3.5.76 \(\int \frac {\tanh ^{-1}(a x)^3}{(1-a^2 x^2)^{5/2}} \, dx\) [476]

Optimal. Leaf size=191 \[ -\frac {2}{27 a \left (1-a^2 x^2\right )^{3/2}}-\frac {40}{9 a \sqrt {1-a^2 x^2}}+\frac {2 x \tanh ^{-1}(a x)}{9 \left (1-a^2 x^2\right )^{3/2}}+\frac {40 x \tanh ^{-1}(a x)}{9 \sqrt {1-a^2 x^2}}-\frac {\tanh ^{-1}(a x)^2}{3 a \left (1-a^2 x^2\right )^{3/2}}-\frac {2 \tanh ^{-1}(a x)^2}{a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{3 \left (1-a^2 x^2\right )^{3/2}}+\frac {2 x \tanh ^{-1}(a x)^3}{3 \sqrt {1-a^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 191, 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, 6105, 6107} \begin {gather*} -\frac {40}{9 a \sqrt {1-a^2 x^2}}-\frac {2}{27 a \left (1-a^2 x^2\right )^{3/2}}+\frac {2 x \tanh ^{-1}(a x)^3}{3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {2 \tanh ^{-1}(a x)^2}{a \sqrt {1-a^2 x^2}}-\frac {\tanh ^{-1}(a x)^2}{3 a \left (1-a^2 x^2\right )^{3/2}}+\frac {40 x \tanh ^{-1}(a x)}{9 \sqrt {1-a^2 x^2}}+\frac {2 x \tanh ^{-1}(a x)}{9 \left (1-a^2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^3/(1 - a^2*x^2)^(5/2),x]

[Out]

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

Rule 6105

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

Rule 6107

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

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

________________________________________________________________________________________

Mathematica [A]
time = 0.06, size = 87, normalized size = 0.46 \begin {gather*} \frac {-122+120 a^2 x^2-6 a x \left (-21+20 a^2 x^2\right ) \tanh ^{-1}(a x)+9 \left (-7+6 a^2 x^2\right ) \tanh ^{-1}(a x)^2-9 a x \left (-3+2 a^2 x^2\right ) \tanh ^{-1}(a x)^3}{27 a \left (1-a^2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[a*x]^3/(1 - a^2*x^2)^(5/2),x]

[Out]

(-122 + 120*a^2*x^2 - 6*a*x*(-21 + 20*a^2*x^2)*ArcTanh[a*x] + 9*(-7 + 6*a^2*x^2)*ArcTanh[a*x]^2 - 9*a*x*(-3 +
2*a^2*x^2)*ArcTanh[a*x]^3)/(27*a*(1 - a^2*x^2)^(3/2))

________________________________________________________________________________________

Maple [A]
time = 0.68, size = 105, normalized size = 0.55

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (18 \arctanh \left (a x \right )^{3} a^{3} x^{3}+120 a^{3} x^{3} \arctanh \left (a x \right )-54 a^{2} x^{2} \arctanh \left (a x \right )^{2}-27 \arctanh \left (a x \right )^{3} a x -120 a^{2} x^{2}-126 a x \arctanh \left (a x \right )+63 \arctanh \left (a x \right )^{2}+122\right )}{27 a \left (a^{2} x^{2}-1\right )^{2}}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)^3/(-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)^3*a^3*x^3+120*a^3*x^3*arctanh(a*x)-54*a^2*x^2*arctanh(a*x)^2-27*ar
ctanh(a*x)^3*a*x-120*a^2*x^2-126*a*x*arctanh(a*x)+63*arctanh(a*x)^2+122)/(a^2*x^2-1)^2

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^3/(-a^2*x^2+1)^(5/2),x, algorithm="maxima")

[Out]

integrate(arctanh(a*x)^3/(-a^2*x^2 + 1)^(5/2), x)

________________________________________________________________________________________

Fricas [A]
time = 0.42, size = 134, normalized size = 0.70 \begin {gather*} \frac {{\left (960 \, a^{2} x^{2} - 9 \, {\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 18 \, {\left (6 \, a^{2} x^{2} - 7\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 24 \, {\left (20 \, a^{3} x^{3} - 21 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 976\right )} \sqrt {-a^{2} x^{2} + 1}}{216 \, {\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)^3/(-a^2*x^2+1)^(5/2),x, algorithm="fricas")

[Out]

1/216*(960*a^2*x^2 - 9*(2*a^3*x^3 - 3*a*x)*log(-(a*x + 1)/(a*x - 1))^3 + 18*(6*a^2*x^2 - 7)*log(-(a*x + 1)/(a*
x - 1))^2 - 24*(20*a^3*x^3 - 21*a*x)*log(-(a*x + 1)/(a*x - 1)) - 976)*sqrt(-a^2*x^2 + 1)/(a^5*x^4 - 2*a^3*x^2
+ a)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {atanh}^{3}{\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)**3/(-a**2*x**2+1)**(5/2),x)

[Out]

Integral(atanh(a*x)**3/(-(a*x - 1)*(a*x + 1))**(5/2), x)

________________________________________________________________________________________

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)^3/(-a^2*x^2+1)^(5/2),x, algorithm="giac")

[Out]

integrate(arctanh(a*x)^3/(-a^2*x^2 + 1)^(5/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {atanh}\left (a\,x\right )}^3}{{\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)^3/(1 - a^2*x^2)^(5/2),x)

[Out]

int(atanh(a*x)^3/(1 - a^2*x^2)^(5/2), x)

________________________________________________________________________________________