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

Optimal. Leaf size=77 \[ -\frac {3 \tanh ^{-1}(a x)}{32 a^4}+\frac {x^4 \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}-\frac {x^3}{16 a \left (1-a^2 x^2\right )^2}+\frac {3 x}{32 a^3 \left (1-a^2 x^2\right )} \]

[Out]

-1/16*x^3/a/(-a^2*x^2+1)^2+3/32*x/a^3/(-a^2*x^2+1)-3/32*arctanh(a*x)/a^4+1/4*x^4*arctanh(a*x)/(-a^2*x^2+1)^2

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Rubi [A]  time = 0.06, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6008, 288, 206} \[ -\frac {x^3}{16 a \left (1-a^2 x^2\right )^2}+\frac {3 x}{32 a^3 \left (1-a^2 x^2\right )}+\frac {x^4 \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )^2}-\frac {3 \tanh ^{-1}(a x)}{32 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^3/(16*a*(1 - a^2*x^2)^2) + (3*x)/(32*a^3*(1 - a^2*x^2)) - (3*ArcTanh[a*x])/(32*a^4) + (x^4*ArcTanh[a*x])/(4
*(1 - a^2*x^2)^2)

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

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Mathematica [A]  time = 0.09, size = 98, normalized size = 1.27 \[ -\frac {5 \log (1-a x)}{64 a^4}+\frac {5 \log (a x+1)}{64 a^4}+\frac {\left (2 a^2 x^2-1\right ) \tanh ^{-1}(a x)}{4 a^4 \left (a^2 x^2-1\right )^2}-\frac {5 x}{32 a^3 \left (a^2 x^2-1\right )}-\frac {x}{16 a^3 \left (a^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*x/(a^3*(-1 + a^2*x^2)^2) - (5*x)/(32*a^3*(-1 + a^2*x^2)) + ((-1 + 2*a^2*x^2)*ArcTanh[a*x])/(4*a^4*(-1 +
a^2*x^2)^2) - (5*Log[1 - a*x])/(64*a^4) + (5*Log[1 + a*x])/(64*a^4)

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fricas [A]  time = 0.80, size = 71, normalized size = 0.92 \[ -\frac {10 \, a^{3} x^{3} - 6 \, a x - {\left (5 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 3\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{64 \, {\left (a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(10*a^3*x^3 - 6*a*x - (5*a^4*x^4 + 6*a^2*x^2 - 3)*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 = 0.19, size = 239, normalized size = 3.10 \[ \frac {1}{256} \, {\left (2 \, {\left (\frac {{\left (a x - 1\right )}^{2} {\left (\frac {4 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )}}{{\left (a x + 1\right )}^{2} a^{5}} + \frac {\frac {{\left (a x + 1\right )}^{2} a^{5}}{{\left (a x - 1\right )}^{2}} + \frac {4 \, {\left (a x + 1\right )} a^{5}}{a x - 1}}{a^{10}}\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right ) + \frac {{\left (a x - 1\right )}^{2} {\left (\frac {8 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )}}{{\left (a x + 1\right )}^{2} a^{5}} - \frac {\frac {{\left (a x + 1\right )}^{2} a^{5}}{{\left (a x - 1\right )}^{2}} + \frac {8 \, {\left (a x + 1\right )} a^{5}}{a x - 1}}{a^{10}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(2*((a*x - 1)^2*(4*(a*x + 1)/(a*x - 1) + 1)/((a*x + 1)^2*a^5) + ((a*x + 1)^2*a^5/(a*x - 1)^2 + 4*(a*x +
1)*a^5/(a*x - 1))/a^10)*log(-(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) + 1)/(a*((a*x + 1)/(a*x
- 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) - 1)) + (a*x - 1)^2*(8*(a*x + 1)/(a*x - 1) + 1)/((a*x + 1)^2*a^5) - ((a*
x + 1)^2*a^5/(a*x - 1)^2 + 8*(a*x + 1)*a^5/(a*x - 1))/a^10)*a

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maple [A]  time = 0.04, size = 136, normalized size = 1.77 \[ \frac {\arctanh \left (a x \right )}{16 a^{4} \left (a x -1\right )^{2}}+\frac {3 \arctanh \left (a x \right )}{16 a^{4} \left (a x -1\right )}+\frac {\arctanh \left (a x \right )}{16 a^{4} \left (a x +1\right )^{2}}-\frac {3 \arctanh \left (a x \right )}{16 a^{4} \left (a x +1\right )}-\frac {1}{64 a^{4} \left (a x -1\right )^{2}}-\frac {5}{64 a^{4} \left (a x -1\right )}-\frac {5 \ln \left (a x -1\right )}{64 a^{4}}+\frac {1}{64 a^{4} \left (a x +1\right )^{2}}-\frac {5}{64 a^{4} \left (a x +1\right )}+\frac {5 \ln \left (a x +1\right )}{64 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/a^4*arctanh(a*x)/(a*x-1)^2+3/16/a^4*arctanh(a*x)/(a*x-1)+1/16/a^4*arctanh(a*x)/(a*x+1)^2-3/16/a^4*arctanh
(a*x)/(a*x+1)-1/64/a^4/(a*x-1)^2-5/64/a^4/(a*x-1)-5/64/a^4*ln(a*x-1)+1/64/a^4/(a*x+1)^2-5/64/a^4/(a*x+1)+5/64/
a^4*ln(a*x+1)

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maxima [A]  time = 0.31, size = 99, normalized size = 1.29 \[ -\frac {1}{64} \, a {\left (\frac {2 \, {\left (5 \, a^{2} x^{3} - 3 \, x\right )}}{a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}} - \frac {5 \, \log \left (a x + 1\right )}{a^{5}} + \frac {5 \, \log \left (a x - 1\right )}{a^{5}}\right )} + \frac {{\left (2 \, a^{2} x^{2} - 1\right )} \operatorname {artanh}\left (a x\right )}{4 \, {\left (a^{8} x^{4} - 2 \, a^{6} x^{2} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*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) + 1/4*(2*a
^2*x^2 - 1)*arctanh(a*x)/(a^8*x^4 - 2*a^6*x^2 + a^4)

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mupad [B]  time = 1.39, size = 83, normalized size = 1.08 \[ \frac {5\,\mathrm {atanh}\left (a\,x\right )}{32\,a^4}+\frac {\frac {\ln \left (1-a\,x\right )}{8}-\frac {\ln \left (a\,x+1\right )}{8}+\frac {3\,a\,x}{32}+x^2\,\left (\frac {a^2\,\ln \left (a\,x+1\right )}{4}-\frac {a^2\,\ln \left (1-a\,x\right )}{4}\right )-\frac {5\,a^3\,x^3}{32}}{a^4\,{\left (a^2\,x^2-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*atanh(a*x))/(32*a^4) + (log(1 - a*x)/8 - log(a*x + 1)/8 + (3*a*x)/32 + x^2*((a^2*log(a*x + 1))/4 - (a^2*log
(1 - a*x))/4) - (5*a^3*x^3)/32)/(a^4*(a^2*x^2 - 1)^2)

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sympy [A]  time = 2.41, size = 158, normalized size = 2.05 \[ \begin {cases} \frac {5 a^{4} x^{4} \operatorname {atanh}{\left (a x \right )}}{32 a^{8} x^{4} - 64 a^{6} x^{2} + 32 a^{4}} - \frac {5 a^{3} x^{3}}{32 a^{8} x^{4} - 64 a^{6} x^{2} + 32 a^{4}} + \frac {6 a^{2} x^{2} \operatorname {atanh}{\left (a x \right )}}{32 a^{8} x^{4} - 64 a^{6} x^{2} + 32 a^{4}} + \frac {3 a x}{32 a^{8} x^{4} - 64 a^{6} x^{2} + 32 a^{4}} - \frac {3 \operatorname {atanh}{\left (a x \right )}}{32 a^{8} x^{4} - 64 a^{6} x^{2} + 32 a^{4}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((5*a**4*x**4*atanh(a*x)/(32*a**8*x**4 - 64*a**6*x**2 + 32*a**4) - 5*a**3*x**3/(32*a**8*x**4 - 64*a**
6*x**2 + 32*a**4) + 6*a**2*x**2*atanh(a*x)/(32*a**8*x**4 - 64*a**6*x**2 + 32*a**4) + 3*a*x/(32*a**8*x**4 - 64*
a**6*x**2 + 32*a**4) - 3*atanh(a*x)/(32*a**8*x**4 - 64*a**6*x**2 + 32*a**4), Ne(a, 0)), (0, True))

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