3.1 \(\int x^3 \tanh ^{-1}(a+b x)^2 \, dx\)

Optimal. Leaf size=263 \[ \frac {a \left (a^2+1\right ) \text {Li}_2\left (-\frac {a+b x+1}{-a-b x+1}\right )}{b^4}+\frac {\left (6 a^2+1\right ) \log \left (1-(a+b x)^2\right )}{4 b^4}-\frac {a \left (a^2+1\right ) \tanh ^{-1}(a+b x)^2}{b^4}+\frac {\left (6 a^2+1\right ) (a+b x) \tanh ^{-1}(a+b x)}{2 b^4}+\frac {2 a \left (a^2+1\right ) \log \left (\frac {2}{-a-b x+1}\right ) \tanh ^{-1}(a+b x)}{b^4}-\frac {\left (a^4+6 a^2+1\right ) \tanh ^{-1}(a+b x)^2}{4 b^4}+\frac {(a+b x)^2}{12 b^4}+\frac {\log \left (1-(a+b x)^2\right )}{12 b^4}+\frac {(a+b x)^3 \tanh ^{-1}(a+b x)}{6 b^4}-\frac {a (a+b x)^2 \tanh ^{-1}(a+b x)}{b^4}+\frac {a \tanh ^{-1}(a+b x)}{b^4}-\frac {a x}{b^3}+\frac {1}{4} x^4 \tanh ^{-1}(a+b x)^2 \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 15, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {6111, 5928, 5910, 260, 5916, 321, 206, 266, 43, 6048, 5948, 5984, 5918, 2402, 2315} \[ \frac {a \left (a^2+1\right ) \text {PolyLog}\left (2,-\frac {a+b x+1}{-a-b x+1}\right )}{b^4}+\frac {\left (6 a^2+1\right ) \log \left (1-(a+b x)^2\right )}{4 b^4}-\frac {a \left (a^2+1\right ) \tanh ^{-1}(a+b x)^2}{b^4}-\frac {\left (a^4+6 a^2+1\right ) \tanh ^{-1}(a+b x)^2}{4 b^4}+\frac {\left (6 a^2+1\right ) (a+b x) \tanh ^{-1}(a+b x)}{2 b^4}+\frac {2 a \left (a^2+1\right ) \log \left (\frac {2}{-a-b x+1}\right ) \tanh ^{-1}(a+b x)}{b^4}-\frac {a x}{b^3}+\frac {(a+b x)^2}{12 b^4}+\frac {\log \left (1-(a+b x)^2\right )}{12 b^4}+\frac {(a+b x)^3 \tanh ^{-1}(a+b x)}{6 b^4}-\frac {a (a+b x)^2 \tanh ^{-1}(a+b x)}{b^4}+\frac {a \tanh ^{-1}(a+b x)}{b^4}+\frac {1}{4} x^4 \tanh ^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcTanh[a + b*x]^2,x]

[Out]

-((a*x)/b^3) + (a + b*x)^2/(12*b^4) + (a*ArcTanh[a + b*x])/b^4 + ((1 + 6*a^2)*(a + b*x)*ArcTanh[a + b*x])/(2*b
^4) - (a*(a + b*x)^2*ArcTanh[a + b*x])/b^4 + ((a + b*x)^3*ArcTanh[a + b*x])/(6*b^4) - (a*(1 + a^2)*ArcTanh[a +
b*x]^2)/b^4 - ((1 + 6*a^2 + a^4)*ArcTanh[a + b*x]^2)/(4*b^4) + (x^4*ArcTanh[a + b*x]^2)/4 + (2*a*(1 + a^2)*Ar
cTanh[a + b*x]*Log[2/(1 - a - b*x)])/b^4 + Log[1 - (a + b*x)^2]/(12*b^4) + ((1 + 6*a^2)*Log[1 - (a + b*x)^2])/
(4*b^4) + (a*(1 + a^2)*PolyLog[2, -((1 + a + b*x)/(1 - a - b*x))])/b^4

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 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 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 5928

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(
a + b*ArcTanh[c*x])^p)/(e*(q + 1)), x] - Dist[(b*c*p)/(e*(q + 1)), Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^(p
- 1), (d + e*x)^(q + 1)/(1 - c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] &
& NeQ[q, -1]

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 5984

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 6048

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :>
Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x]
&& IGtQ[p, 0] && EqQ[c^2*d + e, 0] && IGtQ[m, 0]

Rule 6111

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &
& IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 1.26, size = 187, normalized size = 0.71 \[ -\frac {12 \left (a^3+a\right ) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(a+b x)}\right )+36 a^2 \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )+11 a^2-2 \tanh ^{-1}(a+b x) \left (12 \left (a^3+a\right ) \log \left (e^{-2 \tanh ^{-1}(a+b x)}+1\right )+13 a^3+9 a^2 b x-3 a b^2 x^2+9 a+b^3 x^3+3 b x\right )+3 \left (a^4-4 a^3+6 a^2-4 a-b^4 x^4+1\right ) \tanh ^{-1}(a+b x)^2+10 a b x+8 \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )-b^2 x^2+1}{12 b^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3*ArcTanh[a + b*x]^2,x]

[Out]

-1/12*(1 + 11*a^2 + 10*a*b*x - b^2*x^2 + 3*(1 - 4*a + 6*a^2 - 4*a^3 + a^4 - b^4*x^4)*ArcTanh[a + b*x]^2 - 2*Ar
cTanh[a + b*x]*(9*a + 13*a^3 + 3*b*x + 9*a^2*b*x - 3*a*b^2*x^2 + b^3*x^3 + 12*(a + a^3)*Log[1 + E^(-2*ArcTanh[
a + b*x])]) + 8*Log[1/Sqrt[1 - (a + b*x)^2]] + 36*a^2*Log[1/Sqrt[1 - (a + b*x)^2]] + 12*(a + a^3)*PolyLog[2, -
E^(-2*ArcTanh[a + b*x])])/b^4

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3} \operatorname {artanh}\left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3*arctanh(b*x + a)^2, x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.06, size = 967, normalized size = 3.68 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b^4*ln(-1/2*b*x-1/2*a+1/2)*ln(1/2+1/2*b*x+1/2*a)*a^3+3/4/b^4*ln(-1/2*b*x-1/2*a+1/2)*ln(1/2+1/2*b*x+1/2*a)*
a^2-3/4/b^4*ln(b*x+a+1)*ln(-1/2*b*x-1/2*a+1/2)*a^2-1/b^4*arctanh(b*x+a)*ln(b*x+a-1)*a+1/2/b^4*ln(-1/2*b*x-1/2*
a+1/2)*ln(1/2+1/2*b*x+1/2*a)*a-1/8/b^4*ln(b*x+a+1)*ln(-1/2*b*x-1/2*a+1/2)*a^4+1/4/b^4*arctanh(b*x+a)*ln(b*x+a-
1)*a^4+1/2/b^4*ln(b*x+a-1)*ln(1/2+1/2*b*x+1/2*a)*a-1/8/b^4*ln(b*x+a-1)*ln(1/2+1/2*b*x+1/2*a)*a^4+3/2/b^4*arcta
nh(b*x+a)*ln(b*x+a-1)*a^2+1/2/b^4*ln(b*x+a-1)*ln(1/2+1/2*b*x+1/2*a)*a^3-3/4/b^4*ln(b*x+a-1)*ln(1/2+1/2*b*x+1/2
*a)*a^2-1/b^4*arctanh(b*x+a)*ln(b*x+a+1)*a^3-3/2/b^4*arctanh(b*x+a)*ln(b*x+a+1)*a^2-1/b^4*arctanh(b*x+a)*ln(b*
x+a+1)*a-1/b^4*arctanh(b*x+a)*ln(b*x+a-1)*a^3-1/4/b^4*arctanh(b*x+a)*ln(b*x+a+1)*a^4-1/2/b^2*arctanh(b*x+a)*x^
2*a+3/2/b^3*arctanh(b*x+a)*x*a^2-1/2/b^4*ln(b*x+a+1)*ln(-1/2*b*x-1/2*a+1/2)*a^3-1/2/b^4*ln(b*x+a+1)*ln(-1/2*b*
x-1/2*a+1/2)*a+1/8/b^4*ln(-1/2*b*x-1/2*a+1/2)*ln(1/2+1/2*b*x+1/2*a)*a^4+1/2*a*arctanh(b*x+a)/b^4-5/6*a*x/b^3-1
/4/b^4*arctanh(b*x+a)*ln(b*x+a+1)-1/8/b^4*ln(-1/2*b*x-1/2*a+1/2)*ln(b*x+a+1)+1/8/b^4*ln(-1/2*b*x-1/2*a+1/2)*ln
(1/2+1/2*b*x+1/2*a)+13/6/b^4*arctanh(b*x+a)*a^3+3/2/b^4*ln(b*x+a-1)*a^2+3/2/b^4*ln(b*x+a+1)*a^2-1/8/b^4*ln(b*x
+a-1)*ln(1/2+1/2*b*x+1/2*a)+1/4/b^4*arctanh(b*x+a)*ln(b*x+a-1)+3/8/b^4*ln(b*x+a+1)^2*a^2+1/4/b^4*ln(b*x+a+1)^2
*a+1/16/b^4*ln(b*x+a-1)^2*a^4-1/4/b^4*ln(b*x+a-1)^2*a^3+3/8/b^4*ln(b*x+a-1)^2*a^2-1/4/b^4*ln(b*x+a-1)^2*a+1/b^
4*dilog(1/2+1/2*b*x+1/2*a)*a^3+1/b^4*dilog(1/2+1/2*b*x+1/2*a)*a-1/2/b^4*a*ln(b*x+a-1)+1/2/b^4*a*ln(b*x+a+1)+1/
6/b*arctanh(b*x+a)*x^3+1/2/b^3*arctanh(b*x+a)*x+1/16/b^4*ln(b*x+a+1)^2*a^4+1/4/b^4*ln(b*x+a+1)^2*a^3-11/12/b^4
*a^2+1/12/b^2*x^2+1/3/b^4*ln(b*x+a+1)+1/3/b^4*ln(b*x+a-1)+1/16/b^4*ln(b*x+a-1)^2+1/16/b^4*ln(b*x+a+1)^2+1/4*x^
4*arctanh(b*x+a)^2

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maxima [A]  time = 0.33, size = 320, normalized size = 1.22 \[ \frac {1}{4} \, x^{4} \operatorname {artanh}\left (b x + a\right )^{2} + \frac {1}{48} \, b^{2} {\left (\frac {48 \, {\left (a^{3} + a\right )} {\left (\log \left (b x + a - 1\right ) \log \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, b x - \frac {1}{2} \, a + \frac {1}{2}\right )\right )}}{b^{6}} + \frac {4 \, {\left (13 \, a^{3} + 18 \, a^{2} + 9 \, a + 4\right )} \log \left (b x + a + 1\right )}{b^{6}} + \frac {4 \, b^{2} x^{2} - 40 \, a b x + 3 \, {\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right )^{2} - 6 \, {\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right ) \log \left (b x + a - 1\right ) + 3 \, {\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left (b x + a - 1\right )^{2} - 4 \, {\left (13 \, a^{3} - 18 \, a^{2} + 9 \, a - 4\right )} \log \left (b x + a - 1\right )}{b^{6}}\right )} + \frac {1}{12} \, b {\left (\frac {2 \, {\left (b^{2} x^{3} - 3 \, a b x^{2} + 3 \, {\left (3 \, a^{2} + 1\right )} x\right )}}{b^{4}} - \frac {3 \, {\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{5}} + \frac {3 \, {\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left (b x + a - 1\right )}{b^{5}}\right )} \operatorname {artanh}\left (b x + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arctanh(b*x + a)^2 + 1/48*b^2*(48*(a^3 + a)*(log(b*x + a - 1)*log(1/2*b*x + 1/2*a + 1/2) + dilog(-1/2*
b*x - 1/2*a + 1/2))/b^6 + 4*(13*a^3 + 18*a^2 + 9*a + 4)*log(b*x + a + 1)/b^6 + (4*b^2*x^2 - 40*a*b*x + 3*(a^4
+ 4*a^3 + 6*a^2 + 4*a + 1)*log(b*x + a + 1)^2 - 6*(a^4 + 4*a^3 + 6*a^2 + 4*a + 1)*log(b*x + a + 1)*log(b*x + a
- 1) + 3*(a^4 - 4*a^3 + 6*a^2 - 4*a + 1)*log(b*x + a - 1)^2 - 4*(13*a^3 - 18*a^2 + 9*a - 4)*log(b*x + a - 1))
/b^6) + 1/12*b*(2*(b^2*x^3 - 3*a*b*x^2 + 3*(3*a^2 + 1)*x)/b^4 - 3*(a^4 + 4*a^3 + 6*a^2 + 4*a + 1)*log(b*x + a
+ 1)/b^5 + 3*(a^4 - 4*a^3 + 6*a^2 - 4*a + 1)*log(b*x + a - 1)/b^5)*arctanh(b*x + a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*atanh(a + b*x)^2,x)

[Out]

int(x^3*atanh(a + b*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*atanh(a + b*x)**2, x)

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