\(\int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx\) [506]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F(-2)]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 121 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx=4 a^4 \sqrt {c-\frac {c}{a x}}-\frac {14 a^4 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}+\frac {18 a^4 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}-\frac {10 a^4 \left (c-\frac {c}{a x}\right )^{7/2}}{7 c^3}+\frac {2 a^4 \left (c-\frac {c}{a x}\right )^{9/2}}{9 c^4} \]

[Out]

-14/3*a^4*(c-c/a/x)^(3/2)/c+18/5*a^4*(c-c/a/x)^(5/2)/c^2-10/7*a^4*(c-c/a/x)^(7/2)/c^3+2/9*a^4*(c-c/a/x)^(9/2)/
c^4+4*a^4*(c-c/a/x)^(1/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 121, 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} \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx=\frac {2 a^4 \left (c-\frac {c}{a x}\right )^{9/2}}{9 c^4}-\frac {10 a^4 \left (c-\frac {c}{a x}\right )^{7/2}}{7 c^3}+\frac {18 a^4 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}-\frac {14 a^4 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}+4 a^4 \sqrt {c-\frac {c}{a x}} \]

[In]

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

[Out]

4*a^4*Sqrt[c - c/(a*x)] - (14*a^4*(c - c/(a*x))^(3/2))/(3*c) + (18*a^4*(c - c/(a*x))^(5/2))/(5*c^2) - (10*a^4*
(c - c/(a*x))^(7/2))/(7*c^3) + (2*a^4*(c - c/(a*x))^(9/2))/(9*c^4)

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^5} \, dx \\ & = -\int \frac {\sqrt {c-\frac {c}{a x}} (1+a x)}{x^5 (1-a x)} \, dx \\ & = \frac {c \int \frac {1+a x}{\sqrt {c-\frac {c}{a x}} x^6} \, dx}{a} \\ & = \frac {c \int \frac {a+\frac {1}{x}}{\sqrt {c-\frac {c}{a x}} x^5} \, dx}{a} \\ & = -\frac {c \text {Subst}\left (\int \frac {x^3 (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^4}{\sqrt {c-\frac {c x}{a}}}-\frac {7 a^4 \sqrt {c-\frac {c x}{a}}}{c}+\frac {9 a^4 \left (c-\frac {c x}{a}\right )^{3/2}}{c^2}-\frac {5 a^4 \left (c-\frac {c x}{a}\right )^{5/2}}{c^3}+\frac {a^4 \left (c-\frac {c x}{a}\right )^{7/2}}{c^4}\right ) \, dx,x,\frac {1}{x}\right )}{a} \\ & = 4 a^4 \sqrt {c-\frac {c}{a x}}-\frac {14 a^4 \left (c-\frac {c}{a x}\right )^{3/2}}{3 c}+\frac {18 a^4 \left (c-\frac {c}{a x}\right )^{5/2}}{5 c^2}-\frac {10 a^4 \left (c-\frac {c}{a x}\right )^{7/2}}{7 c^3}+\frac {2 a^4 \left (c-\frac {c}{a x}\right )^{9/2}}{9 c^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.43 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx=\frac {2 \sqrt {c-\frac {c}{a x}} \left (35+85 a x+102 a^2 x^2+136 a^3 x^3+272 a^4 x^4\right )}{315 x^4} \]

[In]

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

[Out]

(2*Sqrt[c - c/(a*x)]*(35 + 85*a*x + 102*a^2*x^2 + 136*a^3*x^3 + 272*a^4*x^4))/(315*x^4)

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.42

method result size
gosper \(\frac {2 \left (272 a^{4} x^{4}+136 a^{3} x^{3}+102 a^{2} x^{2}+85 a x +35\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{315 x^{4}}\) \(51\)
trager \(\frac {2 \left (272 a^{4} x^{4}+136 a^{3} x^{3}+102 a^{2} x^{2}+85 a x +35\right ) \sqrt {-\frac {-a c x +c}{a x}}}{315 x^{4}}\) \(53\)
risch \(\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (272 a^{5} x^{5}-136 a^{4} x^{4}-34 a^{3} x^{3}-17 a^{2} x^{2}-50 a x -35\right )}{315 \left (a x -1\right ) x^{4}}\) \(66\)
default \(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (-630 \sqrt {a \,x^{2}-x}\, a^{\frac {11}{2}} x^{6}-630 a^{\frac {11}{2}} \sqrt {\left (a x -1\right ) x}\, x^{6}+1260 \left (a \,x^{2}-x \right )^{\frac {3}{2}} a^{\frac {9}{2}} x^{4}+315 \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{5} x^{6}-315 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{5} x^{6}+716 a^{\frac {7}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x^{3}+444 a^{\frac {5}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x^{2}+240 a^{\frac {3}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x +70 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {a}\right )}{315 x^{5} \sqrt {\left (a x -1\right ) x}\, \sqrt {a}}\) \(230\)

[In]

int(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(1/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

2/315*(272*a^4*x^4+136*a^3*x^3+102*a^2*x^2+85*a*x+35)*(c*(a*x-1)/a/x)^(1/2)/x^4

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.43 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx=\frac {2 \, {\left (272 \, a^{4} x^{4} + 136 \, a^{3} x^{3} + 102 \, a^{2} x^{2} + 85 \, a x + 35\right )} \sqrt {\frac {a c x - c}{a x}}}{315 \, x^{4}} \]

[In]

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

[Out]

2/315*(272*a^4*x^4 + 136*a^3*x^3 + 102*a^2*x^2 + 85*a*x + 35)*sqrt((a*c*x - c)/(a*x))/x^4

Sympy [F]

\[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx=\int \frac {\sqrt {- c \left (-1 + \frac {1}{a x}\right )} \left (a x + 1\right )}{x^{5} \left (a x - 1\right )}\, dx \]

[In]

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

[Out]

Integral(sqrt(-c*(-1 + 1/(a*x)))*(a*x + 1)/(x**5*(a*x - 1)), x)

Maxima [F]

\[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx=\int { \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a x}}}{{\left (a x - 1\right )} x^{5}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a*x-1)*(a*x+1)*(c-c/a/x)^(1/2)/x^5,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%%%{%%%{%%{[-315,0]:[1,0,%%%{-1,[1]%%%}]%%},[0,9]%%%},[10]%%
%}+%%%{%%{[

Mupad [B] (verification not implemented)

Time = 4.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx=\frac {544\,a^4\,\sqrt {c-\frac {c}{a\,x}}}{315}+\frac {2\,\sqrt {c-\frac {c}{a\,x}}}{9\,x^4}+\frac {34\,a\,\sqrt {c-\frac {c}{a\,x}}}{63\,x^3}+\frac {68\,a^2\,\sqrt {c-\frac {c}{a\,x}}}{105\,x^2}+\frac {272\,a^3\,\sqrt {c-\frac {c}{a\,x}}}{315\,x} \]

[In]

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

[Out]

(544*a^4*(c - c/(a*x))^(1/2))/315 + (2*(c - c/(a*x))^(1/2))/(9*x^4) + (34*a*(c - c/(a*x))^(1/2))/(63*x^3) + (6
8*a^2*(c - c/(a*x))^(1/2))/(105*x^2) + (272*a^3*(c - c/(a*x))^(1/2))/(315*x)