\(\int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^6} \, dx\) [218]

Optimal result

Integrand size = 18, antiderivative size = 119 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^6} \, dx=\frac {8 x}{35 c^6 \sqrt {1-a^2 x^2}}+\frac {1}{7 a c^6 (1-a x)^3 \sqrt {1-a^2 x^2}}+\frac {4}{35 a c^6 (1-a x)^2 \sqrt {1-a^2 x^2}}+\frac {4}{35 a c^6 (1-a x) \sqrt {1-a^2 x^2}} \] Output:

8/35*x/c^6/(-a^2*x^2+1)^(1/2)+1/7/a/c^6/(-a*x+1)^3/(-a^2*x^2+1)^(1/2)+4/35 
/a/c^6/(-a*x+1)^2/(-a^2*x^2+1)^(1/2)+4/35/a/c^6/(-a*x+1)/(-a^2*x^2+1)^(1/2 
)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.51 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^6} \, dx=\frac {-13+4 a x+20 a^2 x^2-24 a^3 x^3+8 a^4 x^4}{35 a c^6 (-1+a x)^3 \sqrt {1-a^2 x^2}} \] Input:

Integrate[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^6),x]
 

Output:

(-13 + 4*a*x + 20*a^2*x^2 - 24*a^3*x^3 + 8*a^4*x^4)/(35*a*c^6*(-1 + a*x)^3 
*Sqrt[1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6677, 27, 461, 461, 470, 208}

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.

\(\displaystyle \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^6} \, dx\)

\(\Big \downarrow \) 6677

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {1}{(1-a x)^3 \left (1-a^2 x^2\right )^{3/2}}dx}{c^6}\)

\(\Big \downarrow \) 461

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

\(\Big \downarrow \) 461

\(\displaystyle \frac {\frac {4}{7} \left (\frac {3}{5} \int \frac {1}{(1-a x) \left (1-a^2 x^2\right )^{3/2}}dx+\frac {1}{5 a (1-a x)^2 \sqrt {1-a^2 x^2}}\right )+\frac {1}{7 a (1-a x)^3 \sqrt {1-a^2 x^2}}}{c^6}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {\frac {4}{7} \left (\frac {3}{5} \left (\frac {2}{3} \int \frac {1}{\left (1-a^2 x^2\right )^{3/2}}dx+\frac {1}{3 a (1-a x) \sqrt {1-a^2 x^2}}\right )+\frac {1}{5 a (1-a x)^2 \sqrt {1-a^2 x^2}}\right )+\frac {1}{7 a (1-a x)^3 \sqrt {1-a^2 x^2}}}{c^6}\)

\(\Big \downarrow \) 208

\(\displaystyle \frac {\frac {4}{7} \left (\frac {3}{5} \left (\frac {2 x}{3 \sqrt {1-a^2 x^2}}+\frac {1}{3 a (1-a x) \sqrt {1-a^2 x^2}}\right )+\frac {1}{5 a (1-a x)^2 \sqrt {1-a^2 x^2}}\right )+\frac {1}{7 a (1-a x)^3 \sqrt {1-a^2 x^2}}}{c^6}\)

Input:

Int[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^6),x]
 

Output:

(1/(7*a*(1 - a*x)^3*Sqrt[1 - a^2*x^2]) + (4*(1/(5*a*(1 - a*x)^2*Sqrt[1 - a 
^2*x^2]) + (3*((2*x)/(3*Sqrt[1 - a^2*x^2]) + 1/(3*a*(1 - a*x)*Sqrt[1 - a^2 
*x^2])))/5))/7)/c^6
 

Defintions of rubi rules used

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 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

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 470

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[(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, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + 
 p + 1, 0] && IntegerQ[2*p]
 

rule 6677

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

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.55

method	result	size
gosper	\(\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} \left (8 a^{4} x^{4}-24 a^{3} x^{3}+20 a^{2} x^{2}+4 a x -13\right )}{35 \left (a x -1\right )^{5} c^{6} a \left (a x +1\right )^{2}}\)	\(65\)
trager	\(-\frac {\left (8 a^{4} x^{4}-24 a^{3} x^{3}+20 a^{2} x^{2}+4 a x -13\right ) \sqrt {-a^{2} x^{2}+1}}{35 c^{6} \left (a x -1\right )^{4} a \left (a x +1\right )}\)	\(65\)
orering	\(\frac {\left (8 a^{4} x^{4}-24 a^{3} x^{3}+20 a^{2} x^{2}+4 a x -13\right ) \left (a x -1\right ) \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{35 a \left (a x +1\right )^{2} \left (-a c x +c \right )^{6}}\)	\(69\)
default	\(\text {Expression too large to display}\)	\(1373\)

Input:

int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^6,x,method=_RETURNVERBOSE)
 

Output:

1/35*(-a^2*x^2+1)^(3/2)*(8*a^4*x^4-24*a^3*x^3+20*a^2*x^2+4*a*x-13)/(a*x-1) 
^5/c^6/a/(a*x+1)^2
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^6} \, dx=\frac {13 \, a^{5} x^{5} - 39 \, a^{4} x^{4} + 26 \, a^{3} x^{3} + 26 \, a^{2} x^{2} - 39 \, a x - {\left (8 \, a^{4} x^{4} - 24 \, a^{3} x^{3} + 20 \, a^{2} x^{2} + 4 \, a x - 13\right )} \sqrt {-a^{2} x^{2} + 1} + 13}{35 \, {\left (a^{6} c^{6} x^{5} - 3 \, a^{5} c^{6} x^{4} + 2 \, a^{4} c^{6} x^{3} + 2 \, a^{3} c^{6} x^{2} - 3 \, a^{2} c^{6} x + a c^{6}\right )}} \] Input:

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^6,x, algorithm="fricas 
")
 

Output:

1/35*(13*a^5*x^5 - 39*a^4*x^4 + 26*a^3*x^3 + 26*a^2*x^2 - 39*a*x - (8*a^4* 
x^4 - 24*a^3*x^3 + 20*a^2*x^2 + 4*a*x - 13)*sqrt(-a^2*x^2 + 1) + 13)/(a^6* 
c^6*x^5 - 3*a^5*c^6*x^4 + 2*a^4*c^6*x^3 + 2*a^3*c^6*x^2 - 3*a^2*c^6*x + a* 
c^6)
 

Sympy [F]

\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^6} \, dx=\frac {\int \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{9} x^{9} - 3 a^{8} x^{8} + 8 a^{6} x^{6} - 6 a^{5} x^{5} - 6 a^{4} x^{4} + 8 a^{3} x^{3} - 3 a x + 1}\, dx + \int \left (- \frac {a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1}}{a^{9} x^{9} - 3 a^{8} x^{8} + 8 a^{6} x^{6} - 6 a^{5} x^{5} - 6 a^{4} x^{4} + 8 a^{3} x^{3} - 3 a x + 1}\right )\, dx}{c^{6}} \] Input:

integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(-a*c*x+c)**6,x)
 

Output:

(Integral(sqrt(-a**2*x**2 + 1)/(a**9*x**9 - 3*a**8*x**8 + 8*a**6*x**6 - 6* 
a**5*x**5 - 6*a**4*x**4 + 8*a**3*x**3 - 3*a*x + 1), x) + Integral(-a**2*x* 
*2*sqrt(-a**2*x**2 + 1)/(a**9*x**9 - 3*a**8*x**8 + 8*a**6*x**6 - 6*a**5*x* 
*5 - 6*a**4*x**4 + 8*a**3*x**3 - 3*a*x + 1), x))/c**6
 

Maxima [F]

\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^6} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a c x - c\right )}^{6} {\left (a x + 1\right )}^{3}} \,d x } \] Input:

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^6,x, algorithm="maxima 
")
 

Output:

integrate((-a^2*x^2 + 1)^(3/2)/((a*c*x - c)^6*(a*x + 1)^3), x)
 

Giac [F]

\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^6} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a c x - c\right )}^{6} {\left (a x + 1\right )}^{3}} \,d x } \] Input:

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^6,x, algorithm="giac")
 

Output:

integrate((-a^2*x^2 + 1)^(3/2)/((a*c*x - c)^6*(a*x + 1)^3), x)
 

Mupad [B] (verification not implemented)

Time = 14.09 (sec) , antiderivative size = 347, normalized size of antiderivative = 2.92 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^6} \, dx=\frac {3\,a\,\sqrt {1-a^2\,x^2}}{40\,\left (a^4\,c^6\,x^2-2\,a^3\,c^6\,x+a^2\,c^6\right )}+\frac {a^3\,\sqrt {1-a^2\,x^2}}{35\,\left (a^6\,c^6\,x^2-2\,a^5\,c^6\,x+a^4\,c^6\right )}+\frac {a\,\sqrt {1-a^2\,x^2}}{14\,\left (a^6\,c^6\,x^4-4\,a^5\,c^6\,x^3+6\,a^4\,c^6\,x^2-4\,a^3\,c^6\,x+a^2\,c^6\right )}+\frac {\sqrt {1-a^2\,x^2}}{16\,\sqrt {-a^2}\,\left (c^6\,x\,\sqrt {-a^2}+\frac {c^6\,\sqrt {-a^2}}{a}\right )}+\frac {93\,\sqrt {1-a^2\,x^2}}{560\,\sqrt {-a^2}\,\left (c^6\,x\,\sqrt {-a^2}-\frac {c^6\,\sqrt {-a^2}}{a}\right )}+\frac {13\,\sqrt {1-a^2\,x^2}}{140\,\sqrt {-a^2}\,\left (3\,c^6\,x\,\sqrt {-a^2}-\frac {c^6\,\sqrt {-a^2}}{a}+a^2\,c^6\,x^3\,\sqrt {-a^2}-3\,a\,c^6\,x^2\,\sqrt {-a^2}\right )} \] Input:

int((1 - a^2*x^2)^(3/2)/((c - a*c*x)^6*(a*x + 1)^3),x)
 

Output:

(3*a*(1 - a^2*x^2)^(1/2))/(40*(a^2*c^6 - 2*a^3*c^6*x + a^4*c^6*x^2)) + (a^ 
3*(1 - a^2*x^2)^(1/2))/(35*(a^4*c^6 - 2*a^5*c^6*x + a^6*c^6*x^2)) + (a*(1 
- a^2*x^2)^(1/2))/(14*(a^2*c^6 - 4*a^3*c^6*x + 6*a^4*c^6*x^2 - 4*a^5*c^6*x 
^3 + a^6*c^6*x^4)) + (1 - a^2*x^2)^(1/2)/(16*(-a^2)^(1/2)*(c^6*x*(-a^2)^(1 
/2) + (c^6*(-a^2)^(1/2))/a)) + (93*(1 - a^2*x^2)^(1/2))/(560*(-a^2)^(1/2)* 
(c^6*x*(-a^2)^(1/2) - (c^6*(-a^2)^(1/2))/a)) + (13*(1 - a^2*x^2)^(1/2))/(1 
40*(-a^2)^(1/2)*(3*c^6*x*(-a^2)^(1/2) - (c^6*(-a^2)^(1/2))/a + a^2*c^6*x^3 
*(-a^2)^(1/2) - 3*a*c^6*x^2*(-a^2)^(1/2)))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a c x)^6} \, dx=\frac {4 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-12 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+12 \sqrt {-a^{2} x^{2}+1}\, a x -4 \sqrt {-a^{2} x^{2}+1}+24 a^{4} x^{4}-72 a^{3} x^{3}+60 a^{2} x^{2}+12 a x -39}{105 \sqrt {-a^{2} x^{2}+1}\, a \,c^{6} \left (a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:

int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^6,x)
 

Output:

(4*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 12*sqrt( - a**2*x**2 + 1)*a**2*x**2 
+ 12*sqrt( - a**2*x**2 + 1)*a*x - 4*sqrt( - a**2*x**2 + 1) + 24*a**4*x**4 
- 72*a**3*x**3 + 60*a**2*x**2 + 12*a*x - 39)/(105*sqrt( - a**2*x**2 + 1)*a 
*c**6*(a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))