3.4.77 \(\int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx\) [377]

Optimal. Leaf size=167 \[ -\frac {a \left (1-\frac {1}{a x}\right )^{\frac {2-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}} x^2}{(3+n) (c-a c x)^{5/2}}+\frac {a \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {3+n}{2}} \left (1-\frac {1}{a x}\right )^{\frac {2-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}} x^2 \, _2F_1\left (\frac {1}{2},\frac {3+n}{2};\frac {3}{2};\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{(3+n) (c-a c x)^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6311, 6316, 96, 134} \begin {gather*} \frac {a x^2 \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {n+3}{2}} \left (1-\frac {1}{a x}\right )^{\frac {2-n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \, _2F_1\left (\frac {1}{2},\frac {n+3}{2};\frac {3}{2};\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{(n+3) (c-a c x)^{5/2}}-\frac {a x^2 \left (1-\frac {1}{a x}\right )^{\frac {2-n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{(n+3) (c-a c x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcCoth[a*x])/(c - a*c*x)^(5/2),x]

[Out]

-((a*(1 - 1/(a*x))^((2 - n)/2)*(1 + 1/(a*x))^((2 + n)/2)*x^2)/((3 + n)*(c - a*c*x)^(5/2))) + (a*((a - x^(-1))/
(a + x^(-1)))^((3 + n)/2)*(1 - 1/(a*x))^((2 - n)/2)*(1 + 1/(a*x))^((2 + n)/2)*x^2*Hypergeometric2F1[1/2, (3 +
n)/2, 3/2, 2/((a + x^(-1))*x)])/((3 + n)*(c - a*c*x)^(5/2))

Rule 96

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)/((m + 1)*(b*e - a*f))), x] - Dist[n*((d*e - c*f)/((m + 1)*(b*e - a*f
))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

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 6311

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 6316

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> Dist[(-c^p)*x^m*(1/x)^m, S
ubst[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 \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx &=\frac {\left (\left (1-\frac {1}{a x}\right )^{5/2} x^{5/2}\right ) \int \frac {e^{n \coth ^{-1}(a x)}}{\left (1-\frac {1}{a x}\right )^{5/2} x^{5/2}} \, dx}{(c-a c x)^{5/2}}\\ &=-\frac {\left (1-\frac {1}{a x}\right )^{5/2} \text {Subst}\left (\int \sqrt {x} \left (1-\frac {x}{a}\right )^{-\frac {5}{2}-\frac {n}{2}} \left (1+\frac {x}{a}\right )^{n/2} \, dx,x,\frac {1}{x}\right )}{\left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2}}\\ &=-\frac {a \left (1-\frac {1}{a x}\right )^{\frac {2-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}} x^2}{(3+n) (c-a c x)^{5/2}}+\frac {\left (a \left (1-\frac {1}{a x}\right )^{5/2}\right ) \text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{-\frac {3}{2}-\frac {n}{2}} \left (1+\frac {x}{a}\right )^{n/2}}{\sqrt {x}} \, dx,x,\frac {1}{x}\right )}{2 (3+n) \left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2}}\\ &=-\frac {a \left (1-\frac {1}{a x}\right )^{\frac {2-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}} x^2}{(3+n) (c-a c x)^{5/2}}+\frac {a \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {3+n}{2}} \left (1-\frac {1}{a x}\right )^{\frac {2-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}} x^2 \, _2F_1\left (\frac {1}{2},\frac {3+n}{2};\frac {3}{2};\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )}{(3+n) (c-a c x)^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 117, normalized size = 0.70 \begin {gather*} \frac {\left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{n/2} \left (-1-a x+(-1+a x) \left (\frac {-1+a x}{1+a x}\right )^{\frac {1+n}{2}} \, _2F_1\left (\frac {1}{2},\frac {3+n}{2};\frac {3}{2};\frac {2}{1+a x}\right )\right )}{a c^2 (3+n) (-1+a x) \sqrt {c-a c x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcCoth[a*x])/(c - a*c*x)^(5/2),x]

[Out]

((1 + 1/(a*x))^(n/2)*(-1 - a*x + (-1 + a*x)*((-1 + a*x)/(1 + a*x))^((1 + n)/2)*Hypergeometric2F1[1/2, (3 + n)/
2, 3/2, 2/(1 + a*x)]))/(a*c^2*(3 + n)*(1 - 1/(a*x))^(n/2)*(-1 + a*x)*Sqrt[c - a*c*x])

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {{\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )}}{\left (-a c x +c \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arccoth(a*x))/(-a*c*x+c)^(5/2),x)

[Out]

int(exp(n*arccoth(a*x))/(-a*c*x+c)^(5/2),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(exp(n*arccoth(a*x))/(-a*c*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(((a*x + 1)/(a*x - 1))^(1/2*n)/(-a*c*x + c)^(5/2), 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(exp(n*arccoth(a*x))/(-a*c*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{n \operatorname {acoth}{\left (a x \right )}}}{\left (- c \left (a x - 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*acoth(a*x))/(-a*c*x+c)**(5/2),x)

[Out]

Integral(exp(n*acoth(a*x))/(-c*(a*x - 1))**(5/2), 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(exp(n*arccoth(a*x))/(-a*c*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate(((a*x + 1)/(a*x - 1))^(1/2*n)/(-a*c*x + c)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {e}}^{n\,\mathrm {acoth}\left (a\,x\right )}}{{\left (c-a\,c\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*acoth(a*x))/(c - a*c*x)^(5/2),x)

[Out]

int(exp(n*acoth(a*x))/(c - a*c*x)^(5/2), x)

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