3.632 \(\int \frac{e^{2 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^{9/2}} \, dx\)

Optimal. Leaf size=120 \[ -\frac{16 x}{45 c^4 \sqrt{c-a^2 c x^2}}-\frac{8 x}{45 c^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac{2 x}{15 c^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac{x}{9 c \left (c-a^2 c x^2\right )^{7/2}}-\frac{2 (a x+1)}{9 a \left (c-a^2 c x^2\right )^{9/2}} \]

[Out]

(-2*(1 + a*x))/(9*a*(c - a^2*c*x^2)^(9/2)) - x/(9*c*(c - a^2*c*x^2)^(7/2)) - (2*x)/(15*c^2*(c - a^2*c*x^2)^(5/
2)) - (8*x)/(45*c^3*(c - a^2*c*x^2)^(3/2)) - (16*x)/(45*c^4*Sqrt[c - a^2*c*x^2])

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Rubi [A]  time = 0.131384, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {6167, 6141, 653, 192, 191} \[ -\frac{16 x}{45 c^4 \sqrt{c-a^2 c x^2}}-\frac{8 x}{45 c^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac{2 x}{15 c^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac{x}{9 c \left (c-a^2 c x^2\right )^{7/2}}-\frac{2 (a x+1)}{9 a \left (c-a^2 c x^2\right )^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcCoth[a*x])/(c - a^2*c*x^2)^(9/2),x]

[Out]

(-2*(1 + a*x))/(9*a*(c - a^2*c*x^2)^(9/2)) - x/(9*c*(c - a^2*c*x^2)^(7/2)) - (2*x)/(15*c^2*(c - a^2*c*x^2)^(5/
2)) - (8*x)/(45*c^3*(c - a^2*c*x^2)^(3/2)) - (16*x)/(45*c^4*Sqrt[c - a^2*c*x^2])

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6141

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

Rule 653

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + c*x^2)^(p + 1))/(c*(
p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -1]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{e^{2 \coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{9/2}} \, dx &=-\int \frac{e^{2 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{9/2}} \, dx\\ &=-\left (c \int \frac{(1+a x)^2}{\left (c-a^2 c x^2\right )^{11/2}} \, dx\right )\\ &=-\frac{2 (1+a x)}{9 a \left (c-a^2 c x^2\right )^{9/2}}-\frac{7}{9} \int \frac{1}{\left (c-a^2 c x^2\right )^{9/2}} \, dx\\ &=-\frac{2 (1+a x)}{9 a \left (c-a^2 c x^2\right )^{9/2}}-\frac{x}{9 c \left (c-a^2 c x^2\right )^{7/2}}-\frac{2 \int \frac{1}{\left (c-a^2 c x^2\right )^{7/2}} \, dx}{3 c}\\ &=-\frac{2 (1+a x)}{9 a \left (c-a^2 c x^2\right )^{9/2}}-\frac{x}{9 c \left (c-a^2 c x^2\right )^{7/2}}-\frac{2 x}{15 c^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac{8 \int \frac{1}{\left (c-a^2 c x^2\right )^{5/2}} \, dx}{15 c^2}\\ &=-\frac{2 (1+a x)}{9 a \left (c-a^2 c x^2\right )^{9/2}}-\frac{x}{9 c \left (c-a^2 c x^2\right )^{7/2}}-\frac{2 x}{15 c^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac{8 x}{45 c^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac{16 \int \frac{1}{\left (c-a^2 c x^2\right )^{3/2}} \, dx}{45 c^3}\\ &=-\frac{2 (1+a x)}{9 a \left (c-a^2 c x^2\right )^{9/2}}-\frac{x}{9 c \left (c-a^2 c x^2\right )^{7/2}}-\frac{2 x}{15 c^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac{8 x}{45 c^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac{16 x}{45 c^4 \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0594047, size = 112, normalized size = 0.93 \[ \frac{\sqrt{1-a^2 x^2} \left (-16 a^7 x^7+32 a^6 x^6+24 a^5 x^5-80 a^4 x^4+10 a^3 x^3+60 a^2 x^2-25 a x-10\right )}{45 a c^4 (1-a x)^{9/2} (a x+1)^{5/2} \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcCoth[a*x])/(c - a^2*c*x^2)^(9/2),x]

[Out]

(Sqrt[1 - a^2*x^2]*(-10 - 25*a*x + 60*a^2*x^2 + 10*a^3*x^3 - 80*a^4*x^4 + 24*a^5*x^5 + 32*a^6*x^6 - 16*a^7*x^7
))/(45*a*c^4*(1 - a*x)^(9/2)*(1 + a*x)^(5/2)*Sqrt[c - a^2*c*x^2])

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Maple [A]  time = 0.057, size = 80, normalized size = 0.7 \begin{align*} -{\frac{ \left ( ax+1 \right ) ^{2} \left ( 16\,{a}^{7}{x}^{7}-32\,{x}^{6}{a}^{6}-24\,{x}^{5}{a}^{5}+80\,{x}^{4}{a}^{4}-10\,{x}^{3}{a}^{3}-60\,{a}^{2}{x}^{2}+25\,ax+10 \right ) }{45\,a} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(a*x-1)/(-a^2*c*x^2+c)^(9/2),x)

[Out]

-1/45*(a*x+1)^2*(16*a^7*x^7-32*a^6*x^6-24*a^5*x^5+80*a^4*x^4-10*a^3*x^3-60*a^2*x^2+25*a*x+10)/a/(-a^2*c*x^2+c)
^(9/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.04126, size = 317, normalized size = 2.64 \begin{align*} \frac{{\left (16 \, a^{7} x^{7} - 32 \, a^{6} x^{6} - 24 \, a^{5} x^{5} + 80 \, a^{4} x^{4} - 10 \, a^{3} x^{3} - 60 \, a^{2} x^{2} + 25 \, a x + 10\right )} \sqrt{-a^{2} c x^{2} + c}}{45 \,{\left (a^{9} c^{5} x^{8} - 2 \, a^{8} c^{5} x^{7} - 2 \, a^{7} c^{5} x^{6} + 6 \, a^{6} c^{5} x^{5} - 6 \, a^{4} c^{5} x^{3} + 2 \, a^{3} c^{5} x^{2} + 2 \, a^{2} c^{5} x - a c^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(9/2),x, algorithm="fricas")

[Out]

1/45*(16*a^7*x^7 - 32*a^6*x^6 - 24*a^5*x^5 + 80*a^4*x^4 - 10*a^3*x^3 - 60*a^2*x^2 + 25*a*x + 10)*sqrt(-a^2*c*x
^2 + c)/(a^9*c^5*x^8 - 2*a^8*c^5*x^7 - 2*a^7*c^5*x^6 + 6*a^6*c^5*x^5 - 6*a^4*c^5*x^3 + 2*a^3*c^5*x^2 + 2*a^2*c
^5*x - a*c^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a**2*c*x**2+c)**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{9}{2}}{\left (a x - 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(9/2),x, algorithm="giac")

[Out]

integrate((a*x + 1)/((-a^2*c*x^2 + c)^(9/2)*(a*x - 1)), x)