\(\int \frac {e^{-\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx\) [522]

Optimal result

Integrand size = 27, antiderivative size = 113 \[ \int \frac {e^{-\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {8 a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}{5 \sqrt {c-\frac {c}{a x}}}+\frac {2}{5} a^2 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}+\frac {2 a^2 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}}{5 c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6313, 809, 671, 663} \[ \int \frac {e^{-\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2 a^2 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}}{5 c}+\frac {2}{5} a^2 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}+\frac {8 a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}{5 \sqrt {c-\frac {c}{a x}}} \]

[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 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 809

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 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 {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) \text {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) \text {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] (verified)

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.51 \[ \int \frac {e^{-\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2 a \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}} \left (1-3 a x+6 a^2 x^2\right )}{5 x (-1+a x)} \]

[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))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.55

method	result	size
gosper	\(\frac {2 \left (a x +1\right ) \left (6 a^{2} x^{2}-3 a x +1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {a x -1}{a x +1}}}{5 \left (a x -1\right ) x^{2}}\)	\(62\)
default	\(\frac {2 \left (a x +1\right ) \left (6 a^{2} x^{2}-3 a x +1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {a x -1}{a x +1}}}{5 \left (a x -1\right ) x^{2}}\)	\(62\)
risch	\(\frac {2 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (6 a^{3} x^{3}+3 a^{2} x^{2}-2 a x +1\right )}{5 \left (a x -1\right ) x^{2}}\)	\(65\)

[In]

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

[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)/(a*x-1)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.61 \[ \int \frac {e^{-\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2 \, {\left (6 \, a^{3} x^{3} + 3 \, a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{5 \, {\left (a x^{3} - x^{2}\right )}} \]

[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)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [B] (verification not implemented)

Time = 4.00 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.55 \[ \int \frac {e^{-\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2\,\sqrt {c-\frac {c}{a\,x}}\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (6\,a^3\,x^3+3\,a^2\,x^2-2\,a\,x+1\right )}{5\,x^2\,\left (a\,x-1\right )} \]

[In]

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

[Out]

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