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

Optimal. Leaf size=116 \[ \frac {32 \left (c-a^2 c x^2\right )^{3/2} \left (1-\frac {1}{a x}\right )^{\frac {5-n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n-5}{2}} \, _2F_1\left (5,\frac {5-n}{2};\frac {7-n}{2};\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a^4 (5-n) x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}} \]

[Out]

32*(1-1/a/x)^(5/2-1/2*n)*(1+1/a/x)^(-5/2+1/2*n)*(-a^2*c*x^2+c)^(3/2)*hypergeom([5, 5/2-1/2*n],[7/2-1/2*n],(a-1
/x)/(a+1/x))/a^4/(5-n)/(1-1/a^2/x^2)^(3/2)/x^3

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Rubi [A]  time = 0.21, antiderivative size = 116, 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 = {6192, 6195, 131} \[ \frac {32 \left (c-a^2 c x^2\right )^{3/2} \left (1-\frac {1}{a x}\right )^{\frac {5-n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n-5}{2}} \, _2F_1\left (5,\frac {5-n}{2};\frac {7-n}{2};\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a^4 (5-n) x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*(1 - 1/(a*x))^((5 - n)/2)*(1 + 1/(a*x))^((-5 + n)/2)*(c - a^2*c*x^2)^(3/2)*Hypergeometric2F1[5, (5 - n)/2,
(7 - n)/2, (a - x^(-1))/(a + x^(-1))])/(a^4*(5 - n)*(1 - 1/(a^2*x^2))^(3/2)*x^3)

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 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 6195

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

Rubi steps

\begin {align*} \int e^{n \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx &=\frac {\left (c-a^2 c x^2\right )^{3/2} \int e^{n \coth ^{-1}(a x)} \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \, dx}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}\\ &=-\frac {\left (c-a^2 c x^2\right )^{3/2} \operatorname {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{\frac {3}{2}-\frac {n}{2}} \left (1+\frac {x}{a}\right )^{\frac {3}{2}+\frac {n}{2}}}{x^5} \, dx,x,\frac {1}{x}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}\\ &=\frac {32 \left (1-\frac {1}{a x}\right )^{\frac {5-n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {1}{2} (-5+n)} \left (c-a^2 c x^2\right )^{3/2} \, _2F_1\left (5,\frac {5-n}{2};\frac {7-n}{2};\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a^4 (5-n) \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}\\ \end {align*}

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Mathematica [B]  time = 2.41, size = 280, normalized size = 2.41 \[ \frac {c^2 \left (96 a^3 c x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \left (a x \sqrt {1-\frac {1}{a^2 x^2}} (a x+n) e^{n \coth ^{-1}(a x)}+2 (n-1) e^{(n+1) \coth ^{-1}(a x)} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};e^{2 \coth ^{-1}(a x)}\right )\right )-c \left (a^2 x^2-1\right ) \left (16 a \left (n^3-n^2+3 n-3\right ) x \sqrt {1-\frac {1}{a^2 x^2}} e^{(n+1) \coth ^{-1}(a x)} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};e^{2 \coth ^{-1}(a x)}\right )+2 \left (a^2 x^2-1\right )^2 e^{n \coth ^{-1}(a x)} \left (a \left (n^2+3\right ) x \sqrt {1-\frac {1}{a^2 x^2}} \cosh \left (3 \coth ^{-1}(a x)\right )-a \left (n^2-21\right ) x+2 n \left (\left (n^2+3\right ) \cosh \left (2 \coth ^{-1}(a x)\right )-n^2+1\right )\right )\right )\right )}{192 a \left (c-a^2 c x^2\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^2*(96*a^3*c*(1 - 1/(a^2*x^2))^(3/2)*x^3*(a*E^(n*ArcCoth[a*x])*Sqrt[1 - 1/(a^2*x^2)]*x*(n + a*x) + 2*E^((1 +
n)*ArcCoth[a*x])*(-1 + n)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, E^(2*ArcCoth[a*x])]) - c*(-1 + a^2*x^2)*
(2*E^(n*ArcCoth[a*x])*(-1 + a^2*x^2)^2*(-(a*(-21 + n^2)*x) + 2*n*(1 - n^2 + (3 + n^2)*Cosh[2*ArcCoth[a*x]]) +
a*(3 + n^2)*Sqrt[1 - 1/(a^2*x^2)]*x*Cosh[3*ArcCoth[a*x]]) + 16*a*E^((1 + n)*ArcCoth[a*x])*(-3 + 3*n - n^2 + n^
3)*Sqrt[1 - 1/(a^2*x^2)]*x*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, E^(2*ArcCoth[a*x])])))/(192*a*(c - a^2*c
*x^2)^(3/2))

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{2} c x^{2} - c\right )} \sqrt {-a^{2} c x^{2} + c} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a^2*c*x^2 - c)*sqrt(-a^2*c*x^2 + 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^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*acoth(a*x))*(-a**2*c*x**2+c)**(3/2),x)

[Out]

Integral((-c*(a*x - 1)*(a*x + 1))**(3/2)*exp(n*acoth(a*x)), x)

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