3.10.33 \(\int e^{n \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^p \, dx\) [933]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 116, normalized size of antiderivative = 1.00, 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 = {6332, 6329, 141} \begin {gather*} -\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)} \end {gather*}

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 141

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)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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 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)} \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 ) \text {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*}

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Mathematica [F]
time = 0.56, size = 0, normalized size = 0.00 \begin {gather*} \int e^{n \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^p \, dx \end {gather*}

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 = 0.01, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )} \left (c -\frac {c}{a^{2} x^{2}}\right )^{p}\, dx\]

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 = 0.00, 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)^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 = 0.00, 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)^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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, 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)^p,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {e}}^{n\,\mathrm {acoth}\left (a\,x\right )}\,{\left (c-\frac {c}{a^2\,x^2}\right )}^p \,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))^p,x)

[Out]

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

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