\(\int \frac {e^{n \text {arctanh}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\) [1374]

Optimal result

Integrand size = 24, antiderivative size = 102 \[ \int \frac {e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {e^{n \text {arctanh}(a x)} (n-3 a x)}{a c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}-\frac {6 e^{n \text {arctanh}(a x)} (n-a x)}{a c^2 \left (1-n^2\right ) \left (9-n^2\right ) \sqrt {c-a^2 c x^2}} \] Output:

-exp(n*arctanh(a*x))*(-3*a*x+n)/a/c/(-n^2+9)/(-a^2*c*x^2+c)^(3/2)-6*exp(n* 
arctanh(a*x))*(-a*x+n)/a/c^2/(-n^2+1)/(-n^2+9)/(-a^2*c*x^2+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.21 \[ \int \frac {e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {(1-a x)^{\frac {1}{2} (-3-n)} (1+a x)^{\frac {1}{2} (-3+n)} \sqrt {1-a^2 x^2} \left (-n^3-9 a x+3 a n^2 x+6 a^3 x^3+n \left (7-6 a^2 x^2\right )\right )}{a c^2 (-3+n) (-1+n) (1+n) (3+n) \sqrt {c-a^2 c x^2}} \] Input:

Integrate[E^(n*ArcTanh[a*x])/(c - a^2*c*x^2)^(5/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6686, 6685}

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.

\(\displaystyle \int \frac {e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 6686

\(\displaystyle \frac {6 \int \frac {e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}}dx}{c \left (9-n^2\right )}-\frac {(n-3 a x) e^{n \text {arctanh}(a x)}}{a c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}\)

\(\Big \downarrow \) 6685

\(\displaystyle -\frac {6 (n-a x) e^{n \text {arctanh}(a x)}}{a c^2 \left (1-n^2\right ) \left (9-n^2\right ) \sqrt {c-a^2 c x^2}}-\frac {(n-3 a x) e^{n \text {arctanh}(a x)}}{a c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}\)

Input:

Int[E^(n*ArcTanh[a*x])/(c - a^2*c*x^2)^(5/2),x]
 

Output:

-((E^(n*ArcTanh[a*x])*(n - 3*a*x))/(a*c*(9 - n^2)*(c - a^2*c*x^2)^(3/2))) 
- (6*E^(n*ArcTanh[a*x])*(n - a*x))/(a*c^2*(1 - n^2)*(9 - n^2)*Sqrt[c - a^2 
*c*x^2])
 

Defintions of rubi rules used

rule 6685

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> 
Simp[(n - a*x)*(E^(n*ArcTanh[a*x])/(a*c*(n^2 - 1)*Sqrt[c + d*x^2])), x] /; 
FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !IntegerQ[n]
 

rule 6686

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> S 
imp[(n + 2*a*(p + 1)*x)*(c + d*x^2)^(p + 1)*(E^(n*ArcTanh[a*x])/(a*c*(n^2 - 
 4*(p + 1)^2))), x] - Simp[2*(p + 1)*((2*p + 3)/(c*(n^2 - 4*(p + 1)^2))) 
Int[(c + d*x^2)^(p + 1)*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n}, x 
] && EqQ[a^2*c + d, 0] && LtQ[p, -1] &&  !IntegerQ[n] && NeQ[n^2 - 4*(p + 1 
)^2, 0] && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82

method	result	size
gosper	\(\frac {\left (a x -1\right ) \left (a x +1\right ) \left (6 a^{3} x^{3}-6 n \,x^{2} a^{2}+3 n^{2} x a -n^{3}-9 a x +7 n \right ) {\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )}}{a \left (n^{4}-10 n^{2}+9\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\)	\(84\)
orering	\(\frac {\left (a x -1\right ) \left (a x +1\right ) \left (6 a^{3} x^{3}-6 n \,x^{2} a^{2}+3 n^{2} x a -n^{3}-9 a x +7 n \right ) {\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )}}{a \left (n^{4}-10 n^{2}+9\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\)	\(84\)

Input:

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

Output:

(a*x-1)*(a*x+1)*(6*a^3*x^3-6*a^2*n*x^2+3*a*n^2*x-n^3-9*a*x+7*n)*exp(n*arct 
anh(a*x))/a/(n^4-10*n^2+9)/(-a^2*c*x^2+c)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.63 \[ \int \frac {e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\left (6 \, a^{3} x^{3} - 6 \, a^{2} n x^{2} - n^{3} + 3 \, {\left (a n^{2} - 3 \, a\right )} x + 7 \, n\right )} \sqrt {-a^{2} c x^{2} + c} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a c^{3} n^{4} - 10 \, a c^{3} n^{2} + {\left (a^{5} c^{3} n^{4} - 10 \, a^{5} c^{3} n^{2} + 9 \, a^{5} c^{3}\right )} x^{4} + 9 \, a c^{3} - 2 \, {\left (a^{3} c^{3} n^{4} - 10 \, a^{3} c^{3} n^{2} + 9 \, a^{3} c^{3}\right )} x^{2}} \] Input:

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

Output:

-(6*a^3*x^3 - 6*a^2*n*x^2 - n^3 + 3*(a*n^2 - 3*a)*x + 7*n)*sqrt(-a^2*c*x^2 
 + c)*(-(a*x + 1)/(a*x - 1))^(1/2*n)/(a*c^3*n^4 - 10*a*c^3*n^2 + (a^5*c^3* 
n^4 - 10*a^5*c^3*n^2 + 9*a^5*c^3)*x^4 + 9*a*c^3 - 2*(a^3*c^3*n^4 - 10*a^3* 
c^3*n^2 + 9*a^3*c^3)*x^2)
 

Sympy [F]

\[ \int \frac {e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {e^{n \operatorname {atanh}{\left (a x \right )}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(exp(n*atanh(a*x))/(-a**2*c*x**2+c)**(5/2),x)
 

Output:

Integral(exp(n*atanh(a*x))/(-c*(a*x - 1)*(a*x + 1))**(5/2), x)
 

Maxima [F]

\[ \int \frac {e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {\left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {\left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 26.71 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.57 \[ \int \frac {e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\mathrm {e}}^{\frac {n\,\ln \left (a\,x+1\right )}{2}-\frac {n\,\ln \left (1-a\,x\right )}{2}}\,\left (\frac {6\,x^3}{c^2\,\left (n^4-10\,n^2+9\right )}+\frac {7\,n-n^3}{a^3\,c^2\,\left (n^4-10\,n^2+9\right )}+\frac {3\,x\,\left (n^2-3\right )}{a^2\,c^2\,\left (n^4-10\,n^2+9\right )}-\frac {6\,n\,x^2}{a\,c^2\,\left (n^4-10\,n^2+9\right )}\right )}{\frac {\sqrt {c-a^2\,c\,x^2}}{a^2}-x^2\,\sqrt {c-a^2\,c\,x^2}} \] Input:

int(exp(n*atanh(a*x))/(c - a^2*c*x^2)^(5/2),x)
 

Output:

-(exp((n*log(a*x + 1))/2 - (n*log(1 - a*x))/2)*((6*x^3)/(c^2*(n^4 - 10*n^2 
 + 9)) + (7*n - n^3)/(a^3*c^2*(n^4 - 10*n^2 + 9)) + (3*x*(n^2 - 3))/(a^2*c 
^2*(n^4 - 10*n^2 + 9)) - (6*n*x^2)/(a*c^2*(n^4 - 10*n^2 + 9))))/((c - a^2* 
c*x^2)^(1/2)/a^2 - x^2*(c - a^2*c*x^2)^(1/2))
 

Reduce [F]

\[ \int \frac {e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\int \frac {e^{\mathit {atanh} \left (a x \right ) n}}{\sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}-2 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+\sqrt {-a^{2} x^{2}+1}}d x}{\sqrt {c}\, c^{2}} \] Input:

int(exp(n*atanh(a*x))/(-a^2*c*x^2+c)^(5/2),x)
 

Output:

int(e**(atanh(a*x)*n)/(sqrt( - a**2*x**2 + 1)*a**4*x**4 - 2*sqrt( - a**2*x 
**2 + 1)*a**2*x**2 + sqrt( - a**2*x**2 + 1)),x)/(sqrt(c)*c**2)