\(\int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx\) [371]

Optimal result

Integrand size = 18, antiderivative size = 224 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {(5+n) \left (1-\frac {1}{a x}\right )^{-3-\frac {n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}}}{a c^4 (6+n)}-\frac {\left (14+8 n+n^2\right ) \left (1-\frac {1}{a x}\right )^{-2-\frac {n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}}}{a c^4 (4+n) (6+n)}-\frac {\left (14+8 n+n^2\right ) \left (1-\frac {1}{a x}\right )^{-1-\frac {n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}}}{a c^4 (6+n) \left (8+6 n+n^2\right )}-\frac {\left (1-\frac {1}{a x}\right )^{-3-\frac {n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {2+n}{2}}}{a^2 c^4 x} \]

[Out]

-(n^2+8*n+14)*(1-1/a/x)^(-2-1/2*n)*(1+1/a/x)^(1+1/2*n)/a/c^4/(n^2+10*n+24)-(n^2+8*n+14)*(1-1/a/x)^(-1-1/2*n)*(
1+1/a/x)^(1+1/2*n)/a/c^4/(n^3+12*n^2+44*n+48)+(5+n)*(1-1/a/x)^(-3-1/2*n)*(1+1/a/x)^(1+1/2*n)/a/c^4/(6+n)-(1-1/
a/x)^(-3-1/2*n)*(1+1/a/x)^(1+1/2*n)/a^2/c^4/x

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6310, 6315, 92, 80, 47, 37} \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {\left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-3}}{a^2 c^4 x}-\frac {\left (n^2+8 n+14\right ) \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-2}}{a c^4 (n+4) (n+6)}-\frac {\left (n^2+8 n+14\right ) \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-1}}{a c^4 (n+6) \left (n^2+6 n+8\right )}+\frac {(n+5) \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-3}}{a c^4 (n+6)} \]

[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 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 47

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 80

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 92

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 6310

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 6315

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*} \text {integral}& = \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 {\text {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 {\text {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 ) \text {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 ) \text {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] (verified)

Time = 0.41 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.37 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {e^{n \coth ^{-1}(a x)} \left (-12-8 n-n^2+(4+n)^2 \cosh \left (2 \coth ^{-1}(a x)\right )-2 (4+n) \sinh \left (2 \coth ^{-1}(a x)\right )\right ) \left (\cosh \left (4 \coth ^{-1}(a x)\right )+\sinh \left (4 \coth ^{-1}(a x)\right )\right )}{2 a c^4 (2+n) (4+n) (6+n)} \]

[In]

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

[Out]

-1/2*(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]]))/(a*c^4*(2 + n)*(4 + n)*(6 + n))

Maple [A] (verified)

Time = 12.14 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.30

method	result	size
gosper	\(-\frac {\left (a x +1\right ) \left (2 a^{2} x^{2}-2 a n x -8 a x +n^{2}+8 n +14\right ) {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )}}{\left (a x -1\right )^{3} c^{4} a \left (n^{2}+8 n +12\right ) \left (4+n \right )}\)	\(68\)
parallelrisch	\(\frac {2 x^{2} {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a^{2} n -x \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a \,n^{2}-6 x \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a -6 x \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a n +6 x^{2} {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a^{2}-8 \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} n -14 \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )}-{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} n^{2}-2 x^{3} {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a^{3}}{c^{4} \left (a x -1\right )^{3} a \left (n^{2}+8 n +12\right ) \left (4+n \right )}\)	\(145\)

[In]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.02 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {{\left (2 \, a^{3} x^{3} - 2 \, {\left (a^{2} n + 3 \, a^{2}\right )} x^{2} + n^{2} + {\left (a n^{2} + 6 \, a n + 6 \, a\right )} x + 8 \, n + 14\right )} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a c^{4} n^{3} + 12 \, a c^{4} n^{2} + 44 \, a c^{4} n + 48 \, a c^{4} - {\left (a^{4} c^{4} n^{3} + 12 \, a^{4} c^{4} n^{2} + 44 \, a^{4} c^{4} n + 48 \, a^{4} c^{4}\right )} x^{3} + 3 \, {\left (a^{3} c^{4} n^{3} + 12 \, a^{3} c^{4} n^{2} + 44 \, a^{3} c^{4} n + 48 \, a^{3} c^{4}\right )} x^{2} - 3 \, {\left (a^{2} c^{4} n^{3} + 12 \, a^{2} c^{4} n^{2} + 44 \, a^{2} c^{4} n + 48 \, a^{2} c^{4}\right )} x} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\text {Timed out} \]

[In]

integrate(exp(n*acoth(a*x))/(-a*c*x+c)**4,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\int { \frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (a c x - c\right )}^{4}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\int { \frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (a c x - c\right )}^{4}} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 4.70 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.80 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {{\left (\frac {a\,x+1}{a\,x}\right )}^{n/2}\,\left (\frac {2\,x^3}{a\,c^4\,\left (n^3+12\,n^2+44\,n+48\right )}+\frac {n^2+8\,n+14}{a^4\,c^4\,\left (n^3+12\,n^2+44\,n+48\right )}-\frac {x^2\,\left (2\,n+6\right )}{a^2\,c^4\,\left (n^3+12\,n^2+44\,n+48\right )}+\frac {x\,\left (n^2+6\,n+6\right )}{a^3\,c^4\,\left (n^3+12\,n^2+44\,n+48\right )}\right )}{{\left (\frac {a\,x-1}{a\,x}\right )}^{n/2}\,\left (\frac {3\,x}{a^2}-\frac {1}{a^3}+x^3-\frac {3\,x^2}{a}\right )} \]

[In]

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

[Out]

-(((a*x + 1)/(a*x))^(n/2)*((2*x^3)/(a*c^4*(44*n + 12*n^2 + n^3 + 48)) + (8*n + n^2 + 14)/(a^4*c^4*(44*n + 12*n
^2 + n^3 + 48)) - (x^2*(2*n + 6))/(a^2*c^4*(44*n + 12*n^2 + n^3 + 48)) + (x*(6*n + n^2 + 6))/(a^3*c^4*(44*n +
12*n^2 + n^3 + 48))))/(((a*x - 1)/(a*x))^(n/2)*((3*x)/a^2 - 1/a^3 + x^3 - (3*x^2)/a))