3.746 \(\int \frac {e^{n \coth ^{-1}(a x)}}{\sqrt {c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=111 \[ \frac {2 x \sqrt {1-\frac {1}{a^2 x^2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n-1}{2}} \, _2F_1\left (1,\frac {1-n}{2};\frac {3-n}{2};\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{(1-n) \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6192, 6195, 131} \[ \frac {2 x \sqrt {1-\frac {1}{a^2 x^2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n-1}{2}} \, _2F_1\left (1,\frac {1-n}{2};\frac {3-n}{2};\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{(1-n) \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 131

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

Rule 6192

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 6195

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_)^2)^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^p, Subst[Int[((
1 - x/a)^(p - n/2)*(1 + x/a)^(p + n/2))/x^(m + 2), x], x, 1/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] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {e^{n \coth ^{-1}(a x)}}{\sqrt {c-a^2 c x^2}} \, dx &=\frac {\left (\sqrt {1-\frac {1}{a^2 x^2}} x\right ) \int \frac {e^{n \coth ^{-1}(a x)}}{\sqrt {1-\frac {1}{a^2 x^2}} x} \, dx}{\sqrt {c-a^2 c x^2}}\\ &=-\frac {\left (\sqrt {1-\frac {1}{a^2 x^2}} x\right ) \operatorname {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (1+\frac {x}{a}\right )^{-\frac {1}{2}+\frac {n}{2}}}{x} \, dx,x,\frac {1}{x}\right )}{\sqrt {c-a^2 c x^2}}\\ &=\frac {2 \sqrt {1-\frac {1}{a^2 x^2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {1}{2} (-1+n)} x \, _2F_1\left (1,\frac {1-n}{2};\frac {3-n}{2};\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{(1-n) \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 81, normalized size = 0.73 \[ -\frac {2 \sqrt {c-a^2 c x^2} e^{(n+1) \coth ^{-1}(a x)} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};e^{2 \coth ^{-1}(a x)}\right )}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a^2 c n x+a^2 c x\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} c x^{2} + c} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{a^{2} c x^{2} - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{\sqrt {-a^{2} c x^{2} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))/(-a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.40, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )}}{\sqrt {-a^{2} c \,x^{2}+c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arccoth(a*x))/(-a^2*c*x^2+c)^(1/2),x)

[Out]

int(exp(n*arccoth(a*x))/(-a^2*c*x^2+c)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{\sqrt {-a^{2} c x^{2} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {e}}^{n\,\mathrm {acoth}\left (a\,x\right )}}{\sqrt {c-a^2\,c\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*acoth(a*x))/(c - a^2*c*x^2)^(1/2),x)

[Out]

int(exp(n*acoth(a*x))/(c - a^2*c*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{n \operatorname {acoth}{\left (a x \right )}}}{\sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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