### 3.371 $$\int \frac{e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx$$

Optimal. Leaf size=224 $-\frac{\left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \left (1-\frac{1}{a x}\right )^{-\frac{n}{2}-3}}{a^2 c^4 x}-\frac{\left (n^2+8 n+14\right ) \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \left (1-\frac{1}{a x}\right )^{-\frac{n}{2}-2}}{a c^4 (n+4) (n+6)}-\frac{\left (n^2+8 n+14\right ) \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \left (1-\frac{1}{a x}\right )^{-\frac{n}{2}-1}}{a c^4 (n+6) \left (n^2+6 n+8\right )}+\frac{(n+5) \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \left (1-\frac{1}{a x}\right )^{-\frac{n}{2}-3}}{a c^4 (n+6)}$

[Out]

((5 + n)*(1 - 1/(a*x))^(-3 - n/2)*(1 + 1/(a*x))^((2 + n)/2))/(a*c^4*(6 + n)) - ((14 + 8*n + n^2)*(1 - 1/(a*x))
^(-2 - n/2)*(1 + 1/(a*x))^((2 + n)/2))/(a*c^4*(4 + n)*(6 + n)) - ((14 + 8*n + n^2)*(1 - 1/(a*x))^(-1 - n/2)*(1
+ 1/(a*x))^((2 + n)/2))/(a*c^4*(6 + n)*(8 + 6*n + n^2)) - ((1 - 1/(a*x))^(-3 - n/2)*(1 + 1/(a*x))^((2 + n)/2)
)/(a^2*c^4*x)

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Rubi [A]  time = 0.261659, antiderivative size = 224, 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, 6180, 90, 79, 45, 37} $-\frac{\left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \left (1-\frac{1}{a x}\right )^{-\frac{n}{2}-3}}{a^2 c^4 x}-\frac{\left (n^2+8 n+14\right ) \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \left (1-\frac{1}{a x}\right )^{-\frac{n}{2}-2}}{a c^4 (n+4) (n+6)}-\frac{\left (n^2+8 n+14\right ) \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \left (1-\frac{1}{a x}\right )^{-\frac{n}{2}-1}}{a c^4 (n+6) \left (n^2+6 n+8\right )}+\frac{(n+5) \left (\frac{1}{a x}+1\right )^{\frac{n+2}{2}} \left (1-\frac{1}{a x}\right )^{-\frac{n}{2}-3}}{a c^4 (n+6)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((5 + n)*(1 - 1/(a*x))^(-3 - n/2)*(1 + 1/(a*x))^((2 + n)/2))/(a*c^4*(6 + n)) - ((14 + 8*n + n^2)*(1 - 1/(a*x))
^(-2 - n/2)*(1 + 1/(a*x))^((2 + n)/2))/(a*c^4*(4 + n)*(6 + n)) - ((14 + 8*n + n^2)*(1 - 1/(a*x))^(-1 - n/2)*(1
+ 1/(a*x))^((2 + n)/2))/(a*c^4*(6 + n)*(8 + 6*n + n^2)) - ((1 - 1/(a*x))^(-3 - n/2)*(1 + 1/(a*x))^((2 + n)/2)
)/(a^2*c^4*x)

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 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

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 45

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] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 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)^4} \, dx &=\frac{\int \frac{e^{n \coth ^{-1}(a x)}}{\left (1-\frac{1}{a x}\right )^4 x^4} \, dx}{a^4 c^4}\\ &=-\frac{\operatorname{Subst}\left (\int x^2 \left (1-\frac{x}{a}\right )^{-4-\frac{n}{2}} \left (1+\frac{x}{a}\right )^{n/2} \, dx,x,\frac{1}{x}\right )}{a^4 c^4}\\ &=-\frac{\left (1-\frac{1}{a x}\right )^{-3-\frac{n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}}}{a^2 c^4 x}-\frac{\operatorname{Subst}\left (\int \left (1-\frac{x}{a}\right )^{-4-\frac{n}{2}} \left (1+\frac{x}{a}\right )^{n/2} \left (-1-\frac{(4+n) x}{a}\right ) \, dx,x,\frac{1}{x}\right )}{a^2 c^4}\\ &=\frac{(5+n) \left (1-\frac{1}{a x}\right )^{-3-\frac{n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}}}{a c^4 (6+n)}-\frac{\left (1-\frac{1}{a x}\right )^{-3-\frac{n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}}}{a^2 c^4 x}-\frac{\left (14+8 n+n^2\right ) \operatorname{Subst}\left (\int \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^2 c^4 (6+n)}\\ &=\frac{(5+n) \left (1-\frac{1}{a x}\right )^{-3-\frac{n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}}}{a c^4 (6+n)}-\frac{\left (14+8 n+n^2\right ) \left (1-\frac{1}{a x}\right )^{-2-\frac{n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}}}{a c^4 (4+n) (6+n)}-\frac{\left (1-\frac{1}{a x}\right )^{-3-\frac{n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}}}{a^2 c^4 x}-\frac{\left (14+8 n+n^2\right ) \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^4 (4+n) (6+n)}\\ &=\frac{(5+n) \left (1-\frac{1}{a x}\right )^{-3-\frac{n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}}}{a c^4 (6+n)}-\frac{\left (14+8 n+n^2\right ) \left (1-\frac{1}{a x}\right )^{-2-\frac{n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}}}{a c^4 (4+n) (6+n)}-\frac{\left (14+8 n+n^2\right ) \left (1-\frac{1}{a x}\right )^{-1-\frac{n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}}}{a c^4 (2+n) (4+n) (6+n)}-\frac{\left (1-\frac{1}{a x}\right )^{-3-\frac{n}{2}} \left (1+\frac{1}{a x}\right )^{\frac{2+n}{2}}}{a^2 c^4 x}\\ \end{align*}

Mathematica [A]  time = 0.274697, size = 83, normalized size = 0.37 $-\frac{e^{n \coth ^{-1}(a x)} \left (\cosh \left (4 \coth ^{-1}(a x)\right )+\sinh \left (4 \coth ^{-1}(a x)\right )\right ) \left ((n+4)^2 \cosh \left (2 \coth ^{-1}(a x)\right )-2 (n+4) \sinh \left (2 \coth ^{-1}(a x)\right )-n^2-8 n-12\right )}{2 a c^4 (n+2) (n+4) (n+6)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(E^(n*ArcCoth[a*x])*(-12 - 8*n - n^2 + (4 + n)^2*Cosh[2*ArcCoth[a*x]] - 2*(4 + n)*Sinh[2*ArcCoth[a*x]])*(Cosh
[4*ArcCoth[a*x]] + Sinh[4*ArcCoth[a*x]]))/(2*a*c^4*(2 + n)*(4 + n)*(6 + n))

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Maple [A]  time = 0.045, size = 68, normalized size = 0.3 \begin{align*} -{\frac{ \left ( 2\,{a}^{2}{x}^{2}-2\,anx-8\,ax+{n}^{2}+8\,n+14 \right ) \left ( ax+1 \right ){{\rm e}^{n{\rm arccoth} \left (ax\right )}}}{ \left ( ax-1 \right ) ^{3}{c}^{4}a \left ({n}^{2}+8\,n+12 \right ) \left ( 4+n \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arccoth(a*x))/(-a*c*x+c)^4,x)

[Out]

-(a*x+1)*(2*a^2*x^2-2*a*n*x-8*a*x+n^2+8*n+14)*exp(n*arccoth(a*x))/(a*x-1)^3/c^4/a/(n^2+8*n+12)/(4+n)

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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 )}^{4}}\,{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)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.79254, size = 485, normalized size = 2.17 \begin{align*} -\frac{{\left (2 \, a^{3} x^{3} + 2 \,{\left (a^{2} n - 3 \, a^{2}\right )} x^{2} + n^{2} +{\left (a n^{2} - 6 \, a n + 6 \, a\right )} x - 8 \, n + 14\right )} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{2} \, n}}{a c^{4} n^{3} - 12 \, a c^{4} n^{2} + 44 \, a c^{4} n - 48 \, a c^{4} -{\left (a^{4} c^{4} n^{3} - 12 \, a^{4} c^{4} n^{2} + 44 \, a^{4} c^{4} n - 48 \, a^{4} c^{4}\right )} x^{3} + 3 \,{\left (a^{3} c^{4} n^{3} - 12 \, a^{3} c^{4} n^{2} + 44 \, a^{3} c^{4} n - 48 \, a^{3} c^{4}\right )} x^{2} - 3 \,{\left (a^{2} c^{4} n^{3} - 12 \, a^{2} c^{4} n^{2} + 44 \, a^{2} c^{4} n - 48 \, a^{2} c^{4}\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)^4,x, algorithm="fricas")

[Out]

-(2*a^3*x^3 + 2*(a^2*n - 3*a^2)*x^2 + n^2 + (a*n^2 - 6*a*n + 6*a)*x - 8*n + 14)*((a*x - 1)/(a*x + 1))^(1/2*n)/
(a*c^4*n^3 - 12*a*c^4*n^2 + 44*a*c^4*n - 48*a*c^4 - (a^4*c^4*n^3 - 12*a^4*c^4*n^2 + 44*a^4*c^4*n - 48*a^4*c^4)
*x^3 + 3*(a^3*c^4*n^3 - 12*a^3*c^4*n^2 + 44*a^3*c^4*n - 48*a^3*c^4)*x^2 - 3*(a^2*c^4*n^3 - 12*a^2*c^4*n^2 + 44
*a^2*c^4*n - 48*a^2*c^4)*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)**4,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 )}^{4}}\,{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)^4,x, algorithm="giac")

[Out]

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