3.630 \(\int \frac{e^{2 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=74 \[ -\frac{2 x}{5 c^2 \sqrt{c-a^2 c x^2}}-\frac{x}{5 c \left (c-a^2 c x^2\right )^{3/2}}-\frac{2 (a x+1)}{5 a \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.110384, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ -\frac{2 x}{5 c^2 \sqrt{c-a^2 c x^2}}-\frac{x}{5 c \left (c-a^2 c x^2\right )^{3/2}}-\frac{2 (a x+1)}{5 a \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1 + a*x))/(5*a*(c - a^2*c*x^2)^(5/2)) - x/(5*c*(c - a^2*c*x^2)^(3/2)) - (2*x)/(5*c^2*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 )^{5/2}} \, dx &=-\int \frac{e^{2 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\\ &=-\left (c \int \frac{(1+a x)^2}{\left (c-a^2 c x^2\right )^{7/2}} \, dx\right )\\ &=-\frac{2 (1+a x)}{5 a \left (c-a^2 c x^2\right )^{5/2}}-\frac{3}{5} \int \frac{1}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\\ &=-\frac{2 (1+a x)}{5 a \left (c-a^2 c x^2\right )^{5/2}}-\frac{x}{5 c \left (c-a^2 c x^2\right )^{3/2}}-\frac{2 \int \frac{1}{\left (c-a^2 c x^2\right )^{3/2}} \, dx}{5 c}\\ &=-\frac{2 (1+a x)}{5 a \left (c-a^2 c x^2\right )^{5/2}}-\frac{x}{5 c \left (c-a^2 c x^2\right )^{3/2}}-\frac{2 x}{5 c^2 \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0381066, size = 53, normalized size = 0.72 \[ -\frac{2 a^3 x^3-4 a^2 x^2+a x+2}{5 a c^2 (a x-1)^2 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2 + a*x - 4*a^2*x^2 + 2*a^3*x^3)/(5*a*c^2*(-1 + a*x)^2*Sqrt[c - a^2*c*x^2])

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Maple [A]  time = 0.043, size = 47, normalized size = 0.6 \begin{align*} -{\frac{ \left ( ax+1 \right ) ^{2} \left ( 2\,{x}^{3}{a}^{3}-4\,{a}^{2}{x}^{2}+ax+2 \right ) }{5\,a} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

-1/5*(a*x+1)^2*(2*a^3*x^3-4*a^2*x^2+a*x+2)/a/(-a^2*c*x^2+c)^(5/2)

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.0755, size = 151, normalized size = 2.04 \begin{align*} \frac{{\left (2 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + a x + 2\right )} \sqrt{-a^{2} c x^{2} + c}}{5 \,{\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\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)^(5/2),x, algorithm="fricas")

[Out]

1/5*(2*a^3*x^3 - 4*a^2*x^2 + a*x + 2)*sqrt(-a^2*c*x^2 + c)/(a^5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a^2*c^3*x - a*c^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a x + 1}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{5}{2}} \left (a x - 1\right )}\, dx \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)**(5/2),x)

[Out]

Integral((a*x + 1)/((-c*(a*x - 1)*(a*x + 1))**(5/2)*(a*x - 1)), x)

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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{5}{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)^(5/2),x, algorithm="giac")

[Out]

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