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

Optimal. Leaf size=100 \[ -\frac{a^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{7 c^4 \left (a-\frac{1}{x}\right )^5}+\frac{12 a^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{35 c^4 \left (a-\frac{1}{x}\right )^4}-\frac{23 a^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{105 c^4 \left (a-\frac{1}{x}\right )^3} \]

[Out]

-(a^4*(1 - 1/(a^2*x^2))^(3/2))/(7*c^4*(a - x^(-1))^5) + (12*a^3*(1 - 1/(a^2*x^2))^(3/2))/(35*c^4*(a - x^(-1))^
4) - (23*a^2*(1 - 1/(a^2*x^2))^(3/2))/(105*c^4*(a - x^(-1))^3)

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Rubi [A]  time = 0.229582, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {6175, 6178, 1639, 793, 659, 651} \[ -\frac{a^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{7 c^4 \left (a-\frac{1}{x}\right )^5}+\frac{12 a^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{35 c^4 \left (a-\frac{1}{x}\right )^4}-\frac{23 a^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{105 c^4 \left (a-\frac{1}{x}\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]/(c - a*c*x)^4,x]

[Out]

-(a^4*(1 - 1/(a^2*x^2))^(3/2))/(7*c^4*(a - x^(-1))^5) + (12*a^3*(1 - 1/(a^2*x^2))^(3/2))/(35*c^4*(a - x^(-1))^
4) - (23*a^2*(1 - 1/(a^2*x^2))^(3/2))/(105*c^4*(a - x^(-1))^3)

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 1639

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 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 659

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*(m + p + 1)), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2
], 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^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx &=\frac{\int \frac{e^{\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 \sqrt{1-\frac{x^2}{a^2}}}{\left (1-\frac{x}{a}\right )^5} \, dx,x,\frac{1}{x}\right )}{a^4 c^4}\\ &=\frac{a^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{c^4 \left (a-\frac{1}{x}\right )^4}-\frac{\operatorname{Subst}\left (\int \frac{\left (\frac{4}{a^2}-\frac{3 x}{a^3}\right ) \sqrt{1-\frac{x^2}{a^2}}}{\left (1-\frac{x}{a}\right )^5} \, dx,x,\frac{1}{x}\right )}{c^4}\\ &=-\frac{a^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{7 c^4 \left (a-\frac{1}{x}\right )^5}+\frac{a^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{c^4 \left (a-\frac{1}{x}\right )^4}-\frac{23 \operatorname{Subst}\left (\int \frac{\sqrt{1-\frac{x^2}{a^2}}}{\left (1-\frac{x}{a}\right )^4} \, dx,x,\frac{1}{x}\right )}{7 a^2 c^4}\\ &=-\frac{a^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{7 c^4 \left (a-\frac{1}{x}\right )^5}+\frac{12 a^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{35 c^4 \left (a-\frac{1}{x}\right )^4}-\frac{23 \operatorname{Subst}\left (\int \frac{\sqrt{1-\frac{x^2}{a^2}}}{\left (1-\frac{x}{a}\right )^3} \, dx,x,\frac{1}{x}\right )}{35 a^2 c^4}\\ &=-\frac{a^4 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{7 c^4 \left (a-\frac{1}{x}\right )^5}+\frac{12 a^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{35 c^4 \left (a-\frac{1}{x}\right )^4}-\frac{23 a^2 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}{105 c^4 \left (a-\frac{1}{x}\right )^3}\\ \end{align*}

Mathematica [A]  time = 0.0562056, size = 51, normalized size = 0.51 \[ -\frac{x \sqrt{1-\frac{1}{a^2 x^2}} \left (2 a^3 x^3-8 a^2 x^2+13 a x+23\right )}{105 c^4 (a x-1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - 1/(a^2*x^2)]*x*(23 + 13*a*x - 8*a^2*x^2 + 2*a^3*x^3))/(105*c^4*(-1 + a*x)^4)

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Maple [A]  time = 0.052, size = 50, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,{a}^{2}{x}^{2}-10\,ax+23 \right ) \left ( ax+1 \right ) }{105\,{c}^{4} \left ( ax-1 \right ) ^{3}a}{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(2*a^2*x^2-10*a*x+23)*(a*x+1)/(a*x-1)^3/c^4/((a*x-1)/(a*x+1))^(1/2)/a

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Maxima [A]  time = 0.995723, size = 74, normalized size = 0.74 \begin{align*} \frac{\frac{42 \,{\left (a x - 1\right )}}{a x + 1} - \frac{35 \,{\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 15}{420 \, a c^{4} \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))^(1/2)/(-a*c*x+c)^4,x, algorithm="maxima")

[Out]

1/420*(42*(a*x - 1)/(a*x + 1) - 35*(a*x - 1)^2/(a*x + 1)^2 - 15)/(a*c^4*((a*x - 1)/(a*x + 1))^(7/2))

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Fricas [A]  time = 1.61714, size = 205, normalized size = 2.05 \begin{align*} -\frac{{\left (2 \, a^{4} x^{4} - 6 \, a^{3} x^{3} + 5 \, a^{2} x^{2} + 36 \, a x + 23\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{105 \,{\left (a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} + 6 \, a^{3} c^{4} x^{2} - 4 \, a^{2} c^{4} x + a c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(2*a^4*x^4 - 6*a^3*x^3 + 5*a^2*x^2 + 36*a*x + 23)*sqrt((a*x - 1)/(a*x + 1))/(a^5*c^4*x^4 - 4*a^4*c^4*x^
3 + 6*a^3*c^4*x^2 - 4*a^2*c^4*x + a*c^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{a^{4} x^{4} \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}} - 4 a^{3} x^{3} \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}} + 6 a^{2} x^{2} \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}} - 4 a x \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}} + \sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}\, dx}{c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(a**4*x**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) - 4*a**3*x**3*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) + 6*a*
*2*x**2*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) - 4*a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) + sqrt(a*x/(a*x + 1) - 1/(
a*x + 1))), x)/c**4

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Giac [A]  time = 1.14222, size = 93, normalized size = 0.93 \begin{align*} \frac{{\left (a x + 1\right )}^{3}{\left (\frac{42 \,{\left (a x - 1\right )}}{a x + 1} - \frac{35 \,{\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 15\right )}}{420 \,{\left (a x - 1\right )}^{3} a c^{4} \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))^(1/2)/(-a*c*x+c)^4,x, algorithm="giac")

[Out]

1/420*(a*x + 1)^3*(42*(a*x - 1)/(a*x + 1) - 35*(a*x - 1)^2/(a*x + 1)^2 - 15)/((a*x - 1)^3*a*c^4*sqrt((a*x - 1)
/(a*x + 1)))