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

Optimal. Leaf size=271 \[ 4 a b c^4 d^3 \log (x)+4 b c^4 d^3 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac {7 b c^3 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x}-\frac {b c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )}{x^2}-\frac {d^3 (c x+1)^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}-\frac {b c d^3 \left (a+b \tanh ^{-1}(c x)\right )}{6 x^3}-2 b^2 c^4 d^3 \text {Li}_2(-c x)+2 b^2 c^4 d^3 \text {Li}_2(c x)+2 b^2 c^4 d^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )+\frac {11}{3} b^2 c^4 d^3 \log (x)+b^2 c^4 d^3 \tanh ^{-1}(c x)-\frac {b^2 c^3 d^3}{x}-\frac {b^2 c^2 d^3}{12 x^2}-\frac {11}{6} b^2 c^4 d^3 \log \left (1-c^2 x^2\right ) \]

[Out]

-1/12*b^2*c^2*d^3/x^2-b^2*c^3*d^3/x+b^2*c^4*d^3*arctanh(c*x)-1/6*b*c*d^3*(a+b*arctanh(c*x))/x^3-b*c^2*d^3*(a+b
*arctanh(c*x))/x^2-7/2*b*c^3*d^3*(a+b*arctanh(c*x))/x-1/4*d^3*(c*x+1)^4*(a+b*arctanh(c*x))^2/x^4+4*a*b*c^4*d^3
*ln(x)+11/3*b^2*c^4*d^3*ln(x)+4*b*c^4*d^3*(a+b*arctanh(c*x))*ln(2/(-c*x+1))-11/6*b^2*c^4*d^3*ln(-c^2*x^2+1)-2*
b^2*c^4*d^3*polylog(2,-c*x)+2*b^2*c^4*d^3*polylog(2,c*x)+2*b^2*c^4*d^3*polylog(2,1-2/(-c*x+1))

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Rubi [A]  time = 0.31, antiderivative size = 271, normalized size of antiderivative = 1.00, 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 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 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 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)^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*}

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Mathematica [A]  time = 0.80, size = 343, normalized size = 1.27 \[ -\frac {d^3 \left (12 a^2 c^3 x^3+18 a^2 c^2 x^2+12 a^2 c x+3 a^2-48 a b c^4 x^4 \log (c x)+21 a b c^4 x^4 \log (1-c x)-21 a b c^4 x^4 \log (c x+1)+42 a b c^3 x^3+12 a b c^2 x^2+24 a b c^4 x^4 \log \left (1-c^2 x^2\right )+2 b \tanh ^{-1}(c x) \left (3 a \left (4 c^3 x^3+6 c^2 x^2+4 c x+1\right )-24 b c^4 x^4 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+b c x \left (-6 c^3 x^3+21 c^2 x^2+6 c x+1\right )\right )+2 a b c x+24 b^2 c^4 x^4 \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )-b^2 c^4 x^4+12 b^2 c^3 x^3+b^2 c^2 x^2-44 b^2 c^4 x^4 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )+3 b^2 \left (-15 c^4 x^4+4 c^3 x^3+6 c^2 x^2+4 c x+1\right ) \tanh ^{-1}(c x)^2\right )}{12 x^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/12*(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])]))/x^4

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} c^{3} d^{3} x^{3} + 3 \, a^{2} c^{2} d^{3} x^{2} + 3 \, a^{2} c d^{3} x + a^{2} d^{3} + {\left (b^{2} c^{3} d^{3} x^{3} + 3 \, b^{2} c^{2} d^{3} x^{2} + 3 \, b^{2} c d^{3} x + b^{2} d^{3}\right )} \operatorname {artanh}\left (c x\right )^{2} + 2 \, {\left (a b c^{3} d^{3} x^{3} + 3 \, a b c^{2} d^{3} x^{2} + 3 \, a b c d^{3} x + a b d^{3}\right )} \operatorname {artanh}\left (c x\right )}{x^{5}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c d x + d\right )}^{3} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{x^{5}}\,{d x} \]

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)

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maple [B]  time = 0.08, size = 646, normalized size = 2.38 \[ -\frac {c^{2} d^{3} a b}{x^{2}}+\frac {c^{4} d^{3} b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{8}+4 c^{4} d^{3} a b \ln \left (c x \right )-\frac {2 c^{3} d^{3} a b \arctanh \left (c x \right )}{x}-\frac {3 c^{2} d^{3} a b \arctanh \left (c x \right )}{x^{2}}-\frac {2 c \,d^{3} a b \arctanh \left (c x \right )}{x^{3}}-\frac {4 c^{4} d^{3} b^{2} \ln \left (c x +1\right )}{3}-2 c^{4} d^{3} b^{2} \dilog \left (c x \right )-2 c^{4} d^{3} b^{2} \dilog \left (c x +1\right )+2 c^{4} d^{3} b^{2} \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )-\frac {d^{3} b^{2} \arctanh \left (c x \right )^{2}}{4 x^{4}}-\frac {c^{3} d^{3} a^{2}}{x}-\frac {c \,d^{3} a^{2}}{x^{3}}-\frac {3 c^{2} d^{3} a^{2}}{2 x^{2}}+\frac {c^{4} d^{3} b^{2} \ln \left (c x +1\right )^{2}}{16}+\frac {11 c^{4} d^{3} b^{2} \ln \left (c x \right )}{3}-\frac {15 c^{4} d^{3} b^{2} \ln \left (c x -1\right )^{2}}{16}-\frac {7 c^{4} d^{3} b^{2} \ln \left (c x -1\right )}{3}-\frac {b^{2} c^{2} d^{3}}{12 x^{2}}-\frac {c \,d^{3} a b}{6 x^{3}}-\frac {b^{2} c^{3} d^{3}}{x}-\frac {d^{3} a^{2}}{4 x^{4}}-\frac {3 c^{2} d^{3} b^{2} \arctanh \left (c x \right )^{2}}{2 x^{2}}-\frac {c \,d^{3} b^{2} \arctanh \left (c x \right )^{2}}{x^{3}}-\frac {c^{2} d^{3} b^{2} \arctanh \left (c x \right )}{x^{2}}-2 c^{4} d^{3} b^{2} \ln \left (c x \right ) \ln \left (c x +1\right )-\frac {c^{4} d^{3} b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{8}-\frac {7 c^{3} d^{3} a b}{2 x}+\frac {15 c^{4} d^{3} b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{8}-\frac {c^{4} d^{3} a b \ln \left (c x +1\right )}{4}-\frac {15 c^{4} d^{3} a b \ln \left (c x -1\right )}{4}-\frac {d^{3} a b \arctanh \left (c x \right )}{2 x^{4}}+4 c^{4} d^{3} b^{2} \arctanh \left (c x \right ) \ln \left (c x \right )-\frac {15 c^{4} d^{3} b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{4}-\frac {c^{3} d^{3} b^{2} \arctanh \left (c x \right )^{2}}{x}-\frac {c \,d^{3} b^{2} \arctanh \left (c x \right )}{6 x^{3}}-\frac {c^{4} d^{3} b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{4}-\frac {7 c^{3} d^{3} b^{2} \arctanh \left (c x \right )}{2 x} \]

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]

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

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maxima [B]  time = 0.96, size = 813, normalized size = 3.00 \[ -2 \, {\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^{4} d^{3} - 2 \, {\left (\log \left (c x\right ) \log \left (-c x + 1\right ) + {\rm Li}_2\left (-c x + 1\right )\right )} b^{2} c^{4} d^{3} + 2 \, {\left (\log \left (c x + 1\right ) \log \left (-c x\right ) + {\rm Li}_2\left (c x + 1\right )\right )} b^{2} c^{4} d^{3} - b^{2} c^{4} d^{3} \log \left (c x + 1\right ) - 2 \, b^{2} c^{4} d^{3} \log \left (c x - 1\right ) + 3 \, b^{2} c^{4} d^{3} \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^{3} d^{3} + \frac {3}{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^{2} d^{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 c d^{3} - \frac {a^{2} c^{3} d^{3}}{x} + \frac {1}{12} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x\right )}{x^{4}}\right )} a b d^{3} + \frac {1}{48} \, {\left ({\left (32 \, c^{2} \log \relax (x) - \frac {3 \, c^{2} x^{2} \log \left (c x + 1\right )^{2} + 3 \, c^{2} x^{2} \log \left (c x - 1\right )^{2} + 16 \, c^{2} x^{2} \log \left (c x - 1\right ) - 2 \, {\left (3 \, c^{2} x^{2} \log \left (c x - 1\right ) - 8 \, c^{2} x^{2}\right )} \log \left (c x + 1\right ) + 4}{x^{2}}\right )} c^{2} + 4 \, {\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c \operatorname {artanh}\left (c x\right )\right )} b^{2} d^{3} - \frac {3 \, a^{2} c^{2} d^{3}}{2 \, x^{2}} - \frac {a^{2} c d^{3}}{x^{3}} - \frac {b^{2} d^{3} \operatorname {artanh}\left (c x\right )^{2}}{4 \, x^{4}} - \frac {a^{2} d^{3}}{4 \, x^{4}} - \frac {8 \, b^{2} c^{3} d^{3} x^{2} + {\left (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}\right )} \log \left (c x + 1\right )^{2} - {\left (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}\right )} \log \left (-c x + 1\right )^{2} + 4 \, {\left (3 \, b^{2} c^{3} d^{3} x^{2} + b^{2} c^{2} d^{3} x\right )} \log \left (c x + 1\right ) - 2 \, {\left (6 \, b^{2} c^{3} d^{3} x^{2} + 2 \, b^{2} c^{2} d^{3} x + {\left (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}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{8 \, x^{3}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\right )}^3}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d^{3} \left (\int \frac {a^{2}}{x^{5}}\, dx + \int \frac {3 a^{2} c}{x^{4}}\, dx + \int \frac {3 a^{2} c^{2}}{x^{3}}\, dx + \int \frac {a^{2} c^{3}}{x^{2}}\, dx + \int \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{x^{5}}\, dx + \int \frac {2 a b \operatorname {atanh}{\left (c x \right )}}{x^{5}}\, dx + \int \frac {3 b^{2} c \operatorname {atanh}^{2}{\left (c x \right )}}{x^{4}}\, dx + \int \frac {3 b^{2} c^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{x^{3}}\, dx + \int \frac {b^{2} c^{3} \operatorname {atanh}^{2}{\left (c x \right )}}{x^{2}}\, dx + \int \frac {6 a b c \operatorname {atanh}{\left (c x \right )}}{x^{4}}\, dx + \int \frac {6 a b c^{2} \operatorname {atanh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac {2 a b c^{3} \operatorname {atanh}{\left (c x \right )}}{x^{2}}\, dx\right ) \]

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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