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

Optimal. Leaf size=192 \[ -\frac {3 \tanh ^{-1}(a x)^3}{32 a^4}+\frac {27 \tanh ^{-1}(a x)}{256 a^4}+\frac {x^4 \tanh ^{-1}(a x)^3}{4 \left (1-a^2 x^2\right )^2}+\frac {3 x^4 \tanh ^{-1}(a x)}{32 \left (1-a^2 x^2\right )^2}-\frac {3 x^3}{128 a \left (1-a^2 x^2\right )^2}-\frac {3 x^3 \tanh ^{-1}(a x)^2}{16 a \left (1-a^2 x^2\right )^2}-\frac {9 \tanh ^{-1}(a x)}{32 a^4 \left (1-a^2 x^2\right )}+\frac {45 x}{256 a^3 \left (1-a^2 x^2\right )}+\frac {9 x \tanh ^{-1}(a x)^2}{32 a^3 \left (1-a^2 x^2\right )} \]

[Out]

-3/128*x^3/a/(-a^2*x^2+1)^2+45/256*x/a^3/(-a^2*x^2+1)+27/256*arctanh(a*x)/a^4+3/32*x^4*arctanh(a*x)/(-a^2*x^2+
1)^2-9/32*arctanh(a*x)/a^4/(-a^2*x^2+1)-3/16*x^3*arctanh(a*x)^2/a/(-a^2*x^2+1)^2+9/32*x*arctanh(a*x)^2/a^3/(-a
^2*x^2+1)-3/32*arctanh(a*x)^3/a^4+1/4*x^4*arctanh(a*x)^3/(-a^2*x^2+1)^2

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Rubi [A]  time = 0.27, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6008, 6004, 6000, 5994, 199, 206, 288} \[ -\frac {3 x^3}{128 a \left (1-a^2 x^2\right )^2}+\frac {45 x}{256 a^3 \left (1-a^2 x^2\right )}+\frac {x^4 \tanh ^{-1}(a x)^3}{4 \left (1-a^2 x^2\right )^2}+\frac {3 x^4 \tanh ^{-1}(a x)}{32 \left (1-a^2 x^2\right )^2}-\frac {3 x^3 \tanh ^{-1}(a x)^2}{16 a \left (1-a^2 x^2\right )^2}+\frac {9 x \tanh ^{-1}(a x)^2}{32 a^3 \left (1-a^2 x^2\right )}-\frac {9 \tanh ^{-1}(a x)}{32 a^4 \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)^3}{32 a^4}+\frac {27 \tanh ^{-1}(a x)}{256 a^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcTanh[a*x]^3)/(1 - a^2*x^2)^3,x]

[Out]

(-3*x^3)/(128*a*(1 - a^2*x^2)^2) + (45*x)/(256*a^3*(1 - a^2*x^2)) + (27*ArcTanh[a*x])/(256*a^4) + (3*x^4*ArcTa
nh[a*x])/(32*(1 - a^2*x^2)^2) - (9*ArcTanh[a*x])/(32*a^4*(1 - a^2*x^2)) - (3*x^3*ArcTanh[a*x]^2)/(16*a*(1 - a^
2*x^2)^2) + (9*x*ArcTanh[a*x]^2)/(32*a^3*(1 - a^2*x^2)) - (3*ArcTanh[a*x]^3)/(32*a^4) + (x^4*ArcTanh[a*x]^3)/(
4*(1 - a^2*x^2)^2)

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 5994

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

Rule 6000

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^2)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> -Simp[(a + b*ArcT
anh[c*x])^(p + 1)/(2*b*c^3*d^2*(p + 1)), x] + (-Dist[(b*p)/(2*c), Int[(x*(a + b*ArcTanh[c*x])^(p - 1))/(d + e*
x^2)^2, x], x] + Simp[(x*(a + b*ArcTanh[c*x])^p)/(2*c^2*d*(d + e*x^2)), x]) /; FreeQ[{a, b, c, d, e}, x] && Eq
Q[c^2*d + e, 0] && GtQ[p, 0]

Rule 6004

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

Rule 6008

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Sim
p[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(m + 1), Int[(f*x)
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d
+ e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {x^3 \tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^3} \, dx &=\frac {x^4 \tanh ^{-1}(a x)^3}{4 \left (1-a^2 x^2\right )^2}-\frac {1}{4} (3 a) \int \frac {x^4 \tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^3} \, dx\\ &=\frac {3 x^4 \tanh ^{-1}(a x)}{32 \left (1-a^2 x^2\right )^2}-\frac {3 x^3 \tanh ^{-1}(a x)^2}{16 a \left (1-a^2 x^2\right )^2}+\frac {x^4 \tanh ^{-1}(a x)^3}{4 \left (1-a^2 x^2\right )^2}+\frac {9 \int \frac {x^2 \tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx}{16 a}-\frac {1}{32} (3 a) \int \frac {x^4}{\left (1-a^2 x^2\right )^3} \, dx\\ &=-\frac {3 x^3}{128 a \left (1-a^2 x^2\right )^2}+\frac {3 x^4 \tanh ^{-1}(a x)}{32 \left (1-a^2 x^2\right )^2}-\frac {3 x^3 \tanh ^{-1}(a x)^2}{16 a \left (1-a^2 x^2\right )^2}+\frac {9 x \tanh ^{-1}(a x)^2}{32 a^3 \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)^3}{32 a^4}+\frac {x^4 \tanh ^{-1}(a x)^3}{4 \left (1-a^2 x^2\right )^2}-\frac {9 \int \frac {x \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx}{16 a^2}+\frac {9 \int \frac {x^2}{\left (1-a^2 x^2\right )^2} \, dx}{128 a}\\ &=-\frac {3 x^3}{128 a \left (1-a^2 x^2\right )^2}+\frac {9 x}{256 a^3 \left (1-a^2 x^2\right )}+\frac {3 x^4 \tanh ^{-1}(a x)}{32 \left (1-a^2 x^2\right )^2}-\frac {9 \tanh ^{-1}(a x)}{32 a^4 \left (1-a^2 x^2\right )}-\frac {3 x^3 \tanh ^{-1}(a x)^2}{16 a \left (1-a^2 x^2\right )^2}+\frac {9 x \tanh ^{-1}(a x)^2}{32 a^3 \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)^3}{32 a^4}+\frac {x^4 \tanh ^{-1}(a x)^3}{4 \left (1-a^2 x^2\right )^2}-\frac {9 \int \frac {1}{1-a^2 x^2} \, dx}{256 a^3}+\frac {9 \int \frac {1}{\left (1-a^2 x^2\right )^2} \, dx}{32 a^3}\\ &=-\frac {3 x^3}{128 a \left (1-a^2 x^2\right )^2}+\frac {45 x}{256 a^3 \left (1-a^2 x^2\right )}-\frac {9 \tanh ^{-1}(a x)}{256 a^4}+\frac {3 x^4 \tanh ^{-1}(a x)}{32 \left (1-a^2 x^2\right )^2}-\frac {9 \tanh ^{-1}(a x)}{32 a^4 \left (1-a^2 x^2\right )}-\frac {3 x^3 \tanh ^{-1}(a x)^2}{16 a \left (1-a^2 x^2\right )^2}+\frac {9 x \tanh ^{-1}(a x)^2}{32 a^3 \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)^3}{32 a^4}+\frac {x^4 \tanh ^{-1}(a x)^3}{4 \left (1-a^2 x^2\right )^2}+\frac {9 \int \frac {1}{1-a^2 x^2} \, dx}{64 a^3}\\ &=-\frac {3 x^3}{128 a \left (1-a^2 x^2\right )^2}+\frac {45 x}{256 a^3 \left (1-a^2 x^2\right )}+\frac {27 \tanh ^{-1}(a x)}{256 a^4}+\frac {3 x^4 \tanh ^{-1}(a x)}{32 \left (1-a^2 x^2\right )^2}-\frac {9 \tanh ^{-1}(a x)}{32 a^4 \left (1-a^2 x^2\right )}-\frac {3 x^3 \tanh ^{-1}(a x)^2}{16 a \left (1-a^2 x^2\right )^2}+\frac {9 x \tanh ^{-1}(a x)^2}{32 a^3 \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)^3}{32 a^4}+\frac {x^4 \tanh ^{-1}(a x)^3}{4 \left (1-a^2 x^2\right )^2}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 135, normalized size = 0.70 \[ \frac {-48 a x \left (5 a^2 x^2-3\right ) \tanh ^{-1}(a x)^2+48 \left (5 a^2 x^2-4\right ) \tanh ^{-1}(a x)+16 \left (5 a^4 x^4+6 a^2 x^2-3\right ) \tanh ^{-1}(a x)^3+3 \left (-34 a^3 x^3-17 \left (a^2 x^2-1\right )^2 \log (1-a x)+17 \left (a^2 x^2-1\right )^2 \log (a x+1)+30 a x\right )}{512 a^4 \left (a^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*ArcTanh[a*x]^3)/(1 - a^2*x^2)^3,x]

[Out]

(48*(-4 + 5*a^2*x^2)*ArcTanh[a*x] - 48*a*x*(-3 + 5*a^2*x^2)*ArcTanh[a*x]^2 + 16*(-3 + 6*a^2*x^2 + 5*a^4*x^4)*A
rcTanh[a*x]^3 + 3*(30*a*x - 34*a^3*x^3 - 17*(-1 + a^2*x^2)^2*Log[1 - a*x] + 17*(-1 + a^2*x^2)^2*Log[1 + a*x]))
/(512*a^4*(-1 + a^2*x^2)^2)

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fricas [A]  time = 1.52, size = 140, normalized size = 0.73 \[ -\frac {102 \, a^{3} x^{3} - 2 \, {\left (5 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 3\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 12 \, {\left (5 \, a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 90 \, a x - 3 \, {\left (17 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 15\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{512 \, {\left (a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(a*x)^3/(-a^2*x^2+1)^3,x, algorithm="fricas")

[Out]

-1/512*(102*a^3*x^3 - 2*(5*a^4*x^4 + 6*a^2*x^2 - 3)*log(-(a*x + 1)/(a*x - 1))^3 + 12*(5*a^3*x^3 - 3*a*x)*log(-
(a*x + 1)/(a*x - 1))^2 - 90*a*x - 3*(17*a^4*x^4 + 6*a^2*x^2 - 15)*log(-(a*x + 1)/(a*x - 1)))/(a^8*x^4 - 2*a^6*
x^2 + a^4)

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giac [B]  time = 0.21, size = 341, normalized size = 1.78 \[ \frac {1}{2048} \, {\left (4 \, {\left (\frac {{\left (a x - 1\right )}^{2} {\left (\frac {4 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )}}{{\left (a x + 1\right )}^{2} a^{5}} + \frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2} a^{5}} + \frac {4 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{5}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 6 \, {\left (\frac {{\left (a x - 1\right )}^{2} {\left (\frac {8 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )}}{{\left (a x + 1\right )}^{2} a^{5}} - \frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2} a^{5}} - \frac {8 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{5}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 6 \, {\left (\frac {{\left (a x - 1\right )}^{2} {\left (\frac {16 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )}}{{\left (a x + 1\right )}^{2} a^{5}} + \frac {{\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2} a^{5}} + \frac {16 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{5}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + \frac {3 \, {\left (a x - 1\right )}^{2} {\left (\frac {32 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )}}{{\left (a x + 1\right )}^{2} a^{5}} - \frac {3 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2} a^{5}} - \frac {96 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{5}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(a*x)^3/(-a^2*x^2+1)^3,x, algorithm="giac")

[Out]

1/2048*(4*((a*x - 1)^2*(4*(a*x + 1)/(a*x - 1) + 1)/((a*x + 1)^2*a^5) + (a*x + 1)^2/((a*x - 1)^2*a^5) + 4*(a*x
+ 1)/((a*x - 1)*a^5))*log(-(a*x + 1)/(a*x - 1))^3 + 6*((a*x - 1)^2*(8*(a*x + 1)/(a*x - 1) + 1)/((a*x + 1)^2*a^
5) - (a*x + 1)^2/((a*x - 1)^2*a^5) - 8*(a*x + 1)/((a*x - 1)*a^5))*log(-(a*x + 1)/(a*x - 1))^2 + 6*((a*x - 1)^2
*(16*(a*x + 1)/(a*x - 1) + 1)/((a*x + 1)^2*a^5) + (a*x + 1)^2/((a*x - 1)^2*a^5) + 16*(a*x + 1)/((a*x - 1)*a^5)
)*log(-(a*x + 1)/(a*x - 1)) + 3*(a*x - 1)^2*(32*(a*x + 1)/(a*x - 1) + 1)/((a*x + 1)^2*a^5) - 3*(a*x + 1)^2/((a
*x - 1)^2*a^5) - 96*(a*x + 1)/((a*x - 1)*a^5))*a

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maple [C]  time = 0.90, size = 2634, normalized size = 13.72 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctanh(a*x)^3/(-a^2*x^2+1)^3,x)

[Out]

15/64*I/a^2/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*x
^2+15/64*I/a^2/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*x^2+15/32*I/a^2/(a*x-1)^2
/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*x^2-15/128*I/(a*x-1)^2/(a*x+1)^2*arctanh(a*x
)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*x^4-15/128*I/(a*x-1)^2/(a*x+1)^2*arc
tanh(a*x)^2*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*x
^4+15/128*I/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*c
sgn(I*(a*x+1)^2/(a^2*x^2-1))*x^4+15/128*I/a^4/(a*x-1)^2/(a*x+1)^2*Pi*arctanh(a*x)^2*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)))^2-15/64*I/a^4/(a*x-1)^2/(a*x+1)^2*Pi*arctanh(a*x)
^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2-15/128*I/a^4/(a*x-1)^2/(a*x+1)^2*Pi*arct
anh(a*x)^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))-15/128*I/a^4/(a*x-1)^2/(a*x+1)^2
*Pi*arctanh(a*x)^2*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)))
^2-15/32*I/a^2/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*x^2+3/16/a^4*arctanh
(a*x)^3/(a*x-1)-3/16/a^4*arctanh(a*x)^3/(a*x+1)+1/16/a^4*arctanh(a*x)^3/(a*x-1)^2+1/16/a^4*arctanh(a*x)^3/(a*x
+1)^2-15/64/a^4*arctanh(a*x)^2*ln(a*x-1)-15/64*I/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x
^2-1))^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*x^4+15/64*I/a^2/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+
1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*x^2+15/64*I/a^2/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*c
sgn(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*x^2-15/64*I/a^2/(
a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*csgn(I*(a*x+1)
^2/(a^2*x^2-1))*x^2+15/128*I/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*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)))*csgn(I*(a*x+1)^2/(a^2*x^2-1))*x^4+15/128*I/a^4/(a*x-1)^2/(a*x
+1)^2*Pi*arctanh(a*x)^2*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)))+15/32*I/a^2/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^
2-1))^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*x^2-3/64/a^4*arctanh(a*x)^2/(a*x-1)^2-15/64/a^4*arctanh(a*x)^2/(a*x
-1)+3/64/a^4*arctanh(a*x)^2/(a*x+1)^2-15/64/a^4*arctanh(a*x)^2/(a*x+1)-15/64*I/a^2/(a*x-1)^2/(a*x+1)^2*arctanh
(a*x)^2*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)))*csgn(I*
(a*x+1)^2/(a^2*x^2-1))*x^2+15/64/a^4*arctanh(a*x)^2*ln(a*x+1)-15/32/a^4*arctanh(a*x)^2*ln((a*x+1)/(-a^2*x^2+1)
^(1/2))+15/64*I/a^4/(a*x-1)^2/(a*x+1)^2*Pi*arctanh(a*x)^2+15/64*I/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*x^4-15
/128*I/a^4/(a*x-1)^2/(a*x+1)^2*Pi*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3-15/64*I/a^4/(a*x-1)^2/(a*x+1)
^2*Pi*arctanh(a*x)^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^2-15/128*I/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(
I*(a*x+1)^2/(a^2*x^2-1))^3*x^4-15/64*I/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1))
)^2*x^4+15/64*I/a^4/(a*x-1)^2/(a*x+1)^2*Pi*arctanh(a*x)^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^3-15/128*I/a^4/(a
*x-1)^2/(a*x+1)^2*Pi*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^3-15/32*I/a^2/(a*
x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*x^2+15/64*I/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I/(1+(a*x+1)^2/(-a^2
*x^2+1)))^3*x^4-15/128*I/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2
*x^2+1)))^3*x^4-5/16/a^2/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^3*x^2+9/128/a^2/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)*x^2
-51/256/a/(a*x-1)^2/(a*x+1)^2*x^3+45/256/a^3/(a*x-1)^2/(a*x+1)^2*x+5/32/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^3*x^4
+51/256/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)*x^4+5/32/a^4/(a*x-1)^2/(a*x+1)^2*arctanh(a*x)^3-45/256/a^4/(a*x-1)^2/
(a*x+1)^2*arctanh(a*x)

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maxima [B]  time = 0.34, size = 437, normalized size = 2.28 \[ -\frac {3}{64} \, a {\left (\frac {2 \, {\left (5 \, a^{2} x^{3} - 3 \, x\right )}}{a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}} - \frac {5 \, \log \left (a x + 1\right )}{a^{5}} + \frac {5 \, \log \left (a x - 1\right )}{a^{5}}\right )} \operatorname {artanh}\left (a x\right )^{2} + \frac {{\left (2 \, a^{2} x^{2} - 1\right )} \operatorname {artanh}\left (a x\right )^{3}}{4 \, {\left (a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}\right )}} - \frac {1}{512} \, {\left (\frac {{\left (102 \, a^{3} x^{3} - 10 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{3} + 30 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} \log \left (a x - 1\right ) + 10 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{3} - 90 \, a x - 3 \, {\left (17 \, a^{4} x^{4} - 34 \, a^{2} x^{2} + 10 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} + 17\right )} \log \left (a x + 1\right ) + 51 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )\right )} a^{2}}{a^{11} x^{4} - 2 \, a^{9} x^{2} + a^{7}} - \frac {12 \, {\left (20 \, a^{2} x^{2} - 5 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 10 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) - 5 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x - 1\right )^{2} - 16\right )} a \operatorname {artanh}\left (a x\right )}{a^{10} x^{4} - 2 \, a^{8} x^{2} + a^{6}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(a*x)^3/(-a^2*x^2+1)^3,x, algorithm="maxima")

[Out]

-3/64*a*(2*(5*a^2*x^3 - 3*x)/(a^8*x^4 - 2*a^6*x^2 + a^4) - 5*log(a*x + 1)/a^5 + 5*log(a*x - 1)/a^5)*arctanh(a*
x)^2 + 1/4*(2*a^2*x^2 - 1)*arctanh(a*x)^3/(a^8*x^4 - 2*a^6*x^2 + a^4) - 1/512*((102*a^3*x^3 - 10*(a^4*x^4 - 2*
a^2*x^2 + 1)*log(a*x + 1)^3 + 30*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^2*log(a*x - 1) + 10*(a^4*x^4 - 2*a^2*x
^2 + 1)*log(a*x - 1)^3 - 90*a*x - 3*(17*a^4*x^4 - 34*a^2*x^2 + 10*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1)^2 + 1
7)*log(a*x + 1) + 51*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1))*a^2/(a^11*x^4 - 2*a^9*x^2 + a^7) - 12*(20*a^2*x^2
- 5*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)^2 + 10*(a^4*x^4 - 2*a^2*x^2 + 1)*log(a*x + 1)*log(a*x - 1) - 5*(a^
4*x^4 - 2*a^2*x^2 + 1)*log(a*x - 1)^2 - 16)*a*arctanh(a*x)/(a^10*x^4 - 2*a^8*x^2 + a^6))*a

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mupad [B]  time = 3.09, size = 414, normalized size = 2.16 \[ \frac {48\,\ln \left (1-a\,x\right )-48\,\ln \left (a\,x+1\right )+51\,\mathrm {atanh}\left (a\,x\right )+45\,a\,x-3\,{\ln \left (a\,x+1\right )}^3+3\,{\ln \left (1-a\,x\right )}^3-9\,\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^2+9\,{\ln \left (a\,x+1\right )}^2\,\ln \left (1-a\,x\right )-51\,a^3\,x^3+6\,a^2\,x^2\,{\ln \left (a\,x+1\right )}^3-6\,a^2\,x^2\,{\ln \left (1-a\,x\right )}^3-30\,a^3\,x^3\,{\ln \left (a\,x+1\right )}^2-30\,a^3\,x^3\,{\ln \left (1-a\,x\right )}^2+5\,a^4\,x^4\,{\ln \left (a\,x+1\right )}^3-5\,a^4\,x^4\,{\ln \left (1-a\,x\right )}^3-102\,a^2\,x^2\,\mathrm {atanh}\left (a\,x\right )+51\,a^4\,x^4\,\mathrm {atanh}\left (a\,x\right )+18\,a\,x\,{\ln \left (a\,x+1\right )}^2+18\,a\,x\,{\ln \left (1-a\,x\right )}^2+60\,a^2\,x^2\,\ln \left (a\,x+1\right )-60\,a^2\,x^2\,\ln \left (1-a\,x\right )-36\,a\,x\,\ln \left (a\,x+1\right )\,\ln \left (1-a\,x\right )+18\,a^2\,x^2\,\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^2-18\,a^2\,x^2\,{\ln \left (a\,x+1\right )}^2\,\ln \left (1-a\,x\right )+15\,a^4\,x^4\,\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^2-15\,a^4\,x^4\,{\ln \left (a\,x+1\right )}^2\,\ln \left (1-a\,x\right )+60\,a^3\,x^3\,\ln \left (a\,x+1\right )\,\ln \left (1-a\,x\right )}{256\,a^4\,{\left (a^2\,x^2-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^3*atanh(a*x)^3)/(a^2*x^2 - 1)^3,x)

[Out]

(48*log(1 - a*x) - 48*log(a*x + 1) + 51*atanh(a*x) + 45*a*x - 3*log(a*x + 1)^3 + 3*log(1 - a*x)^3 - 9*log(a*x
+ 1)*log(1 - a*x)^2 + 9*log(a*x + 1)^2*log(1 - a*x) - 51*a^3*x^3 + 6*a^2*x^2*log(a*x + 1)^3 - 6*a^2*x^2*log(1
- a*x)^3 - 30*a^3*x^3*log(a*x + 1)^2 - 30*a^3*x^3*log(1 - a*x)^2 + 5*a^4*x^4*log(a*x + 1)^3 - 5*a^4*x^4*log(1
- a*x)^3 - 102*a^2*x^2*atanh(a*x) + 51*a^4*x^4*atanh(a*x) + 18*a*x*log(a*x + 1)^2 + 18*a*x*log(1 - a*x)^2 + 60
*a^2*x^2*log(a*x + 1) - 60*a^2*x^2*log(1 - a*x) - 36*a*x*log(a*x + 1)*log(1 - a*x) + 18*a^2*x^2*log(a*x + 1)*l
og(1 - a*x)^2 - 18*a^2*x^2*log(a*x + 1)^2*log(1 - a*x) + 15*a^4*x^4*log(a*x + 1)*log(1 - a*x)^2 - 15*a^4*x^4*l
og(a*x + 1)^2*log(1 - a*x) + 60*a^3*x^3*log(a*x + 1)*log(1 - a*x))/(256*a^4*(a^2*x^2 - 1)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x^{3} \operatorname {atanh}^{3}{\left (a x \right )}}{a^{6} x^{6} - 3 a^{4} x^{4} + 3 a^{2} x^{2} - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atanh(a*x)**3/(-a**2*x**2+1)**3,x)

[Out]

-Integral(x**3*atanh(a*x)**3/(a**6*x**6 - 3*a**4*x**4 + 3*a**2*x**2 - 1), x)

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