\(\int \frac {e^{-3 \text {arctanh}(a x)}}{(c-a^2 c x^2)^{7/2}} \, dx\) [1315]

Optimal result

Integrand size = 24, antiderivative size = 275 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {\sqrt {1-a^2 x^2}}{32 a c^3 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{16 a c^3 (1+a x)^4 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{12 a c^3 (1+a x)^3 \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2}}{32 a c^3 (1+a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^3 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {5 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{32 a c^3 \sqrt {c-a^2 c x^2}} \] Output:

1/32*(-a^2*x^2+1)^(1/2)/a/c^3/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)-1/16*(-a^2*x^2 
+1)^(1/2)/a/c^3/(a*x+1)^4/(-a^2*c*x^2+c)^(1/2)-1/12*(-a^2*x^2+1)^(1/2)/a/c 
^3/(a*x+1)^3/(-a^2*c*x^2+c)^(1/2)-3/32*(-a^2*x^2+1)^(1/2)/a/c^3/(a*x+1)^2/ 
(-a^2*c*x^2+c)^(1/2)-1/8*(-a^2*x^2+1)^(1/2)/a/c^3/(a*x+1)/(-a^2*c*x^2+c)^( 
1/2)+5/32*(-a^2*x^2+1)^(1/2)*arctanh(a*x)/a/c^3/(-a^2*c*x^2+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.37 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (32+15 a x-35 a^2 x^2-45 a^3 x^3-15 a^4 x^4+15 (-1+a x) (1+a x)^4 \text {arctanh}(a x)\right )}{96 a c^3 (-1+a x) (1+a x)^4 \sqrt {c-a^2 c x^2}} \] Input:

Integrate[1/(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^(7/2)),x]
 

Output:

(Sqrt[1 - a^2*x^2]*(32 + 15*a*x - 35*a^2*x^2 - 45*a^3*x^3 - 15*a^4*x^4 + 1 
5*(-1 + a*x)*(1 + a*x)^4*ArcTanh[a*x]))/(96*a*c^3*(-1 + a*x)*(1 + a*x)^4*S 
qrt[c - a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.42, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6693, 6690, 54, 2009}

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.

\(\displaystyle \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 6693

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

\(\Big \downarrow \) 6690

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

\(\Big \downarrow \) 54

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \left (\frac {1}{32 (a x-1)^2}+\frac {1}{8 (a x+1)^2}+\frac {3}{16 (a x+1)^3}+\frac {1}{4 (a x+1)^4}+\frac {1}{4 (a x+1)^5}-\frac {5}{32 \left (a^2 x^2-1\right )}\right )dx}{c^3 \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {5 \text {arctanh}(a x)}{32 a}+\frac {1}{32 a (1-a x)}-\frac {1}{8 a (a x+1)}-\frac {3}{32 a (a x+1)^2}-\frac {1}{12 a (a x+1)^3}-\frac {1}{16 a (a x+1)^4}\right )}{c^3 \sqrt {c-a^2 c x^2}}\)

Input:

Int[1/(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^(7/2)),x]
 

Output:

(Sqrt[1 - a^2*x^2]*(1/(32*a*(1 - a*x)) - 1/(16*a*(1 + a*x)^4) - 1/(12*a*(1 
 + a*x)^3) - 3/(32*a*(1 + a*x)^2) - 1/(8*a*(1 + a*x)) + (5*ArcTanh[a*x])/( 
32*a)))/(c^3*Sqrt[c - a^2*c*x^2])
 

Defintions of rubi rules used

rule 54

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && 
 ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6690

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

rule 6693

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {\left (15 \ln \left (a x +1\right ) x^{5} a^{5}-15 \ln \left (a x -1\right ) x^{5} a^{5}+45 \ln \left (a x +1\right ) x^{4} a^{4}-45 \ln \left (a x -1\right ) x^{4} a^{4}-30 a^{4} x^{4}+30 \ln \left (a x +1\right ) x^{3} a^{3}-30 a^{3} \ln \left (a x -1\right ) x^{3}-90 a^{3} x^{3}-30 \ln \left (a x +1\right ) x^{2} a^{2}+30 a^{2} \ln \left (a x -1\right ) x^{2}-70 a^{2} x^{2}-45 \ln \left (a x +1\right ) x a +45 a \ln \left (a x -1\right ) x +30 a x -15 \ln \left (a x +1\right )+15 \ln \left (a x -1\right )+64\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}}{192 \sqrt {-a^{2} x^{2}+1}\, \left (a x +1\right )^{4} \left (a x -1\right ) c^{4} a}\)	\(227\)

Input:

int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(7/2),x,method=_RETURNVE 
RBOSE)
 

Output:

1/192*(15*ln(a*x+1)*x^5*a^5-15*ln(a*x-1)*x^5*a^5+45*ln(a*x+1)*x^4*a^4-45*l 
n(a*x-1)*x^4*a^4-30*a^4*x^4+30*ln(a*x+1)*x^3*a^3-30*a^3*ln(a*x-1)*x^3-90*a 
^3*x^3-30*ln(a*x+1)*x^2*a^2+30*a^2*ln(a*x-1)*x^2-70*a^2*x^2-45*ln(a*x+1)*x 
*a+45*a*ln(a*x-1)*x+30*a*x-15*ln(a*x+1)+15*ln(a*x-1)+64)/(-a^2*x^2+1)^(1/2 
)/(a*x+1)^4/(a*x-1)*(-c*(a^2*x^2-1))^(1/2)/c^4/a
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 559, normalized size of antiderivative = 2.03 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\left [\frac {15 \, {\left (a^{7} x^{7} + 3 \, a^{6} x^{6} + a^{5} x^{5} - 5 \, a^{4} x^{4} - 5 \, a^{3} x^{3} + a^{2} x^{2} + 3 \, a x + 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right ) - 4 \, {\left (32 \, a^{5} x^{5} + 81 \, a^{4} x^{4} + 19 \, a^{3} x^{3} - 99 \, a^{2} x^{2} - 81 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{384 \, {\left (a^{8} c^{4} x^{7} + 3 \, a^{7} c^{4} x^{6} + a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} - 5 \, a^{4} c^{4} x^{3} + a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x + a c^{4}\right )}}, \frac {15 \, {\left (a^{7} x^{7} + 3 \, a^{6} x^{6} + a^{5} x^{5} - 5 \, a^{4} x^{4} - 5 \, a^{3} x^{3} + a^{2} x^{2} + 3 \, a x + 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right ) - 2 \, {\left (32 \, a^{5} x^{5} + 81 \, a^{4} x^{4} + 19 \, a^{3} x^{3} - 99 \, a^{2} x^{2} - 81 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{192 \, {\left (a^{8} c^{4} x^{7} + 3 \, a^{7} c^{4} x^{6} + a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} - 5 \, a^{4} c^{4} x^{3} + a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x + a c^{4}\right )}}\right ] \] Input:

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(7/2),x, algorithm 
="fricas")
 

Output:

[1/384*(15*(a^7*x^7 + 3*a^6*x^6 + a^5*x^5 - 5*a^4*x^4 - 5*a^3*x^3 + a^2*x^ 
2 + 3*a*x + 1)*sqrt(c)*log(-(a^6*c*x^6 + 5*a^4*c*x^4 - 5*a^2*c*x^2 - 4*(a^ 
3*x^3 + a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - c)/(a^6*x^6 
 - 3*a^4*x^4 + 3*a^2*x^2 - 1)) - 4*(32*a^5*x^5 + 81*a^4*x^4 + 19*a^3*x^3 - 
 99*a^2*x^2 - 81*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^8*c^4*x^ 
7 + 3*a^7*c^4*x^6 + a^6*c^4*x^5 - 5*a^5*c^4*x^4 - 5*a^4*c^4*x^3 + a^3*c^4* 
x^2 + 3*a^2*c^4*x + a*c^4), 1/192*(15*(a^7*x^7 + 3*a^6*x^6 + a^5*x^5 - 5*a 
^4*x^4 - 5*a^3*x^3 + a^2*x^2 + 3*a*x + 1)*sqrt(-c)*arctan(2*sqrt(-a^2*c*x^ 
2 + c)*sqrt(-a^2*x^2 + 1)*a*sqrt(-c)*x/(a^4*c*x^4 - c)) - 2*(32*a^5*x^5 + 
81*a^4*x^4 + 19*a^3*x^3 - 99*a^2*x^2 - 81*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(- 
a^2*x^2 + 1))/(a^8*c^4*x^7 + 3*a^7*c^4*x^6 + a^6*c^4*x^5 - 5*a^5*c^4*x^4 - 
 5*a^4*c^4*x^3 + a^3*c^4*x^2 + 3*a^2*c^4*x + a*c^4)]
 

Sympy [F]

\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}} \left (a x + 1\right )^{3}}\, dx \] Input:

integrate(1/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/(-a**2*c*x**2+c)**(7/2),x)
 

Output:

Integral((-(a*x - 1)*(a*x + 1))**(3/2)/((-c*(a*x - 1)*(a*x + 1))**(7/2)*(a 
*x + 1)**3), x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.44 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=-\frac {15 \, a^{4} x^{4} + 45 \, a^{3} x^{3} + 35 \, a^{2} x^{2} - 15 \, a x - 32}{96 \, {\left (a^{6} c^{\frac {7}{2}} x^{5} + 3 \, a^{5} c^{\frac {7}{2}} x^{4} + 2 \, a^{4} c^{\frac {7}{2}} x^{3} - 2 \, a^{3} c^{\frac {7}{2}} x^{2} - 3 \, a^{2} c^{\frac {7}{2}} x - a c^{\frac {7}{2}}\right )}} + \frac {5 \, \log \left (a x + 1\right )}{64 \, a c^{\frac {7}{2}}} - \frac {5 \, \log \left (a x - 1\right )}{64 \, a c^{\frac {7}{2}}} \] Input:

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(7/2),x, algorithm 
="maxima")
 

Output:

-1/96*(15*a^4*x^4 + 45*a^3*x^3 + 35*a^2*x^2 - 15*a*x - 32)/(a^6*c^(7/2)*x^ 
5 + 3*a^5*c^(7/2)*x^4 + 2*a^4*c^(7/2)*x^3 - 2*a^3*c^(7/2)*x^2 - 3*a^2*c^(7 
/2)*x - a*c^(7/2)) + 5/64*log(a*x + 1)/(a*c^(7/2)) - 5/64*log(a*x - 1)/(a* 
c^(7/2))
 

Giac [F]

\[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int { \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} {\left (a x + 1\right )}^{3}} \,d x } \] Input:

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(7/2),x, algorithm 
="giac")
 

Output:

integrate((-a^2*x^2 + 1)^(3/2)/((-a^2*c*x^2 + c)^(7/2)*(a*x + 1)^3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\int \frac {{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (c-a^2\,c\,x^2\right )}^{7/2}\,{\left (a\,x+1\right )}^3} \,d x \] Input:

int((1 - a^2*x^2)^(3/2)/((c - a^2*c*x^2)^(7/2)*(a*x + 1)^3),x)
 

Output:

int((1 - a^2*x^2)^(3/2)/((c - a^2*c*x^2)^(7/2)*(a*x + 1)^3), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-3 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{7/2}} \, dx=\frac {\sqrt {c}\, \left (-15 \,\mathrm {log}\left (a x -1\right ) a^{5} x^{5}-45 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}-30 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+30 \,\mathrm {log}\left (a x -1\right ) a^{2} x^{2}+45 \,\mathrm {log}\left (a x -1\right ) a x +15 \,\mathrm {log}\left (a x -1\right )+15 \,\mathrm {log}\left (a x +1\right ) a^{5} x^{5}+45 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}+30 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-30 \,\mathrm {log}\left (a x +1\right ) a^{2} x^{2}-45 \,\mathrm {log}\left (a x +1\right ) a x -15 \,\mathrm {log}\left (a x +1\right )+10 a^{5} x^{5}-70 a^{3} x^{3}-90 a^{2} x^{2}+54\right )}{192 a \,c^{4} \left (a^{5} x^{5}+3 a^{4} x^{4}+2 a^{3} x^{3}-2 a^{2} x^{2}-3 a x -1\right )} \] Input:

int(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(7/2),x)
 

Output:

(sqrt(c)*( - 15*log(a*x - 1)*a**5*x**5 - 45*log(a*x - 1)*a**4*x**4 - 30*lo 
g(a*x - 1)*a**3*x**3 + 30*log(a*x - 1)*a**2*x**2 + 45*log(a*x - 1)*a*x + 1 
5*log(a*x - 1) + 15*log(a*x + 1)*a**5*x**5 + 45*log(a*x + 1)*a**4*x**4 + 3 
0*log(a*x + 1)*a**3*x**3 - 30*log(a*x + 1)*a**2*x**2 - 45*log(a*x + 1)*a*x 
 - 15*log(a*x + 1) + 10*a**5*x**5 - 70*a**3*x**3 - 90*a**2*x**2 + 54))/(19 
2*a*c**4*(a**5*x**5 + 3*a**4*x**4 + 2*a**3*x**3 - 2*a**2*x**2 - 3*a*x - 1) 
)