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

Optimal. Leaf size=172 \[ \frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{c \sqrt{c-\frac{c}{a^2 x^2}}}-\frac{\sqrt{1-\frac{1}{a^2 x^2}}}{2 a c (a x+1) \sqrt{c-\frac{c}{a^2 x^2}}}+\frac{\sqrt{1-\frac{1}{a^2 x^2}} \log (1-a x)}{4 a c \sqrt{c-\frac{c}{a^2 x^2}}}-\frac{5 \sqrt{1-\frac{1}{a^2 x^2}} \log (a x+1)}{4 a c \sqrt{c-\frac{c}{a^2 x^2}}} \]

[Out]

(Sqrt[1 - 1/(a^2*x^2)]*x)/(c*Sqrt[c - c/(a^2*x^2)]) - Sqrt[1 - 1/(a^2*x^2)]/(2*a*c*Sqrt[c - c/(a^2*x^2)]*(1 +
a*x)) + (Sqrt[1 - 1/(a^2*x^2)]*Log[1 - a*x])/(4*a*c*Sqrt[c - c/(a^2*x^2)]) - (5*Sqrt[1 - 1/(a^2*x^2)]*Log[1 +
a*x])/(4*a*c*Sqrt[c - c/(a^2*x^2)])

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Rubi [A]  time = 0.1429, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {6197, 6193, 88} \[ \frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{c \sqrt{c-\frac{c}{a^2 x^2}}}-\frac{\sqrt{1-\frac{1}{a^2 x^2}}}{2 a c (a x+1) \sqrt{c-\frac{c}{a^2 x^2}}}+\frac{\sqrt{1-\frac{1}{a^2 x^2}} \log (1-a x)}{4 a c \sqrt{c-\frac{c}{a^2 x^2}}}-\frac{5 \sqrt{1-\frac{1}{a^2 x^2}} \log (a x+1)}{4 a c \sqrt{c-\frac{c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcCoth[a*x]*(c - c/(a^2*x^2))^(3/2)),x]

[Out]

(Sqrt[1 - 1/(a^2*x^2)]*x)/(c*Sqrt[c - c/(a^2*x^2)]) - Sqrt[1 - 1/(a^2*x^2)]/(2*a*c*Sqrt[c - c/(a^2*x^2)]*(1 +
a*x)) + (Sqrt[1 - 1/(a^2*x^2)]*Log[1 - a*x])/(4*a*c*Sqrt[c - c/(a^2*x^2)]) - (5*Sqrt[1 - 1/(a^2*x^2)]*Log[1 +
a*x])/(4*a*c*Sqrt[c - c/(a^2*x^2)])

Rule 6197

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c + d/x^2
)^FracPart[p])/(1 - 1/(a^2*x^2))^FracPart[p], Int[u*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a
, c, d, n, p}, x] && EqQ[c + a^2*d, 0] &&  !IntegerQ[n/2] &&  !(IntegerQ[p] || GtQ[c, 0])

Rule 6193

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u*(-1
+ a*x)^(p - n/2)*(1 + a*x)^(p + n/2))/x^(2*p), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] &&  !
IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{e^{-\coth ^{-1}(a x)}}{\left (c-\frac{c}{a^2 x^2}\right )^{3/2}} \, dx &=\frac{\sqrt{1-\frac{1}{a^2 x^2}} \int \frac{e^{-\coth ^{-1}(a x)}}{\left (1-\frac{1}{a^2 x^2}\right )^{3/2}} \, dx}{c \sqrt{c-\frac{c}{a^2 x^2}}}\\ &=\frac{\left (a^3 \sqrt{1-\frac{1}{a^2 x^2}}\right ) \int \frac{x^3}{(-1+a x) (1+a x)^2} \, dx}{c \sqrt{c-\frac{c}{a^2 x^2}}}\\ &=\frac{\left (a^3 \sqrt{1-\frac{1}{a^2 x^2}}\right ) \int \left (\frac{1}{a^3}+\frac{1}{4 a^3 (-1+a x)}+\frac{1}{2 a^3 (1+a x)^2}-\frac{5}{4 a^3 (1+a x)}\right ) \, dx}{c \sqrt{c-\frac{c}{a^2 x^2}}}\\ &=\frac{\sqrt{1-\frac{1}{a^2 x^2}} x}{c \sqrt{c-\frac{c}{a^2 x^2}}}-\frac{\sqrt{1-\frac{1}{a^2 x^2}}}{2 a c \sqrt{c-\frac{c}{a^2 x^2}} (1+a x)}+\frac{\sqrt{1-\frac{1}{a^2 x^2}} \log (1-a x)}{4 a c \sqrt{c-\frac{c}{a^2 x^2}}}-\frac{5 \sqrt{1-\frac{1}{a^2 x^2}} \log (1+a x)}{4 a c \sqrt{c-\frac{c}{a^2 x^2}}}\\ \end{align*}

Mathematica [A]  time = 0.0645993, size = 65, normalized size = 0.38 \[ \frac{\left (1-\frac{1}{a^2 x^2}\right )^{3/2} \left (4 a x-\frac{2}{a x+1}+\log (1-a x)-5 \log (a x+1)\right )}{4 a \left (c-\frac{c}{a^2 x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcCoth[a*x]*(c - c/(a^2*x^2))^(3/2)),x]

[Out]

((1 - 1/(a^2*x^2))^(3/2)*(4*a*x - 2/(1 + a*x) + Log[1 - a*x] - 5*Log[1 + a*x]))/(4*a*(c - c/(a^2*x^2))^(3/2))

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Maple [A]  time = 0.281, size = 103, normalized size = 0.6 \begin{align*} -{\frac{ \left ( ax+1 \right ) \left ( -4\,{a}^{2}{x}^{2}+5\,ax\ln \left ( ax+1 \right ) -\ln \left ( ax-1 \right ) xa-4\,ax+5\,\ln \left ( ax+1 \right ) -\ln \left ( ax-1 \right ) +2 \right ) \left ( ax-1 \right ) }{4\,{a}^{4}{x}^{3}}\sqrt{{\frac{ax-1}{ax+1}}} \left ({\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((a*x-1)/(a*x+1))^(1/2)*(a*x+1)*(-4*a^2*x^2+5*a*x*ln(a*x+1)-ln(a*x-1)*x*a-4*a*x+5*ln(a*x+1)-ln(a*x-1)+2)*
(a*x-1)/a^4/x^3/(c*(a^2*x^2-1)/a^2/x^2)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{a x - 1}{a x + 1}}}{{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(1/2)/(c-c/a^2/x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt((a*x - 1)/(a*x + 1))/(c - c/(a^2*x^2))^(3/2), x)

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Fricas [A]  time = 1.97078, size = 155, normalized size = 0.9 \begin{align*} \frac{{\left (4 \, a^{2} x^{2} + 4 \, a x - 5 \,{\left (a x + 1\right )} \log \left (a x + 1\right ) +{\left (a x + 1\right )} \log \left (a x - 1\right ) - 2\right )} \sqrt{a^{2} c}}{4 \,{\left (a^{3} c^{2} x + a^{2} c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(1/2)/(c-c/a^2/x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))**(1/2)/(c-c/a**2/x**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{a x - 1}{a x + 1}}}{{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(1/2)/(c-c/a^2/x^2)^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt((a*x - 1)/(a*x + 1))/(c - c/(a^2*x^2))^(3/2), x)