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

Optimal. Leaf size=67 \[ \frac{a^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{7 c^3 \left (a-\frac{1}{x}\right )^6}-\frac{6 a^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{35 c^3 \left (a-\frac{1}{x}\right )^5} \]

[Out]

(a^5*(1 - 1/(a^2*x^2))^(5/2))/(7*c^3*(a - x^(-1))^6) - (6*a^4*(1 - 1/(a^2*x^2))^(5/2))/(35*c^3*(a - x^(-1))^5)

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Rubi [A]  time = 0.134002, antiderivative size = 67, normalized size of antiderivative = 1., 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} \[ \frac{a^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{7 c^3 \left (a-\frac{1}{x}\right )^6}-\frac{6 a^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{35 c^3 \left (a-\frac{1}{x}\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^5*(1 - 1/(a^2*x^2))^(5/2))/(7*c^3*(a - x^(-1))^6) - (6*a^4*(1 - 1/(a^2*x^2))^(5/2))/(35*c^3*(a - x^(-1))^5)

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 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 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]

Rubi steps

\begin{align*} \int \frac{e^{3 \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx &=-\frac{\int \frac{e^{3 \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^2}{a^2}\right )^{3/2}}{\left (1-\frac{x}{a}\right )^6} \, dx,x,\frac{1}{x}\right )}{a^3 c^3}\\ &=\frac{a^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{7 c^3 \left (a-\frac{1}{x}\right )^6}-\frac{6 \operatorname{Subst}\left (\int \frac{\left (1-\frac{x^2}{a^2}\right )^{3/2}}{\left (1-\frac{x}{a}\right )^5} \, dx,x,\frac{1}{x}\right )}{7 a^2 c^3}\\ &=\frac{a^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{7 c^3 \left (a-\frac{1}{x}\right )^6}-\frac{6 a^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}{35 c^3 \left (a-\frac{1}{x}\right )^5}\\ \end{align*}

Mathematica [A]  time = 0.0612204, size = 41, normalized size = 0.61 \[ -\frac{x \sqrt{1-\frac{1}{a^2 x^2}} (a x-6) (a x+1)^2}{35 c^3 (a x-1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - 1/(a^2*x^2)]*x*(-6 + a*x)*(1 + a*x)^2)/(35*c^3*(-1 + a*x)^4)

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Maple [A]  time = 0.125, size = 41, normalized size = 0.6 \begin{align*} -{\frac{ \left ( ax-6 \right ) \left ( ax+1 \right ) }{35\,{c}^{3} \left ( ax-1 \right ) ^{2}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(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^3,x)

[Out]

-1/35*(a*x-6)*(a*x+1)/(a*x-1)^2/c^3/((a*x-1)/(a*x+1))^(3/2)/a

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Maxima [A]  time = 1.03509, size = 53, normalized size = 0.79 \begin{align*} -\frac{\frac{7 \,{\left (a x - 1\right )}}{a x + 1} - 5}{70 \, a c^{3} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/70*(7*(a*x - 1)/(a*x + 1) - 5)/(a*c^3*((a*x - 1)/(a*x + 1))^(7/2))

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Fricas [A]  time = 1.63554, size = 201, normalized size = 3. \begin{align*} -\frac{{\left (a^{4} x^{4} - 3 \, a^{3} x^{3} - 15 \, a^{2} x^{2} - 17 \, a x - 6\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{35 \,{\left (a^{5} c^{3} x^{4} - 4 \, a^{4} c^{3} x^{3} + 6 \, a^{3} c^{3} x^{2} - 4 \, 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))^(3/2)/(-a*c*x+c)^3,x, algorithm="fricas")

[Out]

-1/35*(a^4*x^4 - 3*a^3*x^3 - 15*a^2*x^2 - 17*a*x - 6)*sqrt((a*x - 1)/(a*x + 1))/(a^5*c^3*x^4 - 4*a^4*c^3*x^3 +
 6*a^3*c^3*x^2 - 4*a^2*c^3*x + a*c^3)

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

[Out]

Timed out

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Giac [A]  time = 1.30981, size = 72, normalized size = 1.07 \begin{align*} -\frac{{\left (a x + 1\right )}^{3}{\left (\frac{7 \,{\left (a x - 1\right )}}{a x + 1} - 5\right )}}{70 \,{\left (a x - 1\right )}^{3} a c^{3} \sqrt{\frac{a x - 1}{a x + 1}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/70*(a*x + 1)^3*(7*(a*x - 1)/(a*x + 1) - 5)/((a*x - 1)^3*a*c^3*sqrt((a*x - 1)/(a*x + 1)))