3.462 \(\int \frac {\tanh ^{-1}(a x)}{(1-a^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=89 \[ -\frac {2}{3 a \sqrt {1-a^2 x^2}}-\frac {1}{9 a \left (1-a^2 x^2\right )^{3/2}}+\frac {2 x \tanh ^{-1}(a x)}{3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)}{3 \left (1-a^2 x^2\right )^{3/2}} \]

[Out]

-1/9/a/(-a^2*x^2+1)^(3/2)+1/3*x*arctanh(a*x)/(-a^2*x^2+1)^(3/2)-2/3/a/(-a^2*x^2+1)^(1/2)+2/3*x*arctanh(a*x)/(-
a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {5960, 5958} \[ -\frac {2}{3 a \sqrt {1-a^2 x^2}}-\frac {1}{9 a \left (1-a^2 x^2\right )^{3/2}}+\frac {2 x \tanh ^{-1}(a x)}{3 \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)}{3 \left (1-a^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]/(1 - a^2*x^2)^(5/2),x]

[Out]

-1/(9*a*(1 - a^2*x^2)^(3/2)) - 2/(3*a*Sqrt[1 - a^2*x^2]) + (x*ArcTanh[a*x])/(3*(1 - a^2*x^2)^(3/2)) + (2*x*Arc
Tanh[a*x])/(3*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 )^{5/2}} \, dx &=-\frac {1}{9 a \left (1-a^2 x^2\right )^{3/2}}+\frac {x \tanh ^{-1}(a x)}{3 \left (1-a^2 x^2\right )^{3/2}}+\frac {2}{3} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {1}{9 a \left (1-a^2 x^2\right )^{3/2}}-\frac {2}{3 a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)}{3 \left (1-a^2 x^2\right )^{3/2}}+\frac {2 x \tanh ^{-1}(a x)}{3 \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 49, normalized size = 0.55 \[ -\frac {\left (6 a^3 x^3-9 a x\right ) \tanh ^{-1}(a x)-6 a^2 x^2+7}{9 a \left (1-a^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[a*x]/(1 - a^2*x^2)^(5/2),x]

[Out]

-1/9*(7 - 6*a^2*x^2 + (-9*a*x + 6*a^3*x^3)*ArcTanh[a*x])/(a*(1 - a^2*x^2)^(3/2))

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fricas [A]  time = 0.49, size = 73, normalized size = 0.82 \[ \frac {{\left (12 \, a^{2} x^{2} - 3 \, {\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 14\right )} \sqrt {-a^{2} x^{2} + 1}}{18 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^(5/2),x, algorithm="fricas")

[Out]

1/18*(12*a^2*x^2 - 3*(2*a^3*x^3 - 3*a*x)*log(-(a*x + 1)/(a*x - 1)) - 14)*sqrt(-a^2*x^2 + 1)/(a^5*x^4 - 2*a^3*x
^2 + a)

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giac [A]  time = 0.71, size = 90, normalized size = 1.01 \[ -\frac {{\left (2 \, a^{2} x^{2} - 3\right )} \sqrt {-a^{2} x^{2} + 1} x \log \left (-\frac {a x + 1}{a x - 1}\right )}{6 \, {\left (a^{2} x^{2} - 1\right )}^{2}} - \frac {6 \, a^{2} x^{2} - 7}{9 \, {\left (a^{2} x^{2} - 1\right )} \sqrt {-a^{2} x^{2} + 1} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^(5/2),x, algorithm="giac")

[Out]

-1/6*(2*a^2*x^2 - 3)*sqrt(-a^2*x^2 + 1)*x*log(-(a*x + 1)/(a*x - 1))/(a^2*x^2 - 1)^2 - 1/9*(6*a^2*x^2 - 7)/((a^
2*x^2 - 1)*sqrt(-a^2*x^2 + 1)*a)

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maple [A]  time = 0.40, size = 59, normalized size = 0.66 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \left (6 a^{3} x^{3} \arctanh \left (a x \right )-6 a^{2} x^{2}-9 a x \arctanh \left (a x \right )+7\right )}{9 a \left (a^{2} x^{2}-1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)/(-a^2*x^2+1)^(5/2),x)

[Out]

-1/9/a*(-a^2*x^2+1)^(1/2)*(6*a^3*x^3*arctanh(a*x)-6*a^2*x^2-9*a*x*arctanh(a*x)+7)/(a^2*x^2-1)^2

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maxima [A]  time = 0.31, size = 74, normalized size = 0.83 \[ -\frac {1}{9} \, a {\left (\frac {6}{\sqrt {-a^{2} x^{2} + 1} a^{2}} + \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{2}}\right )} + \frac {1}{3} \, {\left (\frac {2 \, x}{\sqrt {-a^{2} x^{2} + 1}} + \frac {x}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^(5/2),x, algorithm="maxima")

[Out]

-1/9*a*(6/(sqrt(-a^2*x^2 + 1)*a^2) + 1/((-a^2*x^2 + 1)^(3/2)*a^2)) + 1/3*(2*x/sqrt(-a^2*x^2 + 1) + x/(-a^2*x^2
+ 1)^(3/2))*arctanh(a*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atanh}\left (a\,x\right )}{{\left (1-a^2\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)/(1 - a^2*x^2)^(5/2),x)

[Out]

int(atanh(a*x)/(1 - a^2*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\left (a x \right )}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)/(-a**2*x**2+1)**(5/2),x)

[Out]

Integral(atanh(a*x)/(-(a*x - 1)*(a*x + 1))**(5/2), x)

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