3.1.24 \(\int x^5 (a+b \tanh ^{-1}(c x))^3 \, dx\) [24]

Optimal. Leaf size=247 \[ \frac {19 b^3 x}{60 c^5}+\frac {b^3 x^3}{60 c^3}-\frac {19 b^3 \tanh ^{-1}(c x)}{60 c^6}+\frac {4 b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{15 c^4}+\frac {b^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )}{20 c^2}+\frac {23 b \left (a+b \tanh ^{-1}(c x)\right )^2}{30 c^6}+\frac {b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^3}+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{6 c^6}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {23 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{15 c^6}-\frac {23 b^3 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{30 c^6} \]

[Out]

19/60*b^3*x/c^5+1/60*b^3*x^3/c^3-19/60*b^3*arctanh(c*x)/c^6+4/15*b^2*x^2*(a+b*arctanh(c*x))/c^4+1/20*b^2*x^4*(
a+b*arctanh(c*x))/c^2+23/30*b*(a+b*arctanh(c*x))^2/c^6+1/2*b*x*(a+b*arctanh(c*x))^2/c^5+1/6*b*x^3*(a+b*arctanh
(c*x))^2/c^3+1/10*b*x^5*(a+b*arctanh(c*x))^2/c-1/6*(a+b*arctanh(c*x))^3/c^6+1/6*x^6*(a+b*arctanh(c*x))^3-23/15
*b^2*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c^6-23/30*b^3*polylog(2,1-2/(-c*x+1))/c^6

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Rubi [A]
time = 0.67, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 33, number of rules used = 11, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {6037, 6127, 308, 212, 327, 6131, 6055, 2449, 2352, 6021, 6095} \begin {gather*} -\frac {23 b^2 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{15 c^6}+\frac {4 b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{15 c^4}+\frac {b^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )}{20 c^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{6 c^6}+\frac {23 b \left (a+b \tanh ^{-1}(c x)\right )^2}{30 c^6}+\frac {b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^3}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c}-\frac {23 b^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{30 c^6}-\frac {19 b^3 \tanh ^{-1}(c x)}{60 c^6}+\frac {19 b^3 x}{60 c^5}+\frac {b^3 x^3}{60 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*ArcTanh[c*x])^3,x]

[Out]

(19*b^3*x)/(60*c^5) + (b^3*x^3)/(60*c^3) - (19*b^3*ArcTanh[c*x])/(60*c^6) + (4*b^2*x^2*(a + b*ArcTanh[c*x]))/(
15*c^4) + (b^2*x^4*(a + b*ArcTanh[c*x]))/(20*c^2) + (23*b*(a + b*ArcTanh[c*x])^2)/(30*c^6) + (b*x*(a + b*ArcTa
nh[c*x])^2)/(2*c^5) + (b*x^3*(a + b*ArcTanh[c*x])^2)/(6*c^3) + (b*x^5*(a + b*ArcTanh[c*x])^2)/(10*c) - (a + b*
ArcTanh[c*x])^3/(6*c^6) + (x^6*(a + b*ArcTanh[c*x])^3)/6 - (23*b^2*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(15*
c^6) - (23*b^3*PolyLog[2, 1 - 2/(1 - c*x)])/(30*c^6)

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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*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 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 6127

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 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]

Rubi steps

\begin {align*} \int x^5 \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {1}{2} (b c) \int \frac {x^6 \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx\\ &=\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {b \int x^4 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{2 c}-\frac {b \int \frac {x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{2 c}\\ &=\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {1}{5} b^2 \int \frac {x^5 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx+\frac {b \int x^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{2 c^3}-\frac {b \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{2 c^3}\\ &=\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^3}+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {b \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{2 c^5}-\frac {b \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{2 c^5}+\frac {b^2 \int x^3 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{5 c^2}-\frac {b^2 \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{5 c^2}-\frac {b^2 \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{3 c^2}\\ &=\frac {b^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )}{20 c^2}+\frac {b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^3}+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{6 c^6}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {b^2 \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{5 c^4}-\frac {b^2 \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{5 c^4}+\frac {b^2 \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{3 c^4}-\frac {b^2 \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{3 c^4}-\frac {b^2 \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c^4}-\frac {b^3 \int \frac {x^4}{1-c^2 x^2} \, dx}{20 c}\\ &=\frac {4 b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{15 c^4}+\frac {b^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )}{20 c^2}+\frac {23 b \left (a+b \tanh ^{-1}(c x)\right )^2}{30 c^6}+\frac {b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^3}+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{6 c^6}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {b^2 \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{5 c^5}-\frac {b^2 \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{3 c^5}-\frac {b^2 \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{c^5}-\frac {b^3 \int \frac {x^2}{1-c^2 x^2} \, dx}{10 c^3}-\frac {b^3 \int \frac {x^2}{1-c^2 x^2} \, dx}{6 c^3}-\frac {b^3 \int \left (-\frac {1}{c^4}-\frac {x^2}{c^2}+\frac {1}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx}{20 c}\\ &=\frac {19 b^3 x}{60 c^5}+\frac {b^3 x^3}{60 c^3}+\frac {4 b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{15 c^4}+\frac {b^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )}{20 c^2}+\frac {23 b \left (a+b \tanh ^{-1}(c x)\right )^2}{30 c^6}+\frac {b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^3}+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{6 c^6}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {23 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{15 c^6}-\frac {b^3 \int \frac {1}{1-c^2 x^2} \, dx}{20 c^5}-\frac {b^3 \int \frac {1}{1-c^2 x^2} \, dx}{10 c^5}-\frac {b^3 \int \frac {1}{1-c^2 x^2} \, dx}{6 c^5}+\frac {b^3 \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{5 c^5}+\frac {b^3 \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{3 c^5}+\frac {b^3 \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c^5}\\ &=\frac {19 b^3 x}{60 c^5}+\frac {b^3 x^3}{60 c^3}-\frac {19 b^3 \tanh ^{-1}(c x)}{60 c^6}+\frac {4 b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{15 c^4}+\frac {b^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )}{20 c^2}+\frac {23 b \left (a+b \tanh ^{-1}(c x)\right )^2}{30 c^6}+\frac {b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^3}+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{6 c^6}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {23 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{15 c^6}-\frac {b^3 \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{5 c^6}-\frac {b^3 \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{3 c^6}-\frac {b^3 \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{c^6}\\ &=\frac {19 b^3 x}{60 c^5}+\frac {b^3 x^3}{60 c^3}-\frac {19 b^3 \tanh ^{-1}(c x)}{60 c^6}+\frac {4 b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{15 c^4}+\frac {b^2 x^4 \left (a+b \tanh ^{-1}(c x)\right )}{20 c^2}+\frac {23 b \left (a+b \tanh ^{-1}(c x)\right )^2}{30 c^6}+\frac {b x \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^3}+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{6 c^6}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {23 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{15 c^6}-\frac {23 b^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{30 c^6}\\ \end {align*}

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Mathematica [A]
time = 0.51, size = 305, normalized size = 1.23 \begin {gather*} \frac {-19 a b^2+30 a^2 b c x+19 b^3 c x+16 a b^2 c^2 x^2+10 a^2 b c^3 x^3+b^3 c^3 x^3+3 a b^2 c^4 x^4+6 a^2 b c^5 x^5+10 a^3 c^6 x^6+2 b^2 \left (b \left (-23+15 c x+5 c^3 x^3+3 c^5 x^5\right )+15 a \left (-1+c^6 x^6\right )\right ) \tanh ^{-1}(c x)^2+10 b^3 \left (-1+c^6 x^6\right ) \tanh ^{-1}(c x)^3+b \tanh ^{-1}(c x) \left (30 a^2 c^6 x^6+4 a b c x \left (15+5 c^2 x^2+3 c^4 x^4\right )+b^2 \left (-19+16 c^2 x^2+3 c^4 x^4\right )-92 b^2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )+15 a^2 b \log (1-c x)-15 a^2 b \log (1+c x)+46 a b^2 \log \left (1-c^2 x^2\right )+46 b^3 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )}{60 c^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-19*a*b^2 + 30*a^2*b*c*x + 19*b^3*c*x + 16*a*b^2*c^2*x^2 + 10*a^2*b*c^3*x^3 + b^3*c^3*x^3 + 3*a*b^2*c^4*x^4 +
6*a^2*b*c^5*x^5 + 10*a^3*c^6*x^6 + 2*b^2*(b*(-23 + 15*c*x + 5*c^3*x^3 + 3*c^5*x^5) + 15*a*(-1 + c^6*x^6))*Arc
Tanh[c*x]^2 + 10*b^3*(-1 + c^6*x^6)*ArcTanh[c*x]^3 + b*ArcTanh[c*x]*(30*a^2*c^6*x^6 + 4*a*b*c*x*(15 + 5*c^2*x^
2 + 3*c^4*x^4) + b^2*(-19 + 16*c^2*x^2 + 3*c^4*x^4) - 92*b^2*Log[1 + E^(-2*ArcTanh[c*x])]) + 15*a^2*b*Log[1 -
c*x] - 15*a^2*b*Log[1 + c*x] + 46*a*b^2*Log[1 - c^2*x^2] + 46*b^3*PolyLog[2, -E^(-2*ArcTanh[c*x])])/(60*c^6)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 2.72, size = 1242, normalized size = 5.03

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1242\)
default	\(\text {Expression too large to display}\)	\(1242\)
risch	\(\text {Expression too large to display}\)	\(1362\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arctanh(c*x))^3,x,method=_RETURNVERBOSE)

[Out]

1/c^6*(1/6*c^6*x^6*a^3+1/4*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^
2-1))^2-1/3*b^3+1/8*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))
+1/8*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^
2*x^2+1)))^2-1/8*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)
^2/(-c^2*x^2+1)))^2+1/4*a^2*b*ln(c*x-1)-1/4*a^2*b*ln(c*x+1)+1/2*b^3*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/
2))+1/4*b^3*arctanh(c*x)^2*ln(c*x-1)-1/4*b^3*arctanh(c*x)^2*ln(c*x+1)-23/15*b^3*arctanh(c*x)*ln(1+I*(c*x+1)/(-
c^2*x^2+1)^(1/2))-23/15*b^3*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+1/8*a*b^2*ln(c*x-1)^2+1/8*a*b^2*ln
(c*x+1)^2+23/30*a*b^2*ln(c*x-1)+23/30*a*b^2*ln(c*x+1)+19/60*b^3*c*x+1/60*b^3*c^3*x^3-1/2*a*b^2*arctanh(c*x)*ln
(c*x+1)-1/4*a*b^2*ln(c*x-1)*ln(1/2*c*x+1/2)-1/4*I*b^3*arctanh(c*x)^2*Pi+1/10*a^2*b*c^5*x^5+1/6*a^2*b*c^3*x^3+1
/2*a^2*b*c*x+1/10*b^3*c^5*x^5*arctanh(c*x)^2+1/6*b^3*c^6*x^6*arctanh(c*x)^3+1/6*b^3*c^3*x^3*arctanh(c*x)^2+1/2
*b^3*c*x*arctanh(c*x)^2+4/15*b^3*arctanh(c*x)*c^2*x^2+1/20*b^3*arctanh(c*x)*c^4*x^4+1/20*a*b^2*c^4*x^4+4/15*a*
b^2*c^2*x^2-1/4*a*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/4*a*b^2*ln(-1/2*c*x+1/2)*ln(1/2*c*x+1/2)+1/2*a*b^2*arctanh(
c*x)*ln(c*x-1)-19/60*b^3*arctanh(c*x)-23/15*b^3*dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))-23/15*b^3*dilog(1+I*(c*x
+1)/(-c^2*x^2+1)^(1/2))+23/30*b^3*arctanh(c*x)^2-1/6*b^3*arctanh(c*x)^3-1/8*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+
(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))
)+1/8*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3+1/2*a^2*b*c^6*x^6*arctanh(c*x)+1/2*a*b^2*c^6*x^6
*arctanh(c*x)^2+1/5*a*b^2*c^5*x^5*arctanh(c*x)+1/3*a*b^2*c^3*x^3*arctanh(c*x)+a*b^2*c*x*arctanh(c*x)+1/4*I*b^3
*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))^2-1/4*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*
x^2+1)))^3+1/8*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arctanh(c*x))^3,x, algorithm="maxima")

[Out]

1/2*a*b^2*x^6*arctanh(c*x)^2 + 1/6*a^3*x^6 + 1/60*(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^2*b + 1/120*(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)*a*b^2 - 1/1728000*(500*c^7*((2*c^4*x
^6 + 3*c^2*x^4 + 6*x^2)/c^11 + 6*log(c^2*x^2 - 1)/c^13) + 728*c^6*(2*(3*c^4*x^5 + 5*c^2*x^3 + 15*x)/c^11 - 15*
log(c*x + 1)/c^12 + 15*log(c*x - 1)/c^12) + 1485*c^5*((c^2*x^4 + 2*x^2)/c^9 + 2*log(c^2*x^2 - 1)/c^11) - 62208
0000*c^5*integrate(1/3600*x^5*log(c*x + 1)/(c^7*x^2 - c^5), x) + 9750*c^4*(2*(c^2*x^3 + 3*x)/c^9 - 3*log(c*x +
1)/c^10 + 3*log(c*x - 1)/c^10) - 2700*c^3*(x^2/c^7 + log(c^2*x^2 - 1)/c^9) - 1036800000*c^3*integrate(1/3600*
x^3*log(c*x + 1)/(c^7*x^2 - c^5), x) + 227700*c^2*(2*x/c^7 - log(c*x + 1)/c^8 + log(c*x - 1)/c^8) - 5495040000
*c*integrate(1/3600*x*log(c*x + 1)/(c^7*x^2 - c^5), x) + (1000*(36*log(-c*x + 1)^3 - 18*log(-c*x + 1)^2 + 6*lo
g(-c*x + 1) - 1)*(c*x - 1)^6 + 1728*(125*log(-c*x + 1)^3 - 75*log(-c*x + 1)^2 + 30*log(-c*x + 1) - 6)*(c*x - 1
)^5 + 16875*(32*log(-c*x + 1)^3 - 24*log(-c*x + 1)^2 + 12*log(-c*x + 1) - 3)*(c*x - 1)^4 + 80000*(9*log(-c*x +
1)^3 - 9*log(-c*x + 1)^2 + 6*log(-c*x + 1) - 2)*(c*x - 1)^3 + 135000*(4*log(-c*x + 1)^3 - 6*log(-c*x + 1)^2 +
6*log(-c*x + 1) - 3)*(c*x - 1)^2 + 216000*(log(-c*x + 1)^3 - 3*log(-c*x + 1)^2 + 6*log(-c*x + 1) - 6)*(c*x -
1))/c^6 - 60*(600*(c^6*x^6 - 1)*log(c*x + 1)^3 + 240*(3*c^5*x^5 + 5*c^3*x^3 + 15*c*x)*log(c*x + 1)^2 - 30*(10*
c^6*x^6 - 12*c^5*x^5 + 15*c^4*x^4 - 20*c^3*x^3 + 30*c^2*x^2 - 60*c*x - 60*(c^6*x^6 - 1)*log(c*x + 1) + 37)*log
(-c*x + 1)^2 + (100*c^6*x^6 + 264*c^5*x^5 - 165*c^4*x^4 + 1140*c^3*x^3 - 1230*c^2*x^2 - 1800*(c^6*x^6 - 1)*log
(c*x + 1)^2 + 8820*c*x - 480*(3*c^5*x^5 + 5*c^3*x^3 + 15*c*x + 23)*log(c*x + 1))*log(-c*x + 1))/c^6 + 264600*l
og(3600*c^7*x^2 - 3600*c^5)/c^6 - 2384640000*integrate(1/3600*log(c*x + 1)/(c^7*x^2 - c^5), x))*b^3

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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^5*(a+b*arctanh(c*x))^3,x, algorithm="fricas")

[Out]

integral(b^3*x^5*arctanh(c*x)^3 + 3*a*b^2*x^5*arctanh(c*x)^2 + 3*a^2*b*x^5*arctanh(c*x) + a^3*x^5, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{5} \left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*atanh(c*x))**3,x)

[Out]

Integral(x**5*(a + b*atanh(c*x))**3, 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^5*(a+b*arctanh(c*x))^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^3*x^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^5\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*atanh(c*x))^3,x)

[Out]

int(x^5*(a + b*atanh(c*x))^3, x)

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