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

Optimal. Leaf size=295 \[ \frac {2 n \sqrt {c-\frac {c}{a^2 x^2}} \left (\frac {1}{a x}+1\right )^{\frac {n-1}{2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}} \, _2F_1\left (1,\frac {1-n}{2};\frac {3-n}{2};\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a (1-n) \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {2^{\frac {n+1}{2}} \sqrt {c-\frac {c}{a^2 x^2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}} \, _2F_1\left (\frac {1-n}{2},\frac {1-n}{2};\frac {3-n}{2};\frac {a-\frac {1}{x}}{2 a}\right )}{a (1-n) \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x \sqrt {c-\frac {c}{a^2 x^2}} \left (\frac {1}{a x}+1\right )^{\frac {n+1}{2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}}}{\sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

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Rubi [C]  time = 0.15, antiderivative size = 111, normalized size of antiderivative = 0.38, 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 = {6197, 6194, 136} \[ -\frac {2^{\frac {3}{2}-\frac {n}{2}} \sqrt {c-\frac {c}{a^2 x^2}} \left (\frac {1}{a x}+1\right )^{\frac {n+3}{2}} F_1\left (\frac {n+3}{2};\frac {n-1}{2},2;\frac {n+5}{2};\frac {a+\frac {1}{x}}{2 a},1+\frac {1}{a x}\right )}{a (n+3) \sqrt {1-\frac {1}{a^2 x^2}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((2^(3/2 - n/2)*Sqrt[c - c/(a^2*x^2)]*(1 + 1/(a*x))^((3 + n)/2)*AppellF1[(3 + n)/2, (-1 + n)/2, 2, (5 + n)/2,
(a + x^(-1))/(2*a), 1 + 1/(a*x)])/(a*(3 + n)*Sqrt[1 - 1/(a^2*x^2)]))

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 6194

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

Rule 6197

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

Rubi steps

\begin {align*} \int e^{n \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} \, dx &=\frac {\sqrt {c-\frac {c}{a^2 x^2}} \int e^{n \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} \, dx}{\sqrt {1-\frac {1}{a^2 x^2}}}\\ &=-\frac {\sqrt {c-\frac {c}{a^2 x^2}} \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^2} \, dx,x,\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}}}\\ &=-\frac {2^{\frac {3}{2}-\frac {n}{2}} \sqrt {c-\frac {c}{a^2 x^2}} \left (1+\frac {1}{a x}\right )^{\frac {3+n}{2}} F_1\left (\frac {3+n}{2};\frac {1}{2} (-1+n),2;\frac {5+n}{2};\frac {a+\frac {1}{x}}{2 a},1+\frac {1}{a x}\right )}{a (3+n) \sqrt {1-\frac {1}{a^2 x^2}}}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 146, normalized size = 0.49 \[ \frac {a x^2 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a^2 x^2}} e^{n \coth ^{-1}(a x)} \left (a (n+1) x \sqrt {1-\frac {1}{a^2 x^2}}+2 e^{\coth ^{-1}(a x)} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};-e^{2 \coth ^{-1}(a x)}\right )+2 n e^{\coth ^{-1}(a x)} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};e^{2 \coth ^{-1}(a x)}\right )\right )}{(n+1) \left (a^2 x^2-1\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*E^(n*ArcCoth[a*x])*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a^2*x^2)]*x^2*(a*(1 + n)*Sqrt[1 - 1/(a^2*x^2)]*x + 2*E
^ArcCoth[a*x]*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -E^(2*ArcCoth[a*x])] + 2*E^ArcCoth[a*x]*n*Hypergeomet
ric2F1[1, (1 + n)/2, (3 + n)/2, E^(2*ArcCoth[a*x])]))/((1 + n)*(-1 + a^2*x^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )} \sqrt {c -\frac {c}{a^{2} x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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