Li}_3\left (1-\frac {2}{1-c x}\right )-\frac {1}{2} b^2 c^2 d^2 \text {Li}_3\left (-1+\frac {2}{1-c x}\right )+\frac {1}{2} \left (b^2 c^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (b^2 c^4 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {5}{2} c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {2 c d^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+2 c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+b^2 c^2 d^2 \log (x)-\frac {1}{2} b^2 c^2 d^2 \log \left (1-c^2 x^2\right )+4 b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )+b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )-2 b^2 c^2 d^2 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )+\frac {1}{2} b^2 c^2 d^2 \text {Li}_3\left (1-\frac {2}{1-c x}\right )-\frac {1}{2} b^2 c^2 d^2 \text {Li}_3\left (-1+\frac {2}{1-c x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.75, size = 370, normalized size = 1.18 \[ -\frac {d^2 \left (-2 a^2 c^2 x^2 \log (x)+4 a^2 c x+a^2+2 a b c^2 x^2 (\text {Li}_2(-c x)-\text {Li}_2(c x))+4 a b c x \left (c x \left (\log \left (1-c^2 x^2\right )-2 \log (c x)\right )+2 \tanh ^{-1}(c x)\right )+a b \left (c x (c x \log (1-c x)-c x \log (c x+1)+2)+2 \tanh ^{-1}(c x)\right )-2 b^2 c^2 x^2 \left (\tanh ^{-1}(c x) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+\tanh ^{-1}(c x) \text {Li}_2\left (e^{2 \tanh ^{-1}(c x)}\right )+\frac {1}{2} \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c x)}\right )-\frac {1}{2} \text {Li}_3\left (e^{2 \tanh ^{-1}(c x)}\right )-\frac {2}{3} \tanh ^{-1}(c x)^3-\tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+\tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+\frac {i \pi ^3}{24}\right )+b^2 \left (-2 c^2 x^2 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )+\left (1-c^2 x^2\right ) \tanh ^{-1}(c x)^2+2 c x \tanh ^{-1}(c x)\right )+4 b^2 c x \left (c x \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )+\tanh ^{-1}(c x) \left ((1-c x) \tanh ^{-1}(c x)-2 c x \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )\right )\right )}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} c^{2} d^{2} x^{2} + 2 \, a^{2} c d^{2} x + a^{2} d^{2} + {\left (b^{2} c^{2} d^{2} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} d^{2}\right )} \operatorname {artanh}\left (c x\right )^{2} + 2 \, {\left (a b c^{2} d^{2} x^{2} + 2 \, a b c d^{2} x + a b d^{2}\right )} \operatorname {artanh}\left (c x\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.74, size = 1167, normalized size = 3.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*c^2*d^2*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^2*d^2*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+1/2*I*c^2*d^2*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csg
n(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)^2-2
*c*d^2*a^2/x-1/2*d^2*a^2/x^2+4*c^2*d^2*b^2*arctanh(c*x)*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-c^2*d^2*a*b*ln(c*x)*l
n(c*x+1)+2*c^2*d^2*a*b*arctanh(c*x)*ln(c*x)-4*c*d^2*a*b*arctanh(c*x)/x-c^2*d^2*a*b*dilog(c*x+1)-5/2*c^2*d^2*a*
b*ln(c*x-1)-3/2*c^2*d^2*a*b*ln(c*x+1)+c^2*d^2*b^2*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+2*c^2*d^2*b^
2*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))-c*d^2*a*b/x-d^2*a*b*arctanh(c*x)/x^2+4*c^2*d^2*a*b*ln(c*
x)-c^2*d^2*b^2*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+c^2*d^2*b^2*arctanh(c*x)^2*ln(c*x)-c^2*d^2*b^2*
arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+c^2*d^2*b^2*arctanh(c*x)^2*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))+2*c^2*
d^2*b^2*arctanh(c*x)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))-2*c*d^2*b^2*arctanh(c*x)^2/x-c*d^2*b^2*arctanh(c*x)
/x-c^2*d^2*a*b*dilog(c*x)-2*c^2*d^2*b^2*polylog(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))+1/2*c^2*d^2*b^2*polylog(3,-(c*x
+1)^2/(-c^2*x^2+1))-c^2*d^2*b^2*arctanh(c*x)+c^2*d^2*b^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-4*c^2*d^2*b^2*dilog(
(c*x+1)/(-c^2*x^2+1)^(1/2))+4*c^2*d^2*b^2*dilog(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+c^2*d^2*b^2*ln((c*x+1)/(-c^2*x^2
+1)^(1/2)-1)-1/2*d^2*b^2*arctanh(c*x)^2/x^2-3/2*c^2*d^2*b^2*arctanh(c*x)^2+c^2*d^2*a^2*ln(c*x)-2*c^2*d^2*b^2*p
olylog(3,(c*x+1)/(-c^2*x^2+1)^(1/2))+1/2*I*c^2*d^2*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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d^{2} \left (\int \frac {a^{2}}{x^{3}}\, dx + \int \frac {2 a^{2} c}{x^{2}}\, dx + \int \frac {a^{2} c^{2}}{x}\, dx + \int \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{x^{3}}\, dx + \int \frac {2 a b \operatorname {atanh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac {2 b^{2} c \operatorname {atanh}^{2}{\left (c x \right )}}{x^{2}}\, dx + \int \frac {b^{2} c^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{x}\, dx + \int \frac {4 a b c \operatorname {atanh}{\left (c x \right )}}{x^{2}}\, dx + \int \frac {2 a b c^{2} \operatorname {atanh}{\left (c x \right )}}{x}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*(Integral(a**2/x**3, x) + Integral(2*a**2*c/x**2, x) + Integral(a**2*c**2/x, x) + Integral(b**2*atanh(c*x
)**2/x**3, x) + Integral(2*a*b*atanh(c*x)/x**3, x) + Integral(2*b**2*c*atanh(c*x)**2/x**2, x) + Integral(b**2*
c**2*atanh(c*x)**2/x, x) + Integral(4*a*b*c*atanh(c*x)/x**2, x) + Integral(2*a*b*c**2*atanh(c*x)/x, x))

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