### 3.370 $$\int \frac{e^{n \coth ^{-1}(a x)}}{(c-a c x)^3} \, dx$$

Optimal. Leaf size=104 $\frac{\left (1-\frac{1}{a x}\right )^{-\frac{n}{2}-2} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}}}{a c^3 (n+4)}-\frac{(n+3) \left (1-\frac{1}{a x}\right )^{-\frac{n}{2}-1} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}}}{a c^3 (n+2) (n+4)}$

[Out]

((1 - 1/(a*x))^(-2 - n/2)*(1 + 1/(a*x))^((2 + n)/2))/(a*c^3*(4 + n)) - ((3 + n)*(1 - 1/(a*x))^(-1 - n/2)*(1 +
1/(a*x))^((2 + n)/2))/(a*c^3*(2 + n)*(4 + n))

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Rubi [A]  time = 0.149331, antiderivative size = 104, 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, 6180, 79, 37} $\frac{\left (1-\frac{1}{a x}\right )^{-\frac{n}{2}-2} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}}}{a c^3 (n+4)}-\frac{(n+3) \left (1-\frac{1}{a x}\right )^{-\frac{n}{2}-1} \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}}}{a c^3 (n+2) (n+4)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((1 - 1/(a*x))^(-2 - n/2)*(1 + 1/(a*x))^((2 + n)/2))/(a*c^3*(4 + n)) - ((3 + n)*(1 - 1/(a*x))^(-1 - n/2)*(1 +
1/(a*x))^((2 + n)/2))/(a*c^3*(2 + n)*(4 + n))

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 6180

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^p, Subst[Int[((1
+ (d*x)/c)^p*(1 + x/a)^(n/2))/(x^(m + 2)*(1 - x/a)^(n/2)), x], x, 1/x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ
[c^2 - a^2*d^2, 0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegerQ[m]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.216626, size = 64, normalized size = 0.62 $\frac{(-a x+n+3) e^{n \coth ^{-1}(a x)} \left (\cosh \left (3 \coth ^{-1}(a x)\right )+\sinh \left (3 \coth ^{-1}(a x)\right )\right )}{a^2 c^3 (n+2) (n+4) x \sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(E^(n*ArcCoth[a*x])*(3 + n - a*x)*(Cosh[3*ArcCoth[a*x]] + Sinh[3*ArcCoth[a*x]]))/(a^2*c^3*(2 + n)*(4 + n)*Sqrt
[1 - 1/(a^2*x^2)]*x)

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Maple [A]  time = 0.049, size = 46, normalized size = 0.4 \begin{align*} -{\frac{{{\rm e}^{n{\rm arccoth} \left (ax\right )}} \left ( ax-n-3 \right ) \left ( ax+1 \right ) }{ \left ( ax-1 \right ) ^{2}{c}^{3} \left ({n}^{2}+6\,n+8 \right ) a}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arccoth(a*x))/(-a*c*x+c)^3,x)

[Out]

-exp(n*arccoth(a*x))*(a*x-n-3)*(a*x+1)/(a*x-1)^2/c^3/(n^2+6*n+8)/a

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n}}{{\left (a c x - c\right )}^{3}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))/(-a*c*x+c)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.62983, size = 259, normalized size = 2.49 \begin{align*} -\frac{{\left (a^{2} x^{2} +{\left (a n - 2 \, a\right )} x + n - 3\right )} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n}}{a c^{3} n^{2} - 6 \, a c^{3} n + 8 \, a c^{3} +{\left (a^{3} c^{3} n^{2} - 6 \, a^{3} c^{3} n + 8 \, a^{3} c^{3}\right )} x^{2} - 2 \,{\left (a^{2} c^{3} n^{2} - 6 \, a^{2} c^{3} n + 8 \, a^{2} c^{3}\right )} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))/(-a*c*x+c)^3,x, algorithm="fricas")

[Out]

-(a^2*x^2 + (a*n - 2*a)*x + n - 3)*((a*x - 1)/(a*x + 1))^(1/2*n)/(a*c^3*n^2 - 6*a*c^3*n + 8*a*c^3 + (a^3*c^3*n
^2 - 6*a^3*c^3*n + 8*a^3*c^3)*x^2 - 2*(a^2*c^3*n^2 - 6*a^2*c^3*n + 8*a^2*c^3)*x)

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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(exp(n*acoth(a*x))/(-a*c*x+c)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n}}{{\left (a c x - c\right )}^{3}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))/(-a*c*x+c)^3,x, algorithm="giac")

[Out]

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