### 3.186 $$\int \frac{e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx$$

Optimal. Leaf size=125 $\frac{\left (a+\frac{1}{x}\right )^8}{11 a^9 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{11/2}}-\frac{10 \left (a+\frac{1}{x}\right )^7}{33 a^8 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{9/2}}+\frac{79 \left (a+\frac{1}{x}\right )^6}{231 a^7 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}-\frac{152 \left (a+\frac{1}{x}\right )^5}{1155 a^6 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}$

[Out]

(-152*(a + x^(-1))^5)/(1155*a^6*c^5*(1 - 1/(a^2*x^2))^(5/2)) + (79*(a + x^(-1))^6)/(231*a^7*c^5*(1 - 1/(a^2*x^
2))^(7/2)) - (10*(a + x^(-1))^7)/(33*a^8*c^5*(1 - 1/(a^2*x^2))^(9/2)) + (a + x^(-1))^8/(11*a^9*c^5*(1 - 1/(a^2
*x^2))^(11/2))

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Rubi [A]  time = 0.389062, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 7, 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} $\frac{\left (a+\frac{1}{x}\right )^8}{11 a^9 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{11/2}}-\frac{10 \left (a+\frac{1}{x}\right )^7}{33 a^8 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{9/2}}+\frac{79 \left (a+\frac{1}{x}\right )^6}{231 a^7 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}-\frac{152 \left (a+\frac{1}{x}\right )^5}{1155 a^6 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(3*ArcCoth[a*x])/(c - a*c*x)^5,x]

[Out]

(-152*(a + x^(-1))^5)/(1155*a^6*c^5*(1 - 1/(a^2*x^2))^(5/2)) + (79*(a + x^(-1))^6)/(231*a^7*c^5*(1 - 1/(a^2*x^
2))^(7/2)) - (10*(a + x^(-1))^7)/(33*a^8*c^5*(1 - 1/(a^2*x^2))^(9/2)) + (a + x^(-1))^8/(11*a^9*c^5*(1 - 1/(a^2
*x^2))^(11/2))

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)^5} \, dx &=-\frac{\int \frac{e^{3 \coth ^{-1}(a x)}}{\left (1-\frac{1}{a x}\right )^5 x^5} \, dx}{a^5 c^5}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (1-\frac{x^2}{a^2}\right )^{3/2}}{\left (1-\frac{x}{a}\right )^8} \, dx,x,\frac{1}{x}\right )}{a^5 c^5}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (1+\frac{x}{a}\right )^8}{\left (1-\frac{x^2}{a^2}\right )^{13/2}} \, dx,x,\frac{1}{x}\right )}{a^5 c^5}\\ &=\frac{\left (a+\frac{1}{x}\right )^8}{11 a^9 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{11/2}}-\frac{\operatorname{Subst}\left (\int \frac{\left (1+\frac{x}{a}\right )^7 \left (8 a^3+11 a^2 x+11 a x^2\right )}{\left (1-\frac{x^2}{a^2}\right )^{11/2}} \, dx,x,\frac{1}{x}\right )}{11 a^5 c^5}\\ &=-\frac{10 \left (a+\frac{1}{x}\right )^7}{33 a^8 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{9/2}}+\frac{\left (a+\frac{1}{x}\right )^8}{11 a^9 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{11/2}}+\frac{\operatorname{Subst}\left (\int \frac{\left (1+\frac{x}{a}\right )^6 \left (138 a^3+99 a^2 x\right )}{\left (1-\frac{x^2}{a^2}\right )^{9/2}} \, dx,x,\frac{1}{x}\right )}{99 a^5 c^5}\\ &=\frac{79 \left (a+\frac{1}{x}\right )^6}{231 a^7 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}-\frac{10 \left (a+\frac{1}{x}\right )^7}{33 a^8 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{9/2}}+\frac{\left (a+\frac{1}{x}\right )^8}{11 a^9 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{11/2}}-\frac{152 \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 )}{231 a^2 c^5}\\ &=-\frac{152 \left (a+\frac{1}{x}\right )^5}{1155 a^6 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}+\frac{79 \left (a+\frac{1}{x}\right )^6}{231 a^7 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{7/2}}-\frac{10 \left (a+\frac{1}{x}\right )^7}{33 a^8 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{9/2}}+\frac{\left (a+\frac{1}{x}\right )^8}{11 a^9 c^5 \left (1-\frac{1}{a^2 x^2}\right )^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.0680123, size = 58, normalized size = 0.46 $-\frac{x \sqrt{1-\frac{1}{a^2 x^2}} (a x+1)^2 \left (2 a^3 x^3-16 a^2 x^2+61 a x-152\right )}{1155 c^5 (a x-1)^6}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(3*ArcCoth[a*x])/(c - a*c*x)^5,x]

[Out]

-(Sqrt[1 - 1/(a^2*x^2)]*x*(1 + a*x)^2*(-152 + 61*a*x - 16*a^2*x^2 + 2*a^3*x^3))/(1155*c^5*(-1 + a*x)^6)

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Maple [A]  time = 0.125, size = 58, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,{x}^{3}{a}^{3}-16\,{a}^{2}{x}^{2}+61\,ax-152 \right ) \left ( ax+1 \right ) }{1155\,{c}^{5} \left ( ax-1 \right ) ^{4}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)^5,x)

[Out]

-1/1155*(2*a^3*x^3-16*a^2*x^2+61*a*x-152)*(a*x+1)/(a*x-1)^4/c^5/((a*x-1)/(a*x+1))^(3/2)/a

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Maxima [A]  time = 1.03058, size = 96, normalized size = 0.77 \begin{align*} -\frac{\frac{385 \,{\left (a x - 1\right )}}{a x + 1} - \frac{495 \,{\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac{231 \,{\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 105}{9240 \, a c^{5} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{11}{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)^5,x, algorithm="maxima")

[Out]

-1/9240*(385*(a*x - 1)/(a*x + 1) - 495*(a*x - 1)^2/(a*x + 1)^2 + 231*(a*x - 1)^3/(a*x + 1)^3 - 105)/(a*c^5*((a
*x - 1)/(a*x + 1))^(11/2))

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Fricas [A]  time = 1.54921, size = 296, normalized size = 2.37 \begin{align*} -\frac{{\left (2 \, a^{6} x^{6} - 10 \, a^{5} x^{5} + 19 \, a^{4} x^{4} - 15 \, a^{3} x^{3} - 289 \, a^{2} x^{2} - 395 \, a x - 152\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{1155 \,{\left (a^{7} c^{5} x^{6} - 6 \, a^{6} c^{5} x^{5} + 15 \, a^{5} c^{5} x^{4} - 20 \, a^{4} c^{5} x^{3} + 15 \, a^{3} c^{5} x^{2} - 6 \, a^{2} c^{5} x + a c^{5}\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)^5,x, algorithm="fricas")

[Out]

-1/1155*(2*a^6*x^6 - 10*a^5*x^5 + 19*a^4*x^4 - 15*a^3*x^3 - 289*a^2*x^2 - 395*a*x - 152)*sqrt((a*x - 1)/(a*x +
1))/(a^7*c^5*x^6 - 6*a^6*c^5*x^5 + 15*a^5*c^5*x^4 - 20*a^4*c^5*x^3 + 15*a^3*c^5*x^2 - 6*a^2*c^5*x + a*c^5)

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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)**5,x)

[Out]

Timed out

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Giac [A]  time = 1.53506, size = 115, normalized size = 0.92 \begin{align*} -\frac{{\left (a x + 1\right )}^{5}{\left (\frac{385 \,{\left (a x - 1\right )}}{a x + 1} - \frac{495 \,{\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac{231 \,{\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 105\right )}}{9240 \,{\left (a x - 1\right )}^{5} a c^{5} \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)^5,x, algorithm="giac")

[Out]

-1/9240*(a*x + 1)^5*(385*(a*x - 1)/(a*x + 1) - 495*(a*x - 1)^2/(a*x + 1)^2 + 231*(a*x - 1)^3/(a*x + 1)^3 - 105
)/((a*x - 1)^5*a*c^5*sqrt((a*x - 1)/(a*x + 1)))