3.13 \(\int x^5 (a+b \tanh ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=145 \[ \frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{9 c^3}+\frac{a b x}{3 c^5}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^6}+\frac{1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}+\frac{b^2 x^4}{60 c^2}+\frac{4 b^2 x^2}{45 c^4}+\frac{23 b^2 \log \left (1-c^2 x^2\right )}{90 c^6}+\frac{b^2 x \tanh ^{-1}(c x)}{3 c^5} \]

[Out]

(a*b*x)/(3*c^5) + (4*b^2*x^2)/(45*c^4) + (b^2*x^4)/(60*c^2) + (b^2*x*ArcTanh[c*x])/(3*c^5) + (b*x^3*(a + b*Arc
Tanh[c*x]))/(9*c^3) + (b*x^5*(a + b*ArcTanh[c*x]))/(15*c) - (a + b*ArcTanh[c*x])^2/(6*c^6) + (x^6*(a + b*ArcTa
nh[c*x])^2)/6 + (23*b^2*Log[1 - c^2*x^2])/(90*c^6)

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Rubi [A]  time = 0.326833, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5916, 5980, 266, 43, 5910, 260, 5948} \[ \frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{9 c^3}+\frac{a b x}{3 c^5}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^6}+\frac{1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}+\frac{b^2 x^4}{60 c^2}+\frac{4 b^2 x^2}{45 c^4}+\frac{23 b^2 \log \left (1-c^2 x^2\right )}{90 c^6}+\frac{b^2 x \tanh ^{-1}(c x)}{3 c^5} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*ArcTanh[c*x])^2,x]

[Out]

(a*b*x)/(3*c^5) + (4*b^2*x^2)/(45*c^4) + (b^2*x^4)/(60*c^2) + (b^2*x*ArcTanh[c*x])/(3*c^5) + (b*x^3*(a + b*Arc
Tanh[c*x]))/(9*c^3) + (b*x^5*(a + b*ArcTanh[c*x]))/(15*c) - (a + b*ArcTanh[c*x])^2/(6*c^6) + (x^6*(a + b*ArcTa
nh[c*x])^2)/6 + (23*b^2*Log[1 - c^2*x^2])/(90*c^6)

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 5980

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

Rubi steps

\begin{align*} \int x^5 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx &=\frac{1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{1}{3} (b c) \int \frac{x^6 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac{1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{b \int x^4 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{3 c}-\frac{b \int \frac{x^4 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{3 c}\\ &=\frac{b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}+\frac{1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{1}{15} b^2 \int \frac{x^5}{1-c^2 x^2} \, dx+\frac{b \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{3 c^3}-\frac{b \int \frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{3 c^3}\\ &=\frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{9 c^3}+\frac{b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}+\frac{1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{1}{30} b^2 \operatorname{Subst}\left (\int \frac{x^2}{1-c^2 x} \, dx,x,x^2\right )+\frac{b \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{3 c^5}-\frac{b \int \frac{a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{3 c^5}-\frac{b^2 \int \frac{x^3}{1-c^2 x^2} \, dx}{9 c^2}\\ &=\frac{a b x}{3 c^5}+\frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{9 c^3}+\frac{b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^6}+\frac{1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{1}{30} b^2 \operatorname{Subst}\left (\int \left (-\frac{1}{c^4}-\frac{x}{c^2}-\frac{1}{c^4 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )+\frac{b^2 \int \tanh ^{-1}(c x) \, dx}{3 c^5}-\frac{b^2 \operatorname{Subst}\left (\int \frac{x}{1-c^2 x} \, dx,x,x^2\right )}{18 c^2}\\ &=\frac{a b x}{3 c^5}+\frac{b^2 x^2}{30 c^4}+\frac{b^2 x^4}{60 c^2}+\frac{b^2 x \tanh ^{-1}(c x)}{3 c^5}+\frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{9 c^3}+\frac{b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^6}+\frac{1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{b^2 \log \left (1-c^2 x^2\right )}{30 c^6}-\frac{b^2 \int \frac{x}{1-c^2 x^2} \, dx}{3 c^4}-\frac{b^2 \operatorname{Subst}\left (\int \left (-\frac{1}{c^2}-\frac{1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{18 c^2}\\ &=\frac{a b x}{3 c^5}+\frac{4 b^2 x^2}{45 c^4}+\frac{b^2 x^4}{60 c^2}+\frac{b^2 x \tanh ^{-1}(c x)}{3 c^5}+\frac{b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{9 c^3}+\frac{b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^6}+\frac{1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{23 b^2 \log \left (1-c^2 x^2\right )}{90 c^6}\\ \end{align*}

Mathematica [A]  time = 0.0678579, size = 164, normalized size = 1.13 \[ \frac{30 a^2 c^6 x^6+12 a b c^5 x^5+20 a b c^3 x^3+4 b c x \tanh ^{-1}(c x) \left (15 a c^5 x^5+b \left (3 c^4 x^4+5 c^2 x^2+15\right )\right )+60 a b c x+2 b (15 a+23 b) \log (1-c x)-30 a b \log (c x+1)+3 b^2 c^4 x^4+16 b^2 c^2 x^2+30 b^2 \left (c^6 x^6-1\right ) \tanh ^{-1}(c x)^2+46 b^2 \log (c x+1)}{180 c^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*a*b*c*x + 16*b^2*c^2*x^2 + 20*a*b*c^3*x^3 + 3*b^2*c^4*x^4 + 12*a*b*c^5*x^5 + 30*a^2*c^6*x^6 + 4*b*c*x*(15*
a*c^5*x^5 + b*(15 + 5*c^2*x^2 + 3*c^4*x^4))*ArcTanh[c*x] + 30*b^2*(-1 + c^6*x^6)*ArcTanh[c*x]^2 + 2*b*(15*a +
23*b)*Log[1 - c*x] - 30*a*b*Log[1 + c*x] + 46*b^2*Log[1 + c*x])/(180*c^6)

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Maple [B]  time = 0.023, size = 314, normalized size = 2.2 \begin{align*}{\frac{{x}^{6}{a}^{2}}{6}}+{\frac{{b}^{2}{x}^{6} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{6}}+{\frac{{b}^{2}{\it Artanh} \left ( cx \right ){x}^{5}}{15\,c}}+{\frac{{b}^{2}{\it Artanh} \left ( cx \right ){x}^{3}}{9\,{c}^{3}}}+{\frac{{b}^{2}x{\it Artanh} \left ( cx \right ) }{3\,{c}^{5}}}+{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx-1 \right ) }{6\,{c}^{6}}}-{\frac{{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) }{6\,{c}^{6}}}+{\frac{{b}^{2} \left ( \ln \left ( cx-1 \right ) \right ) ^{2}}{24\,{c}^{6}}}-{\frac{{b}^{2}\ln \left ( cx-1 \right ) }{12\,{c}^{6}}\ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{{b}^{2}\ln \left ( cx+1 \right ) }{12\,{c}^{6}}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) }+{\frac{{b}^{2}}{12\,{c}^{6}}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{{b}^{2} \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{24\,{c}^{6}}}+{\frac{{b}^{2}{x}^{4}}{60\,{c}^{2}}}+{\frac{4\,{b}^{2}{x}^{2}}{45\,{c}^{4}}}+{\frac{23\,{b}^{2}\ln \left ( cx-1 \right ) }{90\,{c}^{6}}}+{\frac{23\,{b}^{2}\ln \left ( cx+1 \right ) }{90\,{c}^{6}}}+{\frac{ab{x}^{6}{\it Artanh} \left ( cx \right ) }{3}}+{\frac{ab{x}^{5}}{15\,c}}+{\frac{ab{x}^{3}}{9\,{c}^{3}}}+{\frac{xab}{3\,{c}^{5}}}+{\frac{ab\ln \left ( cx-1 \right ) }{6\,{c}^{6}}}-{\frac{ab\ln \left ( cx+1 \right ) }{6\,{c}^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^6*a^2+1/6*b^2*x^6*arctanh(c*x)^2+1/15/c*b^2*arctanh(c*x)*x^5+1/9/c^3*b^2*arctanh(c*x)*x^3+1/3*b^2*x*arct
anh(c*x)/c^5+1/6/c^6*b^2*arctanh(c*x)*ln(c*x-1)-1/6/c^6*b^2*arctanh(c*x)*ln(c*x+1)+1/24/c^6*b^2*ln(c*x-1)^2-1/
12/c^6*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)-1/12/c^6*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/12/c^6*b^2*ln(-1/2*c*x+1/2)*ln(
1/2+1/2*c*x)+1/24/c^6*b^2*ln(c*x+1)^2+1/60*b^2*x^4/c^2+4/45*b^2*x^2/c^4+23/90/c^6*b^2*ln(c*x-1)+23/90/c^6*b^2*
ln(c*x+1)+1/3*a*b*x^6*arctanh(c*x)+1/15/c*x^5*a*b+1/9*a*b*x^3/c^3+1/3*a*b*x/c^5+1/6/c^6*a*b*ln(c*x-1)-1/6/c^6*
a*b*ln(c*x+1)

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Maxima [A]  time = 1.01312, size = 290, normalized size = 2. \begin{align*} \frac{1}{6} \, b^{2} x^{6} \operatorname{artanh}\left (c x\right )^{2} + \frac{1}{6} \, a^{2} x^{6} + \frac{1}{90} \,{\left (30 \, x^{6} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{2 \,{\left (3 \, c^{4} x^{5} + 5 \, c^{2} x^{3} + 15 \, x\right )}}{c^{6}} - \frac{15 \, \log \left (c x + 1\right )}{c^{7}} + \frac{15 \, \log \left (c x - 1\right )}{c^{7}}\right )}\right )} a b + \frac{1}{360} \,{\left (4 \, c{\left (\frac{2 \,{\left (3 \, c^{4} x^{5} + 5 \, c^{2} x^{3} + 15 \, x\right )}}{c^{6}} - \frac{15 \, \log \left (c x + 1\right )}{c^{7}} + \frac{15 \, \log \left (c x - 1\right )}{c^{7}}\right )} \operatorname{artanh}\left (c x\right ) + \frac{6 \, c^{4} x^{4} + 32 \, c^{2} x^{2} - 2 \,{\left (15 \, \log \left (c x - 1\right ) - 46\right )} \log \left (c x + 1\right ) + 15 \, \log \left (c x + 1\right )^{2} + 15 \, \log \left (c x - 1\right )^{2} + 92 \, \log \left (c x - 1\right )}{c^{6}}\right )} b^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^2*x^6*arctanh(c*x)^2 + 1/6*a^2*x^6 + 1/90*(30*x^6*arctanh(c*x) + c*(2*(3*c^4*x^5 + 5*c^2*x^3 + 15*x)/c^6
- 15*log(c*x + 1)/c^7 + 15*log(c*x - 1)/c^7))*a*b + 1/360*(4*c*(2*(3*c^4*x^5 + 5*c^2*x^3 + 15*x)/c^6 - 15*log
(c*x + 1)/c^7 + 15*log(c*x - 1)/c^7)*arctanh(c*x) + (6*c^4*x^4 + 32*c^2*x^2 - 2*(15*log(c*x - 1) - 46)*log(c*x
+ 1) + 15*log(c*x + 1)^2 + 15*log(c*x - 1)^2 + 92*log(c*x - 1))/c^6)*b^2

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Fricas [A]  time = 2.18829, size = 437, normalized size = 3.01 \begin{align*} \frac{60 \, a^{2} c^{6} x^{6} + 24 \, a b c^{5} x^{5} + 6 \, b^{2} c^{4} x^{4} + 40 \, a b c^{3} x^{3} + 32 \, b^{2} c^{2} x^{2} + 120 \, a b c x + 15 \,{\left (b^{2} c^{6} x^{6} - b^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )^{2} - 4 \,{\left (15 \, a b - 23 \, b^{2}\right )} \log \left (c x + 1\right ) + 4 \,{\left (15 \, a b + 23 \, b^{2}\right )} \log \left (c x - 1\right ) + 4 \,{\left (15 \, a b c^{6} x^{6} + 3 \, b^{2} c^{5} x^{5} + 5 \, b^{2} c^{3} x^{3} + 15 \, b^{2} c x\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{360 \, c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/360*(60*a^2*c^6*x^6 + 24*a*b*c^5*x^5 + 6*b^2*c^4*x^4 + 40*a*b*c^3*x^3 + 32*b^2*c^2*x^2 + 120*a*b*c*x + 15*(b
^2*c^6*x^6 - b^2)*log(-(c*x + 1)/(c*x - 1))^2 - 4*(15*a*b - 23*b^2)*log(c*x + 1) + 4*(15*a*b + 23*b^2)*log(c*x
- 1) + 4*(15*a*b*c^6*x^6 + 3*b^2*c^5*x^5 + 5*b^2*c^3*x^3 + 15*b^2*c*x)*log(-(c*x + 1)/(c*x - 1)))/c^6

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Sympy [A]  time = 4.11455, size = 211, normalized size = 1.46 \begin{align*} \begin{cases} \frac{a^{2} x^{6}}{6} + \frac{a b x^{6} \operatorname{atanh}{\left (c x \right )}}{3} + \frac{a b x^{5}}{15 c} + \frac{a b x^{3}}{9 c^{3}} + \frac{a b x}{3 c^{5}} - \frac{a b \operatorname{atanh}{\left (c x \right )}}{3 c^{6}} + \frac{b^{2} x^{6} \operatorname{atanh}^{2}{\left (c x \right )}}{6} + \frac{b^{2} x^{5} \operatorname{atanh}{\left (c x \right )}}{15 c} + \frac{b^{2} x^{4}}{60 c^{2}} + \frac{b^{2} x^{3} \operatorname{atanh}{\left (c x \right )}}{9 c^{3}} + \frac{4 b^{2} x^{2}}{45 c^{4}} + \frac{b^{2} x \operatorname{atanh}{\left (c x \right )}}{3 c^{5}} + \frac{23 b^{2} \log{\left (x - \frac{1}{c} \right )}}{45 c^{6}} - \frac{b^{2} \operatorname{atanh}^{2}{\left (c x \right )}}{6 c^{6}} + \frac{23 b^{2} \operatorname{atanh}{\left (c x \right )}}{45 c^{6}} & \text{for}\: c \neq 0 \\\frac{a^{2} x^{6}}{6} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x**6/6 + a*b*x**6*atanh(c*x)/3 + a*b*x**5/(15*c) + a*b*x**3/(9*c**3) + a*b*x/(3*c**5) - a*b*at
anh(c*x)/(3*c**6) + b**2*x**6*atanh(c*x)**2/6 + b**2*x**5*atanh(c*x)/(15*c) + b**2*x**4/(60*c**2) + b**2*x**3*
atanh(c*x)/(9*c**3) + 4*b**2*x**2/(45*c**4) + b**2*x*atanh(c*x)/(3*c**5) + 23*b**2*log(x - 1/c)/(45*c**6) - b*
*2*atanh(c*x)**2/(6*c**6) + 23*b**2*atanh(c*x)/(45*c**6), Ne(c, 0)), (a**2*x**6/6, True))

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Giac [A]  time = 1.30206, size = 259, normalized size = 1.79 \begin{align*} \frac{1}{6} \, a^{2} x^{6} + \frac{a b x^{5}}{15 \, c} + \frac{b^{2} x^{4}}{60 \, c^{2}} + \frac{1}{24} \,{\left (b^{2} x^{6} - \frac{b^{2}}{c^{6}}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )^{2} + \frac{a b x^{3}}{9 \, c^{3}} + \frac{1}{90} \,{\left (15 \, a b x^{6} + \frac{3 \, b^{2} x^{5}}{c} + \frac{5 \, b^{2} x^{3}}{c^{3}} + \frac{15 \, b^{2} x}{c^{5}}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right ) + \frac{4 \, b^{2} x^{2}}{45 \, c^{4}} + \frac{a b x}{3 \, c^{5}} - \frac{{\left (15 \, a b - 23 \, b^{2}\right )} \log \left (c x + 1\right )}{90 \, c^{6}} + \frac{{\left (15 \, a b + 23 \, b^{2}\right )} \log \left (c x - 1\right )}{90 \, c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a^2*x^6 + 1/15*a*b*x^5/c + 1/60*b^2*x^4/c^2 + 1/24*(b^2*x^6 - b^2/c^6)*log(-(c*x + 1)/(c*x - 1))^2 + 1/9*a
*b*x^3/c^3 + 1/90*(15*a*b*x^6 + 3*b^2*x^5/c + 5*b^2*x^3/c^3 + 15*b^2*x/c^5)*log(-(c*x + 1)/(c*x - 1)) + 4/45*b
^2*x^2/c^4 + 1/3*a*b*x/c^5 - 1/90*(15*a*b - 23*b^2)*log(c*x + 1)/c^6 + 1/90*(15*a*b + 23*b^2)*log(c*x - 1)/c^6