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

Optimal. Leaf size=116 \[ -\frac{2^{-\frac{n}{2}+p+1} \left (1-\frac{1}{a^2 x^2}\right )^{-p} \left (c-\frac{c}{a^2 x^2}\right )^p \left (\frac{1}{a x}+1\right )^{\frac{n}{2}+p+1} F_1\left (\frac{n}{2}+p+1;\frac{1}{2} (n-2 p),2;\frac{n}{2}+p+2;\frac{a+\frac{1}{x}}{2 a},1+\frac{1}{a x}\right )}{a (n+2 p+2)} \]

[Out]

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

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Rubi [A]  time = 0.114846, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6197, 6194, 136} \[ -\frac{2^{-\frac{n}{2}+p+1} \left (1-\frac{1}{a^2 x^2}\right )^{-p} \left (c-\frac{c}{a^2 x^2}\right )^p \left (\frac{1}{a x}+1\right )^{\frac{n}{2}+p+1} F_1\left (\frac{n}{2}+p+1;\frac{1}{2} (n-2 p),2;\frac{n}{2}+p+2;\frac{a+\frac{1}{x}}{2 a},1+\frac{1}{a x}\right )}{a (n+2 p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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])

Rubi steps

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

Mathematica [F]  time = 0.473303, size = 0, normalized size = 0. \[ \int e^{n \coth ^{-1}(a x)} \left (c-\frac{c}{a^2 x^2}\right )^p \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.202, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n{\rm arccoth} \left (ax\right )}} \left ( c-{\frac{c}{{a}^{2}{x}^{2}}} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{p} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n} \left (\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}\right )^{p}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (- c \left (-1 + \frac{1}{a x}\right ) \left (1 + \frac{1}{a x}\right )\right )^{p} e^{n \operatorname{acoth}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c - \frac{c}{a^{2} x^{2}}\right )}^{p} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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