\(\int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx\) [530]

Optimal result

Integrand size = 27, antiderivative size = 113 \[ \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=4 a^2 \sqrt {c-\frac {c}{a x}}+\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}+\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}-4 \sqrt {2} a^2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right ) \]

[Out]

2/3*a^2*(c-c/a/x)^(3/2)/c+2/5*a^2*(c-c/a/x)^(5/2)/c^2-4*a^2*arctanh(1/2*(c-c/a/x)^(1/2)*2^(1/2)/c^(1/2))*2^(1/
2)*c^(1/2)+4*a^2*(c-c/a/x)^(1/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6302, 6268, 25, 528, 457, 81, 52, 65, 214} \[ \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=-4 \sqrt {2} a^2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )+\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}+\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}+4 a^2 \sqrt {c-\frac {c}{a x}} \]

[In]

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

[Out]

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

Rule 25

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(q_.))^(p_.), x_Symbol] :> Dist[(d/a)^p, Int[u*((
a + b*x^n)^(m + p)/x^(n*p)), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[q, -n] && IntegerQ[p] && EqQ[a*c -
b*d, 0] &&  !(IntegerQ[m] && NegQ[n])

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 528

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 6268

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

Rule 6302

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.70 \[ \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2 \sqrt {c-\frac {c}{a x}} \left (3-11 a x+38 a^2 x^2\right )}{15 x^2}-4 \sqrt {2} a^2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right ) \]

[In]

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

[Out]

(2*Sqrt[c - c/(a*x)]*(3 - 11*a*x + 38*a^2*x^2))/(15*x^2) - 4*Sqrt[2]*a^2*Sqrt[c]*ArcTanh[Sqrt[c - c/(a*x)]/(Sq
rt[2]*Sqrt[c])]

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.35

method	result	size
risch	\(\frac {2 \left (38 a^{3} x^{3}-49 a^{2} x^{2}+14 a x -3\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{15 x^{2} \left (a x -1\right )}+\frac {2 a^{2} \sqrt {2}\, \ln \left (\frac {4 c -3 \left (x +\frac {1}{a}\right ) a c +2 \sqrt {2}\, \sqrt {c}\, \sqrt {a^{2} c \left (x +\frac {1}{a}\right )^{2}-3 \left (x +\frac {1}{a}\right ) a c +2 c}}{x +\frac {1}{a}}\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {c \left (a x -1\right ) a x}}{\sqrt {c}\, \left (a x -1\right )}\)	\(152\)
default	\(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (30 \sqrt {\left (a x -1\right ) x}\, a^{\frac {7}{2}} \sqrt {\frac {1}{a}}\, x^{4}-90 a^{\frac {7}{2}} \sqrt {\frac {1}{a}}\, \sqrt {a \,x^{2}-x}\, x^{4}+60 a^{\frac {5}{2}} \sqrt {\frac {1}{a}}\, \left (a \,x^{2}-x \right )^{\frac {3}{2}} x^{2}+45 \sqrt {\frac {1}{a}}\, \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{3} x^{4}-30 a^{\frac {5}{2}} \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) x^{4}-45 \sqrt {\frac {1}{a}}\, \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{3} x^{4}-16 a^{\frac {3}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x \sqrt {\frac {1}{a}}+6 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {a}\, \sqrt {\frac {1}{a}}\right )}{15 x^{3} \sqrt {\left (a x -1\right ) x}\, \sqrt {a}\, \sqrt {\frac {1}{a}}}\)	\(278\)

[In]

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

[Out]

2/15*(38*a^3*x^3-49*a^2*x^2+14*a*x-3)/x^2/(a*x-1)*(c*(a*x-1)/a/x)^(1/2)+2*a^2*2^(1/2)/c^(1/2)*ln((4*c-3*(x+1/a
)*a*c+2*2^(1/2)*c^(1/2)*(a^2*c*(x+1/a)^2-3*(x+1/a)*a*c+2*c)^(1/2))/(x+1/a))/(a*x-1)*(c*(a*x-1)/a/x)^(1/2)*(c*(
a*x-1)*a*x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.60 \[ \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\left [\frac {2 \, {\left (15 \, \sqrt {2} a^{2} \sqrt {c} x^{2} \log \left (\frac {2 \, \sqrt {2} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} - 3 \, a c x + c}{a x + 1}\right ) + {\left (38 \, a^{2} x^{2} - 11 \, a x + 3\right )} \sqrt {\frac {a c x - c}{a x}}\right )}}{15 \, x^{2}}, \frac {2 \, {\left (30 \, \sqrt {2} a^{2} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{2 \, c}\right ) + {\left (38 \, a^{2} x^{2} - 11 \, a x + 3\right )} \sqrt {\frac {a c x - c}{a x}}\right )}}{15 \, x^{2}}\right ] \]

[In]

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

[Out]

[2/15*(15*sqrt(2)*a^2*sqrt(c)*x^2*log((2*sqrt(2)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) - 3*a*c*x + c)/(a*x + 1))
+ (38*a^2*x^2 - 11*a*x + 3)*sqrt((a*c*x - c)/(a*x)))/x^2, 2/15*(30*sqrt(2)*a^2*sqrt(-c)*x^2*arctan(1/2*sqrt(2
)*sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) + (38*a^2*x^2 - 11*a*x + 3)*sqrt((a*c*x - c)/(a*x)))/x^2]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (94) = 188\).

Time = 0.64 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.46 \[ \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=-\frac {4 \, \sqrt {2} a^{3} c \arctan \left (-\frac {\sqrt {2} {\left ({\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )} a + \sqrt {c} {\left | a \right |}\right )}}{2 \, a \sqrt {-c}}\right )}{\sqrt {-c} {\left | a \right |} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (60 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{4} a^{5} c - 45 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{3} a^{4} c^{\frac {3}{2}} {\left | a \right |} + 35 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{2} a^{5} c^{2} - 15 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )} a^{4} c^{\frac {5}{2}} {\left | a \right |} + 3 \, a^{5} c^{3}\right )}}{15 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )}^{5} a^{2} {\left | a \right |} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-4*sqrt(2)*a^3*c*arctan(-1/2*sqrt(2)*((sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*a + sqrt(c)*abs(a))/(a*sqrt(-c
)))/(sqrt(-c)*abs(a)*sgn(x)) + 2/15*(60*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^4*a^5*c - 45*(sqrt(a^2*c)*x
- sqrt(a^2*c*x^2 - a*c*x))^3*a^4*c^(3/2)*abs(a) + 35*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))^2*a^5*c^2 - 15*
(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*a^4*c^(5/2)*abs(a) + 3*a^5*c^3)/((sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a
*c*x))^5*a^2*abs(a)*sgn(x))

Mupad [F(-1)]

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

[In]

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

[Out]

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