3.132 \(\int \frac {\tanh ^{-1}(a x)^3}{x^2 (c+a c x)} \, dx\)

Optimal. Leaf size=191 \[ -\frac {3 a \text {Li}_3\left (\frac {2}{a x+1}-1\right )}{2 c}+\frac {3 a \text {Li}_4\left (\frac {2}{a x+1}-1\right )}{4 c}+\frac {3 a \text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)^2}{2 c}-\frac {3 a \text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)}{c}+\frac {3 a \text {Li}_3\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)}{2 c}+\frac {a \tanh ^{-1}(a x)^3}{c}-\frac {\tanh ^{-1}(a x)^3}{c x}-\frac {a \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c}+\frac {3 a \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^2}{c} \]

[Out]

a*arctanh(a*x)^3/c-arctanh(a*x)^3/c/x+3*a*arctanh(a*x)^2*ln(2-2/(a*x+1))/c-a*arctanh(a*x)^3*ln(2-2/(a*x+1))/c-
3*a*arctanh(a*x)*polylog(2,-1+2/(a*x+1))/c+3/2*a*arctanh(a*x)^2*polylog(2,-1+2/(a*x+1))/c-3/2*a*polylog(3,-1+2
/(a*x+1))/c+3/2*a*arctanh(a*x)*polylog(3,-1+2/(a*x+1))/c+3/4*a*polylog(4,-1+2/(a*x+1))/c

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Rubi [A]  time = 0.46, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {5934, 5916, 5988, 5932, 5948, 6056, 6610, 6060} \[ -\frac {3 a \text {PolyLog}\left (3,\frac {2}{a x+1}-1\right )}{2 c}+\frac {3 a \text {PolyLog}\left (4,\frac {2}{a x+1}-1\right )}{4 c}+\frac {3 a \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )}{2 c}-\frac {3 a \tanh ^{-1}(a x) \text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )}{c}+\frac {3 a \tanh ^{-1}(a x) \text {PolyLog}\left (3,\frac {2}{a x+1}-1\right )}{2 c}+\frac {a \tanh ^{-1}(a x)^3}{c}-\frac {\tanh ^{-1}(a x)^3}{c x}-\frac {a \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c}+\frac {3 a \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^2}{c} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^3/(x^2*(c + a*c*x)),x]

[Out]

(a*ArcTanh[a*x]^3)/c - ArcTanh[a*x]^3/(c*x) + (3*a*ArcTanh[a*x]^2*Log[2 - 2/(1 + a*x)])/c - (a*ArcTanh[a*x]^3*
Log[2 - 2/(1 + a*x)])/c - (3*a*ArcTanh[a*x]*PolyLog[2, -1 + 2/(1 + a*x)])/c + (3*a*ArcTanh[a*x]^2*PolyLog[2, -
1 + 2/(1 + a*x)])/(2*c) - (3*a*PolyLog[3, -1 + 2/(1 + a*x)])/(2*c) + (3*a*ArcTanh[a*x]*PolyLog[3, -1 + 2/(1 +
a*x)])/(2*c) + (3*a*PolyLog[4, -1 + 2/(1 + a*x)])/(4*c)

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 5932

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2 - 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 5934

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/(d*f), Int[((f*x)^(m + 1)*(a + b*ArcTanh[c*x])^p)/(d + e*x
), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0] && LtQ[m, -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]

Rule 5988

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

Rule 6056

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcTa
nh[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 6060

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

Rule 6610

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 \frac {\tanh ^{-1}(a x)^3}{x^2 (c+a c x)} \, dx &=-\left (a \int \frac {\tanh ^{-1}(a x)^3}{x (c+a c x)} \, dx\right )+\frac {\int \frac {\tanh ^{-1}(a x)^3}{x^2} \, dx}{c}\\ &=-\frac {\tanh ^{-1}(a x)^3}{c x}-\frac {a \tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )}{c}+\frac {(3 a) \int \frac {\tanh ^{-1}(a x)^2}{x \left (1-a^2 x^2\right )} \, dx}{c}+\frac {\left (3 a^2\right ) \int \frac {\tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac {a \tanh ^{-1}(a x)^3}{c}-\frac {\tanh ^{-1}(a x)^3}{c x}-\frac {a \tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )}{c}+\frac {3 a \tanh ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{2 c}+\frac {(3 a) \int \frac {\tanh ^{-1}(a x)^2}{x (1+a x)} \, dx}{c}-\frac {\left (3 a^2\right ) \int \frac {\tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac {a \tanh ^{-1}(a x)^3}{c}-\frac {\tanh ^{-1}(a x)^3}{c x}+\frac {3 a \tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )}{c}-\frac {a \tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )}{c}+\frac {3 a \tanh ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{2 c}+\frac {3 a \tanh ^{-1}(a x) \text {Li}_3\left (-1+\frac {2}{1+a x}\right )}{2 c}-\frac {\left (3 a^2\right ) \int \frac {\text {Li}_3\left (-1+\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx}{2 c}-\frac {\left (6 a^2\right ) \int \frac {\tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac {a \tanh ^{-1}(a x)^3}{c}-\frac {\tanh ^{-1}(a x)^3}{c x}+\frac {3 a \tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )}{c}-\frac {a \tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )}{c}-\frac {3 a \tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{c}+\frac {3 a \tanh ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{2 c}+\frac {3 a \tanh ^{-1}(a x) \text {Li}_3\left (-1+\frac {2}{1+a x}\right )}{2 c}+\frac {3 a \text {Li}_4\left (-1+\frac {2}{1+a x}\right )}{4 c}+\frac {\left (3 a^2\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac {a \tanh ^{-1}(a x)^3}{c}-\frac {\tanh ^{-1}(a x)^3}{c x}+\frac {3 a \tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )}{c}-\frac {a \tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )}{c}-\frac {3 a \tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{c}+\frac {3 a \tanh ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{2 c}-\frac {3 a \text {Li}_3\left (-1+\frac {2}{1+a x}\right )}{2 c}+\frac {3 a \tanh ^{-1}(a x) \text {Li}_3\left (-1+\frac {2}{1+a x}\right )}{2 c}+\frac {3 a \text {Li}_4\left (-1+\frac {2}{1+a x}\right )}{4 c}\\ \end {align*}

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Mathematica [C]  time = 0.29, size = 154, normalized size = 0.81 \[ \frac {a \left (-\frac {3}{2} \left (\tanh ^{-1}(a x)-2\right ) \tanh ^{-1}(a x) \text {Li}_2\left (e^{2 \tanh ^{-1}(a x)}\right )+\frac {3}{2} \left (\tanh ^{-1}(a x)-1\right ) \text {Li}_3\left (e^{2 \tanh ^{-1}(a x)}\right )-\frac {3}{4} \text {Li}_4\left (e^{2 \tanh ^{-1}(a x)}\right )+\frac {1}{2} \tanh ^{-1}(a x)^4-\frac {\tanh ^{-1}(a x)^3}{a x}-\tanh ^{-1}(a x)^3-\tanh ^{-1}(a x)^3 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x)^2 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )-\frac {\pi ^4}{64}+\frac {i \pi ^3}{8}\right )}{c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*((I/8)*Pi^3 - Pi^4/64 - ArcTanh[a*x]^3 - ArcTanh[a*x]^3/(a*x) + ArcTanh[a*x]^4/2 + 3*ArcTanh[a*x]^2*Log[1 -
E^(2*ArcTanh[a*x])] - ArcTanh[a*x]^3*Log[1 - E^(2*ArcTanh[a*x])] - (3*(-2 + ArcTanh[a*x])*ArcTanh[a*x]*PolyLo
g[2, E^(2*ArcTanh[a*x])])/2 + (3*(-1 + ArcTanh[a*x])*PolyLog[3, E^(2*ArcTanh[a*x])])/2 - (3*PolyLog[4, E^(2*Ar
cTanh[a*x])])/4))/c

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (a x\right )^{3}}{a c x^{3} + c x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^3/x^2/(a*c*x+c),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (a x\right )^{3}}{{\left (a c x + c\right )} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.78, size = 1451, normalized size = 7.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)^3/x^2/(a*c*x+c),x)

[Out]

-I*a/c*arctanh(a*x)^3*Pi*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2-1/2*I*a/c*Pi*csgn(
I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*arctanh(a*x)^3-1/2*I*
a/c*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1))*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)
/(1+(a*x+1)^2/(-a^2*x^2+1)))*arctanh(a*x)^3+1/2*I*a/c*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)^2/(
a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))*arctanh(a*x)^3+1/2*I*a/c*Pi*csgn(I/(1+(a*
x+1)^2/(-a^2*x^2+1)))*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*arctanh(a*x)^3+1/2*I*a/c
*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1))*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*arctanh
(a*x)^3+1/2*I*a/c*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*
arctanh(a*x)^3-1/2*I*a/c*Pi*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*arctanh(a*x)^3-
arctanh(a*x)^3/c/x-6*a/c*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))-6*a/c*polylog(4,(a*x+1)/(-a^2*x^2+1)^(1/2))-6*a
/c*polylog(4,-(a*x+1)/(-a^2*x^2+1)^(1/2))-6*a/c*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))+1/2*a*arctanh(a*x)^4/c-
a*arctanh(a*x)^3/c-1/2*I*a/c*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*arctanh(a*x)^3
-1/2*I*a/c*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*arctanh(a*x)^3-1/2*I*a/c*arctanh(a*x)
^3*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3-3*a/c*arctanh(a*x)^2*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))+6*a/c*arctanh
(a*x)*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))-2*a/c*arctanh(a*x)^3*ln((a*x+1)/(-a^2*x^2+1)^(1/2))+a/c*arctanh(a*
x)^3*ln((a*x+1)^2/(-a^2*x^2+1)-1)-a/c*arctanh(a*x)^3*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))-3*a/c*arctanh(a*x)^2*pol
ylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))+6*a/c*arctanh(a*x)*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))-a/c*arctanh(a*x)
^3*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))-a/c*ln(2)*arctanh(a*x)^3+3*a/c*arctanh(a*x)^2*ln(1-(a*x+1)/(-a^2*x^2+1)^(1
/2))+6*a/c*arctanh(a*x)*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))+3*a/c*arctanh(a*x)^2*ln(1+(a*x+1)/(-a^2*x^2+1)^(
1/2))+6*a/c*arctanh(a*x)*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))-a/c*arctanh(a*x)^3*ln(a*x)+a/c*arctanh(a*x)^3*
ln(a*x+1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (a x \log \left (a x + 1\right ) - 1\right )} \log \left (-a x + 1\right )^{3}}{8 \, c x} + \frac {1}{8} \, \int \frac {{\left (a x - 1\right )} \log \left (a x + 1\right )^{3} - 3 \, {\left (a x - 1\right )} \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right ) - 3 \, {\left (a^{2} x^{2} + a x - {\left (a^{3} x^{3} + a^{2} x^{2} + a x - 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{2}}{a^{2} c x^{4} - c x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atanh}\left (a\,x\right )}^3}{x^2\,\left (c+a\,c\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^3/(x^2*(c + a*c*x)),x)

[Out]

int(atanh(a*x)^3/(x^2*(c + a*c*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atanh}^{3}{\left (a x \right )}}{a x^{3} + x^{2}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)**3/x**2/(a*c*x+c),x)

[Out]

Integral(atanh(a*x)**3/(a*x**3 + x**2), x)/c

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