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

Optimal. Leaf size=206 \[ -\frac{1}{3} b^2 c^3 d \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )+\frac{5}{6} c^3 d \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{b c^2 d \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac{2}{3} b c^3 d \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac{c d \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac{b c d \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-\frac{1}{2} b^2 c^3 d \log \left (1-c^2 x^2\right )-\frac{b^2 c^2 d}{3 x}+b^2 c^3 d \log (x)+\frac{1}{3} b^2 c^3 d \tanh ^{-1}(c x) \]

[Out]

-(b^2*c^2*d)/(3*x) + (b^2*c^3*d*ArcTanh[c*x])/3 - (b*c*d*(a + b*ArcTanh[c*x]))/(3*x^2) - (b*c^2*d*(a + b*ArcTa
nh[c*x]))/x + (5*c^3*d*(a + b*ArcTanh[c*x])^2)/6 - (d*(a + b*ArcTanh[c*x])^2)/(3*x^3) - (c*d*(a + b*ArcTanh[c*
x])^2)/(2*x^2) + b^2*c^3*d*Log[x] - (b^2*c^3*d*Log[1 - c^2*x^2])/2 + (2*b*c^3*d*(a + b*ArcTanh[c*x])*Log[2 - 2
/(1 + c*x)])/3 - (b^2*c^3*d*PolyLog[2, -1 + 2/(1 + c*x)])/3

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Rubi [A]  time = 0.453118, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 13, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.65, Rules used = {5940, 5916, 5982, 325, 206, 5988, 5932, 2447, 266, 36, 29, 31, 5948} \[ -\frac{1}{3} b^2 c^3 d \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right )+\frac{5}{6} c^3 d \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{b c^2 d \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac{2}{3} b c^3 d \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac{c d \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac{b c d \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-\frac{1}{2} b^2 c^3 d \log \left (1-c^2 x^2\right )-\frac{b^2 c^2 d}{3 x}+b^2 c^3 d \log (x)+\frac{1}{3} b^2 c^3 d \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*c^2*d)/(3*x) + (b^2*c^3*d*ArcTanh[c*x])/3 - (b*c*d*(a + b*ArcTanh[c*x]))/(3*x^2) - (b*c^2*d*(a + b*ArcTa
nh[c*x]))/x + (5*c^3*d*(a + b*ArcTanh[c*x])^2)/6 - (d*(a + b*ArcTanh[c*x])^2)/(3*x^3) - (c*d*(a + b*ArcTanh[c*
x])^2)/(2*x^2) + b^2*c^3*d*Log[x] - (b^2*c^3*d*Log[1 - c^2*x^2])/2 + (2*b*c^3*d*(a + b*ArcTanh[c*x])*Log[2 - 2
/(1 + c*x)])/3 - (b^2*c^3*d*PolyLog[2, -1 + 2/(1 + c*x)])/3

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 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 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 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 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 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 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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{(d+c d x) \left (a+b \tanh ^{-1}(c x)\right )^2}{x^4} \, dx &=\int \left (\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{x^4}+\frac{c d \left (a+b \tanh ^{-1}(c x)\right )^2}{x^3}\right ) \, dx\\ &=d \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^4} \, dx+(c d) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x^3} \, dx\\ &=-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-\frac{c d \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}+\frac{1}{3} (2 b c d) \int \frac{a+b \tanh ^{-1}(c x)}{x^3 \left (1-c^2 x^2\right )} \, dx+\left (b c^2 d\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^2 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-\frac{c d \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}+\frac{1}{3} (2 b c d) \int \frac{a+b \tanh ^{-1}(c x)}{x^3} \, dx+\left (b c^2 d\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^2} \, dx+\frac{1}{3} \left (2 b c^3 d\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx+\left (b c^4 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 )}{3 x^2}-\frac{b c^2 d \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac{5}{6} c^3 d \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-\frac{c d \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}+\frac{1}{3} \left (b^2 c^2 d\right ) \int \frac{1}{x^2 \left (1-c^2 x^2\right )} \, dx+\frac{1}{3} \left (2 b c^3 d\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx+\left (b^2 c^3 d\right ) \int \frac{1}{x \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{b^2 c^2 d}{3 x}-\frac{b c d \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac{b c^2 d \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac{5}{6} c^3 d \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-\frac{c d \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}+\frac{2}{3} b c^3 d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )+\frac{1}{2} \left (b^2 c^3 d\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )+\frac{1}{3} \left (b^2 c^4 d\right ) \int \frac{1}{1-c^2 x^2} \, dx-\frac{1}{3} \left (2 b^2 c^4 d\right ) \int \frac{\log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac{b^2 c^2 d}{3 x}+\frac{1}{3} b^2 c^3 d \tanh ^{-1}(c x)-\frac{b c d \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac{b c^2 d \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac{5}{6} c^3 d \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-\frac{c d \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}+\frac{2}{3} b c^3 d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )-\frac{1}{3} b^2 c^3 d \text{Li}_2\left (-1+\frac{2}{1+c x}\right )+\frac{1}{2} \left (b^2 c^3 d\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{2} \left (b^2 c^5 d\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac{b^2 c^2 d}{3 x}+\frac{1}{3} b^2 c^3 d \tanh ^{-1}(c x)-\frac{b c d \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac{b c^2 d \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac{5}{6} c^3 d \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-\frac{c d \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}+b^2 c^3 d \log (x)-\frac{1}{2} b^2 c^3 d \log \left (1-c^2 x^2\right )+\frac{2}{3} b c^3 d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+c x}\right )-\frac{1}{3} b^2 c^3 d \text{Li}_2\left (-1+\frac{2}{1+c x}\right )\\ \end{align*}

Mathematica [A]  time = 0.472233, size = 246, normalized size = 1.19 \[ -\frac{d \left (2 b^2 c^3 x^3 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+3 a^2 c x+2 a^2+6 a b c^2 x^2-4 a b c^3 x^3 \log (c x)+3 a b c^3 x^3 \log (1-c x)-3 a b c^3 x^3 \log (c x+1)+2 a b c^3 x^3 \log \left (1-c^2 x^2\right )+2 b \tanh ^{-1}(c x) \left (a (3 c x+2)+b c x \left (-c^2 x^2+3 c x+1\right )-2 b c^3 x^3 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )+2 a b c x+2 b^2 c^2 x^2-6 b^2 c^3 x^3 \log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )+b^2 \left (-5 c^3 x^3+3 c x+2\right ) \tanh ^{-1}(c x)^2\right )}{6 x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(d*(2*a^2 + 3*a^2*c*x + 2*a*b*c*x + 6*a*b*c^2*x^2 + 2*b^2*c^2*x^2 + b^2*(2 + 3*c*x - 5*c^3*x^3)*ArcTanh[c*x]^
2 + 2*b*ArcTanh[c*x]*(a*(2 + 3*c*x) + b*c*x*(1 + 3*c*x - c^2*x^2) - 2*b*c^3*x^3*Log[1 - E^(-2*ArcTanh[c*x])])
- 4*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[1 + c*x] - 6*b^2*c^3*x^3*Log[(c*x)/S
qrt[1 - c^2*x^2]] + 2*a*b*c^3*x^3*Log[1 - c^2*x^2] + 2*b^2*c^3*x^3*PolyLog[2, E^(-2*ArcTanh[c*x])]))/(6*x^3)

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Maple [B]  time = 0.07, size = 440, normalized size = 2.1 \begin{align*}{\frac{2\,{c}^{3}d{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx \right ) }{3}}-{\frac{cd{b}^{2}{\it Artanh} \left ( cx \right ) }{3\,{x}^{2}}}+{\frac{2\,{c}^{3}dab\ln \left ( cx \right ) }{3}}+{\frac{{c}^{3}dab\ln \left ( cx+1 \right ) }{6}}+{\frac{5\,{c}^{3}d{b}^{2}\ln \left ( cx-1 \right ) }{12}\ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{{c}^{3}d{b}^{2}}{12}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{{c}^{3}d{b}^{2}\ln \left ( cx+1 \right ) }{12}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) }-{\frac{{c}^{3}d{b}^{2}\ln \left ( cx \right ) \ln \left ( cx+1 \right ) }{3}}-{\frac{5\,{c}^{3}d{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx-1 \right ) }{6}}+{\frac{{c}^{3}d{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) }{6}}-{\frac{{c}^{2}d{b}^{2}{\it Artanh} \left ( cx \right ) }{x}}-{\frac{cd{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{2\,{x}^{2}}}-{\frac{2\,dab{\it Artanh} \left ( cx \right ) }{3\,{x}^{3}}}-{\frac{cdab}{3\,{x}^{2}}}-{\frac{a{c}^{2}db}{x}}-{\frac{{c}^{2}d{b}^{2}}{3\,x}}-{\frac{5\,{c}^{3}dab\ln \left ( cx-1 \right ) }{6}}-{\frac{{a}^{2}d}{3\,{x}^{3}}}-{\frac{cdab{\it Artanh} \left ( cx \right ) }{{x}^{2}}}-{\frac{5\,{c}^{3}d{b}^{2} \left ( \ln \left ( cx-1 \right ) \right ) ^{2}}{24}}-{\frac{d{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{3\,{x}^{3}}}+{c}^{3}d{b}^{2}\ln \left ( cx \right ) +{\frac{{c}^{3}d{b}^{2}}{3}{\it dilog} \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{{c}^{3}d{b}^{2} \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{24}}-{\frac{{c}^{3}d{b}^{2}{\it dilog} \left ( cx \right ) }{3}}-{\frac{2\,{c}^{3}d{b}^{2}\ln \left ( cx-1 \right ) }{3}}-{\frac{{a}^{2}cd}{2\,{x}^{2}}}-{\frac{{c}^{3}d{b}^{2}\ln \left ( cx+1 \right ) }{3}}-{\frac{{c}^{3}d{b}^{2}{\it dilog} \left ( cx+1 \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*d*a*b*arctanh(c*x)/x^2+2/3*c^3*d*a*b*ln(c*x)+1/6*c^3*d*a*b*ln(c*x+1)+2/3*c^3*d*b^2*arctanh(c*x)*ln(c*x)-5/6
*c^3*d*b^2*arctanh(c*x)*ln(c*x-1)+1/6*c^3*d*b^2*arctanh(c*x)*ln(c*x+1)+5/12*c^3*d*b^2*ln(c*x-1)*ln(1/2+1/2*c*x
)-1/12*c^3*d*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+1/12*c^3*d*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)-1/3*c^3*d*b^2*ln(c
*x)*ln(c*x+1)-c^2*d*b^2*arctanh(c*x)/x-1/2*c*d*b^2*arctanh(c*x)^2/x^2-1/3*c*d*a*b/x^2-c^2*d*a*b/x-2/3*d*a*b*ar
ctanh(c*x)/x^3-1/3*c*d*b^2*arctanh(c*x)/x^2-1/3*b^2*c^2*d/x-5/6*c^3*d*a*b*ln(c*x-1)-1/3*a^2*d/x^3-5/24*c^3*d*b
^2*ln(c*x-1)^2-1/3*d*b^2*arctanh(c*x)^2/x^3+c^3*d*b^2*ln(c*x)+1/3*c^3*d*b^2*dilog(1/2+1/2*c*x)-1/24*c^3*d*b^2*
ln(c*x+1)^2-1/3*c^3*d*b^2*dilog(c*x)-2/3*c^3*d*b^2*ln(c*x-1)-1/2*c*a^2*d/x^2-1/3*c^3*d*b^2*ln(c*x+1)-1/3*c^3*d
*b^2*dilog(c*x+1)

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Maxima [B]  time = 3.06758, size = 563, normalized size = 2.73 \begin{align*} -\frac{1}{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 - \frac{1}{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 + \frac{1}{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 - \frac{1}{3} \, b^{2} c^{3} d \log \left (c x + 1\right ) - \frac{2}{3} \, b^{2} c^{3} d \log \left (c x - 1\right ) + b^{2} c^{3} d \log \left (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 c d - \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 - \frac{a^{2} c d}{2 \, x^{2}} - \frac{a^{2} d}{3 \, x^{3}} - \frac{8 \, b^{2} c^{2} d x^{2} -{\left (b^{2} c^{3} d x^{3} - 3 \, b^{2} c d x - 2 \, b^{2} d\right )} \log \left (c x + 1\right )^{2} -{\left (5 \, b^{2} c^{3} d x^{3} - 3 \, b^{2} c d x - 2 \, b^{2} d\right )} \log \left (-c x + 1\right )^{2} + 4 \,{\left (3 \, b^{2} c^{2} d x^{2} + b^{2} c d x\right )} \log \left (c x + 1\right ) - 2 \,{\left (6 \, b^{2} c^{2} d x^{2} + 2 \, b^{2} c d x -{\left (b^{2} c^{3} d x^{3} - 3 \, b^{2} c d x - 2 \, b^{2} d\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{24 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(log(c*x + 1)*log(-1/2*c*x + 1/2) + dilog(1/2*c*x + 1/2))*b^2*c^3*d - 1/3*(log(c*x)*log(-c*x + 1) + dilog
(-c*x + 1))*b^2*c^3*d + 1/3*(log(c*x + 1)*log(-c*x) + dilog(c*x + 1))*b^2*c^3*d - 1/3*b^2*c^3*d*log(c*x + 1) -
2/3*b^2*c^3*d*log(c*x - 1) + b^2*c^3*d*log(x) + 1/2*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*
x)/x^2)*a*b*c*d - 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 - 1/2*a^2*c
*d/x^2 - 1/3*a^2*d/x^3 - 1/24*(8*b^2*c^2*d*x^2 - (b^2*c^3*d*x^3 - 3*b^2*c*d*x - 2*b^2*d)*log(c*x + 1)^2 - (5*b
^2*c^3*d*x^3 - 3*b^2*c*d*x - 2*b^2*d)*log(-c*x + 1)^2 + 4*(3*b^2*c^2*d*x^2 + b^2*c*d*x)*log(c*x + 1) - 2*(6*b^
2*c^2*d*x^2 + 2*b^2*c*d*x - (b^2*c^3*d*x^3 - 3*b^2*c*d*x - 2*b^2*d)*log(c*x + 1))*log(-c*x + 1))/x^3

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d \left (\int \frac{a^{2}}{x^{4}}\, dx + \int \frac{a^{2} c}{x^{3}}\, 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{b^{2} c \operatorname{atanh}^{2}{\left (c x \right )}}{x^{3}}\, dx + \int \frac{2 a b c \operatorname{atanh}{\left (c x \right )}}{x^{3}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c d x + d\right )}{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{2}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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