3.5.77 \(\int \frac {\tanh ^{-1}(a x)^3}{(1-a^2 x^2)^{7/2}} \, dx\) [477]

Optimal. Leaf size=289 \[ -\frac {6}{625 a \left (1-a^2 x^2\right )^{5/2}}-\frac {272}{3375 a \left (1-a^2 x^2\right )^{3/2}}-\frac {4144}{1125 a \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{125 \left (1-a^2 x^2\right )^{5/2}}+\frac {272 x \tanh ^{-1}(a x)}{1125 \left (1-a^2 x^2\right )^{3/2}}+\frac {4144 x \tanh ^{-1}(a x)}{1125 \sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{25 a \left (1-a^2 x^2\right )^{5/2}}-\frac {4 \tanh ^{-1}(a x)^2}{15 a \left (1-a^2 x^2\right )^{3/2}}-\frac {8 \tanh ^{-1}(a x)^2}{5 a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{5 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 x \tanh ^{-1}(a x)^3}{15 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 x \tanh ^{-1}(a x)^3}{15 \sqrt {1-a^2 x^2}} \]

[Out]

-6/625/a/(-a^2*x^2+1)^(5/2)-272/3375/a/(-a^2*x^2+1)^(3/2)+6/125*x*arctanh(a*x)/(-a^2*x^2+1)^(5/2)+272/1125*x*a
rctanh(a*x)/(-a^2*x^2+1)^(3/2)-3/25*arctanh(a*x)^2/a/(-a^2*x^2+1)^(5/2)-4/15*arctanh(a*x)^2/a/(-a^2*x^2+1)^(3/
2)+1/5*x*arctanh(a*x)^3/(-a^2*x^2+1)^(5/2)+4/15*x*arctanh(a*x)^3/(-a^2*x^2+1)^(3/2)-4144/1125/a/(-a^2*x^2+1)^(
1/2)+4144/1125*x*arctanh(a*x)/(-a^2*x^2+1)^(1/2)-8/5*arctanh(a*x)^2/a/(-a^2*x^2+1)^(1/2)+8/15*x*arctanh(a*x)^3
/(-a^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 289, normalized size of antiderivative = 1.00, number of steps used = 9, 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 {4144}{1125 a \sqrt {1-a^2 x^2}}-\frac {272}{3375 a \left (1-a^2 x^2\right )^{3/2}}-\frac {6}{625 a \left (1-a^2 x^2\right )^{5/2}}+\frac {8 x \tanh ^{-1}(a x)^3}{15 \sqrt {1-a^2 x^2}}+\frac {4 x \tanh ^{-1}(a x)^3}{15 \left (1-a^2 x^2\right )^{3/2}}+\frac {x \tanh ^{-1}(a x)^3}{5 \left (1-a^2 x^2\right )^{5/2}}-\frac {8 \tanh ^{-1}(a x)^2}{5 a \sqrt {1-a^2 x^2}}-\frac {4 \tanh ^{-1}(a x)^2}{15 a \left (1-a^2 x^2\right )^{3/2}}-\frac {3 \tanh ^{-1}(a x)^2}{25 a \left (1-a^2 x^2\right )^{5/2}}+\frac {4144 x \tanh ^{-1}(a x)}{1125 \sqrt {1-a^2 x^2}}+\frac {272 x \tanh ^{-1}(a x)}{1125 \left (1-a^2 x^2\right )^{3/2}}+\frac {6 x \tanh ^{-1}(a x)}{125 \left (1-a^2 x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^3/(1 - a^2*x^2)^(7/2),x]

[Out]

-6/(625*a*(1 - a^2*x^2)^(5/2)) - 272/(3375*a*(1 - a^2*x^2)^(3/2)) - 4144/(1125*a*Sqrt[1 - a^2*x^2]) + (6*x*Arc
Tanh[a*x])/(125*(1 - a^2*x^2)^(5/2)) + (272*x*ArcTanh[a*x])/(1125*(1 - a^2*x^2)^(3/2)) + (4144*x*ArcTanh[a*x])
/(1125*Sqrt[1 - a^2*x^2]) - (3*ArcTanh[a*x]^2)/(25*a*(1 - a^2*x^2)^(5/2)) - (4*ArcTanh[a*x]^2)/(15*a*(1 - a^2*
x^2)^(3/2)) - (8*ArcTanh[a*x]^2)/(5*a*Sqrt[1 - a^2*x^2]) + (x*ArcTanh[a*x]^3)/(5*(1 - a^2*x^2)^(5/2)) + (4*x*A
rcTanh[a*x]^3)/(15*(1 - a^2*x^2)^(3/2)) + (8*x*ArcTanh[a*x]^3)/(15*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 )^{7/2}} \, dx &=-\frac {3 \tanh ^{-1}(a x)^2}{25 a \left (1-a^2 x^2\right )^{5/2}}+\frac {x \tanh ^{-1}(a x)^3}{5 \left (1-a^2 x^2\right )^{5/2}}+\frac {6}{25} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{7/2}} \, dx+\frac {4}{5} \int \frac {\tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx\\ &=-\frac {6}{625 a \left (1-a^2 x^2\right )^{5/2}}+\frac {6 x \tanh ^{-1}(a x)}{125 \left (1-a^2 x^2\right )^{5/2}}-\frac {3 \tanh ^{-1}(a x)^2}{25 a \left (1-a^2 x^2\right )^{5/2}}-\frac {4 \tanh ^{-1}(a x)^2}{15 a \left (1-a^2 x^2\right )^{3/2}}+\frac {x \tanh ^{-1}(a x)^3}{5 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 x \tanh ^{-1}(a x)^3}{15 \left (1-a^2 x^2\right )^{3/2}}+\frac {24}{125} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{5/2}} \, dx+\frac {8}{15} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{5/2}} \, dx+\frac {8}{15} \int \frac {\tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {6}{625 a \left (1-a^2 x^2\right )^{5/2}}-\frac {272}{3375 a \left (1-a^2 x^2\right )^{3/2}}+\frac {6 x \tanh ^{-1}(a x)}{125 \left (1-a^2 x^2\right )^{5/2}}+\frac {272 x \tanh ^{-1}(a x)}{1125 \left (1-a^2 x^2\right )^{3/2}}-\frac {3 \tanh ^{-1}(a x)^2}{25 a \left (1-a^2 x^2\right )^{5/2}}-\frac {4 \tanh ^{-1}(a x)^2}{15 a \left (1-a^2 x^2\right )^{3/2}}-\frac {8 \tanh ^{-1}(a x)^2}{5 a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{5 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 x \tanh ^{-1}(a x)^3}{15 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 x \tanh ^{-1}(a x)^3}{15 \sqrt {1-a^2 x^2}}+\frac {16}{125} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx+\frac {16}{45} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx+\frac {16}{5} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {6}{625 a \left (1-a^2 x^2\right )^{5/2}}-\frac {272}{3375 a \left (1-a^2 x^2\right )^{3/2}}-\frac {4144}{1125 a \sqrt {1-a^2 x^2}}+\frac {6 x \tanh ^{-1}(a x)}{125 \left (1-a^2 x^2\right )^{5/2}}+\frac {272 x \tanh ^{-1}(a x)}{1125 \left (1-a^2 x^2\right )^{3/2}}+\frac {4144 x \tanh ^{-1}(a x)}{1125 \sqrt {1-a^2 x^2}}-\frac {3 \tanh ^{-1}(a x)^2}{25 a \left (1-a^2 x^2\right )^{5/2}}-\frac {4 \tanh ^{-1}(a x)^2}{15 a \left (1-a^2 x^2\right )^{3/2}}-\frac {8 \tanh ^{-1}(a x)^2}{5 a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)^3}{5 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 x \tanh ^{-1}(a x)^3}{15 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 x \tanh ^{-1}(a x)^3}{15 \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 119, normalized size = 0.41 \begin {gather*} \frac {-63682+125680 a^2 x^2-62160 a^4 x^4+30 a x \left (2235-4280 a^2 x^2+2072 a^4 x^4\right ) \tanh ^{-1}(a x)-225 \left (149-260 a^2 x^2+120 a^4 x^4\right ) \tanh ^{-1}(a x)^2+1125 a x \left (15-20 a^2 x^2+8 a^4 x^4\right ) \tanh ^{-1}(a x)^3}{16875 a \left (1-a^2 x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[a*x]^3/(1 - a^2*x^2)^(7/2),x]

[Out]

(-63682 + 125680*a^2*x^2 - 62160*a^4*x^4 + 30*a*x*(2235 - 4280*a^2*x^2 + 2072*a^4*x^4)*ArcTanh[a*x] - 225*(149
- 260*a^2*x^2 + 120*a^4*x^4)*ArcTanh[a*x]^2 + 1125*a*x*(15 - 20*a^2*x^2 + 8*a^4*x^4)*ArcTanh[a*x]^3)/(16875*a
*(1 - a^2*x^2)^(5/2))

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Maple [A]
time = 0.71, size = 153, normalized size = 0.53

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (9000 \arctanh \left (a x \right )^{3} a^{5} x^{5}+62160 \arctanh \left (a x \right ) a^{5} x^{5}-27000 a^{4} x^{4} \arctanh \left (a x \right )^{2}-22500 \arctanh \left (a x \right )^{3} a^{3} x^{3}-62160 a^{4} x^{4}-128400 a^{3} x^{3} \arctanh \left (a x \right )+58500 a^{2} x^{2} \arctanh \left (a x \right )^{2}+16875 \arctanh \left (a x \right )^{3} a x +125680 a^{2} x^{2}+67050 a x \arctanh \left (a x \right )-33525 \arctanh \left (a x \right )^{2}-63682\right )}{16875 a \left (a^{2} x^{2}-1\right )^{3}}\)	\(153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)^3/(-a^2*x^2+1)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/16875/a*(-a^2*x^2+1)^(1/2)*(9000*arctanh(a*x)^3*a^5*x^5+62160*arctanh(a*x)*a^5*x^5-27000*a^4*x^4*arctanh(a*
x)^2-22500*arctanh(a*x)^3*a^3*x^3-62160*a^4*x^4-128400*a^3*x^3*arctanh(a*x)+58500*a^2*x^2*arctanh(a*x)^2+16875
*arctanh(a*x)^3*a*x+125680*a^2*x^2+67050*a*x*arctanh(a*x)-33525*arctanh(a*x)^2-63682)/(a^2*x^2-1)^3

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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)^(7/2),x, algorithm="maxima")

[Out]

integrate(arctanh(a*x)^3/(-a^2*x^2 + 1)^(7/2), x)

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Fricas [A]
time = 0.37, size = 176, normalized size = 0.61 \begin {gather*} \frac {{\left (497280 \, a^{4} x^{4} - 1005440 \, a^{2} x^{2} - 1125 \, {\left (8 \, a^{5} x^{5} - 20 \, a^{3} x^{3} + 15 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 450 \, {\left (120 \, a^{4} x^{4} - 260 \, a^{2} x^{2} + 149\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 120 \, {\left (2072 \, a^{5} x^{5} - 4280 \, a^{3} x^{3} + 2235 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + 509456\right )} \sqrt {-a^{2} x^{2} + 1}}{135000 \, {\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, 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)^(7/2),x, algorithm="fricas")

[Out]

1/135000*(497280*a^4*x^4 - 1005440*a^2*x^2 - 1125*(8*a^5*x^5 - 20*a^3*x^3 + 15*a*x)*log(-(a*x + 1)/(a*x - 1))^
3 + 450*(120*a^4*x^4 - 260*a^2*x^2 + 149)*log(-(a*x + 1)/(a*x - 1))^2 - 120*(2072*a^5*x^5 - 4280*a^3*x^3 + 223
5*a*x)*log(-(a*x + 1)/(a*x - 1)) + 509456)*sqrt(-a^2*x^2 + 1)/(a^7*x^6 - 3*a^5*x^4 + 3*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 {7}{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)**(7/2),x)

[Out]

Integral(atanh(a*x)**3/(-(a*x - 1)*(a*x + 1))**(7/2), 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)^(7/2),x, algorithm="giac")

[Out]

integrate(arctanh(a*x)^3/(-a^2*x^2 + 1)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {atanh}\left (a\,x\right )}^3}{{\left (1-a^2\,x^2\right )}^{7/2}} \,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)^(7/2),x)

[Out]

int(atanh(a*x)^3/(1 - a^2*x^2)^(7/2), x)

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