3.477 \(\int \frac {\tanh ^{-1}(a x)^3}{(1-a^2 x^2)^{7/2}} \, dx\)

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

[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 = 0.30, 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 = {5964, 5962, 5958, 5960} \[ -\frac {4144}{1125 a \sqrt {1-a^2 x^2}}-\frac {272}{3375 a \left (1-a^2 x^2\right )^{3/2}}-\frac {6}{625 a \left (1-a^2 x^2\right )^{5/2}}+\frac {8 x \tanh ^{-1}(a x)^3}{15 \sqrt {1-a^2 x^2}}+\frac {4 x \tanh ^{-1}(a x)^3}{15 \left (1-a^2 x^2\right )^{3/2}}+\frac {x \tanh ^{-1}(a x)^3}{5 \left (1-a^2 x^2\right )^{5/2}}-\frac {8 \tanh ^{-1}(a x)^2}{5 a \sqrt {1-a^2 x^2}}-\frac {4 \tanh ^{-1}(a x)^2}{15 a \left (1-a^2 x^2\right )^{3/2}}-\frac {3 \tanh ^{-1}(a x)^2}{25 a \left (1-a^2 x^2\right )^{5/2}}+\frac {4144 x \tanh ^{-1}(a x)}{1125 \sqrt {1-a^2 x^2}}+\frac {272 x \tanh ^{-1}(a x)}{1125 \left (1-a^2 x^2\right )^{3/2}}+\frac {6 x \tanh ^{-1}(a x)}{125 \left (1-a^2 x^2\right )^{5/2}} \]

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 5958

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 5960

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 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)^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 = 0.11, size = 119, normalized size = 0.41 \[ \frac {-62160 a^4 x^4+125680 a^2 x^2+1125 a x \left (8 a^4 x^4-20 a^2 x^2+15\right ) \tanh ^{-1}(a x)^3+30 a x \left (2072 a^4 x^4-4280 a^2 x^2+2235\right ) \tanh ^{-1}(a x)-225 \left (120 a^4 x^4-260 a^2 x^2+149\right ) \tanh ^{-1}(a x)^2-63682}{16875 a \left (1-a^2 x^2\right )^{5/2}} \]

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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fricas [A]  time = 0.47, size = 176, normalized size = 0.61 \[ \frac {{\left (497280 \, a^{4} x^{4} - 1005440 \, a^{2} x^{2} - 1125 \, {\left (8 \, a^{5} x^{5} - 20 \, a^{3} x^{3} + 15 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 450 \, {\left (120 \, a^{4} x^{4} - 260 \, a^{2} x^{2} + 149\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 120 \, {\left (2072 \, a^{5} x^{5} - 4280 \, a^{3} x^{3} + 2235 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + 509456\right )} \sqrt {-a^{2} x^{2} + 1}}{135000 \, {\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (a x\right )^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {7}{2}}}\,{d x} \]

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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maple [A]  time = 0.46, size = 153, normalized size = 0.53 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \left (9000 \arctanh \left (a x \right )^{3} x^{5} a^{5}+62160 \arctanh \left (a x \right ) x^{5} a^{5}-27000 a^{4} x^{4} \arctanh \left (a x \right )^{2}-22500 \arctanh \left (a x \right )^{3} x^{3} a^{3}-62160 x^{4} a^{4}-128400 a^{3} x^{3} \arctanh \left (a x \right )+58500 a^{2} x^{2} \arctanh \left (a x \right )^{2}+16875 \arctanh \left (a x \right )^{3} a x +125680 a^{2} x^{2}+67050 a x \arctanh \left (a x \right )-33525 \arctanh \left (a x \right )^{2}-63682\right )}{16875 a \left (a^{2} x^{2}-1\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16875/a*(-a^2*x^2+1)^(1/2)*(9000*arctanh(a*x)^3*x^5*a^5+62160*arctanh(a*x)*x^5*a^5-27000*a^4*x^4*arctanh(a*
x)^2-22500*arctanh(a*x)^3*x^3*a^3-62160*x^4*a^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 \[ \int \frac {\operatorname {artanh}\left (a x\right )^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {7}{2}}}\,{d x} \]

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

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}^{3}{\left (a x \right )}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}}}\, dx \]

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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