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

Optimal. Leaf size=147 \[ \frac{c x \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-\frac{1}{a^2 x^2}}}+\frac{c \sqrt{c-\frac{c}{a^2 x^2}}}{a^2 x \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{c \sqrt{c-\frac{c}{a^2 x^2}}}{2 a^3 x^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{c \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{a \sqrt{1-\frac{1}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.124496, antiderivative size = 147, 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, 75} \[ \frac{c x \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-\frac{1}{a^2 x^2}}}+\frac{c \sqrt{c-\frac{c}{a^2 x^2}}}{a^2 x \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{c \sqrt{c-\frac{c}{a^2 x^2}}}{2 a^3 x^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{c \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{a \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*Sqrt[c - c/(a^2*x^2)])/(2*a^3*Sqrt[1 - 1/(a^2*x^2)]*x^2) + (c*Sqrt[c - c/(a^2*x^2)])/(a^2*Sqrt[1 - 1/(a^2*
x^2)]*x) + (c*Sqrt[c - c/(a^2*x^2)]*x)/Sqrt[1 - 1/(a^2*x^2)] - (c*Sqrt[c - c/(a^2*x^2)]*Log[x])/(a*Sqrt[1 - 1/
(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 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

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

Mathematica [A]  time = 0.0442541, size = 65, normalized size = 0.44 \[ \frac{\left (c-\frac{c}{a^2 x^2}\right )^{3/2} \left (a^3 x-a^2 \log (x)-\frac{3 a^2}{2}+\frac{a}{x}-\frac{1}{2 x^2}\right )}{a^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.226, size = 80, normalized size = 0.5 \begin{align*} -{\frac{x \left ( -2\,{x}^{3}{a}^{3}+2\,{a}^{2}\ln \left ( x \right ){x}^{2}-2\,ax+1 \right ) }{ \left ( 2\,ax-2 \right ) \left ({a}^{2}{x}^{2}-1 \right ) } \left ({\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}} \right ) ^{{\frac{3}{2}}}\sqrt{{\frac{ax-1}{ax+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.99713, size = 103, normalized size = 0.7 \begin{align*} \frac{{\left (2 \, a^{3} c x^{3} - 2 \, a^{2} c x^{2} \log \left (x\right ) + 2 \, a c x - c\right )} \sqrt{a^{2} c}}{2 \, a^{4} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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