\(\int \frac {e^{3 \text {arctanh}(a x)}}{(c-a^2 c x^2)^3} \, dx\) [1176]

Optimal result

Integrand size = 22, antiderivative size = 105 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {2 (1+a x)^2}{7 a c^3 \left (1-a^2 x^2\right )^{7/2}}+\frac {3 (1+a x)}{35 a c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 x}{35 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 x}{35 c^3 \sqrt {1-a^2 x^2}} \] Output:

2/7*(a*x+1)^2/a/c^3/(-a^2*x^2+1)^(7/2)+3/35*(a*x+1)/a/c^3/(-a^2*x^2+1)^(5/ 
2)+4/35*x/c^3/(-a^2*x^2+1)^(3/2)+8/35*x/c^3/(-a^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.58 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {-13+4 a x+20 a^2 x^2-24 a^3 x^3+8 a^4 x^4}{35 a c^3 (-1+a x)^3 \sqrt {1-a^2 x^2}} \] Input:

Integrate[E^(3*ArcTanh[a*x])/(c - a^2*c*x^2)^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6688, 464, 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.

\(\displaystyle \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx\)

\(\Big \downarrow \) 6688

\(\displaystyle \frac {\int \frac {(a x+1)^3}{\left (1-a^2 x^2\right )^{9/2}}dx}{c^3}\)

\(\Big \downarrow \) 464

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

\(\Big \downarrow \) 461

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

\(\Big \downarrow \) 461

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

\(\Big \downarrow \) 470

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

\(\Big \downarrow \) 208

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

Input:

Int[E^(3*ArcTanh[a*x])/(c - a^2*c*x^2)^3,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^3
 

Defintions of rubi rules used

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 464

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

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 6688

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

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.60

method	result	size
gosper	\(-\frac {\left (8 a^{4} x^{4}-24 a^{3} x^{3}+20 a^{2} x^{2}+4 a x -13\right ) \left (a x +1\right )}{35 \left (a x -1\right )^{2} c^{3} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} a}\)	\(63\)
trager	\(-\frac {\left (8 a^{4} x^{4}-24 a^{3} x^{3}+20 a^{2} x^{2}+4 a x -13\right ) \sqrt {-a^{2} x^{2}+1}}{35 c^{3} \left (a x -1\right )^{4} a \left (a x +1\right )}\)	\(65\)
orering	\(\frac {\left (8 a^{4} x^{4}-24 a^{3} x^{3}+20 a^{2} x^{2}+4 a x -13\right ) \left (a x -1\right ) \left (a x +1\right )^{4}}{35 a \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} \left (-a^{2} c \,x^{2}+c \right )^{3}}\)	\(73\)
default	\(-\frac {\frac {1}{7 a \left (x -\frac {1}{a}\right )^{3} \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}-\frac {4 a \left (\frac {1}{5 a \left (x -\frac {1}{a}\right )^{2} \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}-\frac {3 a \left (\frac {1}{3 a \left (x -\frac {1}{a}\right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}+\frac {-2 \left (x -\frac {1}{a}\right ) a^{2}-2 a}{3 a \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{5}\right )}{7}}{c^{3} a^{3}}\)	\(189\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.37 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {13 \, a^{5} x^{5} - 39 \, a^{4} x^{4} + 26 \, a^{3} x^{3} + 26 \, a^{2} x^{2} - 39 \, a x - {\left (8 \, a^{4} x^{4} - 24 \, a^{3} x^{3} + 20 \, a^{2} x^{2} + 4 \, a x - 13\right )} \sqrt {-a^{2} x^{2} + 1} + 13}{35 \, {\left (a^{6} c^{3} x^{5} - 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} + 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x + a c^{3}\right )}} \] Input:

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

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^3*x^5 - 3*a^5*c^3*x^4 + 2*a^4*c^3*x^3 + 2*a^3*c^3*x^2 - 3*a^2*c^3*x + a* 
c^3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 27.97 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.19 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {29\,\sqrt {1-a^2\,x^2}}{280\,a\,c^3\,{\left (a\,x-1\right )}^2}-\frac {13\,\sqrt {1-a^2\,x^2}}{140\,a\,c^3\,{\left (a\,x-1\right )}^3}+\frac {\sqrt {1-a^2\,x^2}}{14\,a\,c^3\,{\left (a\,x-1\right )}^4}-\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {8\,x}{35\,c^3}+\frac {29}{280\,a\,c^3}\right )}{\left (a\,x-1\right )\,\left (a\,x+1\right )} \] Input:

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

Output:

(29*(1 - a^2*x^2)^(1/2))/(280*a*c^3*(a*x - 1)^2) - (13*(1 - a^2*x^2)^(1/2) 
)/(140*a*c^3*(a*x - 1)^3) + (1 - a^2*x^2)^(1/2)/(14*a*c^3*(a*x - 1)^4) - ( 
(1 - a^2*x^2)^(1/2)*((8*x)/(35*c^3) + 29/(280*a*c^3)))/((a*x - 1)*(a*x + 1 
))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.33 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {4 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-12 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+12 \sqrt {-a^{2} x^{2}+1}\, a x -4 \sqrt {-a^{2} x^{2}+1}+24 a^{4} x^{4}-72 a^{3} x^{3}+60 a^{2} x^{2}+12 a x -39}{105 \sqrt {-a^{2} x^{2}+1}\, a \,c^{3} \left (a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:

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