\(\int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx\) [190]

Optimal result

Integrand size = 18, antiderivative size = 100 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {a^6 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{9 c^4 \left (a-\frac {1}{x}\right )^7}+\frac {16 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{63 c^4 \left (a-\frac {1}{x}\right )^6}-\frac {47 a^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{315 c^4 \left (a-\frac {1}{x}\right )^5} \] Output:

-1/9*a^6*(1-1/a^2/x^2)^(5/2)/c^4/(a-1/x)^7+16/63*a^5*(1-1/a^2/x^2)^(5/2)/c 
^4/(a-1/x)^6-47/315*a^4*(1-1/a^2/x^2)^(5/2)/c^4/(a-1/x)^5
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.50 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x (1+a x)^2 \left (47-14 a x+2 a^2 x^2\right )}{315 c^4 (-1+a x)^5} \] Input:

Integrate[E^(3*ArcCoth[a*x])/(c - a*c*x)^4,x]
 

Output:

-1/315*(Sqrt[1 - 1/(a^2*x^2)]*x*(1 + a*x)^2*(47 - 14*a*x + 2*a^2*x^2))/(c^ 
4*(-1 + a*x)^5)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6724, 25, 27, 570, 529, 27, 669, 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.

\(\displaystyle \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx\)

\(\Big \downarrow \) 6724

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 570

\(\displaystyle -\frac {\int \frac {\left (a+\frac {1}{x}\right )^7}{\left (1-\frac {1}{a^2 x^2}\right )^{11/2} x^2}d\frac {1}{x}}{a^{11} c^4}\)

\(\Big \downarrow \) 529

\(\displaystyle -\frac {\frac {a^3 \left (a+\frac {1}{x}\right )^7}{9 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}-\frac {1}{9} a \int \frac {a \left (a+\frac {1}{x}\right )^6 \left (7 a+\frac {9}{x}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{9/2}}d\frac {1}{x}}{a^{11} c^4}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {a^3 \left (a+\frac {1}{x}\right )^7}{9 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}-\frac {1}{9} a^2 \int \frac {\left (a+\frac {1}{x}\right )^6 \left (7 a+\frac {9}{x}\right )}{\left (1-\frac {1}{a^2 x^2}\right )^{9/2}}d\frac {1}{x}}{a^{11} c^4}\)

\(\Big \downarrow \) 669

\(\displaystyle -\frac {\frac {a^3 \left (a+\frac {1}{x}\right )^7}{9 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}-\frac {1}{9} a^2 \left (\frac {16 a^2 \left (a+\frac {1}{x}\right )^6}{7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}-\frac {47}{7} a^2 \int \frac {\left (a+\frac {1}{x}\right )^5}{\left (1-\frac {1}{a^2 x^2}\right )^{7/2}}d\frac {1}{x}\right )}{a^{11} c^4}\)

\(\Big \downarrow \) 460

\(\displaystyle -\frac {\frac {a^3 \left (a+\frac {1}{x}\right )^7}{9 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}-\frac {1}{9} a^2 \left (\frac {16 a^2 \left (a+\frac {1}{x}\right )^6}{7 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}-\frac {47 a^3 \left (a+\frac {1}{x}\right )^5}{35 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}\right )}{a^{11} c^4}\)

Input:

Int[E^(3*ArcCoth[a*x])/(c - a*c*x)^4,x]
 

Output:

-((-1/9*(a^2*((-47*a^3*(a + x^(-1))^5)/(35*(1 - 1/(a^2*x^2))^(5/2)) + (16* 
a^2*(a + x^(-1))^6)/(7*(1 - 1/(a^2*x^2))^(7/2)))) + (a^3*(a + x^(-1))^7)/( 
9*(1 - 1/(a^2*x^2))^(9/2)))/(a^11*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 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 529

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem 
ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ 
(2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b* 
x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; 
 FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* 
c^2 + a*d^2, 0]
 

rule 570

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

rule 669

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* 
(p + 1))), x] - Simp[e*((m*(d*g + e*f) + 2*e*f*(p + 1))/(2*c*d*(p + 1))) 
Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, 
g}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 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 = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.50

method	result	size
gosper	\(-\frac {\left (2 a^{2} x^{2}-14 a x +47\right ) \left (a x +1\right )}{315 \left (a x -1\right )^{3} c^{4} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\)	\(50\)
default	\(-\frac {\left (2 a^{2} x^{2}-14 a x +47\right ) \left (a x +1\right )}{315 \left (a x -1\right )^{3} c^{4} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\)	\(50\)
orering	\(-\frac {\left (2 a^{2} x^{2}-14 a x +47\right ) \left (a x -1\right ) \left (a x +1\right )}{315 a \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (-a c x +c \right )^{4}}\)	\(54\)
trager	\(-\frac {\left (a x +1\right ) \left (2 a^{4} x^{4}-10 a^{3} x^{3}+21 a^{2} x^{2}+80 a x +47\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{315 a \,c^{4} \left (a x -1\right )^{5}}\)	\(68\)

Input:

int(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^4,x,method=_RETURNVERBOSE)
 

Output:

-1/315*(2*a^2*x^2-14*a*x+47)*(a*x+1)/(a*x-1)^3/c^4/((a*x-1)/(a*x+1))^(3/2) 
/a
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.16 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {{\left (2 \, a^{5} x^{5} - 8 \, a^{4} x^{4} + 11 \, a^{3} x^{3} + 101 \, a^{2} x^{2} + 127 \, a x + 47\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{315 \, {\left (a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} + 10 \, a^{4} c^{4} x^{3} - 10 \, a^{3} c^{4} x^{2} + 5 \, a^{2} c^{4} x - a c^{4}\right )}} \] Input:

integrate(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^4,x, algorithm="fricas")
 

Output:

-1/315*(2*a^5*x^5 - 8*a^4*x^4 + 11*a^3*x^3 + 101*a^2*x^2 + 127*a*x + 47)*s 
qrt((a*x - 1)/(a*x + 1))/(a^6*c^4*x^5 - 5*a^5*c^4*x^4 + 10*a^4*c^4*x^3 - 1 
0*a^3*c^4*x^2 + 5*a^2*c^4*x - a*c^4)
 

Sympy [F]

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

integrate(1/((a*x-1)/(a*x+1))**(3/2)/(-a*c*x+c)**4,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.55 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {\frac {90 \, {\left (a x - 1\right )}}{a x + 1} - \frac {63 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 35}{1260 \, a c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{2}}} \] Input:

integrate(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^4,x, algorithm="maxima")
 

Output:

1/1260*(90*(a*x - 1)/(a*x + 1) - 63*(a*x - 1)^2/(a*x + 1)^2 - 35)/(a*c^4*( 
(a*x - 1)/(a*x + 1))^(9/2))
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.45 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {4 \, {\left (210 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{6} x^{6} + 315 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{5} x^{5} + 441 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{4} x^{4} + 126 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{3} x^{3} + 36 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{2} x^{2} - 9 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x + 1\right )}}{315 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{9} a c^{4}} \] Input:

integrate(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^4,x, algorithm="giac")
 

Output:

-4/315*(210*(a + sqrt(a^2 - 1/x^2))^6*x^6 + 315*(a + sqrt(a^2 - 1/x^2))^5* 
x^5 + 441*(a + sqrt(a^2 - 1/x^2))^4*x^4 + 126*(a + sqrt(a^2 - 1/x^2))^3*x^ 
3 + 36*(a + sqrt(a^2 - 1/x^2))^2*x^2 - 9*(a + sqrt(a^2 - 1/x^2))*x + 1)/(( 
(a + sqrt(a^2 - 1/x^2))*x - 1)^9*a*c^4)
 

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.56 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {\frac {{\left (a\,x-1\right )}^2}{5\,{\left (a\,x+1\right )}^2}-\frac {2\,\left (a\,x-1\right )}{7\,\left (a\,x+1\right )}+\frac {1}{9}}{4\,a\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/2}} \] Input:

int(1/((c - a*c*x)^4*((a*x - 1)/(a*x + 1))^(3/2)),x)
                                                                                    
                                                                                    
 

Output:

-((a*x - 1)^2/(5*(a*x + 1)^2) - (2*(a*x - 1))/(7*(a*x + 1)) + 1/9)/(4*a*c^ 
4*((a*x - 1)/(a*x + 1))^(9/2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.68 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {2 \sqrt {a x -1}\, a^{4} x^{4}-8 \sqrt {a x -1}\, a^{3} x^{3}+12 \sqrt {a x -1}\, a^{2} x^{2}-8 \sqrt {a x -1}\, a x +2 \sqrt {a x -1}-2 \sqrt {a x +1}\, a^{4} x^{4}+10 \sqrt {a x +1}\, a^{3} x^{3}-21 \sqrt {a x +1}\, a^{2} x^{2}-80 \sqrt {a x +1}\, a x -47 \sqrt {a x +1}}{315 \sqrt {a x -1}\, a \,c^{4} \left (a^{4} x^{4}-4 a^{3} x^{3}+6 a^{2} x^{2}-4 a x +1\right )} \] Input:

int(1/((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^4,x)
 

Output:

(2*sqrt(a*x - 1)*a**4*x**4 - 8*sqrt(a*x - 1)*a**3*x**3 + 12*sqrt(a*x - 1)* 
a**2*x**2 - 8*sqrt(a*x - 1)*a*x + 2*sqrt(a*x - 1) - 2*sqrt(a*x + 1)*a**4*x 
**4 + 10*sqrt(a*x + 1)*a**3*x**3 - 21*sqrt(a*x + 1)*a**2*x**2 - 80*sqrt(a* 
x + 1)*a*x - 47*sqrt(a*x + 1))/(315*sqrt(a*x - 1)*a*c**4*(a**4*x**4 - 4*a* 
*3*x**3 + 6*a**2*x**2 - 4*a*x + 1))