3.1.27 \(\int x^2 (a+b \tanh ^{-1}(c x))^3 \, dx\) [27]

Optimal. Leaf size=197 \[ \frac {a b^2 x}{c^2}+\frac {b^3 x \tanh ^{-1}(c x)}{c^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c^3}+\frac {b^3 \log \left (1-c^2 x^2\right )}{2 c^3}-\frac {b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^3}+\frac {b^3 \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 c^3} \]

[Out]

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

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Rubi [A]
time = 0.36, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6037, 6127, 6021, 266, 6095, 6131, 6055, 6205, 6745} \begin {gather*} -\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3}+\frac {a b^2 x}{c^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c^3}-\frac {b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 c^3}+\frac {b^3 x \tanh ^{-1}(c x)}{c^2}+\frac {b^3 \log \left (1-c^2 x^2\right )}{2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcTanh[c*x])^3,x]

[Out]

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

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 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]

Rule 6205

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

Rule 6745

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int x^2 \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^3-(b c) \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx\\ &=\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {b \int x \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c}-\frac {b \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{c}\\ &=\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^3-b^2 \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx-\frac {b \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c x} \, dx}{c^2}\\ &=\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c^3}+\frac {b^2 \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c^2}-\frac {b^2 \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{c^2}+\frac {\left (2 b^2\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c^2}\\ &=\frac {a b^2 x}{c^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c^3}-\frac {b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^3}+\frac {b^3 \int \tanh ^{-1}(c x) \, dx}{c^2}+\frac {b^3 \int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c^2}\\ &=\frac {a b^2 x}{c^2}+\frac {b^3 x \tanh ^{-1}(c x)}{c^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c^3}-\frac {b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^3}+\frac {b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 c^3}-\frac {b^3 \int \frac {x}{1-c^2 x^2} \, dx}{c}\\ &=\frac {a b^2 x}{c^2}+\frac {b^3 x \tanh ^{-1}(c x)}{c^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{c^3}+\frac {b^3 \log \left (1-c^2 x^2\right )}{2 c^3}-\frac {b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^3}+\frac {b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.37, size = 250, normalized size = 1.27 \begin {gather*} \frac {3 a^2 b c^2 x^2+2 a^3 c^3 x^3+6 a^2 b c^3 x^3 \tanh ^{-1}(c x)+3 a^2 b \log \left (1-c^2 x^2\right )+6 a b^2 \left (c x+\left (-1+c^3 x^3\right ) \tanh ^{-1}(c x)^2+\tanh ^{-1}(c x) \left (-1+c^2 x^2-2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )+\text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )\right )+b^3 \left (6 c x \tanh ^{-1}(c x)-3 \tanh ^{-1}(c x)^2+3 c^2 x^2 \tanh ^{-1}(c x)^2-2 \tanh ^{-1}(c x)^3+2 c^3 x^3 \tanh ^{-1}(c x)^3-6 \tanh ^{-1}(c x)^2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+3 \log \left (1-c^2 x^2\right )+6 \tanh ^{-1}(c x) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+3 \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )\right )}{6 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcTanh[c*x])^3,x]

[Out]

(3*a^2*b*c^2*x^2 + 2*a^3*c^3*x^3 + 6*a^2*b*c^3*x^3*ArcTanh[c*x] + 3*a^2*b*Log[1 - c^2*x^2] + 6*a*b^2*(c*x + (-
1 + c^3*x^3)*ArcTanh[c*x]^2 + ArcTanh[c*x]*(-1 + c^2*x^2 - 2*Log[1 + E^(-2*ArcTanh[c*x])]) + PolyLog[2, -E^(-2
*ArcTanh[c*x])]) + b^3*(6*c*x*ArcTanh[c*x] - 3*ArcTanh[c*x]^2 + 3*c^2*x^2*ArcTanh[c*x]^2 - 2*ArcTanh[c*x]^3 +
2*c^3*x^3*ArcTanh[c*x]^3 - 6*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + 3*Log[1 - c^2*x^2] + 6*ArcTanh[c*x]
*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 3*PolyLog[3, -E^(-2*ArcTanh[c*x])]))/(6*c^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.21, size = 1093, normalized size = 5.55

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1093\)
default	\(\text {Expression too large to display}\)	\(1093\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(1/2*a^2*b*c^2*x^2+1/2*b^3*arctanh(c*x)^2*c^2*x^2+b^3*arctanh(c*x)*c*x-a*b^2*dilog(1/2*c*x+1/2)-b^3*arct
anh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))-b^3*ln(2)*arctanh(c*x)^2+a*b^2*c^2*x^2*arctanh(c*x)-b^3*ln(1+(c*x+
1)^2/(-c^2*x^2+1))+1/2*b^3*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))+1/2*a^2*b*ln(c*x-1)+1/2*a^2*b*ln(c*x+1)-b^3*arct
anh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))+1/2*b^3*arctanh(c*x)^2*ln(c*x-1)+1/2*b^3*arctanh(c*x)^2*ln(c*x+1)+1/
4*a*b^2*ln(c*x-1)^2-1/4*a*b^2*ln(c*x+1)^2+1/2*a*b^2*ln(c*x-1)-1/2*a*b^2*ln(c*x+1)+a*b^2*arctanh(c*x)*ln(c*x+1)
-1/2*a*b^2*ln(c*x-1)*ln(1/2*c*x+1/2)+1/2*a*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)-1/2*a*b^2*ln(-1/2*c*x+1/2)*ln(1/2*c*
x+1/2)+a*b^2*arctanh(c*x)*ln(c*x-1)+1/4*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)))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))-1/2*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*csgn(I*
(c*x+1)/(-c^2*x^2+1)^(1/2))-1/4*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)/(-c^2*x^2
+1)^(1/2))^2+b^3*arctanh(c*x)-1/2*b^3*arctanh(c*x)^2+1/3*b^3*arctanh(c*x)^3+a*b^2*c^3*x^3*arctanh(c*x)^2+a^2*b
*c^3*x^3*arctanh(c*x)-1/4*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)))^3-1
/2*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))^3-1/4*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(
c^2*x^2-1))^3+1/2*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*csg
n(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))+1/4*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)))*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c
*x+1)^2/(c^2*x^2-1))-1/2*I*b^3*arctanh(c*x)^2*Pi+1/3*b^3*c^3*x^3*arctanh(c*x)^3+1/3*a^3*c^3*x^3+b^2*c*x*a)

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

[Out]

1/3*a^3*x^3 + 1/2*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*a^2*b - 1/24*((b^3*c^3*x^3 - b^3)*
log(-c*x + 1)^3 - 3*(2*a*b^2*c^3*x^3 + b^3*c^2*x^2 + (b^3*c^3*x^3 + b^3)*log(c*x + 1))*log(-c*x + 1)^2)/c^3 -
integrate(-1/8*((b^3*c^3*x^3 - b^3*c^2*x^2)*log(c*x + 1)^3 + 6*(a*b^2*c^3*x^3 - a*b^2*c^2*x^2)*log(c*x + 1)^2
- (4*a*b^2*c^3*x^3 + 2*b^3*c^2*x^2 + 3*(b^3*c^3*x^3 - b^3*c^2*x^2)*log(c*x + 1)^2 - 2*(6*a*b^2*c^2*x^2 - (6*a*
b^2*c^3 + b^3*c^3)*x^3 - b^3)*log(c*x + 1))*log(-c*x + 1))/(c^3*x - c^2), x)

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

[Out]

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

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\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^2*(a + b*atanh(c*x))^3,x)

[Out]

int(x^2*(a + b*atanh(c*x))^3, x)

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