\(\int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-\frac {c}{a^2 x^2})^{5/2}} \, dx\) [878]

Optimal result

Integrand size = 24, antiderivative size = 264 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{6 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1+a x)^3}+\frac {9 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1+a x)^2}-\frac {31 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1+a x)}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} \log (1-a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {49 \sqrt {1-\frac {1}{a^2 x^2}} \log (1+a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}} \]

[Out]

x*(1-1/a^2/x^2)^(1/2)/c^2/(c-c/a^2/x^2)^(1/2)-1/6*(1-1/a^2/x^2)^(1/2)/a/c^2/(a*x+1)^3/(c-c/a^2/x^2)^(1/2)+9/8*
(1-1/a^2/x^2)^(1/2)/a/c^2/(a*x+1)^2/(c-c/a^2/x^2)^(1/2)-31/8*(1-1/a^2/x^2)^(1/2)/a/c^2/(a*x+1)/(c-c/a^2/x^2)^(
1/2)+1/16*ln(-a*x+1)*(1-1/a^2/x^2)^(1/2)/a/c^2/(c-c/a^2/x^2)^(1/2)-49/16*ln(a*x+1)*(1-1/a^2/x^2)^(1/2)/a/c^2/(
c-c/a^2/x^2)^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6332, 6328, 90} \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {x \sqrt {1-\frac {1}{a^2 x^2}}}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {31 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 (a x+1) \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {9 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 (a x+1)^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{6 a c^2 (a x+1)^3 \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} \log (1-a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {49 \sqrt {1-\frac {1}{a^2 x^2}} \log (a x+1)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}} \]

[In]

Int[1/(E^(3*ArcCoth[a*x])*(c - c/(a^2*x^2))^(5/2)),x]

[Out]

(Sqrt[1 - 1/(a^2*x^2)]*x)/(c^2*Sqrt[c - c/(a^2*x^2)]) - Sqrt[1 - 1/(a^2*x^2)]/(6*a*c^2*Sqrt[c - c/(a^2*x^2)]*(
1 + a*x)^3) + (9*Sqrt[1 - 1/(a^2*x^2)])/(8*a*c^2*Sqrt[c - c/(a^2*x^2)]*(1 + a*x)^2) - (31*Sqrt[1 - 1/(a^2*x^2)
])/(8*a*c^2*Sqrt[c - c/(a^2*x^2)]*(1 + a*x)) + (Sqrt[1 - 1/(a^2*x^2)]*Log[1 - a*x])/(16*a*c^2*Sqrt[c - c/(a^2*
x^2)]) - (49*Sqrt[1 - 1/(a^2*x^2)]*Log[1 + a*x])/(16*a*c^2*Sqrt[c - c/(a^2*x^2)])

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6328

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

Rule 6332

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1-\frac {1}{a^2 x^2}} \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2}} \, dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}} \\ & = \frac {\left (a^5 \sqrt {1-\frac {1}{a^2 x^2}}\right ) \int \frac {x^5}{(-1+a x) (1+a x)^4} \, dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}} \\ & = \frac {\left (a^5 \sqrt {1-\frac {1}{a^2 x^2}}\right ) \int \left (\frac {1}{a^5}+\frac {1}{16 a^5 (-1+a x)}+\frac {1}{2 a^5 (1+a x)^4}-\frac {9}{4 a^5 (1+a x)^3}+\frac {31}{8 a^5 (1+a x)^2}-\frac {49}{16 a^5 (1+a x)}\right ) \, dx}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}} \\ & = \frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{c^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{6 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1+a x)^3}+\frac {9 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1+a x)^2}-\frac {31 \sqrt {1-\frac {1}{a^2 x^2}}}{8 a c^2 \sqrt {c-\frac {c}{a^2 x^2}} (1+a x)}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} \log (1-a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {49 \sqrt {1-\frac {1}{a^2 x^2}} \log (1+a x)}{16 a c^2 \sqrt {c-\frac {c}{a^2 x^2}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.36 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {\left (1-\frac {1}{a^2 x^2}\right )^{5/2} \left (48 x-\frac {8}{a (1+a x)^3}+\frac {54}{a (1+a x)^2}-\frac {186}{a+a^2 x}+\frac {3 \log (1-a x)}{a}-\frac {147 \log (1+a x)}{a}\right )}{48 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \]

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*(c - c/(a^2*x^2))^(5/2)),x]

[Out]

((1 - 1/(a^2*x^2))^(5/2)*(48*x - 8/(a*(1 + a*x)^3) + 54/(a*(1 + a*x)^2) - 186/(a + a^2*x) + (3*Log[1 - a*x])/a
- (147*Log[1 + a*x])/a))/(48*(c - c/(a^2*x^2))^(5/2))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.66

method	result	size
default	\(-\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x +1\right ) \left (a x -1\right ) \left (-48 a^{4} x^{4}+147 a^{3} \ln \left (a x +1\right ) x^{3}-3 a^{3} \ln \left (a x -1\right ) x^{3}-144 a^{3} x^{3}+441 a^{2} \ln \left (a x +1\right ) x^{2}-9 a^{2} \ln \left (a x -1\right ) x^{2}+42 a^{2} x^{2}+441 a \ln \left (a x +1\right ) x -9 a \ln \left (a x -1\right ) x +270 a x +147 \ln \left (a x +1\right )-3 \ln \left (a x -1\right )+140\right )}{48 a^{6} x^{5} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\)	\(175\)

[In]

int(((a*x-1)/(a*x+1))^(3/2)/(c-c/a^2/x^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/48*((a*x-1)/(a*x+1))^(3/2)*(a*x+1)*(a*x-1)*(-48*a^4*x^4+147*a^3*ln(a*x+1)*x^3-3*a^3*ln(a*x-1)*x^3-144*a^3*x
^3+441*a^2*ln(a*x+1)*x^2-9*a^2*ln(a*x-1)*x^2+42*a^2*x^2+441*a*ln(a*x+1)*x-9*a*ln(a*x-1)*x+270*a*x+147*ln(a*x+1
)-3*ln(a*x-1)+140)/a^6/x^5/(c*(a^2*x^2-1)/a^2/x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.52 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {{\left (48 \, a^{4} x^{4} + 144 \, a^{3} x^{3} - 42 \, a^{2} x^{2} - 270 \, a x - 147 \, {\left (a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1\right )} \log \left (a x - 1\right ) - 140\right )} \sqrt {a^{2} c}}{48 \, {\left (a^{5} c^{3} x^{3} + 3 \, a^{4} c^{3} x^{2} + 3 \, a^{3} c^{3} x + a^{2} c^{3}\right )}} \]

[In]

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

[Out]

1/48*(48*a^4*x^4 + 144*a^3*x^3 - 42*a^2*x^2 - 270*a*x - 147*(a^3*x^3 + 3*a^2*x^2 + 3*a*x + 1)*log(a*x + 1) + 3
*(a^3*x^3 + 3*a^2*x^2 + 3*a*x + 1)*log(a*x - 1) - 140)*sqrt(a^2*c)/(a^5*c^3*x^3 + 3*a^4*c^3*x^2 + 3*a^3*c^3*x
+ a^2*c^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(((a*x-1)/(a*x+1))**(3/2)/(c-c/a**2/x**2)**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int { \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(c-c/a^2/x^2)^(5/2),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 [F(-1)]

Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int \frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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