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

Optimal. Leaf size=124 \[ -\frac {8}{15 a^3 \sqrt {a-a x^2}}+\frac {8 x \tanh ^{-1}(x)}{15 a^3 \sqrt {a-a x^2}}-\frac {4}{45 a^2 \left (a-a x^2\right )^{3/2}}+\frac {4 x \tanh ^{-1}(x)}{15 a^2 \left (a-a x^2\right )^{3/2}}-\frac {1}{25 a \left (a-a x^2\right )^{5/2}}+\frac {x \tanh ^{-1}(x)}{5 a \left (a-a x^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5960, 5958} \[ -\frac {8}{15 a^3 \sqrt {a-a x^2}}-\frac {4}{45 a^2 \left (a-a x^2\right )^{3/2}}+\frac {8 x \tanh ^{-1}(x)}{15 a^3 \sqrt {a-a x^2}}+\frac {4 x \tanh ^{-1}(x)}{15 a^2 \left (a-a x^2\right )^{3/2}}-\frac {1}{25 a \left (a-a x^2\right )^{5/2}}+\frac {x \tanh ^{-1}(x)}{5 a \left (a-a x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[x]/(a - a*x^2)^(7/2),x]

[Out]

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

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Mathematica [A]  time = 0.06, size = 55, normalized size = 0.44 \[ \frac {\sqrt {a-a x^2} \left (120 x^4-260 x^2-15 \left (8 x^4-20 x^2+15\right ) x \tanh ^{-1}(x)+149\right )}{225 a^4 \left (x^2-1\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a - a*x^2]*(149 - 260*x^2 + 120*x^4 - 15*x*(15 - 20*x^2 + 8*x^4)*ArcTanh[x]))/(225*a^4*(-1 + x^2)^3)

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fricas [A]  time = 0.56, size = 82, normalized size = 0.66 \[ \frac {{\left (240 \, x^{4} - 520 \, x^{2} - 15 \, {\left (8 \, x^{5} - 20 \, x^{3} + 15 \, x\right )} \log \left (-\frac {x + 1}{x - 1}\right ) + 298\right )} \sqrt {-a x^{2} + a}}{450 \, {\left (a^{4} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{4} x^{2} - a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/450*(240*x^4 - 520*x^2 - 15*(8*x^5 - 20*x^3 + 15*x)*log(-(x + 1)/(x - 1)) + 298)*sqrt(-a*x^2 + a)/(a^4*x^6 -
3*a^4*x^4 + 3*a^4*x^2 - a^4)

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giac [A]  time = 0.19, size = 118, normalized size = 0.95 \[ -\frac {\sqrt {-a x^{2} + a} {\left (4 \, x^{2} {\left (\frac {2 \, x^{2}}{a} - \frac {5}{a}\right )} + \frac {15}{a}\right )} x \log \left (-\frac {x + 1}{x - 1}\right )}{30 \, {\left (a x^{2} - a\right )}^{3}} - \frac {120 \, {\left (a x^{2} - a\right )}^{2} - 20 \, {\left (a x^{2} - a\right )} a + 9 \, a^{2}}{225 \, {\left (a x^{2} - a\right )}^{2} \sqrt {-a x^{2} + a} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*sqrt(-a*x^2 + a)*(4*x^2*(2*x^2/a - 5/a) + 15/a)*x*log(-(x + 1)/(x - 1))/(a*x^2 - a)^3 - 1/225*(120*(a*x^
2 - a)^2 - 20*(a*x^2 - a)*a + 9*a^2)/((a*x^2 - a)^2*sqrt(-a*x^2 + a)*a^3)

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maple [A]  time = 0.48, size = 176, normalized size = 1.42 \[ -\frac {\left (1+x \right )^{2} \left (-1+5 \arctanh \relax (x )\right ) \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}}{800 \left (-1+x \right )^{3} a^{4}}+\frac {5 \left (1+x \right ) \left (-1+3 \arctanh \relax (x )\right ) \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}}{288 \left (-1+x \right )^{2} a^{4}}-\frac {5 \left (\arctanh \relax (x )-1\right ) \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}}{16 \left (-1+x \right ) a^{4}}-\frac {5 \left (1+\arctanh \relax (x )\right ) \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}}{16 \left (1+x \right ) a^{4}}+\frac {5 \left (-1+x \right ) \left (1+3 \arctanh \relax (x )\right ) \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}}{288 \left (1+x \right )^{2} a^{4}}-\frac {\left (-1+x \right )^{2} \left (1+5 \arctanh \relax (x )\right ) \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}}{800 \left (1+x \right )^{3} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/800*(1+x)^2*(-1+5*arctanh(x))*(-(-1+x)*(1+x)*a)^(1/2)/(-1+x)^3/a^4+5/288*(1+x)*(-1+3*arctanh(x))*(-(-1+x)*(
1+x)*a)^(1/2)/(-1+x)^2/a^4-5/16*(arctanh(x)-1)*(-(-1+x)*(1+x)*a)^(1/2)/(-1+x)/a^4-5/16*(1+arctanh(x))*(-(-1+x)
*(1+x)*a)^(1/2)/(1+x)/a^4+5/288*(-1+x)*(1+3*arctanh(x))*(-(-1+x)*(1+x)*a)^(1/2)/(1+x)^2/a^4-1/800*(-1+x)^2*(1+
5*arctanh(x))*(-(-1+x)*(1+x)*a)^(1/2)/(1+x)^3/a^4

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maxima [A]  time = 0.34, size = 99, normalized size = 0.80 \[ \frac {1}{15} \, {\left (\frac {8 \, x}{\sqrt {-a x^{2} + a} a^{3}} + \frac {4 \, x}{{\left (-a x^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {3 \, x}{{\left (-a x^{2} + a\right )}^{\frac {5}{2}} a}\right )} \operatorname {artanh}\relax (x) - \frac {8}{15 \, \sqrt {-a x^{2} + a} a^{3}} - \frac {4}{45 \, {\left (-a x^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {1}{25 \, {\left (-a x^{2} + a\right )}^{\frac {5}{2}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(x)/(a - a*x^2)^(7/2),x)

[Out]

int(atanh(x)/(a - a*x^2)^(7/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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