\(\int \frac {e^{n \text {arctanh}(a x)} x}{(c-a^2 c x^2)^{5/2}} \, dx\) [1373]

Optimal result

Integrand size = 25, antiderivative size = 133 \[ \int \frac {e^{n \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {e^{n \text {arctanh}(a x)}}{3 a^2 c \left (c-a^2 c x^2\right )^{3/2}}+\frac {e^{n \text {arctanh}(a x)} n (n-3 a x)}{3 a^2 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}+\frac {2 e^{n \text {arctanh}(a x)} n (n-a x)}{a^2 c^2 \left (9-10 n^2+n^4\right ) \sqrt {c-a^2 c x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.86 \[ \int \frac {e^{n \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {(1-a x)^{\frac {1}{2} (-3-n)} (1+a x)^{\frac {1}{2} (-3+n)} \sqrt {1-a^2 x^2} \left (3+a n^3 x+a n x \left (-3+2 a^2 x^2\right )-n^2 \left (1+2 a^2 x^2\right )\right )}{a^2 c^2 \left (9-10 n^2+n^4\right ) \sqrt {c-a^2 c x^2}} \] Input:

Integrate[(E^(n*ArcTanh[a*x])*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]*(3 + a*n^ 
3*x + a*n*x*(-3 + 2*a^2*x^2) - n^2*(1 + 2*a^2*x^2)))/(a^2*c^2*(9 - 10*n^2 
+ n^4)*Sqrt[c - a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6695, 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.

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

\(\Big \downarrow \) 6695

\(\displaystyle \frac {e^{n \text {arctanh}(a x)}}{3 a^2 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {n \int \frac {e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}}dx}{3 a}\)

\(\Big \downarrow \) 6686

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

\(\Big \downarrow \) 6685

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

Input:

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

Output:

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

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]
 

rule 6695

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.65

method	result	size
gosper	\(-\frac {\left (a x -1\right ) \left (a x +1\right ) \left (2 a^{3} x^{3} n -2 a^{2} n^{2} x^{2}+a \,n^{3} x -3 n a x -n^{2}+3\right ) {\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )}}{a^{2} \left (n^{4}-10 n^{2}+9\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\)	\(86\)
orering	\(-\frac {\left (a x -1\right ) \left (a x +1\right ) \left (2 a^{3} x^{3} n -2 a^{2} n^{2} x^{2}+a \,n^{3} x -3 n a x -n^{2}+3\right ) {\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )}}{a^{2} \left (n^{4}-10 n^{2}+9\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\)	\(86\)

Input:

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

Output:

-(a*x-1)*(a*x+1)*(2*a^3*n*x^3-2*a^2*n^2*x^2+a*n^3*x-3*a*n*x-n^2+3)*exp(n*a 
rctanh(a*x))/a^2/(n^4-10*n^2+9)/(-a^2*c*x^2+c)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.29 \[ \int \frac {e^{n \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, a^{3} n x^{3} - 2 \, a^{2} n^{2} x^{2} - n^{2} + {\left (a n^{3} - 3 \, a n\right )} x + 3\right )} \sqrt {-a^{2} c x^{2} + c} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{2} c^{3} n^{4} - 10 \, a^{2} c^{3} n^{2} + 9 \, a^{2} c^{3} + {\left (a^{6} c^{3} n^{4} - 10 \, a^{6} c^{3} n^{2} + 9 \, a^{6} c^{3}\right )} x^{4} - 2 \, {\left (a^{4} c^{3} n^{4} - 10 \, a^{4} c^{3} n^{2} + 9 \, a^{4} c^{3}\right )} x^{2}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 26.75 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.23 \[ \int \frac {e^{n \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\mathrm {e}}^{\frac {n\,\ln \left (a\,x+1\right )}{2}-\frac {n\,\ln \left (1-a\,x\right )}{2}}\,\left (\frac {n^2-3}{a^4\,c^2\,\left (n^4-10\,n^2+9\right )}+\frac {2\,n^2\,x^2}{a^2\,c^2\,\left (n^4-10\,n^2+9\right )}-\frac {2\,n\,x^3}{a\,c^2\,\left (n^4-10\,n^2+9\right )}-\frac {n\,x\,\left (n^2-3\right )}{a^3\,c^2\,\left (n^4-10\,n^2+9\right )}\right )}{\frac {\sqrt {c-a^2\,c\,x^2}}{a^2}-x^2\,\sqrt {c-a^2\,c\,x^2}} \] Input:

int((x*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)*((n^2 - 3)/(a^4*c^2*(n^4 - 
10*n^2 + 9)) + (2*n^2*x^2)/(a^2*c^2*(n^4 - 10*n^2 + 9)) - (2*n*x^3)/(a*c^2 
*(n^4 - 10*n^2 + 9)) - (n*x*(n^2 - 3))/(a^3*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]

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

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

Output:

int((e**(atanh(a*x)*n)*x)/(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)