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

Optimal. Leaf size=133 \[ -\frac {8}{15 a \sqrt {1-a^2 x^2}}-\frac {4}{45 a \left (1-a^2 x^2\right )^{3/2}}-\frac {1}{25 a \left (1-a^2 x^2\right )^{5/2}}+\frac {8 x \tanh ^{-1}(a x)}{15 \sqrt {1-a^2 x^2}}+\frac {4 x \tanh ^{-1}(a x)}{15 \left (1-a^2 x^2\right )^{3/2}}+\frac {x \tanh ^{-1}(a x)}{5 \left (1-a^2 x^2\right )^{5/2}} \]

[Out]

-1/25/a/(-a^2*x^2+1)^(5/2)-4/45/a/(-a^2*x^2+1)^(3/2)+1/5*x*arctanh(a*x)/(-a^2*x^2+1)^(5/2)+4/15*x*arctanh(a*x)
/(-a^2*x^2+1)^(3/2)-8/15/a/(-a^2*x^2+1)^(1/2)+8/15*x*arctanh(a*x)/(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {5960, 5958} \[ -\frac {8}{15 a \sqrt {1-a^2 x^2}}-\frac {4}{45 a \left (1-a^2 x^2\right )^{3/2}}-\frac {1}{25 a \left (1-a^2 x^2\right )^{5/2}}+\frac {8 x \tanh ^{-1}(a x)}{15 \sqrt {1-a^2 x^2}}+\frac {4 x \tanh ^{-1}(a x)}{15 \left (1-a^2 x^2\right )^{3/2}}+\frac {x \tanh ^{-1}(a x)}{5 \left (1-a^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(25*a*(1 - a^2*x^2)^(5/2)) - 4/(45*a*(1 - a^2*x^2)^(3/2)) - 8/(15*a*Sqrt[1 - a^2*x^2]) + (x*ArcTanh[a*x])/(
5*(1 - a^2*x^2)^(5/2)) + (4*x*ArcTanh[a*x])/(15*(1 - a^2*x^2)^(3/2)) + (8*x*ArcTanh[a*x])/(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]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{7/2}} \, dx &=-\frac {1}{25 a \left (1-a^2 x^2\right )^{5/2}}+\frac {x \tanh ^{-1}(a x)}{5 \left (1-a^2 x^2\right )^{5/2}}+\frac {4}{5} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{5/2}} \, dx\\ &=-\frac {1}{25 a \left (1-a^2 x^2\right )^{5/2}}-\frac {4}{45 a \left (1-a^2 x^2\right )^{3/2}}+\frac {x \tanh ^{-1}(a x)}{5 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 x \tanh ^{-1}(a x)}{15 \left (1-a^2 x^2\right )^{3/2}}+\frac {8}{15} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {1}{25 a \left (1-a^2 x^2\right )^{5/2}}-\frac {4}{45 a \left (1-a^2 x^2\right )^{3/2}}-\frac {8}{15 a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)}{5 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 x \tanh ^{-1}(a x)}{15 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 x \tanh ^{-1}(a x)}{15 \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 65, normalized size = 0.49 \[ \frac {-120 a^4 x^4+260 a^2 x^2+15 a x \left (8 a^4 x^4-20 a^2 x^2+15\right ) \tanh ^{-1}(a x)-149}{225 a \left (1-a^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-149 + 260*a^2*x^2 - 120*a^4*x^4 + 15*a*x*(15 - 20*a^2*x^2 + 8*a^4*x^4)*ArcTanh[a*x])/(225*a*(1 - a^2*x^2)^(5
/2))

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fricas [A]  time = 0.44, size = 99, normalized size = 0.74 \[ \frac {{\left (240 \, a^{4} x^{4} - 520 \, a^{2} x^{2} - 15 \, {\left (8 \, a^{5} x^{5} - 20 \, a^{3} x^{3} + 15 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + 298\right )} \sqrt {-a^{2} x^{2} + 1}}{450 \, {\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^(7/2),x, algorithm="fricas")

[Out]

1/450*(240*a^4*x^4 - 520*a^2*x^2 - 15*(8*a^5*x^5 - 20*a^3*x^3 + 15*a*x)*log(-(a*x + 1)/(a*x - 1)) + 298)*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 [A]  time = 0.25, size = 114, normalized size = 0.86 \[ -\frac {\sqrt {-a^{2} x^{2} + 1} {\left (4 \, {\left (2 \, a^{4} x^{2} - 5 \, a^{2}\right )} x^{2} + 15\right )} x \log \left (-\frac {a x + 1}{a x - 1}\right )}{30 \, {\left (a^{2} x^{2} - 1\right )}^{3}} + \frac {20 \, a^{2} x^{2} - 120 \, {\left (a^{2} x^{2} - 1\right )}^{2} - 29}{225 \, {\left (a^{2} x^{2} - 1\right )}^{2} \sqrt {-a^{2} x^{2} + 1} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^(7/2),x, algorithm="giac")

[Out]

-1/30*sqrt(-a^2*x^2 + 1)*(4*(2*a^4*x^2 - 5*a^2)*x^2 + 15)*x*log(-(a*x + 1)/(a*x - 1))/(a^2*x^2 - 1)^3 + 1/225*
(20*a^2*x^2 - 120*(a^2*x^2 - 1)^2 - 29)/((a^2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1)*a)

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maple [A]  time = 0.42, size = 79, normalized size = 0.59 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \left (120 \arctanh \left (a x \right ) x^{5} a^{5}-120 x^{4} a^{4}-300 a^{3} x^{3} \arctanh \left (a x \right )+260 a^{2} x^{2}+225 a x \arctanh \left (a x \right )-149\right )}{225 a \left (a^{2} x^{2}-1\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/225/a*(-a^2*x^2+1)^(1/2)*(120*arctanh(a*x)*x^5*a^5-120*x^4*a^4-300*a^3*x^3*arctanh(a*x)+260*a^2*x^2+225*a*x
*arctanh(a*x)-149)/(a^2*x^2-1)^3

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maxima [A]  time = 0.31, size = 108, normalized size = 0.81 \[ -\frac {1}{225} \, a {\left (\frac {120}{\sqrt {-a^{2} x^{2} + 1} a^{2}} + \frac {20}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{2}} + \frac {9}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} a^{2}}\right )} + \frac {1}{15} \, {\left (\frac {8 \, x}{\sqrt {-a^{2} x^{2} + 1}} + \frac {4 \, x}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} + \frac {3 \, x}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}}}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^(7/2),x, algorithm="maxima")

[Out]

-1/225*a*(120/(sqrt(-a^2*x^2 + 1)*a^2) + 20/((-a^2*x^2 + 1)^(3/2)*a^2) + 9/((-a^2*x^2 + 1)^(5/2)*a^2)) + 1/15*
(8*x/sqrt(-a^2*x^2 + 1) + 4*x/(-a^2*x^2 + 1)^(3/2) + 3*x/(-a^2*x^2 + 1)^(5/2))*arctanh(a*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atanh}\left (a\,x\right )}{{\left (1-a^2\,x^2\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)/(1 - a^2*x^2)^(7/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)/(-a**2*x**2+1)**(7/2),x)

[Out]

Integral(atanh(a*x)/(-(a*x - 1)*(a*x + 1))**(7/2), x)

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