3.522 \(\int \frac{e^{-\coth ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^3} \, dx\)

Optimal. Leaf size=113 \[ \frac{2 a^2 \sqrt{1-\frac{1}{a^2 x^2}} \left (c-\frac{c}{a x}\right )^{3/2}}{5 c}+\frac{2}{5} a^2 \sqrt{1-\frac{1}{a^2 x^2}} \sqrt{c-\frac{c}{a x}}+\frac{8 a^2 c \sqrt{1-\frac{1}{a^2 x^2}}}{5 \sqrt{c-\frac{c}{a x}}} \]

[Out]

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

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Rubi [A]  time = 0.223833, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {6178, 795, 657, 649} \[ \frac{2 a^2 \sqrt{1-\frac{1}{a^2 x^2}} \left (c-\frac{c}{a x}\right )^{3/2}}{5 c}+\frac{2}{5} a^2 \sqrt{1-\frac{1}{a^2 x^2}} \sqrt{c-\frac{c}{a x}}+\frac{8 a^2 c \sqrt{1-\frac{1}{a^2 x^2}}}{5 \sqrt{c-\frac{c}{a x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rule 795

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(g*(d + e*x)^m
*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

Rule 657

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[(2*c*d*Simplify[m + p])/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^
2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p]
, 0]

Rule 649

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
0]

Rubi steps

\begin{align*} \int \frac{e^{-\coth ^{-1}(a x)} \sqrt{c-\frac{c}{a x}}}{x^3} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x \left (c-\frac{c x}{a}\right )^{3/2}}{\sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{c}\\ &=\frac{2 a^2 \sqrt{1-\frac{1}{a^2 x^2}} \left (c-\frac{c}{a x}\right )^{3/2}}{5 c}+\frac{(3 a) \operatorname{Subst}\left (\int \frac{\left (c-\frac{c x}{a}\right )^{3/2}}{\sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{5 c}\\ &=\frac{2}{5} a^2 \sqrt{1-\frac{1}{a^2 x^2}} \sqrt{c-\frac{c}{a x}}+\frac{2 a^2 \sqrt{1-\frac{1}{a^2 x^2}} \left (c-\frac{c}{a x}\right )^{3/2}}{5 c}+\frac{1}{5} (4 a) \operatorname{Subst}\left (\int \frac{\sqrt{c-\frac{c x}{a}}}{\sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{8 a^2 c \sqrt{1-\frac{1}{a^2 x^2}}}{5 \sqrt{c-\frac{c}{a x}}}+\frac{2}{5} a^2 \sqrt{1-\frac{1}{a^2 x^2}} \sqrt{c-\frac{c}{a x}}+\frac{2 a^2 \sqrt{1-\frac{1}{a^2 x^2}} \left (c-\frac{c}{a x}\right )^{3/2}}{5 c}\\ \end{align*}

Mathematica [A]  time = 0.105657, size = 58, normalized size = 0.51 \[ \frac{2 a \sqrt{1-\frac{1}{a^2 x^2}} \left (6 a^2 x^2-3 a x+1\right ) \sqrt{c-\frac{c}{a x}}}{5 x (a x-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)]*(1 - 3*a*x + 6*a^2*x^2))/(5*x*(-1 + a*x))

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Maple [A]  time = 0.118, size = 62, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,ax+2 \right ) \left ( 6\,{a}^{2}{x}^{2}-3\,ax+1 \right ) }{5\,{x}^{2} \left ( ax-1 \right ) }\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}}\sqrt{{\frac{ax-1}{ax+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(a*x+1)*(6*a^2*x^2-3*a*x+1)*(c*(a*x-1)/a/x)^(1/2)*((a*x-1)/(a*x+1))^(1/2)/x^2/(a*x-1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.5235, size = 142, normalized size = 1.26 \begin{align*} \frac{2 \,{\left (6 \, a^{3} x^{3} + 3 \, a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt{\frac{a x - 1}{a x + 1}} \sqrt{\frac{a c x - c}{a x}}}{5 \,{\left (a x^{3} - x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(6*a^3*x^3 + 3*a^2*x^2 - 2*a*x + 1)*sqrt((a*x - 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x))/(a*x^3 - 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/x)**(1/2)*((a*x-1)/(a*x+1))**(1/2)/x**3,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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