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

Optimal. Leaf size=94 \[ -\frac{\left (a+\frac{1}{x}\right )^7}{9 a^8 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{9/2}}+\frac{16 \left (a+\frac{1}{x}\right )^6}{63 a^7 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}-\frac{47 \left (a+\frac{1}{x}\right )^5}{315 a^6 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}} \]

[Out]

(-47*(a + x^(-1))^5)/(315*a^6*c^4*(1 - 1/(a^2*x^2))^(5/2)) + (16*(a + x^(-1))^6)/(63*a^7*c^4*(1 - 1/(a^2*x^2))
^(7/2)) - (a + x^(-1))^7/(9*a^8*c^4*(1 - 1/(a^2*x^2))^(9/2))

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Rubi [A]  time = 0.275223, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6175, 6178, 852, 1635, 789, 651} \[ -\frac{\left (a+\frac{1}{x}\right )^7}{9 a^8 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{9/2}}+\frac{16 \left (a+\frac{1}{x}\right )^6}{63 a^7 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}-\frac{47 \left (a+\frac{1}{x}\right )^5}{315 a^6 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-47*(a + x^(-1))^5)/(315*a^6*c^4*(1 - 1/(a^2*x^2))^(5/2)) + (16*(a + x^(-1))^6)/(63*a^7*c^4*(1 - 1/(a^2*x^2))
^(7/2)) - (a + x^(-1))^7/(9*a^8*c^4*(1 - 1/(a^2*x^2))^(9/2))

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 852

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

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 789

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*(p + 1)), x] - Dist[(e*(m*(d*g + e*f) + 2*e*f*(p + 1)))/(2*c*d*(p + 1)
), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0]
&& LtQ[p, -1] && GtQ[m, 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)^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^2}{a^2}\right )^{3/2}}{\left (1-\frac{x}{a}\right )^7} \, dx,x,\frac{1}{x}\right )}{a^4 c^4}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (1+\frac{x}{a}\right )^7}{\left (1-\frac{x^2}{a^2}\right )^{11/2}} \, dx,x,\frac{1}{x}\right )}{a^4 c^4}\\ &=-\frac{\left (a+\frac{1}{x}\right )^7}{9 a^8 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{9/2}}+\frac{\operatorname{Subst}\left (\int \frac{\left (1+\frac{x}{a}\right )^6 \left (7 a^2+9 a x\right )}{\left (1-\frac{x^2}{a^2}\right )^{9/2}} \, dx,x,\frac{1}{x}\right )}{9 a^4 c^4}\\ &=\frac{16 \left (a+\frac{1}{x}\right )^6}{63 a^7 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}-\frac{\left (a+\frac{1}{x}\right )^7}{9 a^8 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{9/2}}-\frac{47 \operatorname{Subst}\left (\int \frac{\left (1+\frac{x}{a}\right )^5}{\left (1-\frac{x^2}{a^2}\right )^{7/2}} \, dx,x,\frac{1}{x}\right )}{63 a^2 c^4}\\ &=-\frac{47 \left (a+\frac{1}{x}\right )^5}{315 a^6 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}+\frac{16 \left (a+\frac{1}{x}\right )^6}{63 a^7 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}-\frac{\left (a+\frac{1}{x}\right )^7}{9 a^8 c^4 \left (1-\frac{1}{a^2 x^2}\right )^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0615454, size = 50, normalized size = 0.53 \[ -\frac{x \sqrt{1-\frac{1}{a^2 x^2}} (a x+1)^2 \left (2 a^2 x^2-14 a x+47\right )}{315 c^4 (a x-1)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - 1/(a^2*x^2)]*x*(1 + a*x)^2*(47 - 14*a*x + 2*a^2*x^2))/(315*c^4*(-1 + a*x)^5)

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Maple [A]  time = 0.123, size = 50, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,{a}^{2}{x}^{2}-14\,ax+47 \right ) \left ( ax+1 \right ) }{315\,{c}^{4} \left ( ax-1 \right ) ^{3}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)^4,x)

[Out]

-1/315*(2*a^2*x^2-14*a*x+47)*(a*x+1)/(a*x-1)^3/c^4/((a*x-1)/(a*x+1))^(3/2)/a

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Maxima [A]  time = 1.02957, size = 74, normalized size = 0.79 \begin{align*} \frac{\frac{90 \,{\left (a x - 1\right )}}{a x + 1} - \frac{63 \,{\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 35}{1260 \, a c^{4} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{9}{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)^4,x, algorithm="maxima")

[Out]

1/1260*(90*(a*x - 1)/(a*x + 1) - 63*(a*x - 1)^2/(a*x + 1)^2 - 35)/(a*c^4*((a*x - 1)/(a*x + 1))^(9/2))

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Fricas [A]  time = 1.57475, size = 251, normalized size = 2.67 \begin{align*} -\frac{{\left (2 \, a^{5} x^{5} - 8 \, a^{4} x^{4} + 11 \, a^{3} x^{3} + 101 \, a^{2} x^{2} + 127 \, a x + 47\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{315 \,{\left (a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} + 10 \, a^{4} c^{4} x^{3} - 10 \, a^{3} c^{4} x^{2} + 5 \, 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))^(3/2)/(-a*c*x+c)^4,x, algorithm="fricas")

[Out]

-1/315*(2*a^5*x^5 - 8*a^4*x^4 + 11*a^3*x^3 + 101*a^2*x^2 + 127*a*x + 47)*sqrt((a*x - 1)/(a*x + 1))/(a^6*c^4*x^
5 - 5*a^5*c^4*x^4 + 10*a^4*c^4*x^3 - 10*a^3*c^4*x^2 + 5*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(1/((a*x-1)/(a*x+1))**(3/2)/(-a*c*x+c)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.38352, size = 93, normalized size = 0.99 \begin{align*} \frac{{\left (a x + 1\right )}^{4}{\left (\frac{90 \,{\left (a x - 1\right )}}{a x + 1} - \frac{63 \,{\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 35\right )}}{1260 \,{\left (a x - 1\right )}^{4} 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))^(3/2)/(-a*c*x+c)^4,x, algorithm="giac")

[Out]

1/1260*(a*x + 1)^4*(90*(a*x - 1)/(a*x + 1) - 63*(a*x - 1)^2/(a*x + 1)^2 - 35)/((a*x - 1)^4*a*c^4*sqrt((a*x - 1
)/(a*x + 1)))