3.223 \(\int \frac{e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx\)

Optimal. Leaf size=61 \[ \frac{2}{3 a c^4 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{1}{3 a^2 c^4 x^2 \sqrt{1-\frac{1}{a^2 x^2}} \left (a-\frac{1}{x}\right )} \]

[Out]

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

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Rubi [A]  time = 0.136697, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {6175, 6178, 855, 12, 261} \[ \frac{2}{3 a c^4 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{1}{3 a^2 c^4 x^2 \sqrt{1-\frac{1}{a^2 x^2}} \left (a-\frac{1}{x}\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)^4),x]

[Out]

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

Rule 6175

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

Rule 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rule 855

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 261

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

Rubi steps

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

Mathematica [A]  time = 0.0609999, size = 50, normalized size = 0.82 \[ \frac{x \sqrt{1-\frac{1}{a^2 x^2}} \left (2 a^2 x^2-2 a x-1\right )}{3 c^4 (a x-1)^2 (a x+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)^4),x]

[Out]

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

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Maple [A]  time = 0.041, size = 50, normalized size = 0.8 \begin{align*}{\frac{ \left ( 2\,{a}^{2}{x}^{2}-2\,ax-1 \right ) \left ( ax+1 \right ) }{3\, \left ( ax-1 \right ) ^{3}{c}^{4}a} \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01646, size = 88, normalized size = 1.44 \begin{align*} \frac{1}{12} \, a{\left (\frac{3 \, \sqrt{\frac{a x - 1}{a x + 1}}}{a^{2} c^{4}} + \frac{\frac{6 \,{\left (a x - 1\right )}}{a x + 1} - 1}{a^{2} c^{4} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^4,x, algorithm="maxima")

[Out]

1/12*a*(3*sqrt((a*x - 1)/(a*x + 1))/(a^2*c^4) + (6*(a*x - 1)/(a*x + 1) - 1)/(a^2*c^4*((a*x - 1)/(a*x + 1))^(3/
2)))

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Fricas [A]  time = 1.55957, size = 123, normalized size = 2.02 \begin{align*} \frac{{\left (2 \, a^{2} x^{2} - 2 \, a x - 1\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{3 \,{\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^4,x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}{{\left (a c x - c\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^4,x, algorithm="giac")

[Out]

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