3.10.31 \(\int e^{n \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} \, dx\) [931]

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

[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 [A]
time = 0.15, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6332, 6329, 130, 71, 98, 133} \begin {gather*} \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}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 130

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

Rule 133

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)/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2,
(-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) &&  !ILtQ[m, 0]

Rule 6329

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 6332

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}} \text {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 = 0.34, size = 146, normalized size = 0.49 \begin {gather*} \frac {a e^{n \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a^2 x^2}} x^2 \left (a (1+n) \sqrt {1-\frac {1}{a^2 x^2}} x+2 e^{\coth ^{-1}(a x)} \, _2F_1\left (1,\frac {1+n}{2};\frac {3+n}{2};-e^{2 \coth ^{-1}(a x)}\right )+2 e^{\coth ^{-1}(a x)} n \, _2F_1\left (1,\frac {1+n}{2};\frac {3+n}{2};e^{2 \coth ^{-1}(a x)}\right )\right )}{(1+n) \left (-1+a^2 x^2\right )} \end {gather*}

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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Maple [F]
time = 0.02, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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

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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