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

Optimal. Leaf size=204 \[ \frac {x}{3 b^2}-\frac {\tanh ^{-1}(a+b x)}{3 b^3}-\frac {2 a (a+b x) \tanh ^{-1}(a+b x)}{b^3}+\frac {(a+b x)^2 \tanh ^{-1}(a+b x)}{3 b^3}+\frac {a \left (3+a^2\right ) \tanh ^{-1}(a+b x)^2}{3 b^3}+\frac {\left (1+3 a^2\right ) \tanh ^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \tanh ^{-1}(a+b x)^2-\frac {2 \left (1+3 a^2\right ) \tanh ^{-1}(a+b x) \log \left (\frac {2}{1-a-b x}\right )}{3 b^3}-\frac {a \log \left (1-(a+b x)^2\right )}{b^3}-\frac {\left (1+3 a^2\right ) \text {PolyLog}\left (2,-\frac {1+a+b x}{1-a-b x}\right )}{3 b^3} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 13, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.083, Rules used = {6246, 6065, 6021, 266, 6037, 327, 212, 6195, 6095, 6131, 6055, 2449, 2352} \begin {gather*} -\frac {\left (3 a^2+1\right ) \text {Li}_2\left (-\frac {a+b x+1}{-a-b x+1}\right )}{3 b^3}+\frac {a \left (a^2+3\right ) \tanh ^{-1}(a+b x)^2}{3 b^3}+\frac {\left (3 a^2+1\right ) \tanh ^{-1}(a+b x)^2}{3 b^3}-\frac {2 \left (3 a^2+1\right ) \log \left (\frac {2}{-a-b x+1}\right ) \tanh ^{-1}(a+b x)}{3 b^3}-\frac {a \log \left (1-(a+b x)^2\right )}{b^3}+\frac {(a+b x)^2 \tanh ^{-1}(a+b x)}{3 b^3}-\frac {2 a (a+b x) \tanh ^{-1}(a+b x)}{b^3}-\frac {\tanh ^{-1}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \tanh ^{-1}(a+b x)^2+\frac {x}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTanh[a + b*x]^2,x]

[Out]

x/(3*b^2) - ArcTanh[a + b*x]/(3*b^3) - (2*a*(a + b*x)*ArcTanh[a + b*x])/b^3 + ((a + b*x)^2*ArcTanh[a + b*x])/(
3*b^3) + (a*(3 + a^2)*ArcTanh[a + b*x]^2)/(3*b^3) + ((1 + 3*a^2)*ArcTanh[a + b*x]^2)/(3*b^3) + (x^3*ArcTanh[a
+ b*x]^2)/3 - (2*(1 + 3*a^2)*ArcTanh[a + b*x]*Log[2/(1 - a - b*x)])/(3*b^3) - (a*Log[1 - (a + b*x)^2])/b^3 - (
(1 + 3*a^2)*PolyLog[2, -((1 + a + b*x)/(1 - a - b*x))])/(3*b^3)

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(463\) vs. \(2(204)=408\).
time = 1.60, size = 463, normalized size = 2.27 \begin {gather*} -\frac {\left (1-(a+b x)^2\right )^{3/2} \left (-\frac {a+b x}{\sqrt {1-(a+b x)^2}}+\frac {6 a (a+b x) \tanh ^{-1}(a+b x)}{\sqrt {1-(a+b x)^2}}+\frac {3 (a+b x) \tanh ^{-1}(a+b x)^2}{\sqrt {1-(a+b x)^2}}-\frac {3 a^2 (a+b x) \tanh ^{-1}(a+b x)^2}{\sqrt {1-(a+b x)^2}}+\tanh ^{-1}(a+b x)^2 \cosh \left (3 \tanh ^{-1}(a+b x)\right )+3 a^2 \tanh ^{-1}(a+b x)^2 \cosh \left (3 \tanh ^{-1}(a+b x)\right )+2 \tanh ^{-1}(a+b x) \cosh \left (3 \tanh ^{-1}(a+b x)\right ) \log \left (1+e^{-2 \tanh ^{-1}(a+b x)}\right )+6 a^2 \tanh ^{-1}(a+b x) \cosh \left (3 \tanh ^{-1}(a+b x)\right ) \log \left (1+e^{-2 \tanh ^{-1}(a+b x)}\right )-6 a \cosh \left (3 \tanh ^{-1}(a+b x)\right ) \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )+\frac {3 \left (1-4 a+3 a^2\right ) \tanh ^{-1}(a+b x)^2+2 \tanh ^{-1}(a+b x) \left (2+\left (3+9 a^2\right ) \log \left (1+e^{-2 \tanh ^{-1}(a+b x)}\right )\right )-18 a \log \left (\frac {1}{\sqrt {1-(a+b x)^2}}\right )}{\sqrt {1-(a+b x)^2}}-\frac {4 \left (1+3 a^2\right ) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a+b x)}\right )}{\left (1-(a+b x)^2\right )^{3/2}}-\sinh \left (3 \tanh ^{-1}(a+b x)\right )+6 a \tanh ^{-1}(a+b x) \sinh \left (3 \tanh ^{-1}(a+b x)\right )-\tanh ^{-1}(a+b x)^2 \sinh \left (3 \tanh ^{-1}(a+b x)\right )-3 a^2 \tanh ^{-1}(a+b x)^2 \sinh \left (3 \tanh ^{-1}(a+b x)\right )\right )}{12 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcTanh[a + b*x]^2,x]

[Out]

-1/12*((1 - (a + b*x)^2)^(3/2)*(-((a + b*x)/Sqrt[1 - (a + b*x)^2]) + (6*a*(a + b*x)*ArcTanh[a + b*x])/Sqrt[1 -
(a + b*x)^2] + (3*(a + b*x)*ArcTanh[a + b*x]^2)/Sqrt[1 - (a + b*x)^2] - (3*a^2*(a + b*x)*ArcTanh[a + b*x]^2)/
Sqrt[1 - (a + b*x)^2] + ArcTanh[a + b*x]^2*Cosh[3*ArcTanh[a + b*x]] + 3*a^2*ArcTanh[a + b*x]^2*Cosh[3*ArcTanh[
a + b*x]] + 2*ArcTanh[a + b*x]*Cosh[3*ArcTanh[a + b*x]]*Log[1 + E^(-2*ArcTanh[a + b*x])] + 6*a^2*ArcTanh[a + b
*x]*Cosh[3*ArcTanh[a + b*x]]*Log[1 + E^(-2*ArcTanh[a + b*x])] - 6*a*Cosh[3*ArcTanh[a + b*x]]*Log[1/Sqrt[1 - (a
+ b*x)^2]] + (3*(1 - 4*a + 3*a^2)*ArcTanh[a + b*x]^2 + 2*ArcTanh[a + b*x]*(2 + (3 + 9*a^2)*Log[1 + E^(-2*ArcT
anh[a + b*x])]) - 18*a*Log[1/Sqrt[1 - (a + b*x)^2]])/Sqrt[1 - (a + b*x)^2] - (4*(1 + 3*a^2)*PolyLog[2, -E^(-2*
ArcTanh[a + b*x])])/(1 - (a + b*x)^2)^(3/2) - Sinh[3*ArcTanh[a + b*x]] + 6*a*ArcTanh[a + b*x]*Sinh[3*ArcTanh[a
+ b*x]] - ArcTanh[a + b*x]^2*Sinh[3*ArcTanh[a + b*x]] - 3*a^2*ArcTanh[a + b*x]^2*Sinh[3*ArcTanh[a + b*x]]))/b
^3

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(665\) vs. \(2(190)=380\).
time = 0.08, size = 666, normalized size = 3.26 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 259, normalized size = 1.27 \begin {gather*} \frac {1}{3} \, x^{3} \operatorname {artanh}\left (b x + a\right )^{2} - \frac {1}{12} \, b^{2} {\left (\frac {4 \, {\left (3 \, a^{2} + 1\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^{5}} + \frac {2 \, {\left (5 \, a^{2} + 6 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{5}} + \frac {{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right )^{2} - 2 \, {\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right ) \log \left (b x + a - 1\right ) + {\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (b x + a - 1\right )^{2} - 4 \, b x - 2 \, {\left (5 \, a^{2} - 6 \, a + 1\right )} \log \left (b x + a - 1\right )}{b^{5}}\right )} + \frac {1}{3} \, b {\left (\frac {b x^{2} - 4 \, a x}{b^{3}} + \frac {{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{4}} - \frac {{\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (b x + a - 1\right )}{b^{4}}\right )} \operatorname {artanh}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arctanh(b*x + a)^2 - 1/12*b^2*(4*(3*a^2 + 1)*(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^5 + 2*(5*a^2 + 6*a + 1)*log(b*x + a + 1)/b^5 + ((a^3 + 3*a^2 + 3*a + 1)*log(b*x + a + 1
)^2 - 2*(a^3 + 3*a^2 + 3*a + 1)*log(b*x + a + 1)*log(b*x + a - 1) + (a^3 - 3*a^2 + 3*a - 1)*log(b*x + a - 1)^2
- 4*b*x - 2*(5*a^2 - 6*a + 1)*log(b*x + a - 1))/b^5) + 1/3*b*((b*x^2 - 4*a*x)/b^3 + (a^3 + 3*a^2 + 3*a + 1)*l
og(b*x + a + 1)/b^4 - (a^3 - 3*a^2 + 3*a - 1)*log(b*x + a - 1)/b^4)*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^2*arctanh(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(x^2*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^{2} \operatorname {atanh}^{2}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(x^2*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^2\,{\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^2*atanh(a + b*x)^2,x)

[Out]

int(x^2*atanh(a + b*x)^2, x)

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