### 3.554 $$\int e^{-\coth ^{-1}(a x)} (c-\frac{c}{a x})^p \, dx$$

Optimal. Leaf size=88 $-\frac{2^{p+\frac{3}{2}} \sqrt{\frac{1}{a x}+1} \left (1-\frac{1}{a x}\right )^{-p} F_1\left (\frac{1}{2};-p-\frac{1}{2},2;\frac{3}{2};\frac{a+\frac{1}{x}}{2 a},1+\frac{1}{a x}\right ) \left (c-\frac{c}{a x}\right )^p}{a}$

[Out]

-((2^(3/2 + p)*Sqrt[1 + 1/(a*x)]*(c - c/(a*x))^p*AppellF1[1/2, -1/2 - p, 2, 3/2, (a + x^(-1))/(2*a), 1 + 1/(a*
x)])/(a*(1 - 1/(a*x))^p))

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Rubi [A]  time = 0.0853336, antiderivative size = 88, 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 = {6182, 6179, 136} $-\frac{2^{p+\frac{3}{2}} \sqrt{\frac{1}{a x}+1} \left (1-\frac{1}{a x}\right )^{-p} F_1\left (\frac{1}{2};-p-\frac{1}{2},2;\frac{3}{2};\frac{a+\frac{1}{x}}{2 a},1+\frac{1}{a x}\right ) \left (c-\frac{c}{a x}\right )^p}{a}$

Antiderivative was successfully veriﬁed.

[In]

Int[(c - c/(a*x))^p/E^ArcCoth[a*x],x]

[Out]

-((2^(3/2 + p)*Sqrt[1 + 1/(a*x)]*(c - c/(a*x))^p*AppellF1[1/2, -1/2 - p, 2, 3/2, (a + x^(-1))/(2*a), 1 + 1/(a*
x)])/(a*(1 - 1/(a*x))^p))

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])

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 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^{-\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^{-\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{1}{2}+p}}{x^2 \sqrt{1+\frac{x}{a}}} \, dx,x,\frac{1}{x}\right )\right )\\ &=-\frac{2^{\frac{3}{2}+p} \left (1-\frac{1}{a x}\right )^{-p} \sqrt{1+\frac{1}{a x}} \left (c-\frac{c}{a x}\right )^p F_1\left (\frac{1}{2};-\frac{1}{2}-p,2;\frac{3}{2};\frac{a+\frac{1}{x}}{2 a},1+\frac{1}{a x}\right )}{a}\\ \end{align*}

Mathematica [F]  time = 1.08491, size = 0, normalized size = 0. $\int e^{-\coth ^{-1}(a x)} \left (c-\frac{c}{a x}\right )^p \, dx$

Veriﬁcation is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.174, size = 0, normalized size = 0. \begin{align*} \int \left ( c-{\frac{c}{ax}} \right ) ^{p}\sqrt{{\frac{ax-1}{ax+1}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^p*((a*x-1)/(a*x+1))^(1/2),x, algorithm="maxima")

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^p*((a*x-1)/(a*x+1))^(1/2),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^p*((a*x-1)/(a*x+1))^(1/2),x, algorithm="giac")

[Out]

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