\(\int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\) [666]

Optimal result

Integrand size = 24, antiderivative size = 182 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{6 (1+a x)^3 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \text {arctanh}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

1/6*a^4*(1-1/a^2/x^2)^(5/2)*x^5/(a*x+1)^3/(-a^2*c*x^2+c)^(5/2)+1/8*a^4*(1-1/a^2/x^2)^(5/2)*x^5/(a*x+1)^2/(-a^2
*c*x^2+c)^(5/2)+1/8*a^4*(1-1/a^2/x^2)^(5/2)*x^5/(a*x+1)/(-a^2*c*x^2+c)^(5/2)-1/8*a^4*(1-1/a^2/x^2)^(5/2)*x^5*a
rctanh(a*x)/(-a^2*c*x^2+c)^(5/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6327, 6328, 46, 213} \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \text {arctanh}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{6 (a x+1)^3 \left (c-a^2 c x^2\right )^{5/2}} \]

[In]

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

[Out]

(a^4*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(6*(1 + a*x)^3*(c - a^2*c*x^2)^(5/2)) + (a^4*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8
*(1 + a*x)^2*(c - a^2*c*x^2)^(5/2)) + (a^4*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) -
(a^4*(1 - 1/(a^2*x^2))^(5/2)*x^5*ArcTanh[a*x])/(8*(c - a^2*c*x^2)^(5/2))

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 6327

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

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]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.39 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-10-9 a x-3 a^2 x^2+3 (1+a x)^3 \text {arctanh}(a x)\right )}{24 c^2 (1+a x)^3 \sqrt {c-a^2 c x^2}} \]

[In]

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

[Out]

-1/24*(Sqrt[1 - 1/(a^2*x^2)]*x*(-10 - 9*a*x - 3*a^2*x^2 + 3*(1 + a*x)^3*ArcTanh[a*x]))/(c^2*(1 + a*x)^3*Sqrt[c
- a^2*c*x^2])

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (3 a^{3} \ln \left (a x +1\right ) x^{3}-3 a^{3} \ln \left (a x -1\right ) x^{3}+9 a^{2} \ln \left (a x +1\right ) x^{2}-9 a^{2} \ln \left (a x -1\right ) x^{2}-6 a^{2} x^{2}+9 a \ln \left (a x +1\right ) x -9 a \ln \left (a x -1\right ) x -18 a x +3 \ln \left (a x +1\right )-3 \ln \left (a x -1\right )-20\right )}{48 \left (a x +1\right ) \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} a}\)	\(169\)

[In]

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

[Out]

1/48*((a*x-1)/(a*x+1))^(3/2)/(a*x+1)/(a*x-1)*(-c*(a^2*x^2-1))^(1/2)*(3*a^3*ln(a*x+1)*x^3-3*a^3*ln(a*x-1)*x^3+9
*a^2*ln(a*x+1)*x^2-9*a^2*ln(a*x-1)*x^2-6*a^2*x^2+9*a*ln(a*x+1)*x-9*a*ln(a*x-1)*x-18*a*x+3*ln(a*x+1)-3*ln(a*x-1
)-20)/(a^2*x^2-1)/c^3/a

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {3 \, {\left (a^{4} x^{3} + 3 \, a^{3} x^{2} + 3 \, a^{2} x + a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) + 2 \, {\left (3 \, a^{2} x^{2} + 9 \, a x + 10\right )} \sqrt {-a^{2} c}}{48 \, {\left (a^{5} c^{3} x^{3} + 3 \, a^{4} c^{3} x^{2} + 3 \, a^{3} c^{3} x + a^{2} c^{3}\right )}} \]

[In]

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

[Out]

-1/48*(3*(a^4*x^3 + 3*a^3*x^2 + 3*a^2*x + a)*sqrt(-c)*log((a^2*c*x^2 + 2*sqrt(-a^2*c)*sqrt(-c)*x + c)/(a^2*x^2
- 1)) + 2*(3*a^2*x^2 + 9*a*x + 10)*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. \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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