3.50 \(\int \frac {\coth ^{-1}(x)}{(a-a x^2)^{3/2}} \, dx\)

Optimal. Leaf size=37 \[ \frac {x \coth ^{-1}(x)}{a \sqrt {a-a x^2}}-\frac {1}{a \sqrt {a-a x^2}} \]

[Out]

-1/a/(-a*x^2+a)^(1/2)+x*arccoth(x)/a/(-a*x^2+a)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {5959} \[ \frac {x \coth ^{-1}(x)}{a \sqrt {a-a x^2}}-\frac {1}{a \sqrt {a-a x^2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[x]/(a - a*x^2)^(3/2),x]

[Out]

-(1/(a*Sqrt[a - a*x^2])) + (x*ArcCoth[x])/(a*Sqrt[a - a*x^2])

Rule 5959

Int[((a_.) + ArcCoth[(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*ArcCoth[c*x]))/(d*Sqrt[d + e*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0
]

Rubi steps

\begin {align*} \int \frac {\coth ^{-1}(x)}{\left (a-a x^2\right )^{3/2}} \, dx &=-\frac {1}{a \sqrt {a-a x^2}}+\frac {x \coth ^{-1}(x)}{a \sqrt {a-a x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 30, normalized size = 0.81 \[ \frac {\sqrt {a-a x^2} \left (1-x \coth ^{-1}(x)\right )}{a^2 \left (x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[x]/(a - a*x^2)^(3/2),x]

[Out]

(Sqrt[a - a*x^2]*(1 - x*ArcCoth[x]))/(a^2*(-1 + x^2))

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fricas [A]  time = 0.47, size = 41, normalized size = 1.11 \[ -\frac {\sqrt {-a x^{2} + a} {\left (x \log \left (\frac {x + 1}{x - 1}\right ) - 2\right )}}{2 \, {\left (a^{2} x^{2} - a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(x)/(-a*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

-1/2*sqrt(-a*x^2 + a)*(x*log((x + 1)/(x - 1)) - 2)/(a^2*x^2 - a^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(x)/(-a*x^2+a)^(3/2),x, algorithm="giac")

[Out]

integrate(arccoth(x)/(-a*x^2 + a)^(3/2), x)

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maple [A]  time = 0.48, size = 52, normalized size = 1.41 \[ -\frac {\left (\mathrm {arccoth}\relax (x )-1\right ) \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}}{2 \left (-1+x \right ) a^{2}}-\frac {\left (\mathrm {arccoth}\relax (x )+1\right ) \sqrt {-\left (-1+x \right ) \left (1+x \right ) a}}{2 \left (1+x \right ) a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(x)/(-a*x^2+a)^(3/2),x)

[Out]

-1/2*(arccoth(x)-1)*(-(-1+x)*(1+x)*a)^(1/2)/(-1+x)/a^2-1/2*(arccoth(x)+1)*(-(-1+x)*(1+x)*a)^(1/2)/(1+x)/a^2

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maxima [A]  time = 0.41, size = 63, normalized size = 1.70 \[ \frac {x \operatorname {arcoth}\relax (x)}{\sqrt {-a x^{2} + a} a} - \frac {\frac {\sqrt {-a x^{2} + a}}{a x + a} - \frac {\sqrt {-a x^{2} + a}}{a x - a}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(x)/(-a*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

x*arccoth(x)/(sqrt(-a*x^2 + a)*a) - 1/2*(sqrt(-a*x^2 + a)/(a*x + a) - sqrt(-a*x^2 + a)/(a*x - a))/a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\mathrm {acoth}\relax (x)}{{\left (a-a\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(x)/(a - a*x^2)^(3/2),x)

[Out]

int(acoth(x)/(a - a*x^2)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(x)/(-a*x**2+a)**(3/2),x)

[Out]

Integral(acoth(x)/(-a*(x - 1)*(x + 1))**(3/2), x)

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