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

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Rubi [A]  time = 0.14, antiderivative size = 61, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.41, size = 58, normalized size = 0.95 \[ \frac {{\left (2 \, a^{2} x^{2} - 2 \, a x - 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{3 \, {\left (a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (a c x - c\right )}^{4}}\,{d x} \]

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)

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maple [A]  time = 0.04, size = 50, normalized size = 0.82 \[ \frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (2 a^{2} x^{2}-2 a x -1\right ) \left (a x +1\right )}{3 \left (a x -1\right )^{3} c^{4} a} \]

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 = 0.31, size = 65, normalized size = 1.07 \[ \frac {1}{12} \, a {\left (\frac {3 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c^{4}} + \frac {\frac {6 \, {\left (a x - 1\right )}}{a x + 1} - 1}{a^{2} c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\right )} \]

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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mupad [B]  time = 1.23, size = 50, normalized size = 0.82 \[ \frac {-2\,a^2\,x^2+2\,a\,x+1}{\left (3\,a\,c^4-3\,a^3\,c^4\,x^2\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \left (- \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{5} x^{5} - 3 a^{4} x^{4} + 2 a^{3} x^{3} + 2 a^{2} x^{2} - 3 a x + 1}\right )\, dx + \int \frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{5} x^{5} - 3 a^{4} x^{4} + 2 a^{3} x^{3} + 2 a^{2} x^{2} - 3 a x + 1}\, dx}{c^{4}} \]

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]

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

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