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

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

[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 = 0.24, 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 = {6246, 6065, 6021, 266, 6037, 327, 212, 272, 45, 6195, 6095, 6131, 6055, 2449, 2352} \begin {gather*} \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 \end {gather*}

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 45

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 212

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

Rule 266

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 272

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 327

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 2352

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

Rule 2449

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 6021

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

Rule 6037

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

Rule 6055

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 6065

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 6095

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 6131

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 6195

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 6246

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 {\text {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} \text {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 {\text {Subst}\left (\int x^2 \tanh ^{-1}(x) \, dx,x,a+b x\right )}{2 b^4}-\frac {\text {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) \text {Subst}\left (\int x \tanh ^{-1}(x) \, dx,x,a+b x\right )}{b^4}+\frac {\left (1+6 a^2\right ) \text {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 {\text {Subst}\left (\int \frac {x^3}{1-x^2} \, dx,x,a+b x\right )}{6 b^4}-\frac {\text {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 \text {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,a+b x\right )}{b^4}-\frac {\left (1+6 a^2\right ) \text {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 {\text {Subst}\left (\int \frac {x}{1-x} \, dx,x,(a+b x)^2\right )}{12 b^4}+\frac {a \text {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 ) \text {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 ) \text {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 {\text {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 ) \text {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 ) \text {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 ) \text {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 = 1.28, size = 187, normalized size = 0.71 \begin {gather*} -\frac {1+11 a^2+10 a b x-b^2 x^2+3 \left (1-4 a+6 a^2-4 a^3+a^4-b^4 x^4\right ) \tanh ^{-1}(a+b x)^2-2 \tanh ^{-1}(a+b x) \left (9 a+13 a^3+3 b x+9 a^2 b x-3 a b^2 x^2+b^3 x^3+12 \left (a+a^3\right ) \log \left (1+e^{-2 \tanh ^{-1}(a+b x)}\right )\right )+8 \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )+36 a^2 \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )+12 \left (a+a^3\right ) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a+b x)}\right )}{12 b^4} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(886\) vs. \(2(251)=502\).
time = 2.18, size = 887, normalized size = 3.37

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(887\)
default	\(\text {Expression too large to display}\)	\(887\)
risch	\(-\frac {1}{12 b^{4}}-\frac {5 a x}{6 b^{3}}-\frac {\ln \left (-b x -a +1\right ) \left (-b x -a +1\right )^{3}}{12 b^{4}}+\frac {\ln \left (-b x -a +1\right ) \left (-b x -a +1\right )^{2}}{8 b^{4}}+\frac {13 \ln \left (-b x -a -1\right ) a^{3}}{12 b^{4}}-\frac {\left (-b^{4} x^{4}+a^{4}+4 a^{3}+6 a^{2}+4 a +1\right ) \ln \left (b x +a +1\right )^{2}}{16 b^{4}}+\frac {3 \ln \left (-b x -a -1\right ) a^{2}}{2 b^{4}}+\frac {3 \ln \left (-b x -a -1\right ) a}{4 b^{4}}+\frac {\ln \left (-b x -a +1\right ) \left (-b x -a +1\right )^{4}}{32 b^{4}}+\frac {\ln \left (-b x -a -1\right )}{3 b^{4}}+\left (-\frac {x^{4} \ln \left (-b x -a +1\right )}{8}-\frac {-2 x^{3} b^{3}-3 \ln \left (-b x -a +1\right ) a^{4}+6 a \,b^{2} x^{2}+12 \ln \left (-b x -a +1\right ) a^{3}-18 a^{2} b x -18 \ln \left (-b x -a +1\right ) a^{2}+12 \ln \left (-b x -a +1\right ) a -6 b x -3 \ln \left (-b x -a +1\right )}{24 b^{4}}\right ) \ln \left (b x +a +1\right )+\frac {a}{b^{4}}-\frac {11 a^{2}}{12 b^{4}}+\frac {x^{2}}{12 b^{2}}+\frac {25 \ln \left (-b x -a +1\right )}{96 b^{4}}-\frac {\ln \left (-b x -a +1\right )^{2}}{16 b^{4}}+\frac {\ln \left (-b x -a +1\right )^{2} x^{4}}{16}+\frac {\ln \left (-b x -a +1\right ) \left (-b x -a +1\right )^{3} a}{6 b^{4}}+\frac {3 \ln \left (-b x -a +1\right ) \left (-b x -a +1\right )^{2} a^{2}}{8 b^{4}}-\frac {\ln \left (-b x -a +1\right ) \left (-b x -a +1\right )^{2} a}{4 b^{4}}+\frac {\ln \left (-b x -a +1\right ) \left (-b x -a +1\right ) a^{3}}{2 b^{4}}+\frac {\ln \left (-b x -a +1\right ) \left (-b x -a +1\right ) a}{2 b^{4}}+\frac {\ln \left (-b x -a +1\right ) \ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right ) a^{3}}{b^{4}}-\frac {\ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right ) \ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right ) a^{3}}{b^{4}}+\frac {\ln \left (-b x -a +1\right ) \ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right ) a}{b^{4}}-\frac {\ln \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right ) \ln \left (\frac {b x}{2}+\frac {a}{2}+\frac {1}{2}\right ) a}{b^{4}}+\frac {\ln \left (-b x -a +1\right ) x^{2} a}{8 b^{2}}-\frac {3 \ln \left (-b x -a +1\right ) x \,a^{2}}{8 b^{3}}+\frac {3 \ln \left (-b x -a +1\right ) x a}{8 b^{3}}-\frac {\ln \left (-b x -a +1\right ) x^{2} a^{2}}{16 b^{2}}+\frac {\ln \left (-b x -a +1\right ) x \,a^{3}}{8 b^{3}}+\frac {a \ln \left (-b x -a +1\right ) x^{3}}{24 b}-\frac {\ln \left (-b x -a +1\right ) x^{3}}{24 b}-\frac {\ln \left (-b x -a +1\right ) x^{2}}{16 b^{2}}-\frac {\ln \left (-b x -a +1\right ) x}{8 b^{3}}+\frac {\ln \left (-b x -a +1\right )^{2} a^{3}}{4 b^{4}}-\frac {3 \ln \left (-b x -a +1\right )^{2} a^{2}}{8 b^{4}}+\frac {\ln \left (-b x -a +1\right )^{2} a}{4 b^{4}}-\frac {25 \ln \left (-b x -a +1\right ) a^{3}}{24 b^{4}}+\frac {25 \ln \left (-b x -a +1\right ) a^{2}}{16 b^{4}}-\frac {25 \ln \left (-b x -a +1\right ) a}{24 b^{4}}-\frac {\ln \left (-b x -a +1\right )^{2} a^{4}}{16 b^{4}}+\frac {25 \ln \left (-b x -a +1\right ) a^{4}}{96 b^{4}}-\frac {x^{4} \ln \left (-b x -a +1\right )}{32}-\frac {\dilog \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right ) a^{3}}{b^{4}}-\frac {\dilog \left (-\frac {b x}{2}-\frac {a}{2}+\frac {1}{2}\right ) a}{b^{4}}\)	\(1001\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^4*(1/3*ln(b*x+a+1)+1/12*(b*x+a)^2-arctanh(b*x+a)*ln(b*x+a-1)*a+3/2*arctanh(b*x+a)*ln(b*x+a-1)*a^2+1/4*arct
anh(b*x+a)*ln(b*x+a-1)*a^4-arctanh(b*x+a)*ln(b*x+a-1)*a^3+3*arctanh(b*x+a)*a^2*(b*x+a)-arctanh(b*x+a)*(b*x+a)^
2*a+1/3*ln(b*x+a-1)+1/16*ln(b*x+a+1)^2+1/16*ln(b*x+a-1)^2-(b*x+a)*a+3/2*ln(b*x+a-1)*a^2+3/2*ln(b*x+a+1)*a^2-1/
2*ln(b*x+a-1)*a+1/2*ln(b*x+a+1)*a+1/6*arctanh(b*x+a)*(b*x+a)^3+1/2*arctanh(b*x+a)*(b*x+a)+1/4*arctanh(b*x+a)*l
n(b*x+a-1)-1/4*arctanh(b*x+a)*ln(b*x+a+1)+1/4*arctanh(b*x+a)^2*a^4+1/4*arctanh(b*x+a)^2*(b*x+a)^4-1/8*ln(-1/2*
b*x-1/2*a+1/2)*ln(b*x+a+1)+1/8*ln(-1/2*b*x-1/2*a+1/2)*ln(1/2*b*x+1/2*a+1/2)+1/16*ln(b*x+a+1)^2*a^4+1/4*ln(b*x+
a+1)^2*a^3+3/8*ln(b*x+a+1)^2*a^2+1/4*ln(b*x+a+1)^2*a-1/8*ln(b*x+a-1)*ln(1/2*b*x+1/2*a+1/2)+1/16*ln(b*x+a-1)^2*
a^4-1/4*ln(b*x+a-1)^2*a^3+3/8*ln(b*x+a-1)^2*a^2-1/4*ln(b*x+a-1)^2*a+dilog(1/2*b*x+1/2*a+1/2)*a^3+dilog(1/2*b*x
+1/2*a+1/2)*a-1/4*arctanh(b*x+a)*ln(b*x+a+1)*a^4-arctanh(b*x+a)*ln(b*x+a+1)*a^3-3/2*arctanh(b*x+a)*ln(b*x+a+1)
*a^2-arctanh(b*x+a)*ln(b*x+a+1)*a-arctanh(b*x+a)^2*a^3*(b*x+a)+3/2*arctanh(b*x+a)^2*a^2*(b*x+a)^2-arctanh(b*x+
a)^2*a*(b*x+a)^3-1/8*ln(b*x+a+1)*ln(-1/2*b*x-1/2*a+1/2)*a^4-1/2*ln(b*x+a+1)*ln(-1/2*b*x-1/2*a+1/2)*a^3-3/4*ln(
b*x+a+1)*ln(-1/2*b*x-1/2*a+1/2)*a^2-1/2*ln(b*x+a+1)*ln(-1/2*b*x-1/2*a+1/2)*a+1/8*ln(-1/2*b*x-1/2*a+1/2)*ln(1/2
*b*x+1/2*a+1/2)*a^4+1/2*ln(-1/2*b*x-1/2*a+1/2)*ln(1/2*b*x+1/2*a+1/2)*a^3+3/4*ln(-1/2*b*x-1/2*a+1/2)*ln(1/2*b*x
+1/2*a+1/2)*a^2+1/2*ln(-1/2*b*x-1/2*a+1/2)*ln(1/2*b*x+1/2*a+1/2)*a-1/8*ln(b*x+a-1)*ln(1/2*b*x+1/2*a+1/2)*a^4+1
/2*ln(b*x+a-1)*ln(1/2*b*x+1/2*a+1/2)*a^3-3/4*ln(b*x+a-1)*ln(1/2*b*x+1/2*a+1/2)*a^2+1/2*ln(b*x+a-1)*ln(1/2*b*x+
1/2*a+1/2)*a)

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Maxima [A]
time = 0.27, size = 320, normalized size = 1.22 \begin {gather*} \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 ) \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arctanh(b*x + a)^2, x)

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

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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