\(\int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx\) [538]

Optimal result

Integrand size = 27, antiderivative size = 109 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\frac {64 a \sqrt {c-\frac {c}{a x}}}{3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {16 a \left (c-\frac {c}{a x}\right )^{3/2}}{3 c \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {2 a \left (c-\frac {c}{a x}\right )^{5/2}}{3 c^2 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

-16/3*a*(c-c/a/x)^(3/2)/c/(1-1/a^2/x^2)^(1/2)-2/3*a*(c-c/a/x)^(5/2)/c^2/(1-1/a^2/x^2)^(1/2)+64/3*a*(c-c/a/x)^(
1/2)/(1-1/a^2/x^2)^(1/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6313, 671, 663} \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=-\frac {2 a \left (c-\frac {c}{a x}\right )^{5/2}}{3 c^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {16 a \left (c-\frac {c}{a x}\right )^{3/2}}{3 c \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {64 a \sqrt {c-\frac {c}{a x}}}{3 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

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

[Out]

(64*a*Sqrt[c - c/(a*x)])/(3*Sqrt[1 - 1/(a^2*x^2)]) - (16*a*(c - c/(a*x))^(3/2))/(3*c*Sqrt[1 - 1/(a^2*x^2)]) -
(2*a*(c - c/(a*x))^(5/2))/(3*c^2*Sqrt[1 - 1/(a^2*x^2)])

Rule 663

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]

Rule 671

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 6313

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{7/2}}{\left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{c^3} \\ & = -\frac {2 a \left (c-\frac {c}{a x}\right )^{5/2}}{3 c^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {8 \text {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{5/2}}{\left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{3 c^2} \\ & = -\frac {16 a \left (c-\frac {c}{a x}\right )^{3/2}}{3 c \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {2 a \left (c-\frac {c}{a x}\right )^{5/2}}{3 c^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {32 \text {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{3/2}}{\left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{3 c} \\ & = \frac {64 a \sqrt {c-\frac {c}{a x}}}{3 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {16 a \left (c-\frac {c}{a x}\right )^{3/2}}{3 c \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {2 a \left (c-\frac {c}{a x}\right )^{5/2}}{3 c^2 \sqrt {1-\frac {1}{a^2 x^2}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.53 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\frac {2 a \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}} \left (-1+10 a x+23 a^2 x^2\right )}{-3+3 a^2 x^2} \]

[In]

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

[Out]

(2*a*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)]*(-1 + 10*a*x + 23*a^2*x^2))/(-3 + 3*a^2*x^2)

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.57

method	result	size
gosper	\(\frac {2 \left (a x +1\right ) \left (23 a^{2} x^{2}+10 a x -1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{3 x \left (a x -1\right )^{2}}\)	\(62\)
default	\(\frac {2 \left (a x +1\right ) \left (23 a^{2} x^{2}+10 a x -1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{3 x \left (a x -1\right )^{2}}\)	\(62\)
risch	\(\frac {2 \left (11 a^{2} x^{2}+10 a x -1\right ) \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\frac {c \left (a x -1\right )}{a x}}}{3 x \left (a x -1\right )}+\frac {8 a^{2} x \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\frac {c \left (a x -1\right )}{a x}}}{a x -1}\)	\(101\)

[In]

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

[Out]

2/3*(a*x+1)*(23*a^2*x^2+10*a*x-1)*(c*(a*x-1)/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x/(a*x-1)^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.54 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\frac {2 \, {\left (23 \, a^{2} x^{2} + 10 \, a x - 1\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{3 \, {\left (a x^{2} - x\right )}} \]

[In]

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

[Out]

2/3*(23*a^2*x^2 + 10*a*x - 1)*sqrt((a*x - 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x))/(a*x^2 - x)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\int { \frac {\sqrt {c - \frac {c}{a x}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{x^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 4.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.50 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx=\frac {2\,\sqrt {c-\frac {c}{a\,x}}\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (23\,a^2\,x^2+10\,a\,x-1\right )}{3\,x\,\left (a\,x-1\right )} \]

[In]

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

[Out]

(2*(c - c/(a*x))^(1/2)*((a*x - 1)/(a*x + 1))^(1/2)*(10*a*x + 23*a^2*x^2 - 1))/(3*x*(a*x - 1))