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

Optimal. Leaf size=96 \[ \frac {2 x \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {n+1}{2}} \left (1-\frac {1}{a x}\right )^{-n/2} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \, _2F_1\left (-\frac {1}{2},\frac {n+1}{2};\frac {1}{2};\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{\sqrt {c-a c x}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6176, 6181, 132} \[ \frac {2 x \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {n+1}{2}} \left (1-\frac {1}{a x}\right )^{-n/2} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \, _2F_1\left (-\frac {1}{2},\frac {n+1}{2};\frac {1}{2};\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{\sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 132

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)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c -
a*d)*(e + f*x)))])/(((b*e - a*f)*(m + 1))*(((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 6176

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 6181

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> -Dist[c^p*x^m*(1/x)^m, Sub
st[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*} \int \frac {e^{n \coth ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx &=\frac {\left (\sqrt {1-\frac {1}{a x}} \sqrt {x}\right ) \int \frac {e^{n \coth ^{-1}(a x)}}{\sqrt {1-\frac {1}{a x}} \sqrt {x}} \, dx}{\sqrt {c-a c x}}\\ &=-\frac {\sqrt {1-\frac {1}{a x}} \operatorname {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{-\frac {1}{2}-\frac {n}{2}} \left (1+\frac {x}{a}\right )^{n/2}}{x^{3/2}} \, dx,x,\frac {1}{x}\right )}{\sqrt {\frac {1}{x}} \sqrt {c-a c x}}\\ &=\frac {2 \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {1+n}{2}} \left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}} x \, _2F_1\left (-\frac {1}{2},\frac {1+n}{2};\frac {1}{2};\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{\sqrt {c-a c x}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 96, normalized size = 1.00 \[ \frac {2 (a x+1) \left (1-\frac {1}{a x}\right )^{-n/2} \left (\frac {1}{a x}+1\right )^{n/2} \left (\frac {a x-1}{a x+1}\right )^{\frac {n+1}{2}} \, _2F_1\left (-\frac {1}{2},\frac {n+1}{2};\frac {1}{2};\frac {2}{a x+1}\right )}{a \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a c x + c} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{a c x - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )}}{\sqrt {-a c x +c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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