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

Optimal. Leaf size=110 \[ -\frac {2^{-\frac {n}{2}+p+1} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (1-\frac {1}{a x}\right )^{-p} \left (c-\frac {c}{a x}\right )^p F_1\left (\frac {n+2}{2};\frac {1}{2} (n-2 p),2;\frac {n+4}{2};\frac {a+\frac {1}{x}}{2 a},1+\frac {1}{a x}\right )}{a (n+2)} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 110, 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 = {6182, 6179, 136} \[ -\frac {2^{-\frac {n}{2}+p+1} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (1-\frac {1}{a x}\right )^{-p} \left (c-\frac {c}{a x}\right )^p F_1\left (\frac {n+2}{2};\frac {1}{2} (n-2 p),2;\frac {n+4}{2};\frac {a+\frac {1}{x}}{2 a},1+\frac {1}{a x}\right )}{a (n+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(1 - n/2 + p)*(1 + 1/(a*x))^((2 + n)/2)*(c - c/(a*x))^p*AppellF1[(2 + n)/2, (n - 2*p)/2, 2, (4 + n)/2, (a
+ x^(-1))/(2*a), 1 + 1/(a*x)])/(a*(2 + n)*(1 - 1/(a*x))^p))

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])

Rule 6179

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

Rule 6182

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] :> Dist[(c + d/x)^p/(1 + d/(c*x))^
p, Int[u*(1 + d/(c*x))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c^2 - a^2*d^2, 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 x}\right )^p \, dx &=\left (\left (1-\frac {1}{a x}\right )^{-p} \left (c-\frac {c}{a x}\right )^p\right ) \int e^{n \coth ^{-1}(a x)} \left (1-\frac {1}{a x}\right )^p \, dx\\ &=-\left (\left (\left (1-\frac {1}{a x}\right )^{-p} \left (c-\frac {c}{a x}\right )^p\right ) \operatorname {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{-\frac {n}{2}+p} \left (1+\frac {x}{a}\right )^{n/2}}{x^2} \, dx,x,\frac {1}{x}\right )\right )\\ &=-\frac {2^{1-\frac {n}{2}+p} \left (1-\frac {1}{a x}\right )^{-p} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}} \left (c-\frac {c}{a x}\right )^p F_1\left (\frac {2+n}{2};\frac {1}{2} (n-2 p),2;\frac {4+n}{2};\frac {a+\frac {1}{x}}{2 a},1+\frac {1}{a x}\right )}{a (2+n)}\\ \end {align*}

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Mathematica [F]  time = 0.97, size = 0, normalized size = 0.00 \[ \int e^{n \coth ^{-1}(a x)} \left (c-\frac {c}{a x}\right )^p \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n} \left (\frac {a c x - c}{a x}\right )^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )} \left (c -\frac {c}{a x}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c - c/(a*x))^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 \[ \int {\mathrm {e}}^{n\,\mathrm {acoth}\left (a\,x\right )}\,{\left (c-\frac {c}{a\,x}\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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