\(\int \frac {e^{n \coth ^{-1}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\) [760]

Optimal result

Integrand size = 24, antiderivative size = 102 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {e^{n \coth ^{-1}(a x)} (n-3 a x)}{a c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}-\frac {6 e^{n \coth ^{-1}(a x)} (n-a x)}{a c^2 \left (1-n^2\right ) \left (9-n^2\right ) \sqrt {c-a^2 c x^2}} \]

[Out]

-exp(n*arccoth(a*x))*(-3*a*x+n)/a/c/(-n^2+9)/(-a^2*c*x^2+c)^(3/2)-6*exp(n*arccoth(a*x))*(-a*x+n)/a/c^2/(n^4-10
*n^2+9)/(-a^2*c*x^2+c)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6320, 6319} \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {6 (n-a x) e^{n \coth ^{-1}(a x)}}{a c^2 \left (1-n^2\right ) \left (9-n^2\right ) \sqrt {c-a^2 c x^2}}-\frac {(n-3 a x) e^{n \coth ^{-1}(a x)}}{a c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}} \]

[In]

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

[Out]

-((E^(n*ArcCoth[a*x])*(n - 3*a*x))/(a*c*(9 - n^2)*(c - a^2*c*x^2)^(3/2))) - (6*E^(n*ArcCoth[a*x])*(n - a*x))/(
a*c^2*(1 - n^2)*(9 - n^2)*Sqrt[c - a^2*c*x^2])

Rule 6319

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(n - a*x)*(E^(n*ArcCoth[a*x])/
(a*c*(n^2 - 1)*Sqrt[c + d*x^2])), x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !IntegerQ[n]

Rule 6320

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(n + 2*a*(p + 1)*x)*(c + d*x^2
)^(p + 1)*(E^(n*ArcCoth[a*x])/(a*c*(n^2 - 4*(p + 1)^2))), x] - Dist[2*(p + 1)*((2*p + 3)/(c*(n^2 - 4*(p + 1)^2
))), Int[(c + d*x^2)^(p + 1)*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !In
tegerQ[n/2] && LtQ[p, -1] && NeQ[p, -3/2] && NeQ[n^2 - 4*(p + 1)^2, 0] && (IntegerQ[p] ||  !IntegerQ[n])

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.08 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {e^{n \coth ^{-1}(a x)} \left (-26 n+2 n^3+27 a x-3 a n^2 x-2 n \left (-1+n^2\right ) \cosh \left (2 \coth ^{-1}(a x)\right )+3 a \left (-1+n^2\right ) \sqrt {1-\frac {1}{a^2 x^2}} x \cosh \left (3 \coth ^{-1}(a x)\right )\right )}{4 a c^2 \left (9-10 n^2+n^4\right ) \sqrt {c-a^2 c x^2}} \]

[In]

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

[Out]

(E^(n*ArcCoth[a*x])*(-26*n + 2*n^3 + 27*a*x - 3*a*n^2*x - 2*n*(-1 + n^2)*Cosh[2*ArcCoth[a*x]] + 3*a*(-1 + n^2)
*Sqrt[1 - 1/(a^2*x^2)]*x*Cosh[3*ArcCoth[a*x]]))/(4*a*c^2*(9 - 10*n^2 + n^4)*Sqrt[c - a^2*c*x^2])

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82

method	result	size
gosper	\(\frac {\left (a x -1\right ) \left (a x +1\right ) \left (6 a^{3} x^{3}-6 n \,x^{2} a^{2}+3 n^{2} x a -n^{3}-9 a x +7 n \right ) {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )}}{a \left (n^{4}-10 n^{2}+9\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\)	\(84\)

[In]

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

[Out]

(a*x-1)*(a*x+1)*(6*a^3*x^3-6*a^2*n*x^2+3*a*n^2*x-n^3-9*a*x+7*n)*exp(n*arccoth(a*x))/a/(n^4-10*n^2+9)/(-a^2*c*x
^2+c)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.62 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\left (6 \, a^{3} x^{3} - 6 \, a^{2} n x^{2} - n^{3} + 3 \, {\left (a n^{2} - 3 \, a\right )} x + 7 \, n\right )} \sqrt {-a^{2} c x^{2} + c} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a c^{3} n^{4} - 10 \, a c^{3} n^{2} + {\left (a^{5} c^{3} n^{4} - 10 \, a^{5} c^{3} n^{2} + 9 \, a^{5} c^{3}\right )} x^{4} + 9 \, a c^{3} - 2 \, {\left (a^{3} c^{3} n^{4} - 10 \, a^{3} c^{3} n^{2} + 9 \, a^{3} c^{3}\right )} x^{2}} \]

[In]

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

[Out]

-(6*a^3*x^3 - 6*a^2*n*x^2 - n^3 + 3*(a*n^2 - 3*a)*x + 7*n)*sqrt(-a^2*c*x^2 + c)*((a*x + 1)/(a*x - 1))^(1/2*n)/
(a*c^3*n^4 - 10*a*c^3*n^2 + (a^5*c^3*n^4 - 10*a^5*c^3*n^2 + 9*a^5*c^3)*x^4 + 9*a*c^3 - 2*(a^3*c^3*n^4 - 10*a^3
*c^3*n^2 + 9*a^3*c^3)*x^2)

Sympy [F]

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

[In]

integrate(exp(n*acoth(a*x))/(-a**2*c*x**2+c)**(5/2),x)

[Out]

Integral(exp(n*acoth(a*x))/(-c*(a*x - 1)*(a*x + 1))**(5/2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

integrate(exp(n*arccoth(a*x))/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.70 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\left (\frac {a\,x+1}{a\,x}\right )}^{n/2}\,\left (\frac {6\,x^3}{c^2\,\left (n^4-10\,n^2+9\right )}+\frac {7\,n-n^3}{a^3\,c^2\,\left (n^4-10\,n^2+9\right )}+\frac {3\,x\,\left (n^2-3\right )}{a^2\,c^2\,\left (n^4-10\,n^2+9\right )}-\frac {6\,n\,x^2}{a\,c^2\,\left (n^4-10\,n^2+9\right )}\right )}{\left (\frac {\sqrt {c-a^2\,c\,x^2}}{a^2}-x^2\,\sqrt {c-a^2\,c\,x^2}\right )\,{\left (\frac {a\,x-1}{a\,x}\right )}^{n/2}} \]

[In]

int(exp(n*acoth(a*x))/(c - a^2*c*x^2)^(5/2),x)

[Out]

-(((a*x + 1)/(a*x))^(n/2)*((6*x^3)/(c^2*(n^4 - 10*n^2 + 9)) + (7*n - n^3)/(a^3*c^2*(n^4 - 10*n^2 + 9)) + (3*x*
(n^2 - 3))/(a^2*c^2*(n^4 - 10*n^2 + 9)) - (6*n*x^2)/(a*c^2*(n^4 - 10*n^2 + 9))))/(((c - a^2*c*x^2)^(1/2)/a^2 -
x^2*(c - a^2*c*x^2)^(1/2))*((a*x - 1)/(a*x))^(n/2))