3.5.7 \(\int \frac {\tanh ^{-1}(a x)^3}{(1-a^2 x^2)^{3/2}} \, dx\) [407]

Optimal. Leaf size=88 \[ -\frac {6}{a \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6109, 6105} \begin {gather*} -\frac {6}{a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{a \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-6/(a*Sqrt[1 - a^2*x^2]) + (6*x*ArcTanh[a*x])/Sqrt[1 - a^2*x^2] - (3*ArcTanh[a*x]^2)/(a*Sqrt[1 - a^2*x^2]) + (
x*ArcTanh[a*x]^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 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]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx &=-\frac {3 \tanh ^{-1}(a x)^2}{a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}+6 \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {6}{a \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{\sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 45, normalized size = 0.51 \begin {gather*} \frac {-6+6 a x \tanh ^{-1}(a x)-3 \tanh ^{-1}(a x)^2+a x \tanh ^{-1}(a x)^3}{a \sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6 + 6*a*x*ArcTanh[a*x] - 3*ArcTanh[a*x]^2 + a*x*ArcTanh[a*x]^3)/(a*Sqrt[1 - a^2*x^2])

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Maple [A]
time = 0.67, size = 56, normalized size = 0.64

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (\arctanh \left (a x \right )^{3} a x +6 a x \arctanh \left (a x \right )-3 \arctanh \left (a x \right )^{2}-6\right )}{a \left (a^{2} x^{2}-1\right )}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 86, normalized size = 0.98 \begin {gather*} \frac {x \operatorname {artanh}\left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1}} + 6 \, a {\left (\frac {x \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1} a} - \frac {1}{\sqrt {-a^{2} x^{2} + 1} a^{2}}\right )} - \frac {3 \, \operatorname {artanh}\left (a x\right )^{2}}{\sqrt {-a^{2} x^{2} + 1} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctanh(a*x)^3/sqrt(-a^2*x^2 + 1) + 6*a*(x*arctanh(a*x)/(sqrt(-a^2*x^2 + 1)*a) - 1/(sqrt(-a^2*x^2 + 1)*a^2))
- 3*arctanh(a*x)^2/(sqrt(-a^2*x^2 + 1)*a)

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Fricas [A]
time = 0.37, size = 87, normalized size = 0.99 \begin {gather*} -\frac {{\left (a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 24 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - 6 \, \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 48\right )} \sqrt {-a^{2} x^{2} + 1}}{8 \, {\left (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)^(3/2),x, algorithm="fricas")

[Out]

-1/8*(a*x*log(-(a*x + 1)/(a*x - 1))^3 + 24*a*x*log(-(a*x + 1)/(a*x - 1)) - 6*log(-(a*x + 1)/(a*x - 1))^2 - 48)
*sqrt(-a^2*x^2 + 1)/(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}^{3}{\left (a x \right )}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{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)**(3/2),x)

[Out]

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

[Out]

integrate(arctanh(a*x)^3/(-a^2*x^2 + 1)^(3/2), 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 )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}} \,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)^(3/2),x)

[Out]

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

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