\(\int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx\) [504]

Optimal result

Integrand size = 27, antiderivative size = 69 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=4 a^2 \sqrt {c-\frac {c}{a x}}-\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{3/2}}{c}+\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6302, 6268, 25, 528, 457, 78} \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}-\frac {2 a^2 \left (c-\frac {c}{a x}\right )^{3/2}}{c}+4 a^2 \sqrt {c-\frac {c}{a x}} \]

[In]

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

[Out]

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

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 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.52 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2 \sqrt {c-\frac {c}{a x}} \left (1+3 a x+6 a^2 x^2\right )}{5 x^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.51

method	result	size
gosper	\(\frac {2 \left (6 a^{2} x^{2}+3 a x +1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{5 x^{2}}\)	\(35\)
trager	\(\frac {2 \left (6 a^{2} x^{2}+3 a x +1\right ) \sqrt {-\frac {-a c x +c}{a x}}}{5 x^{2}}\)	\(37\)
risch	\(\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (6 a^{3} x^{3}-3 a^{2} x^{2}-2 a x -1\right )}{5 \left (a x -1\right ) x^{2}}\)	\(50\)
default	\(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (-10 \sqrt {a \,x^{2}-x}\, a^{\frac {7}{2}} x^{4}-10 a^{\frac {7}{2}} \sqrt {\left (a x -1\right ) x}\, x^{4}+20 a^{\frac {5}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x^{2}+5 \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{3} x^{4}-5 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{3} x^{4}+8 a^{\frac {3}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x +2 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {a}\right )}{5 x^{3} \sqrt {\left (a x -1\right ) x}\, \sqrt {a}}\)	\(192\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.52 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2 \, {\left (6 \, a^{2} x^{2} + 3 \, a x + 1\right )} \sqrt {\frac {a c x - c}{a x}}}{5 \, x^{2}} \]

[In]

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

[Out]

2/5*(6*a^2*x^2 + 3*a*x + 1)*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(1/(a*x-1)*(a*x+1)*(c-c/a/x)**(1/2)/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(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(1/2)/x^3,x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%%{%%{[-5,0]:[1,0,%%%{-1,[1]%%%}]%%},[0,5]%%%},[6]%%%}+
%%%{%%{[%%%

Mupad [B] (verification not implemented)

Time = 3.90 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.46 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2\,\sqrt {c-\frac {c}{a\,x}}\,\left (6\,a^2\,x^2+3\,a\,x+1\right )}{5\,x^2} \]

[In]

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

[Out]

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