\(\int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^5} \, dx\) [208]

Optimal result

Integrand size = 18, antiderivative size = 128 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {a^3 \sqrt {1-\frac {1}{a^2 x^2}}}{7 c^5 \left (a-\frac {1}{x}\right )^4}-\frac {18 a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{35 c^5 \left (a-\frac {1}{x}\right )^3}+\frac {23 a \sqrt {1-\frac {1}{a^2 x^2}}}{35 c^5 \left (a-\frac {1}{x}\right )^2}-\frac {12 \sqrt {1-\frac {1}{a^2 x^2}}}{35 c^5 \left (a-\frac {1}{x}\right )} \] Output:

1/7*a^3*(1-1/a^2/x^2)^(1/2)/c^5/(a-1/x)^4-18/35*a^2*(1-1/a^2/x^2)^(1/2)/c^ 
5/(a-1/x)^3+23/35*a*(1-1/a^2/x^2)^(1/2)/c^5/(a-1/x)^2-12/35*(1-1/a^2/x^2)^ 
(1/2)/c^5/(a-1/x)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.40 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-12+13 a x-8 a^2 x^2+2 a^3 x^3\right )}{35 c^5 (-1+a x)^4} \] Input:

Integrate[1/(E^ArcCoth[a*x]*(c - a*c*x)^5),x]
 

Output:

-1/35*(Sqrt[1 - 1/(a^2*x^2)]*x*(-12 + 13*a*x - 8*a^2*x^2 + 2*a^3*x^3))/(c^ 
5*(-1 + a*x)^4)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.30, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6724, 27, 581, 25, 27, 671, 461, 461, 460}

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.

\(\displaystyle \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^5} \, dx\)

\(\Big \downarrow \) 6724

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {1}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )^4 x^3}d\frac {1}{x}}{a c^5}\)

\(\Big \downarrow \) 581

\(\displaystyle \frac {\int -\frac {a^2 \left (2 a-\frac {3}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )^4}d\frac {1}{x}+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^2}}{a c^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^2}-\int \frac {a^2 \left (2 a-\frac {3}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )^4}d\frac {1}{x}}{a c^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^2}-a^2 \int \frac {2 a-\frac {3}{x}}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )^4}d\frac {1}{x}}{a c^5}\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^2}-a^2 \left (\frac {18}{7} \int \frac {1}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )^3}d\frac {1}{x}-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{7 \left (a-\frac {1}{x}\right )^4}\right )}{a c^5}\)

\(\Big \downarrow \) 461

\(\displaystyle \frac {\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^2}-a^2 \left (\frac {18}{7} \left (\frac {2 \int \frac {1}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )^2}d\frac {1}{x}}{5 a}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{5 \left (a-\frac {1}{x}\right )^3}\right )-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{7 \left (a-\frac {1}{x}\right )^4}\right )}{a c^5}\)

\(\Big \downarrow \) 461

\(\displaystyle \frac {\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^2}-a^2 \left (\frac {18}{7} \left (\frac {2 \left (\frac {\int \frac {1}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a-\frac {1}{x}\right )}d\frac {1}{x}}{3 a}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{3 \left (a-\frac {1}{x}\right )^2}\right )}{5 a}+\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{5 \left (a-\frac {1}{x}\right )^3}\right )-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{7 \left (a-\frac {1}{x}\right )^4}\right )}{a c^5}\)

\(\Big \downarrow \) 460

\(\displaystyle \frac {\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{\left (a-\frac {1}{x}\right )^2}-a^2 \left (\frac {18}{7} \left (\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{5 \left (a-\frac {1}{x}\right )^3}+\frac {2 \left (\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{3 \left (a-\frac {1}{x}\right )^2}+\frac {\sqrt {1-\frac {1}{a^2 x^2}}}{3 \left (a-\frac {1}{x}\right )}\right )}{5 a}\right )-\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{7 \left (a-\frac {1}{x}\right )^4}\right )}{a c^5}\)

Input:

Int[1/(E^ArcCoth[a*x]*(c - a*c*x)^5),x]
 

Output:

(-(a^2*((18*((2*((a*Sqrt[1 - 1/(a^2*x^2)])/(3*(a - x^(-1))^2) + Sqrt[1 - 1 
/(a^2*x^2)]/(3*(a - x^(-1)))))/(5*a) + (a*Sqrt[1 - 1/(a^2*x^2)])/(5*(a - x 
^(-1))^3)))/7 - (a^2*Sqrt[1 - 1/(a^2*x^2)])/(7*(a - x^(-1))^4))) + (a^2*Sq 
rt[1 - 1/(a^2*x^2)])/(a - x^(-1))^2)/(a*c^5)
 

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 460

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

rule 461

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl 
ify[n + 2*p + 2]/(2*c*(n + p + 1))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x 
], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[Simp 
lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
 

rule 581

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

rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m 
 + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m 
+ p + 1))   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, 
e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p 
 + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p 
 + 1, 0]
 

rule 6724

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.45

method	result	size
gosper	\(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (2 a^{3} x^{3}-8 a^{2} x^{2}+13 a x -12\right ) \left (a x +1\right )}{35 \left (a x -1\right )^{4} c^{5} a}\)	\(58\)
default	\(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (2 a^{3} x^{3}-8 a^{2} x^{2}+13 a x -12\right ) \left (a x +1\right )}{35 \left (a x -1\right )^{4} c^{5} a}\)	\(58\)
trager	\(-\frac {\left (2 a^{3} x^{3}-8 a^{2} x^{2}+13 a x -12\right ) \left (a x +1\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{35 a \,c^{5} \left (a x -1\right )^{4}}\)	\(60\)
orering	\(\frac {\left (2 a^{3} x^{3}-8 a^{2} x^{2}+13 a x -12\right ) \left (a x -1\right ) \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}{35 a \left (-a c x +c \right )^{5}}\)	\(62\)

Input:

int(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^5,x,method=_RETURNVERBOSE)
 

Output:

-1/35*((a*x-1)/(a*x+1))^(1/2)*(2*a^3*x^3-8*a^2*x^2+13*a*x-12)*(a*x+1)/(a*x 
-1)^4/c^5/a
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.74 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {{\left (2 \, a^{4} x^{4} - 6 \, a^{3} x^{3} + 5 \, a^{2} x^{2} + a x - 12\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{35 \, {\left (a^{5} c^{5} x^{4} - 4 \, a^{4} c^{5} x^{3} + 6 \, a^{3} c^{5} x^{2} - 4 \, a^{2} c^{5} x + a c^{5}\right )}} \] Input:

integrate(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^5,x, algorithm="fricas")
 

Output:

-1/35*(2*a^4*x^4 - 6*a^3*x^3 + 5*a^2*x^2 + a*x - 12)*sqrt((a*x - 1)/(a*x + 
 1))/(a^5*c^5*x^4 - 4*a^4*c^5*x^3 + 6*a^3*c^5*x^2 - 4*a^2*c^5*x + a*c^5)
 

Sympy [F]

\[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=- \frac {\int \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{5} x^{5} - 5 a^{4} x^{4} + 10 a^{3} x^{3} - 10 a^{2} x^{2} + 5 a x - 1}\, dx}{c^{5}} \] Input:

integrate(((a*x-1)/(a*x+1))**(1/2)/(-a*c*x+c)**5,x)
 

Output:

-Integral(sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**5*x**5 - 5*a**4*x**4 + 10* 
a**3*x**3 - 10*a**2*x**2 + 5*a*x - 1), x)/c**5
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.55 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {\frac {21 \, {\left (a x - 1\right )}}{a x + 1} - \frac {35 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {35 \, {\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 5}{280 \, a c^{5} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}}} \] Input:

integrate(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^5,x, algorithm="maxima")
 

Output:

-1/280*(21*(a*x - 1)/(a*x + 1) - 35*(a*x - 1)^2/(a*x + 1)^2 + 35*(a*x - 1) 
^3/(a*x + 1)^3 - 5)/(a*c^5*((a*x - 1)/(a*x + 1))^(7/2))
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.66 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {4 \, {\left (35 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{3} x^{3} - 21 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{2} x^{2} + 7 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}}{35 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{7} a c^{5}} \] Input:

integrate(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^5,x, algorithm="giac")
 

Output:

4/35*(35*(a + sqrt(a^2 - 1/x^2))^3*x^3 - 21*(a + sqrt(a^2 - 1/x^2))^2*x^2 
+ 7*(a + sqrt(a^2 - 1/x^2))*x - 1)/(((a + sqrt(a^2 - 1/x^2))*x - 1)^7*a*c^ 
5)
 

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.55 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {\frac {{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {3\,\left (a\,x-1\right )}{5\,\left (a\,x+1\right )}+\frac {1}{7}}{8\,a\,c^5\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}} \] Input:

int(((a*x - 1)/(a*x + 1))^(1/2)/(c - a*c*x)^5,x)
                                                                                    
                                                                                    
 

Output:

((a*x - 1)^2/(a*x + 1)^2 - (a*x - 1)^3/(a*x + 1)^3 - (3*(a*x - 1))/(5*(a*x 
 + 1)) + 1/7)/(8*a*c^5*((a*x - 1)/(a*x + 1))^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {2 \sqrt {a x -1}\, a^{3} x^{3}-6 \sqrt {a x -1}\, a^{2} x^{2}+6 \sqrt {a x -1}\, a x -2 \sqrt {a x -1}-2 \sqrt {a x +1}\, a^{3} x^{3}+8 \sqrt {a x +1}\, a^{2} x^{2}-13 \sqrt {a x +1}\, a x +12 \sqrt {a x +1}}{35 \sqrt {a x -1}\, a \,c^{5} \left (a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:

int(((a*x-1)/(a*x+1))^(1/2)/(-a*c*x+c)^5,x)
 

Output:

(2*sqrt(a*x - 1)*a**3*x**3 - 6*sqrt(a*x - 1)*a**2*x**2 + 6*sqrt(a*x - 1)*a 
*x - 2*sqrt(a*x - 1) - 2*sqrt(a*x + 1)*a**3*x**3 + 8*sqrt(a*x + 1)*a**2*x* 
*2 - 13*sqrt(a*x + 1)*a*x + 12*sqrt(a*x + 1))/(35*sqrt(a*x - 1)*a*c**5*(a* 
*3*x**3 - 3*a**2*x**2 + 3*a*x - 1))