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]

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

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Rubi [A]  time = 0.15, antiderivative size = 104, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.25, 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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fricas [A]  time = 0.54, size = 125, normalized size = 1.20 \[ -\frac {{\left (a^{2} x^{2} + {\left (a n - 2 \, a\right )} x + n - 3\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{a c^{3} n^{2} - 6 \, a c^{3} n + 8 \, a c^{3} + {\left (a^{3} c^{3} n^{2} - 6 \, a^{3} c^{3} n + 8 \, a^{3} c^{3}\right )} x^{2} - 2 \, {\left (a^{2} c^{3} n^{2} - 6 \, a^{2} c^{3} n + 8 \, a^{2} c^{3}\right )} x} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{{\left (a c x - c\right )}^{3}}\,{d x} \]

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)

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maple [A]  time = 0.04, size = 46, normalized size = 0.44 \[ -\frac {{\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )} \left (a x -n -3\right ) \left (a x +1\right )}{\left (a x -1\right )^{2} c^{3} \left (n^{2}+6 n +8\right ) a} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{{\left (a c x - c\right )}^{3}}\,{d x} \]

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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mupad [B]  time = 1.65, size = 113, normalized size = 1.09 \[ \frac {{\left (\frac {a\,x+1}{a\,x}\right )}^{n/2}\,\left (\frac {n+3}{a^3\,c^3\,\left (n^2+6\,n+8\right )}-\frac {x^2}{a\,c^3\,\left (n^2+6\,n+8\right )}+\frac {x\,\left (n+2\right )}{a^2\,c^3\,\left (n^2+6\,n+8\right )}\right )}{{\left (\frac {a\,x-1}{a\,x}\right )}^{n/2}\,\left (\frac {1}{a^2}-\frac {2\,x}{a}+x^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*Integral(exp(n*acoth(a*x)), x), Eq(c, 0)), (a**2*x**2*acoth(a*x)/(2*a**3*c**3*x**2*exp(4*acoth(
a*x)) - 4*a**2*c**3*x*exp(4*acoth(a*x)) + 2*a*c**3*exp(4*acoth(a*x))) + 2*a*x*acoth(a*x)/(2*a**3*c**3*x**2*exp
(4*acoth(a*x)) - 4*a**2*c**3*x*exp(4*acoth(a*x)) + 2*a*c**3*exp(4*acoth(a*x))) - a*x/(2*a**3*c**3*x**2*exp(4*a
coth(a*x)) - 4*a**2*c**3*x*exp(4*acoth(a*x)) + 2*a*c**3*exp(4*acoth(a*x))) + acoth(a*x)/(2*a**3*c**3*x**2*exp(
4*acoth(a*x)) - 4*a**2*c**3*x*exp(4*acoth(a*x)) + 2*a*c**3*exp(4*acoth(a*x))) - 1/(2*a**3*c**3*x**2*exp(4*acot
h(a*x)) - 4*a**2*c**3*x*exp(4*acoth(a*x)) + 2*a*c**3*exp(4*acoth(a*x))), Eq(n, -4)), (-a**2*x**2*acoth(a*x)/(2
*a**3*c**3*x**2*exp(2*acoth(a*x)) - 4*a**2*c**3*x*exp(2*acoth(a*x)) + 2*a*c**3*exp(2*acoth(a*x))) + a*x/(2*a**
3*c**3*x**2*exp(2*acoth(a*x)) - 4*a**2*c**3*x*exp(2*acoth(a*x)) + 2*a*c**3*exp(2*acoth(a*x))) + acoth(a*x)/(2*
a**3*c**3*x**2*exp(2*acoth(a*x)) - 4*a**2*c**3*x*exp(2*acoth(a*x)) + 2*a*c**3*exp(2*acoth(a*x))) + 1/(2*a**3*c
**3*x**2*exp(2*acoth(a*x)) - 4*a**2*c**3*x*exp(2*acoth(a*x)) + 2*a*c**3*exp(2*acoth(a*x))), Eq(n, -2)), (-a**2
*x**2*exp(n*acoth(a*x))/(a**3*c**3*n**2*x**2 + 6*a**3*c**3*n*x**2 + 8*a**3*c**3*x**2 - 2*a**2*c**3*n**2*x - 12
*a**2*c**3*n*x - 16*a**2*c**3*x + a*c**3*n**2 + 6*a*c**3*n + 8*a*c**3) + a*n*x*exp(n*acoth(a*x))/(a**3*c**3*n*
*2*x**2 + 6*a**3*c**3*n*x**2 + 8*a**3*c**3*x**2 - 2*a**2*c**3*n**2*x - 12*a**2*c**3*n*x - 16*a**2*c**3*x + a*c
**3*n**2 + 6*a*c**3*n + 8*a*c**3) + 2*a*x*exp(n*acoth(a*x))/(a**3*c**3*n**2*x**2 + 6*a**3*c**3*n*x**2 + 8*a**3
*c**3*x**2 - 2*a**2*c**3*n**2*x - 12*a**2*c**3*n*x - 16*a**2*c**3*x + a*c**3*n**2 + 6*a*c**3*n + 8*a*c**3) + n
*exp(n*acoth(a*x))/(a**3*c**3*n**2*x**2 + 6*a**3*c**3*n*x**2 + 8*a**3*c**3*x**2 - 2*a**2*c**3*n**2*x - 12*a**2
*c**3*n*x - 16*a**2*c**3*x + a*c**3*n**2 + 6*a*c**3*n + 8*a*c**3) + 3*exp(n*acoth(a*x))/(a**3*c**3*n**2*x**2 +
6*a**3*c**3*n*x**2 + 8*a**3*c**3*x**2 - 2*a**2*c**3*n**2*x - 12*a**2*c**3*n*x - 16*a**2*c**3*x + a*c**3*n**2
+ 6*a*c**3*n + 8*a*c**3), True))

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