\(\int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx\) [376]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 20, antiderivative size = 96 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx=-\frac {2 \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {3+n}{2}} \left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}} x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {3}{2},\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{(c-a c x)^{3/2}} \]

[Out]

-2*((a-1/x)/(a+1/x))^(3/2+1/2*n)*(1+1/a/x)^(1+1/2*n)*x*hypergeom([1/2, 3/2+1/2*n],[3/2],2/(a+1/x)/x)/((1-1/a/x
)^(1/2*n))/(-a*c*x+c)^(3/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6311, 6316, 134} \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx=-\frac {2 x \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {n+3}{2}} \left (1-\frac {1}{a x}\right )^{-n/2} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+3}{2},\frac {3}{2},\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{(c-a c x)^{3/2}} \]

[In]

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

[Out]

(-2*((a - x^(-1))/(a + x^(-1)))^((3 + n)/2)*(1 + 1/(a*x))^((2 + n)/2)*x*Hypergeometric2F1[1/2, (3 + n)/2, 3/2,
 2/((a + x^(-1))*x)])/((1 - 1/(a*x))^(n/2)*(c - a*c*x)^(3/2))

Rule 134

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x
)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((b*e - a*f)*(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c
*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)*((c + d*x)/((b*c - a*d)*(e + f*x))))^n, x] /; FreeQ[{a
, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] &&  !IntegerQ[n]

Rule 6311

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

Rule 6316

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\left (1-\frac {1}{a x}\right )^{3/2} x^{3/2}\right ) \int \frac {e^{n \coth ^{-1}(a x)}}{\left (1-\frac {1}{a x}\right )^{3/2} x^{3/2}} \, dx}{(c-a c x)^{3/2}} \\ & = -\frac {\left (1-\frac {1}{a x}\right )^{3/2} \text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{-\frac {3}{2}-\frac {n}{2}} \left (1+\frac {x}{a}\right )^{n/2}}{\sqrt {x}} \, dx,x,\frac {1}{x}\right )}{\left (\frac {1}{x}\right )^{3/2} (c-a c x)^{3/2}} \\ & = -\frac {2 \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {3+n}{2}} \left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}} x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {3}{2},\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{(c-a c x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.98 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx=\frac {2 \left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{n/2} \left (\frac {-1+a x}{1+a x}\right )^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {3}{2},\frac {2}{1+a x}\right )}{a c \sqrt {c-a c x}} \]

[In]

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

[Out]

(2*(1 + 1/(a*x))^(n/2)*((-1 + a*x)/(1 + a*x))^((1 + n)/2)*Hypergeometric2F1[1/2, (3 + n)/2, 3/2, 2/(1 + a*x)])
/(a*c*(1 - 1/(a*x))^(n/2)*Sqrt[c - a*c*x])

Maple [F]

\[\int \frac {{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )}}{\left (-a c x +c \right )^{\frac {3}{2}}}d x\]

[In]

int(exp(n*arccoth(a*x))/(-a*c*x+c)^(3/2),x)

[Out]

int(exp(n*arccoth(a*x))/(-a*c*x+c)^(3/2),x)

Fricas [F]

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

[In]

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

[Out]

integral(sqrt(-a*c*x + c)*((a*x + 1)/(a*x - 1))^(1/2*n)/(a^2*c^2*x^2 - 2*a*c^2*x + c^2), x)

Sympy [F]

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

[In]

integrate(exp(n*acoth(a*x))/(-a*c*x+c)**(3/2),x)

[Out]

Integral(exp(n*acoth(a*x))/(-c*(a*x - 1))**(3/2), x)

Maxima [F]

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

[In]

integrate(exp(n*arccoth(a*x))/(-a*c*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(exp(n*acoth(a*x))/(c - a*c*x)^(3/2),x)

[Out]

int(exp(n*acoth(a*x))/(c - a*c*x)^(3/2), x)