### 3.729 $$\int \frac{e^{-3 \coth ^{-1}(a x)} \sqrt{c-a^2 c x^2}}{x^5} \, dx$$

Optimal. Leaf size=227 $\frac{4 a^2 \sqrt{c-a^2 c x^2}}{x^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{2 a \sqrt{c-a^2 c x^2}}{x^3 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{c-a^2 c x^2}}{x^4 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{\sqrt{c-a^2 c x^2}}{4 a x^5 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 a^3 \log (x) \sqrt{c-a^2 c x^2}}{x \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{4 a^3 \sqrt{c-a^2 c x^2} \log (a x+1)}{x \sqrt{1-\frac{1}{a^2 x^2}}}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.250525, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.111, Rules used = {6192, 6193, 88} $\frac{4 a^2 \sqrt{c-a^2 c x^2}}{x^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{2 a \sqrt{c-a^2 c x^2}}{x^3 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\sqrt{c-a^2 c x^2}}{x^4 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{\sqrt{c-a^2 c x^2}}{4 a x^5 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 a^3 \log (x) \sqrt{c-a^2 c x^2}}{x \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{4 a^3 \sqrt{c-a^2 c x^2} \log (a x+1)}{x \sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[c - a^2*c*x^2]/(E^(3*ArcCoth[a*x])*x^5),x]

[Out]

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

Rule 6192

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

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

Mathematica [A]  time = 0.0517898, size = 83, normalized size = 0.37 $\frac{\sqrt{c-a^2 c x^2} \left (-\frac{2 a^2}{x^2}+\frac{4 a^3}{x}+4 a^4 \log (x)-4 a^4 \log (a x+1)+\frac{a}{x^3}-\frac{1}{4 x^4}\right )}{a x \sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[c - a^2*c*x^2]/(E^(3*ArcCoth[a*x])*x^5),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.133, size = 93, normalized size = 0.4 \begin{align*}{\frac{ \left ( 16\,{a}^{4}\ln \left ( x \right ){x}^{4}-16\,\ln \left ( ax+1 \right ){a}^{4}{x}^{4}+16\,{x}^{3}{a}^{3}-8\,{a}^{2}{x}^{2}+4\,ax-1 \right ) \left ( ax+1 \right ) }{4\,{x}^{4} \left ( ax-1 \right ) ^{2}}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) } \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(16*a^4*ln(x)*x^4-16*ln(a*x+1)*a^4*x^4+16*x^3*a^3-8*a^2*x^2+4*a*x-1)*(-c*(a^2*x^2-1))^(1/2)*(a*x+1)*((a*x-
1)/(a*x+1))^(3/2)/x^4/(a*x-1)^2

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} c x^{2} + c} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}{x^{5}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^5,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.55909, size = 232, normalized size = 1.02 \begin{align*} \frac{16 \, a^{5} \sqrt{-c} x^{4} \log \left (\frac{2 \, a^{3} c x^{2} + 2 \, a^{2} c x + \sqrt{-a^{2} c}{\left (2 \, a x + 1\right )} \sqrt{-c} + a c}{a x^{2} + x}\right ) +{\left (16 \, a^{3} x^{3} - 8 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt{-a^{2} c}}{4 \, a x^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^5,x, algorithm="fricas")

[Out]

1/4*(16*a^5*sqrt(-c)*x^4*log((2*a^3*c*x^2 + 2*a^2*c*x + sqrt(-a^2*c)*(2*a*x + 1)*sqrt(-c) + a*c)/(a*x^2 + x))
+ (16*a^3*x^3 - 8*a^2*x^2 + 4*a*x - 1)*sqrt(-a^2*c))/(a*x^4)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} c x^{2} + c} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}{x^{5}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^5,x, algorithm="giac")

[Out]

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