\(\int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx\) [290]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.37 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {e^{n \coth ^{-1}(a x)} \left (-12-8 n-n^2+(4+n)^2 \cosh \left (2 \coth ^{-1}(a x)\right )-2 (4+n) \sinh \left (2 \coth ^{-1}(a x)\right )\right ) \left (\cosh \left (4 \coth ^{-1}(a x)\right )+\sinh \left (4 \coth ^{-1}(a x)\right )\right )}{2 a c^4 (2+n) (4+n) (6+n)} \] Input:

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

Output:

-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))
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {6725, 101, 25, 27, 88, 55, 48}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 6725

\(\displaystyle -\frac {\int \frac {\left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-4} \left (1+\frac {1}{a x}\right )^{n/2}}{x^2}d\frac {1}{x}}{a^4 c^4}\)

\(\Big \downarrow \) 101

\(\displaystyle -\frac {a^2 \int -\frac {\left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-4} \left (1+\frac {1}{a x}\right )^{n/2} \left (a+\frac {n+4}{x}\right )}{a}d\frac {1}{x}+\frac {a^2 \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-3}}{x}}{a^4 c^4}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {a^2 \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-3} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{x}-a^2 \int \frac {\left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-4} \left (1+\frac {1}{a x}\right )^{n/2} \left (a+\frac {n+4}{x}\right )}{a}d\frac {1}{x}}{a^4 c^4}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {a^2 \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-3} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{x}-a \int \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-4} \left (1+\frac {1}{a x}\right )^{n/2} \left (a+\frac {n+4}{x}\right )d\frac {1}{x}}{a^4 c^4}\)

\(\Big \downarrow \) 88

\(\displaystyle -\frac {\frac {a^2 \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-3} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{x}-a \left (\frac {a^2 (n+5) \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-3} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{n+6}-\frac {a \left (n^2+8 n+14\right ) \int \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-3} \left (1+\frac {1}{a x}\right )^{n/2}d\frac {1}{x}}{n+6}\right )}{a^4 c^4}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {\frac {a^2 \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-3} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{x}-a \left (\frac {a^2 (n+5) \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-3} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{n+6}-\frac {a \left (n^2+8 n+14\right ) \left (\frac {\int \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-2} \left (1+\frac {1}{a x}\right )^{n/2}d\frac {1}{x}}{n+4}+\frac {a \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-2}}{n+4}\right )}{n+6}\right )}{a^4 c^4}\)

\(\Big \downarrow \) 48

\(\displaystyle -\frac {\frac {a^2 \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-3} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{x}-a \left (\frac {a^2 (n+5) \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-3} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}}}{n+6}-\frac {a \left (n^2+8 n+14\right ) \left (\frac {a \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-2}}{n+4}+\frac {a \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} \left (1-\frac {1}{a x}\right )^{-\frac {n}{2}-1}}{(n+2) (n+4)}\right )}{n+6}\right )}{a^4 c^4}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 48

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] && EqQ[m + n + 2, 0] && NeQ[m, -1]
 

rule 55

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] - Simp[d*(S 
implify[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] && 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] ||  !SumSimp 
lerQ[n, 1])
 

rule 88

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(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] && SumSimpl 
erQ[p, 1]
 

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[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 6725

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

Maple [A] (verified)

Time = 13.99 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.30

method	result	size
gosper	\(-\frac {\left (a x +1\right ) \left (2 a^{2} x^{2}-2 n a x -8 a x +n^{2}+8 n +14\right ) {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )}}{\left (a x -1\right )^{3} c^{4} a \left (n^{2}+8 n +12\right ) \left (4+n \right )}\)	\(68\)
orering	\(-\frac {\left (2 a^{2} x^{2}-2 n a x -8 a x +n^{2}+8 n +14\right ) \left (a x -1\right ) \left (a x +1\right ) {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )}}{a \left (n^{3}+12 n^{2}+44 n +48\right ) \left (-a c x +c \right )^{4}}\)	\(72\)
parallelrisch	\(\frac {-14 \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )}+2 x^{2} {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a^{2} n -x \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a \,n^{2}-2 a^{3} {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} x^{3}-8 \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} n -6 x \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a -6 x \,{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a n -{\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} n^{2}+6 x^{2} {\mathrm e}^{n \,\operatorname {arccoth}\left (a x \right )} a^{2}}{c^{4} \left (a x -1\right )^{3} a \left (n^{3}+12 n^{2}+44 n +48\right )}\)	\(145\)

Input:

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

Output:

-(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)

Time = 0.09 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.02 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {{\left (2 \, a^{3} x^{3} - 2 \, {\left (a^{2} n + 3 \, a^{2}\right )} x^{2} + n^{2} + {\left (a n^{2} + 6 \, a n + 6 \, a\right )} x + 8 \, n + 14\right )} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a c^{4} n^{3} + 12 \, a c^{4} n^{2} + 44 \, a c^{4} n + 48 \, a c^{4} - {\left (a^{4} c^{4} n^{3} + 12 \, a^{4} c^{4} n^{2} + 44 \, a^{4} c^{4} n + 48 \, a^{4} c^{4}\right )} x^{3} + 3 \, {\left (a^{3} c^{4} n^{3} + 12 \, a^{3} c^{4} n^{2} + 44 \, a^{3} c^{4} n + 48 \, a^{3} c^{4}\right )} x^{2} - 3 \, {\left (a^{2} c^{4} n^{3} + 12 \, a^{2} c^{4} n^{2} + 44 \, a^{2} c^{4} n + 48 \, a^{2} c^{4}\right )} x} \] Input:

integrate(exp(n*arccoth(a*x))/(-a*c*x+c)^4,x, algorithm="fricas")
 

Output:

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(exp(n*arccoth(a*x))/(-a*c*x+c)^4,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(exp(n*arccoth(a*x))/(-a*c*x+c)^4,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 13.95 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.80 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {{\left (\frac {a\,x+1}{a\,x}\right )}^{n/2}\,\left (\frac {2\,x^3}{a\,c^4\,\left (n^3+12\,n^2+44\,n+48\right )}+\frac {n^2+8\,n+14}{a^4\,c^4\,\left (n^3+12\,n^2+44\,n+48\right )}-\frac {x^2\,\left (2\,n+6\right )}{a^2\,c^4\,\left (n^3+12\,n^2+44\,n+48\right )}+\frac {x\,\left (n^2+6\,n+6\right )}{a^3\,c^4\,\left (n^3+12\,n^2+44\,n+48\right )}\right )}{{\left (\frac {a\,x-1}{a\,x}\right )}^{n/2}\,\left (\frac {3\,x}{a^2}-\frac {1}{a^3}+x^3-\frac {3\,x^2}{a}\right )} \] Input:

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

Output:

-(((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))
 

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.80 \[ \int \frac {e^{n \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {e^{\mathit {acoth} \left (a x \right ) n} \left (2 a^{3} x^{3}+2 a^{2} n \,x^{2}-6 a^{2} x^{2}+a \,n^{2} x -6 a n x +6 a x +n^{2}-8 n +14\right )}{a \,c^{4} \left (a^{3} n^{3} x^{3}-12 a^{3} n^{2} x^{3}+44 a^{3} n \,x^{3}-3 a^{2} n^{3} x^{2}-48 a^{3} x^{3}+36 a^{2} n^{2} x^{2}-132 a^{2} n \,x^{2}+3 a \,n^{3} x +144 a^{2} x^{2}-36 a \,n^{2} x +132 a n x -n^{3}-144 a x +12 n^{2}-44 n +48\right )} \] Input:

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

Output:

(e**(acoth(a*x)*n)*(2*a**3*x**3 + 2*a**2*n*x**2 - 6*a**2*x**2 + a*n**2*x - 
 6*a*n*x + 6*a*x + n**2 - 8*n + 14))/(a*c**4*(a**3*n**3*x**3 - 12*a**3*n** 
2*x**3 + 44*a**3*n*x**3 - 48*a**3*x**3 - 3*a**2*n**3*x**2 + 36*a**2*n**2*x 
**2 - 132*a**2*n*x**2 + 144*a**2*x**2 + 3*a*n**3*x - 36*a*n**2*x + 132*a*n 
*x - 144*a*x - n**3 + 12*n**2 - 44*n + 48))