3.92 \(\int \frac{(d+c d x)^3 (a+b \tanh ^{-1}(c x))^2}{x^5} \, dx\)

Optimal. Leaf size=271 \[ -2 b^2 c^4 d^3 \text{PolyLog}(2,-c x)+2 b^2 c^4 d^3 \text{PolyLog}(2,c x)+2 b^2 c^4 d^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )-\frac{b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}+4 a b c^4 d^3 \log (x)-\frac{7 b c^3 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x}+4 b c^4 d^3 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac{b c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 x^3}-\frac{d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}-\frac{b^2 c^2 d^3}{12 x^2}-\frac{11}{6} b^2 c^4 d^3 \log \left (1-c^2 x^2\right )-\frac{b^2 c^3 d^3}{x}+\frac{11}{3} b^2 c^4 d^3 \log (x)+b^2 c^4 d^3 \tanh ^{-1}(c x) \]

[Out]

-(b^2*c^2*d^3)/(12*x^2) - (b^2*c^3*d^3)/x + b^2*c^4*d^3*ArcTanh[c*x] - (b*c*d^3*(a + b*ArcTanh[c*x]))/(6*x^3)
- (b*c^2*d^3*(a + b*ArcTanh[c*x]))/x^2 - (7*b*c^3*d^3*(a + b*ArcTanh[c*x]))/(2*x) - (d^3*(1 + c*x)^4*(a + b*Ar
cTanh[c*x])^2)/(4*x^4) + 4*a*b*c^4*d^3*Log[x] + (11*b^2*c^4*d^3*Log[x])/3 + 4*b*c^4*d^3*(a + b*ArcTanh[c*x])*L
og[2/(1 - c*x)] - (11*b^2*c^4*d^3*Log[1 - c^2*x^2])/6 - 2*b^2*c^4*d^3*PolyLog[2, -(c*x)] + 2*b^2*c^4*d^3*PolyL
og[2, c*x] + 2*b^2*c^4*d^3*PolyLog[2, 1 - 2/(1 - c*x)]

________________________________________________________________________________________

Rubi [A]  time = 0.307835, antiderivative size = 271, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 14, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.636, Rules used = {37, 5938, 5916, 266, 44, 325, 206, 36, 29, 31, 5912, 5918, 2402, 2315} \[ -2 b^2 c^4 d^3 \text{PolyLog}(2,-c x)+2 b^2 c^4 d^3 \text{PolyLog}(2,c x)+2 b^2 c^4 d^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )-\frac{b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}+4 a b c^4 d^3 \log (x)-\frac{7 b c^3 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x}+4 b c^4 d^3 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac{b c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 x^3}-\frac{d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}-\frac{b^2 c^2 d^3}{12 x^2}-\frac{11}{6} b^2 c^4 d^3 \log \left (1-c^2 x^2\right )-\frac{b^2 c^3 d^3}{x}+\frac{11}{3} b^2 c^4 d^3 \log (x)+b^2 c^4 d^3 \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)^3*(a + b*ArcTanh[c*x])^2)/x^5,x]

[Out]

-(b^2*c^2*d^3)/(12*x^2) - (b^2*c^3*d^3)/x + b^2*c^4*d^3*ArcTanh[c*x] - (b*c*d^3*(a + b*ArcTanh[c*x]))/(6*x^3)
- (b*c^2*d^3*(a + b*ArcTanh[c*x]))/x^2 - (7*b*c^3*d^3*(a + b*ArcTanh[c*x]))/(2*x) - (d^3*(1 + c*x)^4*(a + b*Ar
cTanh[c*x])^2)/(4*x^4) + 4*a*b*c^4*d^3*Log[x] + (11*b^2*c^4*d^3*Log[x])/3 + 4*b*c^4*d^3*(a + b*ArcTanh[c*x])*L
og[2/(1 - c*x)] - (11*b^2*c^4*d^3*Log[1 - c^2*x^2])/6 - 2*b^2*c^4*d^3*PolyLog[2, -(c*x)] + 2*b^2*c^4*d^3*PolyL
og[2, c*x] + 2*b^2*c^4*d^3*PolyLog[2, 1 - 2/(1 - c*x)]

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 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]

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 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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{(d+c d x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x^5} \, dx &=-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}-(2 b c) \int \left (-\frac{d^3 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^4}-\frac{c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^3}-\frac{7 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^2}-\frac{2 c^3 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac{2 c^4 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{-1+c x}\right ) \, dx\\ &=-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{2} \left (b c d^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^4} \, dx+\left (2 b c^2 d^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^3} \, dx+\frac{1}{2} \left (7 b c^3 d^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^2} \, dx+\left (4 b c^4 d^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x} \, dx-\left (4 b c^5 d^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{-1+c x} \, dx\\ &=-\frac{b c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 x^3}-\frac{b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac{7 b c^3 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x}-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}+4 a b c^4 d^3 \log (x)+4 b c^4 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )-2 b^2 c^4 d^3 \text{Li}_2(-c x)+2 b^2 c^4 d^3 \text{Li}_2(c x)+\frac{1}{6} \left (b^2 c^2 d^3\right ) \int \frac{1}{x^3 \left (1-c^2 x^2\right )} \, dx+\left (b^2 c^3 d^3\right ) \int \frac{1}{x^2 \left (1-c^2 x^2\right )} \, dx+\frac{1}{2} \left (7 b^2 c^4 d^3\right ) \int \frac{1}{x \left (1-c^2 x^2\right )} \, dx-\left (4 b^2 c^5 d^3\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac{b^2 c^3 d^3}{x}-\frac{b c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 x^3}-\frac{b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac{7 b c^3 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x}-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}+4 a b c^4 d^3 \log (x)+4 b c^4 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )-2 b^2 c^4 d^3 \text{Li}_2(-c x)+2 b^2 c^4 d^3 \text{Li}_2(c x)+\frac{1}{12} \left (b^2 c^2 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-c^2 x\right )} \, dx,x,x^2\right )+\frac{1}{4} \left (7 b^2 c^4 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )+\left (4 b^2 c^4 d^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )+\left (b^2 c^5 d^3\right ) \int \frac{1}{1-c^2 x^2} \, dx\\ &=-\frac{b^2 c^3 d^3}{x}+b^2 c^4 d^3 \tanh ^{-1}(c x)-\frac{b c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 x^3}-\frac{b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac{7 b c^3 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x}-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}+4 a b c^4 d^3 \log (x)+4 b c^4 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )-2 b^2 c^4 d^3 \text{Li}_2(-c x)+2 b^2 c^4 d^3 \text{Li}_2(c x)+2 b^2 c^4 d^3 \text{Li}_2\left (1-\frac{2}{1-c x}\right )+\frac{1}{12} \left (b^2 c^2 d^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{x^2}+\frac{c^2}{x}-\frac{c^4}{-1+c^2 x}\right ) \, dx,x,x^2\right )+\frac{1}{4} \left (7 b^2 c^4 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{4} \left (7 b^2 c^6 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac{b^2 c^2 d^3}{12 x^2}-\frac{b^2 c^3 d^3}{x}+b^2 c^4 d^3 \tanh ^{-1}(c x)-\frac{b c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 x^3}-\frac{b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac{7 b c^3 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x}-\frac{d^3 (1+c x)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}+4 a b c^4 d^3 \log (x)+\frac{11}{3} b^2 c^4 d^3 \log (x)+4 b c^4 d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )-\frac{11}{6} b^2 c^4 d^3 \log \left (1-c^2 x^2\right )-2 b^2 c^4 d^3 \text{Li}_2(-c x)+2 b^2 c^4 d^3 \text{Li}_2(c x)+2 b^2 c^4 d^3 \text{Li}_2\left (1-\frac{2}{1-c x}\right )\\ \end{align*}

Mathematica [A]  time = 0.771332, size = 343, normalized size = 1.27 \[ -\frac{d^3 \left (24 b^2 c^4 x^4 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+12 a^2 c^3 x^3+18 a^2 c^2 x^2+12 a^2 c x+3 a^2+42 a b c^3 x^3+12 a b c^2 x^2-48 a b c^4 x^4 \log (c x)+21 a b c^4 x^4 \log (1-c x)-21 a b c^4 x^4 \log (c x+1)+24 a b c^4 x^4 \log \left (1-c^2 x^2\right )+2 b \tanh ^{-1}(c x) \left (3 a \left (4 c^3 x^3+6 c^2 x^2+4 c x+1\right )+b c x \left (-6 c^3 x^3+21 c^2 x^2+6 c x+1\right )-24 b c^4 x^4 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )+2 a b c x-b^2 c^4 x^4+12 b^2 c^3 x^3+b^2 c^2 x^2-44 b^2 c^4 x^4 \log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )+3 b^2 \left (-15 c^4 x^4+4 c^3 x^3+6 c^2 x^2+4 c x+1\right ) \tanh ^{-1}(c x)^2\right )}{12 x^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + c*d*x)^3*(a + b*ArcTanh[c*x])^2)/x^5,x]

[Out]

-(d^3*(3*a^2 + 12*a^2*c*x + 2*a*b*c*x + 18*a^2*c^2*x^2 + 12*a*b*c^2*x^2 + b^2*c^2*x^2 + 12*a^2*c^3*x^3 + 42*a*
b*c^3*x^3 + 12*b^2*c^3*x^3 - b^2*c^4*x^4 + 3*b^2*(1 + 4*c*x + 6*c^2*x^2 + 4*c^3*x^3 - 15*c^4*x^4)*ArcTanh[c*x]
^2 + 2*b*ArcTanh[c*x]*(b*c*x*(1 + 6*c*x + 21*c^2*x^2 - 6*c^3*x^3) + 3*a*(1 + 4*c*x + 6*c^2*x^2 + 4*c^3*x^3) -
24*b*c^4*x^4*Log[1 - E^(-2*ArcTanh[c*x])]) - 48*a*b*c^4*x^4*Log[c*x] + 21*a*b*c^4*x^4*Log[1 - c*x] - 21*a*b*c^
4*x^4*Log[1 + c*x] - 44*b^2*c^4*x^4*Log[(c*x)/Sqrt[1 - c^2*x^2]] + 24*a*b*c^4*x^4*Log[1 - c^2*x^2] + 24*b^2*c^
4*x^4*PolyLog[2, E^(-2*ArcTanh[c*x])]))/(12*x^4)

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Maple [B]  time = 0.075, size = 646, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^3*(a+b*arctanh(c*x))^2/x^5,x)

[Out]

-1/12*b^2*c^2*d^3/x^2-1/4*d^3*a^2/x^4-b^2*c^3*d^3/x-2*c^3*d^3*a*b*arctanh(c*x)/x-3*c^2*d^3*a*b*arctanh(c*x)/x^
2-2*c*d^3*a*b*arctanh(c*x)/x^3-15/4*c^4*d^3*a*b*ln(c*x-1)-1/4*c^4*d^3*a*b*ln(c*x+1)-3/2*c^2*d^3*b^2*arctanh(c*
x)^2/x^2-7/2*c^3*d^3*b^2*arctanh(c*x)/x-c^3*d^3*b^2*arctanh(c*x)^2/x-c*d^3*b^2*arctanh(c*x)^2/x^3-c^2*d^3*b^2*
arctanh(c*x)/x^2-1/6*c*d^3*b^2*arctanh(c*x)/x^3-1/2*d^3*a*b*arctanh(c*x)/x^4-1/6*c*d^3*a*b/x^3-7/2*c^3*d^3*a*b
/x-15/4*c^4*d^3*b^2*arctanh(c*x)*ln(c*x-1)-1/4*c^4*d^3*b^2*arctanh(c*x)*ln(c*x+1)-1/8*c^4*d^3*b^2*ln(-1/2*c*x+
1/2)*ln(c*x+1)+1/8*c^4*d^3*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-2*c^4*d^3*b^2*ln(c*x)*ln(c*x+1)+4*c^4*d^3*b^2*
arctanh(c*x)*ln(c*x)+15/8*c^4*d^3*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+4*c^4*d^3*a*b*ln(c*x)-2*c^4*d^3*b^2*dilog(c*x+
1)+1/16*c^4*d^3*b^2*ln(c*x+1)^2-4/3*c^4*d^3*b^2*ln(c*x+1)+11/3*c^4*d^3*b^2*ln(c*x)+2*c^4*d^3*b^2*dilog(1/2+1/2
*c*x)-7/3*c^4*d^3*b^2*ln(c*x-1)-15/16*c^4*d^3*b^2*ln(c*x-1)^2-2*c^4*d^3*b^2*dilog(c*x)-c^3*d^3*a^2/x-3/2*c^2*d
^3*a^2/x^2-c*d^3*a^2/x^3-1/4*d^3*b^2*arctanh(c*x)^2/x^4-c^2*d^3*a*b/x^2

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Maxima [B]  time = 3.10516, size = 1098, normalized size = 4.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^3*(a+b*arctanh(c*x))^2/x^5,x, algorithm="maxima")

[Out]

-2*(log(c*x + 1)*log(-1/2*c*x + 1/2) + dilog(1/2*c*x + 1/2))*b^2*c^4*d^3 - 2*(log(c*x)*log(-c*x + 1) + dilog(-
c*x + 1))*b^2*c^4*d^3 + 2*(log(c*x + 1)*log(-c*x) + dilog(c*x + 1))*b^2*c^4*d^3 - b^2*c^4*d^3*log(c*x + 1) - 2
*b^2*c^4*d^3*log(c*x - 1) + 3*b^2*c^4*d^3*log(x) - (c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*a*b*c^
3*d^3 + 3/2*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*a*b*c^2*d^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*c*d^3 - a^2*c^3*d^3/x + 1/12*((3*c^3*log(c*x + 1) - 3*
c^3*log(c*x - 1) - 2*(3*c^2*x^2 + 1)/x^3)*c - 6*arctanh(c*x)/x^4)*a*b*d^3 + 1/48*((32*c^2*log(x) - (3*c^2*x^2*
log(c*x + 1)^2 + 3*c^2*x^2*log(c*x - 1)^2 + 16*c^2*x^2*log(c*x - 1) - 2*(3*c^2*x^2*log(c*x - 1) - 8*c^2*x^2)*l
og(c*x + 1) + 4)/x^2)*c^2 + 4*(3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x^2 + 1)/x^3)*c*arctanh(c*x)
)*b^2*d^3 - 3/2*a^2*c^2*d^3/x^2 - a^2*c*d^3/x^3 - 1/4*b^2*d^3*arctanh(c*x)^2/x^4 - 1/4*a^2*d^3/x^4 - 1/8*(8*b^
2*c^3*d^3*x^2 + (b^2*c^4*d^3*x^3 + 2*b^2*c^3*d^3*x^2 + 3*b^2*c^2*d^3*x + 2*b^2*c*d^3)*log(c*x + 1)^2 - (7*b^2*
c^4*d^3*x^3 - 2*b^2*c^3*d^3*x^2 - 3*b^2*c^2*d^3*x - 2*b^2*c*d^3)*log(-c*x + 1)^2 + 4*(3*b^2*c^3*d^3*x^2 + b^2*
c^2*d^3*x)*log(c*x + 1) - 2*(6*b^2*c^3*d^3*x^2 + 2*b^2*c^2*d^3*x + (b^2*c^4*d^3*x^3 + 2*b^2*c^3*d^3*x^2 + 3*b^
2*c^2*d^3*x + 2*b^2*c*d^3)*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^{3} d^{3} x^{3} + 3 \, a^{2} c^{2} d^{3} x^{2} + 3 \, a^{2} c d^{3} x + a^{2} d^{3} +{\left (b^{2} c^{3} d^{3} x^{3} + 3 \, b^{2} c^{2} d^{3} x^{2} + 3 \, b^{2} c d^{3} x + b^{2} d^{3}\right )} \operatorname{artanh}\left (c x\right )^{2} + 2 \,{\left (a b c^{3} d^{3} x^{3} + 3 \, a b c^{2} d^{3} x^{2} + 3 \, a b c d^{3} x + a b d^{3}\right )} \operatorname{artanh}\left (c x\right )}{x^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^3*(a+b*arctanh(c*x))^2/x^5,x, algorithm="fricas")

[Out]

integral((a^2*c^3*d^3*x^3 + 3*a^2*c^2*d^3*x^2 + 3*a^2*c*d^3*x + a^2*d^3 + (b^2*c^3*d^3*x^3 + 3*b^2*c^2*d^3*x^2
+ 3*b^2*c*d^3*x + b^2*d^3)*arctanh(c*x)^2 + 2*(a*b*c^3*d^3*x^3 + 3*a*b*c^2*d^3*x^2 + 3*a*b*c*d^3*x + a*b*d^3)
*arctanh(c*x))/x^5, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{3} \left (\int \frac{a^{2}}{x^{5}}\, dx + \int \frac{3 a^{2} c}{x^{4}}\, dx + \int \frac{3 a^{2} c^{2}}{x^{3}}\, dx + \int \frac{a^{2} c^{3}}{x^{2}}\, dx + \int \frac{b^{2} \operatorname{atanh}^{2}{\left (c x \right )}}{x^{5}}\, dx + \int \frac{2 a b \operatorname{atanh}{\left (c x \right )}}{x^{5}}\, dx + \int \frac{3 b^{2} c \operatorname{atanh}^{2}{\left (c x \right )}}{x^{4}}\, dx + \int \frac{3 b^{2} c^{2} \operatorname{atanh}^{2}{\left (c x \right )}}{x^{3}}\, dx + \int \frac{b^{2} c^{3} \operatorname{atanh}^{2}{\left (c x \right )}}{x^{2}}\, dx + \int \frac{6 a b c \operatorname{atanh}{\left (c x \right )}}{x^{4}}\, dx + \int \frac{6 a b c^{2} \operatorname{atanh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac{2 a b c^{3} \operatorname{atanh}{\left (c x \right )}}{x^{2}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)**3*(a+b*atanh(c*x))**2/x**5,x)

[Out]

d**3*(Integral(a**2/x**5, x) + Integral(3*a**2*c/x**4, x) + Integral(3*a**2*c**2/x**3, x) + Integral(a**2*c**3
/x**2, x) + Integral(b**2*atanh(c*x)**2/x**5, x) + Integral(2*a*b*atanh(c*x)/x**5, x) + Integral(3*b**2*c*atan
h(c*x)**2/x**4, x) + Integral(3*b**2*c**2*atanh(c*x)**2/x**3, x) + Integral(b**2*c**3*atanh(c*x)**2/x**2, x) +
Integral(6*a*b*c*atanh(c*x)/x**4, x) + Integral(6*a*b*c**2*atanh(c*x)/x**3, x) + Integral(2*a*b*c**3*atanh(c*
x)/x**2, x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c d x + d\right )}^{3}{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{2}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^3*(a+b*arctanh(c*x))^2/x^5,x, algorithm="giac")

[Out]

integrate((c*d*x + d)^3*(b*arctanh(c*x) + a)^2/x^5, x)