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]

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

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Rubi [A]  time = 0.14, antiderivative size = 172, normalized size of antiderivative = 1.00, 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 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]))

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 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])

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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.82, size = 66, normalized size = 0.38 \[ \frac {{\left (4 \, a^{2} x^{2} + 4 \, a x - 5 \, {\left (a x + 1\right )} \log \left (a x + 1\right ) + {\left (a x + 1\right )} \log \left (a x - 1\right ) - 2\right )} \sqrt {a^{2} c}}{4 \, {\left (a^{3} c^{2} x + a^{2} c^{2}\right )}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(a*x+1)]Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is re
al):Check [abs(t_nostep)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument
Value

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maple [A]  time = 0.07, size = 100, normalized size = 0.58 \[ \frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) \left (4 a^{2} x^{2}+\ln \left (a x -1\right ) x a -5 a x \ln \left (a x +1\right )+4 a x +\ln \left (a x -1\right )-5 \ln \left (a x +1\right )-2\right ) \left (a x -1\right )}{4 a^{4} x^{3} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {3}{2}}} \]

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+ln(a*x-1)*x*a-5*a*x*ln(a*x+1)+4*a*x+ln(a*x-1)-5*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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {a x - 1}{a x + 1}}}{{\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\frac {a\,x-1}{a\,x+1}}}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{3/2}} \,d x \]

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]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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