3.363 \(\int e^{n \coth ^{-1}(a x)} (c-a c x)^{-2+\frac {n}{2}} \, dx\)

Optimal. Leaf size=88 \[ -\frac {2 x \left (1-\frac {1}{a x}\right )^{2-\frac {n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n-2}{2}} (c-a c x)^{\frac {n-4}{2}} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{2-n} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6176, 6181, 131} \[ -\frac {2 x \left (1-\frac {1}{a x}\right )^{2-\frac {n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n-2}{2}} (c-a c x)^{\frac {n-4}{2}} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{2-n} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcCoth[a*x])*(c - a*c*x)^(-2 + n/2),x]

[Out]

(-2*(1 - 1/(a*x))^(2 - n/2)*(1 + 1/(a*x))^((-2 + n)/2)*x*(c - a*c*x)^((-4 + n)/2)*Hypergeometric2F1[2, 1 - n/2
, 2 - n/2, 2/((a + x^(-1))*x)])/(2 - n)

Rule 131

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

Rule 6176

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

Rule 6181

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

Rubi steps

\begin {align*} \int e^{n \coth ^{-1}(a x)} (c-a c x)^{-2+\frac {n}{2}} \, dx &=\left (\left (1-\frac {1}{a x}\right )^{2-\frac {n}{2}} x^{2-\frac {n}{2}} (c-a c x)^{-2+\frac {n}{2}}\right ) \int e^{n \coth ^{-1}(a x)} \left (1-\frac {1}{a x}\right )^{-2+\frac {n}{2}} x^{-2+\frac {n}{2}} \, dx\\ &=-\left (\left (\left (1-\frac {1}{a x}\right )^{2-\frac {n}{2}} \left (\frac {1}{x}\right )^{-2+\frac {n}{2}} (c-a c x)^{-2+\frac {n}{2}}\right ) \operatorname {Subst}\left (\int \frac {x^{-n/2} \left (1+\frac {x}{a}\right )^{n/2}}{\left (1-\frac {x}{a}\right )^2} \, dx,x,\frac {1}{x}\right )\right )\\ &=-\frac {2 \left (1-\frac {1}{a x}\right )^{2-\frac {n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {1}{2} (-2+n)} x (c-a c x)^{\frac {1}{2} (-4+n)} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{2-n}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 89, normalized size = 1.01 \[ \frac {2 \left (1-\frac {1}{a x}\right )^{-n/2} \left (\frac {1}{a x}+1\right )^{n/2} (c-a c x)^{n/2} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {2}{a x+1}\right )}{a c^2 (n-2) (a x+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcCoth[a*x])*(c - a*c*x)^(-2 + n/2),x]

[Out]

(2*(1 + 1/(a*x))^(n/2)*(c - a*c*x)^(n/2)*Hypergeometric2F1[2, 1 - n/2, 2 - n/2, 2/(1 + a*x)])/(a*c^2*(-2 + n)*
(1 - 1/(a*x))^(n/2)*(1 + a*x))

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-a c x + c\right )}^{\frac {1}{2} \, n - 2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )} \left (-a c x +c \right )^{-2+\frac {n}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arccoth(a*x))*(-a*c*x+c)^(-2+1/2*n),x)

[Out]

int(exp(n*arccoth(a*x))*(-a*c*x+c)^(-2+1/2*n),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-a*c*x + c)^(1/2*n - 2)*((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 \[ \int {\mathrm {e}}^{n\,\mathrm {acoth}\left (a\,x\right )}\,{\left (c-a\,c\,x\right )}^{\frac {n}{2}-2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*acoth(a*x))*(-a*c*x+c)**(-2+1/2*n),x)

[Out]

Timed out

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