3.373 \(\int e^{n \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx\)

Optimal. Leaf size=98 \[ \frac {2}{5} x (c-a c x)^{3/2} \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {n-3}{2}} \left (1-\frac {1}{a x}\right )^{-n/2} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \, _2F_1\left (-\frac {5}{2},\frac {n-3}{2};-\frac {3}{2};\frac {2}{\left (a+\frac {1}{x}\right ) x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6176, 6181, 132} \[ \frac {2}{5} x (c-a c x)^{3/2} \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {n-3}{2}} \left (1-\frac {1}{a x}\right )^{-n/2} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \, _2F_1\left (-\frac {5}{2},\frac {n-3}{2};-\frac {3}{2};\frac {2}{\left (a+\frac {1}{x}\right ) x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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)^{3/2} \, dx &=\frac {(c-a c x)^{3/2} \int e^{n \coth ^{-1}(a x)} \left (1-\frac {1}{a x}\right )^{3/2} x^{3/2} \, dx}{\left (1-\frac {1}{a x}\right )^{3/2} x^{3/2}}\\ &=-\frac {\left (\left (\frac {1}{x}\right )^{3/2} (c-a c x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{\frac {3}{2}-\frac {n}{2}} \left (1+\frac {x}{a}\right )^{n/2}}{x^{7/2}} \, dx,x,\frac {1}{x}\right )}{\left (1-\frac {1}{a x}\right )^{3/2}}\\ &=\frac {2}{5} \left (\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )^{\frac {1}{2} (-3+n)} \left (1-\frac {1}{a x}\right )^{-n/2} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}} x (c-a c x)^{3/2} \, _2F_1\left (-\frac {5}{2},\frac {1}{2} (-3+n);-\frac {3}{2};\frac {2}{\left (a+\frac {1}{x}\right ) x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 101, normalized size = 1.03 \[ -\frac {2 c (a x+1)^2 \sqrt {c-a c x} \left (1-\frac {1}{a x}\right )^{-n/2} \left (\frac {1}{a x}+1\right )^{n/2} \left (\frac {a x-1}{a x+1}\right )^{\frac {n-1}{2}} \, _2F_1\left (-\frac {5}{2},\frac {n-3}{2};-\frac {3}{2};\frac {2}{a x+1}\right )}{5 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a c x - c\right )} \sqrt {-a c x + c} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))*(-a*c*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))*(-a*c*x+c)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{1,[0,4,1,0,0]%%%}+%%%{-2,[0,2,1,1,0]%%%}+%%%{1,[0,0,1,2,0]%%%}
/ %%%{1,[0,0,0,2,2]%%%} Error: Bad Argument Value

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )} \left (-a c x +c \right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arccoth(a*x))*(-a*c*x+c)^(3/2),x)

[Out]

int(exp(n*arccoth(a*x))*(-a*c*x+c)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-a c x + c\right )}^{\frac {3}{2}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))*(-a*c*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((-a*c*x + c)^(3/2)*((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 \[ \int {\mathrm {e}}^{n\,\mathrm {acoth}\left (a\,x\right )}\,{\left (c-a\,c\,x\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*acoth(a*x))*(c - a*c*x)^(3/2),x)

[Out]

int(exp(n*acoth(a*x))*(c - a*c*x)^(3/2), 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)**(3/2),x)

[Out]

Timed out

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