3.472 \(\int \frac {\tanh ^{-1}(a x)^2}{(1-a^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=139 \[ \frac {40 x}{27 \sqrt {1-a^2 x^2}}+\frac {2 x}{27 \left (1-a^2 x^2\right )^{3/2}}+\frac {2 x \tanh ^{-1}(a x)^2}{3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^2}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {4 \tanh ^{-1}(a x)}{3 a \sqrt {1-a^2 x^2}}-\frac {2 \tanh ^{-1}(a x)}{9 a \left (1-a^2 x^2\right )^{3/2}} \]

[Out]

2/27*x/(-a^2*x^2+1)^(3/2)-2/9*arctanh(a*x)/a/(-a^2*x^2+1)^(3/2)+1/3*x*arctanh(a*x)^2/(-a^2*x^2+1)^(3/2)+40/27*
x/(-a^2*x^2+1)^(1/2)-4/3*arctanh(a*x)/a/(-a^2*x^2+1)^(1/2)+2/3*x*arctanh(a*x)^2/(-a^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5964, 5962, 191, 192} \[ \frac {40 x}{27 \sqrt {1-a^2 x^2}}+\frac {2 x}{27 \left (1-a^2 x^2\right )^{3/2}}+\frac {2 x \tanh ^{-1}(a x)^2}{3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^2}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {4 \tanh ^{-1}(a x)}{3 a \sqrt {1-a^2 x^2}}-\frac {2 \tanh ^{-1}(a x)}{9 a \left (1-a^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^2/(1 - a^2*x^2)^(5/2),x]

[Out]

(2*x)/(27*(1 - a^2*x^2)^(3/2)) + (40*x)/(27*Sqrt[1 - a^2*x^2]) - (2*ArcTanh[a*x])/(9*a*(1 - a^2*x^2)^(3/2)) -
(4*ArcTanh[a*x])/(3*a*Sqrt[1 - a^2*x^2]) + (x*ArcTanh[a*x]^2)/(3*(1 - a^2*x^2)^(3/2)) + (2*x*ArcTanh[a*x]^2)/(
3*Sqrt[1 - a^2*x^2])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

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, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 5962

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

Rule 5964

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*p*(d + e*x^2)^(
q + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(4*c*d*(q + 1)^2), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q
+ 1)*(a + b*ArcTanh[c*x])^p, x], x] + Dist[(b^2*p*(p - 1))/(4*(q + 1)^2), Int[(d + e*x^2)^q*(a + b*ArcTanh[c*
x])^(p - 2), x], x] - Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p)/(2*d*(q + 1)), x]) /; FreeQ[{a, b, c
, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[q, -1] && GtQ[p, 1] && NeQ[q, -3/2]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^{5/2}} \, dx &=-\frac {2 \tanh ^{-1}(a x)}{9 a \left (1-a^2 x^2\right )^{3/2}}+\frac {x \tanh ^{-1}(a x)^2}{3 \left (1-a^2 x^2\right )^{3/2}}+\frac {2}{9} \int \frac {1}{\left (1-a^2 x^2\right )^{5/2}} \, dx+\frac {2}{3} \int \frac {\tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=\frac {2 x}{27 \left (1-a^2 x^2\right )^{3/2}}-\frac {2 \tanh ^{-1}(a x)}{9 a \left (1-a^2 x^2\right )^{3/2}}-\frac {4 \tanh ^{-1}(a x)}{3 a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^2}{3 \left (1-a^2 x^2\right )^{3/2}}+\frac {2 x \tanh ^{-1}(a x)^2}{3 \sqrt {1-a^2 x^2}}+\frac {4}{27} \int \frac {1}{\left (1-a^2 x^2\right )^{3/2}} \, dx+\frac {4}{3} \int \frac {1}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=\frac {2 x}{27 \left (1-a^2 x^2\right )^{3/2}}+\frac {40 x}{27 \sqrt {1-a^2 x^2}}-\frac {2 \tanh ^{-1}(a x)}{9 a \left (1-a^2 x^2\right )^{3/2}}-\frac {4 \tanh ^{-1}(a x)}{3 a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^2}{3 \left (1-a^2 x^2\right )^{3/2}}+\frac {2 x \tanh ^{-1}(a x)^2}{3 \sqrt {1-a^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 70, normalized size = 0.50 \[ \frac {-40 a^3 x^3-9 a x \left (2 a^2 x^2-3\right ) \tanh ^{-1}(a x)^2+6 \left (6 a^2 x^2-7\right ) \tanh ^{-1}(a x)+42 a x}{27 a \left (1-a^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[a*x]^2/(1 - a^2*x^2)^(5/2),x]

[Out]

(42*a*x - 40*a^3*x^3 + 6*(-7 + 6*a^2*x^2)*ArcTanh[a*x] - 9*a*x*(-3 + 2*a^2*x^2)*ArcTanh[a*x]^2)/(27*a*(1 - a^2
*x^2)^(3/2))

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 105, normalized size = 0.76 \[ -\frac {{\left (160 \, a^{3} x^{3} + 9 \, {\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 168 \, a x - 12 \, {\left (6 \, a^{2} x^{2} - 7\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )\right )} \sqrt {-a^{2} x^{2} + 1}}{108 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^2/(-a^2*x^2+1)^(5/2),x, algorithm="fricas")

[Out]

-1/108*(160*a^3*x^3 + 9*(2*a^3*x^3 - 3*a*x)*log(-(a*x + 1)/(a*x - 1))^2 - 168*a*x - 12*(6*a^2*x^2 - 7)*log(-(a
*x + 1)/(a*x - 1)))*sqrt(-a^2*x^2 + 1)/(a^5*x^4 - 2*a^3*x^2 + a)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (a x\right )^{2}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^2/(-a^2*x^2+1)^(5/2),x, algorithm="giac")

[Out]

integrate(arctanh(a*x)^2/(-a^2*x^2 + 1)^(5/2), x)

________________________________________________________________________________________

maple [A]  time = 0.43, size = 84, normalized size = 0.60 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \left (18 \arctanh \left (a x \right )^{2} x^{3} a^{3}+40 x^{3} a^{3}-36 a^{2} x^{2} \arctanh \left (a x \right )-27 \arctanh \left (a x \right )^{2} a x -42 a x +42 \arctanh \left (a x \right )\right )}{27 a \left (a^{2} x^{2}-1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)^2/(-a^2*x^2+1)^(5/2),x)

[Out]

-1/27/a*(-a^2*x^2+1)^(1/2)*(18*arctanh(a*x)^2*x^3*a^3+40*x^3*a^3-36*a^2*x^2*arctanh(a*x)-27*arctanh(a*x)^2*a*x
-42*a*x+42*arctanh(a*x))/(a^2*x^2-1)^2

________________________________________________________________________________________

maxima [B]  time = 0.50, size = 304, normalized size = 2.19 \[ \frac {1}{3} \, {\left (\frac {2 \, x}{\sqrt {-a^{2} x^{2} + 1}} + \frac {x}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}\right )} \operatorname {artanh}\left (a x\right )^{2} + \frac {1}{27} \, a {\left (\frac {\frac {2 \, x}{\sqrt {-a^{2} x^{2} + 1}} - \frac {1}{\sqrt {-a^{2} x^{2} + 1} a^{2} x + \sqrt {-a^{2} x^{2} + 1} a}}{a} + \frac {\frac {2 \, x}{\sqrt {-a^{2} x^{2} + 1}} - \frac {1}{\sqrt {-a^{2} x^{2} + 1} a^{2} x - \sqrt {-a^{2} x^{2} + 1} a}}{a} - \frac {18 \, \sqrt {-a^{2} x^{2} + 1}}{{\left (a^{2} x + a\right )} a} - \frac {18 \, \sqrt {-a^{2} x^{2} + 1}}{{\left (a^{2} x - a\right )} a} - \frac {18 \, \log \left (a x + 1\right )}{\sqrt {-a^{2} x^{2} + 1} a^{2}} + \frac {18 \, \log \left (-a x + 1\right )}{\sqrt {-a^{2} x^{2} + 1} a^{2}} - \frac {3 \, \log \left (a x + 1\right )}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{2}} + \frac {3 \, \log \left (-a x + 1\right )}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^2/(-a^2*x^2+1)^(5/2),x, algorithm="maxima")

[Out]

1/3*(2*x/sqrt(-a^2*x^2 + 1) + x/(-a^2*x^2 + 1)^(3/2))*arctanh(a*x)^2 + 1/27*a*((2*x/sqrt(-a^2*x^2 + 1) - 1/(sq
rt(-a^2*x^2 + 1)*a^2*x + sqrt(-a^2*x^2 + 1)*a))/a + (2*x/sqrt(-a^2*x^2 + 1) - 1/(sqrt(-a^2*x^2 + 1)*a^2*x - sq
rt(-a^2*x^2 + 1)*a))/a - 18*sqrt(-a^2*x^2 + 1)/((a^2*x + a)*a) - 18*sqrt(-a^2*x^2 + 1)/((a^2*x - a)*a) - 18*lo
g(a*x + 1)/(sqrt(-a^2*x^2 + 1)*a^2) + 18*log(-a*x + 1)/(sqrt(-a^2*x^2 + 1)*a^2) - 3*log(a*x + 1)/((-a^2*x^2 +
1)^(3/2)*a^2) + 3*log(-a*x + 1)/((-a^2*x^2 + 1)^(3/2)*a^2))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atanh}\left (a\,x\right )}^2}{{\left (1-a^2\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^2/(1 - a^2*x^2)^(5/2),x)

[Out]

int(atanh(a*x)^2/(1 - a^2*x^2)^(5/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)**2/(-a**2*x**2+1)**(5/2),x)

[Out]

Integral(atanh(a*x)**2/(-(a*x - 1)*(a*x + 1))**(5/2), x)

________________________________________________________________________________________