### 3.758 $$\int \frac{e^{n \coth ^{-1}(a x)} x^2}{(c-a^2 c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=102 $\frac{\left (3-n^2\right ) (n-a x) e^{n \coth ^{-1}(a x)}}{a^3 c^2 \left (n^4-10 n^2+9\right ) \sqrt{c-a^2 c x^2}}-\frac{(n-3 a x) e^{n \coth ^{-1}(a x)}}{a^3 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}$

[Out]

-((E^(n*ArcCoth[a*x])*(n - 3*a*x))/(a^3*c*(9 - n^2)*(c - a^2*c*x^2)^(3/2))) + (E^(n*ArcCoth[a*x])*(3 - n^2)*(n
- a*x))/(a^3*c^2*(9 - 10*n^2 + n^4)*Sqrt[c - a^2*c*x^2])

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Rubi [A]  time = 0.189252, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 27, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.074, Rules used = {6189, 6184} $\frac{\left (3-n^2\right ) (n-a x) e^{n \coth ^{-1}(a x)}}{a^3 c^2 \left (n^4-10 n^2+9\right ) \sqrt{c-a^2 c x^2}}-\frac{(n-3 a x) e^{n \coth ^{-1}(a x)}}{a^3 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(E^(n*ArcCoth[a*x])*x^2)/(c - a^2*c*x^2)^(5/2),x]

[Out]

-((E^(n*ArcCoth[a*x])*(n - 3*a*x))/(a^3*c*(9 - n^2)*(c - a^2*c*x^2)^(3/2))) + (E^(n*ArcCoth[a*x])*(3 - n^2)*(n
- a*x))/(a^3*c^2*(9 - 10*n^2 + n^4)*Sqrt[c - a^2*c*x^2])

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 6184

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

Rubi steps

\begin{align*} \int \frac{e^{n \coth ^{-1}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx &=-\frac{e^{n \coth ^{-1}(a x)} (n-3 a x)}{a^3 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}-\frac{\left (3-n^2\right ) \int \frac{e^{n \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx}{a^2 c \left (9-n^2\right )}\\ &=-\frac{e^{n \coth ^{-1}(a x)} (n-3 a x)}{a^3 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}+\frac{e^{n \coth ^{-1}(a x)} \left (3-n^2\right ) (n-a x)}{a^3 c^2 \left (9-10 n^2+n^4\right ) \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.688619, size = 109, normalized size = 1.07 $\frac{e^{n \coth ^{-1}(a x)} \left (3 a \left (n^2-1\right ) x \sqrt{1-\frac{1}{a^2 x^2}} \cosh \left (3 \coth ^{-1}(a x)\right )+a n^2 x-2 \left (n^2-1\right ) n \cosh \left (2 \coth ^{-1}(a x)\right )-9 a x-2 n^3+10 n\right )}{4 a^3 c^2 \left (n^4-10 n^2+9\right ) \sqrt{c-a^2 c x^2}}$

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^(n*ArcCoth[a*x])*x^2)/(c - a^2*c*x^2)^(5/2),x]

[Out]

(E^(n*ArcCoth[a*x])*(10*n - 2*n^3 - 9*a*x + a*n^2*x - 2*n*(-1 + n^2)*Cosh[2*ArcCoth[a*x]] + 3*a*(-1 + n^2)*Sqr
t[1 - 1/(a^2*x^2)]*x*Cosh[3*ArcCoth[a*x]]))/(4*a^3*c^2*(9 - 10*n^2 + n^4)*Sqrt[c - a^2*c*x^2])

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Maple [A]  time = 0.043, size = 96, normalized size = 0.9 \begin{align*}{\frac{ \left ({a}^{3}{n}^{2}{x}^{3}-{a}^{2}{n}^{3}{x}^{2}-3\,{x}^{3}{a}^{3}+3\,n{x}^{2}{a}^{2}+2\,{n}^{2}xa-2\,n \right ) \left ( ax-1 \right ) \left ( ax+1 \right ){{\rm e}^{n{\rm arccoth} \left (ax\right )}}}{ \left ({n}^{4}-10\,{n}^{2}+9 \right ){a}^{3}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{-{\frac{5}{2}}}} \end{align*}

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)^(5/2),x)

[Out]

(a*x-1)*(a*x+1)*(a^3*n^2*x^3-a^2*n^3*x^2-3*a^3*x^3+3*a^2*n*x^2+2*a*n^2*x-2*n)*exp(n*arccoth(a*x))/(n^4-10*n^2+
9)/a^3/(-a^2*c*x^2+c)^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{5}{2}}}\,{d x} \end{align*}

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)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^2*((a*x - 1)/(a*x + 1))^(1/2*n)/(-a^2*c*x^2 + c)^(5/2), x)

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Fricas [A]  time = 1.3283, size = 356, normalized size = 3.49 \begin{align*} -\frac{\sqrt{-a^{2} c x^{2} + c}{\left (2 \, a n^{2} x +{\left (a^{3} n^{2} - 3 \, a^{3}\right )} x^{3} +{\left (a^{2} n^{3} - 3 \, a^{2} n\right )} x^{2} + 2 \, n\right )} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n}}{a^{3} c^{3} n^{4} - 10 \, a^{3} c^{3} n^{2} + 9 \, a^{3} c^{3} +{\left (a^{7} c^{3} n^{4} - 10 \, a^{7} c^{3} n^{2} + 9 \, a^{7} c^{3}\right )} x^{4} - 2 \,{\left (a^{5} c^{3} n^{4} - 10 \, a^{5} c^{3} n^{2} + 9 \, a^{5} c^{3}\right )} x^{2}} \end{align*}

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)^(5/2),x, algorithm="fricas")

[Out]

-sqrt(-a^2*c*x^2 + c)*(2*a*n^2*x + (a^3*n^2 - 3*a^3)*x^3 + (a^2*n^3 - 3*a^2*n)*x^2 + 2*n)*((a*x - 1)/(a*x + 1)
)^(1/2*n)/(a^3*c^3*n^4 - 10*a^3*c^3*n^2 + 9*a^3*c^3 + (a^7*c^3*n^4 - 10*a^7*c^3*n^2 + 9*a^7*c^3)*x^4 - 2*(a^5*
c^3*n^4 - 10*a^5*c^3*n^2 + 9*a^5*c^3)*x^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{5}{2}}}\,{d x} \end{align*}

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)^(5/2),x, algorithm="giac")

[Out]

integrate(x^2*((a*x - 1)/(a*x + 1))^(1/2*n)/(-a^2*c*x^2 + c)^(5/2), x)