### 3.766 $$\int e^{3 \coth ^{-1}(a x)} (c-a^2 c x^2)^p \, dx$$

Optimal. Leaf size=118 $\frac{x \left (1-\frac{1}{a^2 x^2}\right )^{-p} \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{\frac{3}{2}-p} \left (1-\frac{1}{a x}\right )^{p-\frac{3}{2}} \left (\frac{1}{a x}+1\right )^{p+\frac{5}{2}} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (-2 p-1,\frac{3}{2}-p,-2 p,\frac{2}{x \left (a+\frac{1}{x}\right )}\right )}{2 p+1}$

[Out]

(((a - x^(-1))/(a + x^(-1)))^(3/2 - p)*(1 - 1/(a*x))^(-3/2 + p)*(1 + 1/(a*x))^(5/2 + p)*x*(c - a^2*c*x^2)^p*Hy
pergeometric2F1[-1 - 2*p, 3/2 - p, -2*p, 2/((a + x^(-1))*x)])/((1 + 2*p)*(1 - 1/(a^2*x^2))^p)

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Rubi [A]  time = 0.145059, antiderivative size = 118, 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 = {6192, 6196, 132} $\frac{x \left (1-\frac{1}{a^2 x^2}\right )^{-p} \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{\frac{3}{2}-p} \left (1-\frac{1}{a x}\right )^{p-\frac{3}{2}} \left (\frac{1}{a x}+1\right )^{p+\frac{5}{2}} \left (c-a^2 c x^2\right )^p \, _2F_1\left (-2 p-1,\frac{3}{2}-p;-2 p;\frac{2}{\left (a+\frac{1}{x}\right ) x}\right )}{2 p+1}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(3*ArcCoth[a*x])*(c - a^2*c*x^2)^p,x]

[Out]

(((a - x^(-1))/(a + x^(-1)))^(3/2 - p)*(1 - 1/(a*x))^(-3/2 + p)*(1 + 1/(a*x))^(5/2 + p)*x*(c - a^2*c*x^2)^p*Hy
pergeometric2F1[-1 - 2*p, 3/2 - p, -2*p, 2/((a + x^(-1))*x)])/((1 + 2*p)*(1 - 1/(a^2*x^2))^p)

Rule 6192

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

Rule 6196

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

Rule 132

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x
)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c -
a*d)*(e + f*x)))])/(((b*e - a*f)*(m + 1))*(((b*e - a*f)*(c + d*x))/((b*c - a*d)*(e + f*x)))^n), x] /; FreeQ[{a
, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 0] &&  !IntegerQ[n]

Rubi steps

\begin{align*} \int e^{3 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx &=\left (\left (1-\frac{1}{a^2 x^2}\right )^{-p} x^{-2 p} \left (c-a^2 c x^2\right )^p\right ) \int e^{3 \coth ^{-1}(a x)} \left (1-\frac{1}{a^2 x^2}\right )^p x^{2 p} \, dx\\ &=-\left (\left (\left (1-\frac{1}{a^2 x^2}\right )^{-p} \left (\frac{1}{x}\right )^{2 p} \left (c-a^2 c x^2\right )^p\right ) \operatorname{Subst}\left (\int x^{-2-2 p} \left (1-\frac{x}{a}\right )^{-\frac{3}{2}+p} \left (1+\frac{x}{a}\right )^{\frac{3}{2}+p} \, dx,x,\frac{1}{x}\right )\right )\\ &=\frac{\left (1-\frac{1}{a^2 x^2}\right )^{-p} \left (\frac{a-\frac{1}{x}}{a+\frac{1}{x}}\right )^{\frac{3}{2}-p} \left (1-\frac{1}{a x}\right )^{-\frac{3}{2}+p} \left (1+\frac{1}{a x}\right )^{\frac{5}{2}+p} x \left (c-a^2 c x^2\right )^p \, _2F_1\left (-1-2 p,\frac{3}{2}-p;-2 p;\frac{2}{\left (a+\frac{1}{x}\right ) x}\right )}{1+2 p}\\ \end{align*}

Mathematica [A]  time = 0.255989, size = 122, normalized size = 1.03 $-\frac{4^{p+1} \left (a x \sqrt{1-\frac{1}{a^2 x^2}}\right )^{-2 p} e^{5 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \left (1-e^{2 \coth ^{-1}(a x)}\right )^{2 p} \left (\frac{e^{\coth ^{-1}(a x)}}{e^{2 \coth ^{-1}(a x)}-1}\right )^{2 p} \text{Hypergeometric2F1}\left (p+\frac{5}{2},2 p+2,p+\frac{7}{2},e^{2 \coth ^{-1}(a x)}\right )}{2 a p+5 a}$

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(3*ArcCoth[a*x])*(c - a^2*c*x^2)^p,x]

[Out]

-((4^(1 + p)*E^(5*ArcCoth[a*x])*(1 - E^(2*ArcCoth[a*x]))^(2*p)*(E^ArcCoth[a*x]/(-1 + E^(2*ArcCoth[a*x])))^(2*p
)*(c - a^2*c*x^2)^p*Hypergeometric2F1[5/2 + p, 2 + 2*p, 7/2 + p, E^(2*ArcCoth[a*x])])/((5*a + 2*a*p)*(a*Sqrt[1
- 1/(a^2*x^2)]*x)^(2*p)))

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Maple [F]  time = 0.372, size = 0, normalized size = 0. \begin{align*} \int{ \left ( -{a}^{2}c{x}^{2}+c \right ) ^{p} \left ({\frac{ax-1}{ax+1}} \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^p,x)

[Out]

int(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} c x^{2} + c\right )}^{p}}{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^p,x, algorithm="maxima")

[Out]

integrate((-a^2*c*x^2 + c)^p/((a*x - 1)/(a*x + 1))^(3/2), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^p,x, algorithm="fricas")

[Out]

integral((a^2*x^2 + 2*a*x + 1)*(-a^2*c*x^2 + c)^p*sqrt((a*x - 1)/(a*x + 1))/(a^2*x^2 - 2*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(1/((a*x-1)/(a*x+1))**(3/2)*(-a**2*c*x**2+c)**p,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} c x^{2} + c\right )}^{p}}{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a^2*c*x^2+c)^p,x, algorithm="giac")

[Out]

integrate((-a^2*c*x^2 + c)^p/((a*x - 1)/(a*x + 1))^(3/2), x)