3.5.72 \(\int \frac {\tanh ^{-1}(a x)^2}{(1-a^2 x^2)^{5/2}} \, dx\) [472]

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

[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)

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Rubi [A]
time = 0.07, 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 = {6111, 6109, 197, 198} \begin {gather*} \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}} \end {gather*}

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 197

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 198

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 6109

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-b)*p*((a + b*Arc
Tanh[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] && E
qQ[c^2*d + e, 0] && GtQ[p, 1]

Rule 6111

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*}

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Mathematica [A]
time = 0.06, size = 70, normalized size = 0.50 \begin {gather*} \frac {42 a x-40 a^3 x^3+6 \left (-7+6 a^2 x^2\right ) \tanh ^{-1}(a x)-9 a x \left (-3+2 a^2 x^2\right ) \tanh ^{-1}(a x)^2}{27 a \left (1-a^2 x^2\right )^{3/2}} \end {gather*}

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))

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Maple [A]
time = 0.70, size = 84, normalized size = 0.60

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (18 \arctanh \left (a x \right )^{2} a^{3} x^{3}+40 a^{3} x^{3}-36 a^{2} x^{2} \arctanh \left (a x \right )-27 \arctanh \left (a x \right )^{2} a x -42 a x +42 \arctanh \left (a x \right )\right )}{27 a \left (a^{2} x^{2}-1\right )^{2}}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (115) = 230\).
time = 0.52, size = 304, normalized size = 2.19 \begin {gather*} \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 )} \end {gather*}

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))

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Fricas [A]
time = 0.36, size = 105, normalized size = 0.76 \begin {gather*} -\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 )}} \end {gather*}

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

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)

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