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

Optimal. Leaf size=119 \[ -\frac {3 x}{8 a \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^2}-\frac {3 \tanh ^{-1}(a x)}{8 a^2} \]

[Out]

-3/8*x/a/(-a^2*x^2+1)-3/8*arctanh(a*x)/a^2+3/4*arctanh(a*x)/a^2/(-a^2*x^2+1)-3/4*x*arctanh(a*x)^2/a/(-a^2*x^2+
1)-1/4*arctanh(a*x)^3/a^2+1/2*arctanh(a*x)^3/a^2/(-a^2*x^2+1)

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Rubi [A]  time = 0.11, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5994, 5956, 199, 206} \[ -\frac {3 x}{8 a \left (1-a^2 x^2\right )}+\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^2}-\frac {3 \tanh ^{-1}(a x)}{8 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x)/(8*a*(1 - a^2*x^2)) - (3*ArcTanh[a*x])/(8*a^2) + (3*ArcTanh[a*x])/(4*a^2*(1 - a^2*x^2)) - (3*x*ArcTanh[
a*x]^2)/(4*a*(1 - a^2*x^2)) - ArcTanh[a*x]^3/(4*a^2) + ArcTanh[a*x]^3/(2*a^2*(1 - a^2*x^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 5956

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

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]

Rubi steps

\begin {align*} \int \frac {x \tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^2} \, dx &=\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {3 \int \frac {\tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx}{2 a}\\ &=-\frac {3 x \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^2}+\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}+\frac {3}{2} \int \frac {x \tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac {3 \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^2}+\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {3 \int \frac {1}{\left (1-a^2 x^2\right )^2} \, dx}{4 a}\\ &=-\frac {3 x}{8 a \left (1-a^2 x^2\right )}+\frac {3 \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^2}+\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}-\frac {3 \int \frac {1}{1-a^2 x^2} \, dx}{8 a}\\ &=-\frac {3 x}{8 a \left (1-a^2 x^2\right )}-\frac {3 \tanh ^{-1}(a x)}{8 a^2}+\frac {3 \tanh ^{-1}(a x)}{4 a^2 \left (1-a^2 x^2\right )}-\frac {3 x \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}-\frac {\tanh ^{-1}(a x)^3}{4 a^2}+\frac {\tanh ^{-1}(a x)^3}{2 a^2 \left (1-a^2 x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 91, normalized size = 0.76 \[ \frac {3 \left (a^2 x^2-1\right ) \log (1-a x)-3 \left (a^2 x^2-1\right ) \log (a x+1)-4 \left (a^2 x^2+1\right ) \tanh ^{-1}(a x)^3+6 a x+12 a x \tanh ^{-1}(a x)^2-12 \tanh ^{-1}(a x)}{16 a^2 \left (a^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*x - 12*ArcTanh[a*x] + 12*a*x*ArcTanh[a*x]^2 - 4*(1 + a^2*x^2)*ArcTanh[a*x]^3 + 3*(-1 + a^2*x^2)*Log[1 - a
*x] - 3*(-1 + a^2*x^2)*Log[1 + a*x])/(16*a^2*(-1 + a^2*x^2))

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fricas [A]  time = 0.46, size = 97, normalized size = 0.82 \[ \frac {6 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 12 \, a x - 6 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{32 \, {\left (a^{4} x^{2} - a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(6*a*x*log(-(a*x + 1)/(a*x - 1))^2 - (a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))^3 + 12*a*x - 6*(a^2*x^2 + 1)
*log(-(a*x + 1)/(a*x - 1)))/(a^4*x^2 - a^2)

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giac [A]  time = 0.19, size = 192, normalized size = 1.61 \[ -\frac {1}{64} \, {\left ({\left (\frac {a x + 1}{{\left (a x - 1\right )} a^{3}} + \frac {a x - 1}{{\left (a x + 1\right )} a^{3}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} - 3 \, {\left (\frac {a x + 1}{{\left (a x - 1\right )} a^{3}} - \frac {a x - 1}{{\left (a x + 1\right )} a^{3}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 6 \, {\left (\frac {a x + 1}{{\left (a x - 1\right )} a^{3}} + \frac {a x - 1}{{\left (a x + 1\right )} a^{3}}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - \frac {6 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} a^{3}} + \frac {6 \, {\left (a x - 1\right )}}{{\left (a x + 1\right )} a^{3}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(((a*x + 1)/((a*x - 1)*a^3) + (a*x - 1)/((a*x + 1)*a^3))*log(-(a*x + 1)/(a*x - 1))^3 - 3*((a*x + 1)/((a*
x - 1)*a^3) - (a*x - 1)/((a*x + 1)*a^3))*log(-(a*x + 1)/(a*x - 1))^2 + 6*((a*x + 1)/((a*x - 1)*a^3) + (a*x - 1
)/((a*x + 1)*a^3))*log(-(a*x + 1)/(a*x - 1)) - 6*(a*x + 1)/((a*x - 1)*a^3) + 6*(a*x - 1)/((a*x + 1)*a^3))*a

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maple [C]  time = 0.81, size = 1708, normalized size = 14.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*I/(a*x-1)/(a*x+1)*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
+3/16*I/(a*x-1)/(a*x+1)*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+3/16*I/a^2/(a*x-1)/(a*x+1)*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))*arctanh(a*x)^2*Pi-3/16*I/a^2/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))
^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*arctanh(a*x)^2*Pi-3/8*I/a^2/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1)
)^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)^2*Pi-3/16*I/a^2/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-
1))*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*arctanh(a*x)^2*Pi-3/16*I/(a*x-1)/(a*x+1)*arctanh(a*x)^2*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*x^2+3/16*I/(a*x-1)/(a*x+1)*arct
anh(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/(1+(a*x+1)^2/(-a^2*x^2+1)))*x^
2-1/2/a^2/(a^2*x^2-1)*arctanh(a*x)^3+3/8/a^2*arctanh(a*x)^2/(a*x-1)+3/8/a^2*arctanh(a*x)^2*ln(a*x-1)+3/8/a^2*a
rctanh(a*x)^2/(a*x+1)-3/8/a^2*arctanh(a*x)^2*ln(a*x+1)+3/4/a^2*arctanh(a*x)^2*ln((a*x+1)/(-a^2*x^2+1)^(1/2))+3
/8/a/(a*x-1)/(a*x+1)*x+1/4/a^2/(a*x-1)/(a*x+1)*arctanh(a*x)^3-3/8/a^2/(a*x-1)/(a*x+1)*arctanh(a*x)+3/16*I/a^2/
(a*x-1)/(a*x+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))*csgn(I/
(1+(a*x+1)^2/(-a^2*x^2+1)))*arctanh(a*x)^2*Pi-3/16*I/(a*x-1)/(a*x+1)*arctanh(a*x)^2*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)))*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*x^2-1/4/(a*
x-1)/(a*x+1)*arctanh(a*x)^3*x^2-3/8/(a*x-1)/(a*x+1)*arctanh(a*x)*x^2+3/8*I/a^2/(a*x-1)/(a*x+1)*arctanh(a*x)^2*
Pi-3/8*I/(a*x-1)/(a*x+1)*arctanh(a*x)^2*Pi*x^2-3/8*I/(a*x-1)/(a*x+1)*arctanh(a*x)^2*Pi*csgn(I/(1+(a*x+1)^2/(-a
^2*x^2+1)))^3*x^2-3/16*I/a^2/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*arctan
h(a*x)^2*Pi-3/16*I/a^2/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*arctanh(a*x)^2*Pi+3/8*I/a^2/(a*x-1)/(a*
x+1)*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*arctanh(a*x)^2*Pi-3/8*I/a^2/(a*x-1)/(a*x+1)*csgn(I/(1+(a*x+1)^2/(-a^
2*x^2+1)))^2*arctanh(a*x)^2*Pi+3/8*I/(a*x-1)/(a*x+1)*arctanh(a*x)^2*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*x^
2+3/16*I/(a*x-1)/(a*x+1)*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+3/16
*I/(a*x-1)/(a*x+1)*arctanh(a*x)^2*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*x^2

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maxima [B]  time = 0.34, size = 298, normalized size = 2.50 \[ \frac {3 \, {\left (\frac {2 \, x}{a^{2} x^{2} - 1} - \frac {\log \left (a x + 1\right )}{a} + \frac {\log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right )^{2}}{8 \, a} - \frac {\frac {{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{3} - 3 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} \log \left (a x - 1\right ) - {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{3} - 12 \, a x + 3 \, {\left (2 \, a^{2} x^{2} + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 2\right )} \log \left (a x + 1\right ) - 6 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )\right )} a^{2}}{a^{5} x^{2} - a^{3}} - \frac {6 \, {\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) + {\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 4\right )} a \operatorname {artanh}\left (a x\right )}{a^{4} x^{2} - a^{2}}}{32 \, a} - \frac {\operatorname {artanh}\left (a x\right )^{3}}{2 \, {\left (a^{2} x^{2} - 1\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(2*x/(a^2*x^2 - 1) - log(a*x + 1)/a + log(a*x - 1)/a)*arctanh(a*x)^2/a - 1/32*(((a^2*x^2 - 1)*log(a*x + 1)
^3 - 3*(a^2*x^2 - 1)*log(a*x + 1)^2*log(a*x - 1) - (a^2*x^2 - 1)*log(a*x - 1)^3 - 12*a*x + 3*(2*a^2*x^2 + (a^2
*x^2 - 1)*log(a*x - 1)^2 - 2)*log(a*x + 1) - 6*(a^2*x^2 - 1)*log(a*x - 1))*a^2/(a^5*x^2 - a^3) - 6*((a^2*x^2 -
1)*log(a*x + 1)^2 - 2*(a^2*x^2 - 1)*log(a*x + 1)*log(a*x - 1) + (a^2*x^2 - 1)*log(a*x - 1)^2 - 4)*a*arctanh(a
*x)/(a^4*x^2 - a^2))/a - 1/2*arctanh(a*x)^3/((a^2*x^2 - 1)*a^2)

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mupad [B]  time = 1.76, size = 239, normalized size = 2.01 \[ -\frac {6\,\ln \left (1-a\,x\right )-6\,\ln \left (a\,x+1\right )+12\,a\,x-{\ln \left (a\,x+1\right )}^3+{\ln \left (1-a\,x\right )}^3-3\,\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^2+3\,{\ln \left (a\,x+1\right )}^2\,\ln \left (1-a\,x\right )-a^2\,x^2\,\left (6\,\ln \left (a\,x+1\right )-6\,\ln \left (1-a\,x\right )\right )-a^2\,x^2\,{\ln \left (a\,x+1\right )}^3+a^2\,x^2\,{\ln \left (1-a\,x\right )}^3+6\,a\,x\,{\ln \left (a\,x+1\right )}^2+6\,a\,x\,{\ln \left (1-a\,x\right )}^2-12\,a\,x\,\ln \left (a\,x+1\right )\,\ln \left (1-a\,x\right )-3\,a^2\,x^2\,\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^2+3\,a^2\,x^2\,{\ln \left (a\,x+1\right )}^2\,\ln \left (1-a\,x\right )}{32\,a^2-32\,a^4\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(6*log(1 - a*x) - 6*log(a*x + 1) + 12*a*x - log(a*x + 1)^3 + log(1 - a*x)^3 - 3*log(a*x + 1)*log(1 - a*x)^2 +
3*log(a*x + 1)^2*log(1 - a*x) - a^2*x^2*(6*log(a*x + 1) - 6*log(1 - a*x)) - a^2*x^2*log(a*x + 1)^3 + a^2*x^2*
log(1 - a*x)^3 + 6*a*x*log(a*x + 1)^2 + 6*a*x*log(1 - a*x)^2 - 12*a*x*log(a*x + 1)*log(1 - a*x) - 3*a^2*x^2*lo
g(a*x + 1)*log(1 - a*x)^2 + 3*a^2*x^2*log(a*x + 1)^2*log(1 - a*x))/(32*a^2 - 32*a^4*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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