3.83 \(\int \frac {(d+c d x)^2 (a+b \tanh ^{-1}(c x))^2}{x^4} \, dx\)

Optimal. Leaf size=244 \[ \frac {8}{3} a b c^3 d^2 \log (x)+\frac {8}{3} b c^3 d^2 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac {2 b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac {d^2 (c x+1)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-\frac {b c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac {4}{3} b^2 c^3 d^2 \text {Li}_2(-c x)+\frac {4}{3} b^2 c^3 d^2 \text {Li}_2(c x)+\frac {4}{3} b^2 c^3 d^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )+2 b^2 c^3 d^2 \log (x)+\frac {1}{3} b^2 c^3 d^2 \tanh ^{-1}(c x)-\frac {b^2 c^2 d^2}{3 x}-b^2 c^3 d^2 \log \left (1-c^2 x^2\right ) \]

[Out]

-1/3*b^2*c^2*d^2/x+1/3*b^2*c^3*d^2*arctanh(c*x)-1/3*b*c*d^2*(a+b*arctanh(c*x))/x^2-2*b*c^2*d^2*(a+b*arctanh(c*
x))/x-1/3*d^2*(c*x+1)^3*(a+b*arctanh(c*x))^2/x^3+8/3*a*b*c^3*d^2*ln(x)+2*b^2*c^3*d^2*ln(x)+8/3*b*c^3*d^2*(a+b*
arctanh(c*x))*ln(2/(-c*x+1))-b^2*c^3*d^2*ln(-c^2*x^2+1)-4/3*b^2*c^3*d^2*polylog(2,-c*x)+4/3*b^2*c^3*d^2*polylo
g(2,c*x)+4/3*b^2*c^3*d^2*polylog(2,1-2/(-c*x+1))

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 13, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {37, 5938, 5916, 325, 206, 266, 36, 29, 31, 5912, 5918, 2402, 2315} \[ -\frac {4}{3} b^2 c^3 d^2 \text {PolyLog}(2,-c x)+\frac {4}{3} b^2 c^3 d^2 \text {PolyLog}(2,c x)+\frac {4}{3} b^2 c^3 d^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )+\frac {8}{3} a b c^3 d^2 \log (x)-\frac {2 b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {8}{3} b c^3 d^2 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac {b c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac {d^2 (c x+1)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-b^2 c^3 d^2 \log \left (1-c^2 x^2\right )-\frac {b^2 c^2 d^2}{3 x}+2 b^2 c^3 d^2 \log (x)+\frac {1}{3} b^2 c^3 d^2 \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^2*(a + b*ArcTanh[c*x])^2)/x^4,x]

[Out]

-(b^2*c^2*d^2)/(3*x) + (b^2*c^3*d^2*ArcTanh[c*x])/3 - (b*c*d^2*(a + b*ArcTanh[c*x]))/(3*x^2) - (2*b*c^2*d^2*(a
+ b*ArcTanh[c*x]))/x - (d^2*(1 + c*x)^3*(a + b*ArcTanh[c*x])^2)/(3*x^3) + (8*a*b*c^3*d^2*Log[x])/3 + 2*b^2*c^
3*d^2*Log[x] + (8*b*c^3*d^2*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/3 - b^2*c^3*d^2*Log[1 - c^2*x^2] - (4*b^2*c
^3*d^2*PolyLog[2, -(c*x)])/3 + (4*b^2*c^3*d^2*PolyLog[2, c*x])/3 + (4*b^2*c^3*d^2*PolyLog[2, 1 - 2/(1 - c*x)])
/3

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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 5912

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b*PolyLog[2, -(c*x)])/2
, x] + Simp[(b*PolyLog[2, c*x])/2, x]) /; FreeQ[{a, b, c}, 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 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[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 5938

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_))^(q_), x_Symbol] :> With[{
u = IntHide[(f*x)^m*(d + e*x)^q, x]}, Dist[(a + b*ArcTanh[c*x])^p, u, x] - Dist[b*c*p, Int[ExpandIntegrand[(a
+ b*ArcTanh[c*x])^(p - 1), u/(1 - c^2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && IGtQ[p, 1] && Eq
Q[c^2*d^2 - e^2, 0] && IntegersQ[m, q] && NeQ[m, -1] && NeQ[q, -1] && ILtQ[m + q + 1, 0] && LtQ[m*q, 0]

Rubi steps

\begin {align*} \int \frac {(d+c d x)^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{x^4} \, dx &=-\frac {d^2 (1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-(2 b c) \int \left (-\frac {d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^3}-\frac {c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac {4 c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x}+\frac {4 c^3 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 (-1+c x)}\right ) \, dx\\ &=-\frac {d^2 (1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}+\frac {1}{3} \left (2 b c d^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^3} \, dx+\left (2 b c^2 d^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx+\frac {1}{3} \left (8 b c^3 d^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx-\frac {1}{3} \left (8 b c^4 d^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{-1+c x} \, dx\\ &=-\frac {b c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac {2 b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac {d^2 (1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}+\frac {8}{3} a b c^3 d^2 \log (x)+\frac {8}{3} b c^3 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )-\frac {4}{3} b^2 c^3 d^2 \text {Li}_2(-c x)+\frac {4}{3} b^2 c^3 d^2 \text {Li}_2(c x)+\frac {1}{3} \left (b^2 c^2 d^2\right ) \int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx+\left (2 b^2 c^3 d^2\right ) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx-\frac {1}{3} \left (8 b^2 c^4 d^2\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac {b^2 c^2 d^2}{3 x}-\frac {b c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac {2 b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac {d^2 (1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}+\frac {8}{3} a b c^3 d^2 \log (x)+\frac {8}{3} b c^3 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )-\frac {4}{3} b^2 c^3 d^2 \text {Li}_2(-c x)+\frac {4}{3} b^2 c^3 d^2 \text {Li}_2(c x)+\left (b^2 c^3 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )+\frac {1}{3} \left (8 b^2 c^3 d^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )+\frac {1}{3} \left (b^2 c^4 d^2\right ) \int \frac {1}{1-c^2 x^2} \, dx\\ &=-\frac {b^2 c^2 d^2}{3 x}+\frac {1}{3} b^2 c^3 d^2 \tanh ^{-1}(c x)-\frac {b c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac {2 b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac {d^2 (1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}+\frac {8}{3} a b c^3 d^2 \log (x)+\frac {8}{3} b c^3 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )-\frac {4}{3} b^2 c^3 d^2 \text {Li}_2(-c x)+\frac {4}{3} b^2 c^3 d^2 \text {Li}_2(c x)+\frac {4}{3} b^2 c^3 d^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )+\left (b^2 c^3 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\left (b^2 c^5 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b^2 c^2 d^2}{3 x}+\frac {1}{3} b^2 c^3 d^2 \tanh ^{-1}(c x)-\frac {b c d^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac {2 b c^2 d^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac {d^2 (1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}+\frac {8}{3} a b c^3 d^2 \log (x)+2 b^2 c^3 d^2 \log (x)+\frac {8}{3} b c^3 d^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )-b^2 c^3 d^2 \log \left (1-c^2 x^2\right )-\frac {4}{3} b^2 c^3 d^2 \text {Li}_2(-c x)+\frac {4}{3} b^2 c^3 d^2 \text {Li}_2(c x)+\frac {4}{3} b^2 c^3 d^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.63, size = 270, normalized size = 1.11 \[ -\frac {d^2 \left (3 a^2 c^2 x^2+3 a^2 c x+a^2-8 a b c^3 x^3 \log (c x)+3 a b c^3 x^3 \log (1-c x)-3 a b c^3 x^3 \log (c x+1)+6 a b c^2 x^2+4 a b c^3 x^3 \log \left (1-c^2 x^2\right )+b \tanh ^{-1}(c x) \left (a \left (6 c^2 x^2+6 c x+2\right )-8 b c^3 x^3 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+b c x \left (-c^2 x^2+6 c x+1\right )\right )+a b c x+4 b^2 c^3 x^3 \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )+b^2 c^2 x^2-6 b^2 c^3 x^3 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )+b^2 \left (-7 c^3 x^3+3 c^2 x^2+3 c x+1\right ) \tanh ^{-1}(c x)^2\right )}{3 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + c*d*x)^2*(a + b*ArcTanh[c*x])^2)/x^4,x]

[Out]

-1/3*(d^2*(a^2 + 3*a^2*c*x + a*b*c*x + 3*a^2*c^2*x^2 + 6*a*b*c^2*x^2 + b^2*c^2*x^2 + b^2*(1 + 3*c*x + 3*c^2*x^
2 - 7*c^3*x^3)*ArcTanh[c*x]^2 + b*ArcTanh[c*x]*(b*c*x*(1 + 6*c*x - c^2*x^2) + a*(2 + 6*c*x + 6*c^2*x^2) - 8*b*
c^3*x^3*Log[1 - E^(-2*ArcTanh[c*x])]) - 8*a*b*c^3*x^3*Log[c*x] + 3*a*b*c^3*x^3*Log[1 - c*x] - 3*a*b*c^3*x^3*Lo
g[1 + c*x] - 6*b^2*c^3*x^3*Log[(c*x)/Sqrt[1 - c^2*x^2]] + 4*a*b*c^3*x^3*Log[1 - c^2*x^2] + 4*b^2*c^3*x^3*PolyL
og[2, E^(-2*ArcTanh[c*x])]))/x^3

________________________________________________________________________________________

fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} c^{2} d^{2} x^{2} + 2 \, a^{2} c d^{2} x + a^{2} d^{2} + {\left (b^{2} c^{2} d^{2} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} d^{2}\right )} \operatorname {artanh}\left (c x\right )^{2} + 2 \, {\left (a b c^{2} d^{2} x^{2} + 2 \, a b c d^{2} x + a b d^{2}\right )} \operatorname {artanh}\left (c x\right )}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^2*(a+b*arctanh(c*x))^2/x^4,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^4, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^2*(a+b*arctanh(c*x))^2/x^4,x, algorithm="giac")

[Out]

integrate((c*d*x + d)^2*(b*arctanh(c*x) + a)^2/x^4, x)

________________________________________________________________________________________

maple [B]  time = 0.08, size = 550, normalized size = 2.25 \[ -\frac {2 c \,d^{2} a b \arctanh \left (c x \right )}{x^{2}}-\frac {2 c^{2} d^{2} a b \arctanh \left (c x \right )}{x}-\frac {5 c^{3} d^{2} b^{2} \ln \left (c x +1\right )}{6}-\frac {7 c^{3} d^{2} b^{2} \ln \left (c x -1\right )^{2}}{12}-\frac {c^{2} d^{2} a^{2}}{x}-\frac {4 c^{3} d^{2} b^{2} \dilog \left (c x \right )}{3}-\frac {4 c^{3} d^{2} b^{2} \dilog \left (c x +1\right )}{3}-\frac {d^{2} b^{2} \arctanh \left (c x \right )^{2}}{3 x^{3}}+\frac {4 c^{3} d^{2} b^{2} \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{3}-\frac {c \,d^{2} a^{2}}{x^{2}}+2 c^{3} d^{2} b^{2} \ln \left (c x \right )+\frac {c^{3} d^{2} b^{2} \ln \left (c x +1\right )^{2}}{12}-\frac {7 c^{3} d^{2} b^{2} \ln \left (c x -1\right )}{6}-\frac {2 c^{2} d^{2} a b}{x}-\frac {b^{2} c^{2} d^{2}}{3 x}-\frac {d^{2} a^{2}}{3 x^{3}}-\frac {c^{3} d^{2} b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{3}+\frac {8 c^{3} d^{2} b^{2} \arctanh \left (c x \right ) \ln \left (c x \right )}{3}-\frac {7 c^{3} d^{2} b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{3}-\frac {2 d^{2} a b \arctanh \left (c x \right )}{3 x^{3}}-\frac {c^{2} d^{2} b^{2} \arctanh \left (c x \right )^{2}}{x}+\frac {7 c^{3} d^{2} b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{6}-\frac {c^{3} d^{2} b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{6}+\frac {c^{3} d^{2} b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{6}+\frac {8 c^{3} d^{2} a b \ln \left (c x \right )}{3}-\frac {7 c^{3} d^{2} a b \ln \left (c x -1\right )}{3}-\frac {c^{3} d^{2} a b \ln \left (c x +1\right )}{3}-\frac {4 c^{3} d^{2} b^{2} \ln \left (c x \right ) \ln \left (c x +1\right )}{3}-\frac {2 c^{2} d^{2} b^{2} \arctanh \left (c x \right )}{x}-\frac {c \,d^{2} b^{2} \arctanh \left (c x \right )^{2}}{x^{2}}-\frac {c \,d^{2} b^{2} \arctanh \left (c x \right )}{3 x^{2}}-\frac {c \,d^{2} a b}{3 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^2*(a+b*arctanh(c*x))^2/x^4,x)

[Out]

-2*c*d^2*a*b*arctanh(c*x)/x^2-2*c^2*d^2*a*b*arctanh(c*x)/x-4/3*c^3*d^2*b^2*dilog(c*x)-4/3*c^3*d^2*b^2*dilog(c*
x+1)-5/6*c^3*d^2*b^2*ln(c*x+1)-7/12*c^3*d^2*b^2*ln(c*x-1)^2-1/3*d^2*b^2*arctanh(c*x)^2/x^3-c^2*d^2*a^2/x-c*d^2
*a^2/x^2+2*c^3*d^2*b^2*ln(c*x)+4/3*c^3*d^2*b^2*dilog(1/2+1/2*c*x)+1/12*c^3*d^2*b^2*ln(c*x+1)^2-7/6*c^3*d^2*b^2
*ln(c*x-1)-2*c^2*d^2*a*b/x-1/3*b^2*c^2*d^2/x-1/3*d^2*a^2/x^3-2*c^2*d^2*b^2*arctanh(c*x)/x-c*d^2*b^2*arctanh(c*
x)^2/x^2-1/3*c*d^2*b^2*arctanh(c*x)/x^2-1/3*c^3*d^2*b^2*arctanh(c*x)*ln(c*x+1)+7/6*c^3*d^2*b^2*ln(c*x-1)*ln(1/
2+1/2*c*x)-1/6*c^3*d^2*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/6*c^3*d^2*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+8/3*c^3
*d^2*a*b*ln(c*x)-7/3*c^3*d^2*a*b*ln(c*x-1)-1/3*c^3*d^2*a*b*ln(c*x+1)-4/3*c^3*d^2*b^2*ln(c*x)*ln(c*x+1)+8/3*c^3
*d^2*b^2*arctanh(c*x)*ln(c*x)-7/3*c^3*d^2*b^2*arctanh(c*x)*ln(c*x-1)-2/3*d^2*a*b*arctanh(c*x)/x^3-1/3*c*d^2*a*
b/x^2-c^2*d^2*b^2*arctanh(c*x)^2/x

________________________________________________________________________________________

maxima [B]  time = 0.94, size = 555, normalized size = 2.27 \[ -\frac {4}{3} \, {\left (\log \left (c x + 1\right ) \log \left (-\frac {1}{2} \, c x + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, c x + \frac {1}{2}\right )\right )} b^{2} c^{3} d^{2} - \frac {4}{3} \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} b^{2} c^{3} d^{2} + \frac {4}{3} \, {\left (\log \left (c x + 1\right ) \log \left (-c x\right ) + {\rm Li}_2\left (c x + 1\right )\right )} b^{2} c^{3} d^{2} - \frac {5}{6} \, b^{2} c^{3} d^{2} \log \left (c x + 1\right ) - \frac {7}{6} \, b^{2} c^{3} d^{2} \log \left (c x - 1\right ) + 2 \, b^{2} c^{3} d^{2} \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 c^{2} d^{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 c d^{2} - \frac {1}{3} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} a b d^{2} - \frac {a^{2} c^{2} d^{2}}{x} - \frac {a^{2} c d^{2}}{x^{2}} - \frac {a^{2} d^{2}}{3 \, x^{3}} - \frac {4 \, b^{2} c^{2} d^{2} x^{2} + {\left (b^{2} c^{3} d^{2} x^{3} + 3 \, b^{2} c^{2} d^{2} x^{2} + 3 \, b^{2} c d^{2} x + b^{2} d^{2}\right )} \log \left (c x + 1\right )^{2} - {\left (7 \, b^{2} c^{3} d^{2} x^{3} - 3 \, b^{2} c^{2} d^{2} x^{2} - 3 \, b^{2} c d^{2} x - b^{2} d^{2}\right )} \log \left (-c x + 1\right )^{2} + 2 \, {\left (6 \, b^{2} c^{2} d^{2} x^{2} + b^{2} c d^{2} x\right )} \log \left (c x + 1\right ) - 2 \, {\left (6 \, b^{2} c^{2} d^{2} x^{2} + b^{2} c d^{2} x + {\left (b^{2} c^{3} d^{2} x^{3} + 3 \, b^{2} c^{2} d^{2} x^{2} + 3 \, b^{2} c d^{2} x + b^{2} d^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^2*(a+b*arctanh(c*x))^2/x^4,x, algorithm="maxima")

[Out]

-4/3*(log(c*x + 1)*log(-1/2*c*x + 1/2) + dilog(1/2*c*x + 1/2))*b^2*c^3*d^2 - 4/3*(log(c*x)*log(-c*x + 1) + dil
og(-c*x + 1))*b^2*c^3*d^2 + 4/3*(log(c*x + 1)*log(-c*x) + dilog(c*x + 1))*b^2*c^3*d^2 - 5/6*b^2*c^3*d^2*log(c*
x + 1) - 7/6*b^2*c^3*d^2*log(c*x - 1) + 2*b^2*c^3*d^2*log(x) - (c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*
x)/x)*a*b*c^2*d^2 + ((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*a*b*c*d^2 - 1/3*((c^2*log
(c^2*x^2 - 1) - c^2*log(x^2) + 1/x^2)*c + 2*arctanh(c*x)/x^3)*a*b*d^2 - a^2*c^2*d^2/x - a^2*c*d^2/x^2 - 1/3*a^
2*d^2/x^3 - 1/12*(4*b^2*c^2*d^2*x^2 + (b^2*c^3*d^2*x^3 + 3*b^2*c^2*d^2*x^2 + 3*b^2*c*d^2*x + b^2*d^2)*log(c*x
+ 1)^2 - (7*b^2*c^3*d^2*x^3 - 3*b^2*c^2*d^2*x^2 - 3*b^2*c*d^2*x - b^2*d^2)*log(-c*x + 1)^2 + 2*(6*b^2*c^2*d^2*
x^2 + b^2*c*d^2*x)*log(c*x + 1) - 2*(6*b^2*c^2*d^2*x^2 + b^2*c*d^2*x + (b^2*c^3*d^2*x^3 + 3*b^2*c^2*d^2*x^2 +
3*b^2*c*d^2*x + b^2*d^2)*log(c*x + 1))*log(-c*x + 1))/x^3

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atanh(c*x))^2*(d + c*d*x)^2)/x^4,x)

[Out]

int(((a + b*atanh(c*x))^2*(d + c*d*x)^2)/x^4, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________