3.8.66 \(\int e^{3 \coth ^{-1}(a x)} (c-a^2 c x^2)^p \, dx\) [766]

Optimal. Leaf size=118 \[ \frac {\left (1-\frac {1}{a^2 x^2}\right )^{-p} \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {3}{2}-p} \left (1-\frac {1}{a x}\right )^{-\frac {3}{2}+p} \left (1+\frac {1}{a x}\right )^{\frac {5}{2}+p} x \left (c-a^2 c x^2\right )^p \, _2F_1\left (-1-2 p,\frac {3}{2}-p;-2 p;\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{1+2 p} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 118, 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 = {6327, 6331, 134} \begin {gather*} \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} \end {gather*}

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 134

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)/((b*e - a*f)*(m + 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)*((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]

Rule 6327

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 6331

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_)^2)^(p_.)*(x_)^(m_), x_Symbol] :> Dist[(-c^p)*x^m*(1/x)^m,
Subst[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] &&  !Inte
gerQ[m]

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 ) \text {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*}

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Mathematica [A]
time = 0.17, size = 122, normalized size = 1.03 \begin {gather*} -\frac {4^{1+p} e^{5 \coth ^{-1}(a x)} \left (1-e^{2 \coth ^{-1}(a x)}\right )^{2 p} \left (\frac {e^{\coth ^{-1}(a x)}}{-1+e^{2 \coth ^{-1}(a x)}}\right )^{2 p} \left (a \sqrt {1-\frac {1}{a^2 x^2}} x\right )^{-2 p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {5}{2}+p,2+2 p;\frac {7}{2}+p;e^{2 \coth ^{-1}(a x)}\right )}{5 a+2 a p} \end {gather*}

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 = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (-a^{2} c \,x^{2}+c \right )^{p}}{\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\, dx\]

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 = 0.00, 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(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 = 0.00, 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(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{p}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c-a^2\,c\,x^2\right )}^p}{{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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