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

Optimal. Leaf size=62 \[ \frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{3 c^3 \left (a-\frac {1}{x}\right )^2}-\frac {2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 c^3 \left (a-\frac {1}{x}\right )} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6175, 6178, 793, 651} \[ \frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{3 c^3 \left (a-\frac {1}{x}\right )^2}-\frac {2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 c^3 \left (a-\frac {1}{x}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 651

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

Rule 793

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d*g - e*f)*(
d + e*x)^m*(a + c*x^2)^(p + 1))/(2*c*d*(m + p + 1)), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 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^{-\coth ^{-1}(a x)}}{(c-a c x)^3} \, dx &=-\frac {\int \frac {e^{-\coth ^{-1}(a x)}}{\left (1-\frac {1}{a x}\right )^3 x^3} \, dx}{a^3 c^3}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x}{\left (1-\frac {x}{a}\right )^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a^3 c^3}\\ &=\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{3 c^3 \left (a-\frac {1}{x}\right )^2}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{\left (1-\frac {x}{a}\right ) \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{3 a^2 c^3}\\ &=\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{3 c^3 \left (a-\frac {1}{x}\right )^2}-\frac {2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 c^3 \left (a-\frac {1}{x}\right )}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 34, normalized size = 0.55 \[ -\frac {x \sqrt {1-\frac {1}{a^2 x^2}} (a x-2)}{3 c^3 (a x-1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.60, size = 57, normalized size = 0.92 \[ -\frac {{\left (a^{2} x^{2} - a x - 2\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{3 \, {\left (a^{3} c^{3} x^{2} - 2 \, a^{2} c^{3} x + a c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 45, normalized size = 0.73 \[ \frac {2 \, {\left (3 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}}{3 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{3} a c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 41, normalized size = 0.66 \[ -\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x -2\right ) \left (a x +1\right )}{3 \left (a x -1\right )^{2} c^{3} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 39, normalized size = 0.63 \[ -\frac {\frac {3 \, {\left (a x - 1\right )}}{a x + 1} - 1}{6 \, a c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 38, normalized size = 0.61 \[ -\frac {\frac {a\,x-1}{a\,x+1}-\frac {1}{3}}{2\,a\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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