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

Optimal. Leaf size=164 \[ -\frac {2 x \sqrt {1-\frac {1}{a^2 x^2}} \left (\frac {1}{a x}+1\right )^{\frac {n-1}{2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}} \, _2F_1\left (1,\frac {1-n}{2};\frac {3-n}{2};\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a^2 c (1-n) \sqrt {c-a^2 c x^2}}-\frac {(n-a x) e^{n \coth ^{-1}(a x)}}{a^3 c \left (1-n^2\right ) \sqrt {c-a^2 c x^2}} \]

[Out]

-exp(n*arccoth(a*x))*(-a*x+n)/a^3/c/(-n^2+1)/(-a^2*c*x^2+c)^(1/2)-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)/a^2/c/(1-n)/(-a^2*c*x^2+c)^(1/2
)

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Rubi [A]  time = 0.34, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6189, 6192, 6195, 131} \[ -\frac {2 x \sqrt {1-\frac {1}{a^2 x^2}} \left (\frac {1}{a x}+1\right )^{\frac {n-1}{2}} \left (1-\frac {1}{a x}\right )^{\frac {1-n}{2}} \, _2F_1\left (1,\frac {1-n}{2};\frac {3-n}{2};\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a^2 c (1-n) \sqrt {c-a^2 c x^2}}-\frac {(n-a x) e^{n \coth ^{-1}(a x)}}{a^3 c \left (1-n^2\right ) \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((E^(n*ArcCoth[a*x])*(n - a*x))/(a^3*c*(1 - n^2)*Sqrt[c - a^2*c*x^2])) - (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
))])/(a^2*c*(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 6189

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(x_)^2*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((n + 2*(p + 1)*a*x)*(c
+ d*x^2)^(p + 1)*E^(n*ArcCoth[a*x]))/(a^3*c*(n^2 - 4*(p + 1)^2)), x] - Dist[(n^2 + 2*(p + 1))/(a^2*c*(n^2 - 4
*(p + 1)^2)), Int[(c + d*x^2)^(p + 1)*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0
] &&  !IntegerQ[n/2] && LeQ[p, -1] && NeQ[n^2 + 2*(p + 1), 0] && NeQ[n^2 - 4*(p + 1)^2, 0] && (IntegerQ[p] ||
!IntegerQ[n])

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

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Mathematica [A]  time = 0.44, size = 127, normalized size = 0.77 \[ -\frac {e^{n \coth ^{-1}(a x)} \left (2 (n-1) \left (a^2 x^2-1\right ) e^{\coth ^{-1}(a x)} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};e^{2 \coth ^{-1}(a x)}\right )+a x \sqrt {1-\frac {1}{a^2 x^2}} (a x-n)\right )}{a^4 c (n-1) (n+1) x \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-a^2 c x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((E^(n*ArcCoth[a*x])*(a*Sqrt[1 - 1/(a^2*x^2)]*x*(-n + a*x) + 2*E^ArcCoth[a*x]*(-1 + n)*(-1 + a^2*x^2)*Hyperge
ometric2F1[1, (1 + n)/2, (3 + n)/2, E^(2*ArcCoth[a*x])]))/(a^4*c*(-1 + n)*(1 + n)*Sqrt[1 - 1/(a^2*x^2)]*x*Sqrt
[c - a^2*c*x^2]))

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} c x^{2} + c} x^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{a^{4} c^{2} x^{4} - 2 \, a^{2} c^{2} x^{2} + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*c*x^2 + c)*x^2*((a*x - 1)/(a*x + 1))^(1/2*n)/(a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )} x^{2}}{\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,{\mathrm {e}}^{n\,\mathrm {acoth}\left (a\,x\right )}}{{\left (c-a^2\,c\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} e^{n \operatorname {acoth}{\left (a x \right )}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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