\(\int \frac {e^{n \text {arctanh}(a x)} x^3}{(c-a^2 c x^2)^{5/2}} \, dx\) [1371]

Optimal result

Integrand size = 27, antiderivative size = 394 \[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {(2+n) (1-a x)^{\frac {1}{2} (-3-n)} (1+a x)^{\frac {1}{2} (-3+n)} \sqrt {1-a^2 x^2}}{a^4 c^2 (3+n) \sqrt {c-a^2 c x^2}}+\frac {x^2 (1-a x)^{\frac {1}{2} (-3-n)} (1+a x)^{\frac {1}{2} (-3+n)} \sqrt {1-a^2 x^2}}{a^2 c^2 \sqrt {c-a^2 c x^2}}+\frac {n (1-a x)^{\frac {1}{2} (-1-n)} (1+a x)^{\frac {1}{2} (-3+n)} \sqrt {1-a^2 x^2}}{a^4 c^2 \left (9-n^2\right ) \sqrt {c-a^2 c x^2}}+\frac {n \left (7-n^2\right ) (1-a x)^{\frac {1}{2} (-1-n)} (1+a x)^{\frac {1}{2} (-1+n)} \sqrt {1-a^2 x^2}}{a^4 c^2 (3-n) (1+n) (3+n) \sqrt {c-a^2 c x^2}}-\frac {n \left (7-n^2\right ) (1-a x)^{\frac {1-n}{2}} (1+a x)^{\frac {1}{2} (-1+n)} \sqrt {1-a^2 x^2}}{a^4 c^2 \left (9-10 n^2+n^4\right ) \sqrt {c-a^2 c x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.28 \[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {(1-a x)^{\frac {1}{2} (-3-n)} (1+a x)^{\frac {1}{2} (-3+n)} \sqrt {1-a^2 x^2} \left (6-6 a n x+3 a^2 \left (-3+n^2\right ) x^2-a^3 n \left (-7+n^2\right ) x^3\right )}{a^4 c^2 \left (9-10 n^2+n^4\right ) \sqrt {c-a^2 c x^2}} \] Input:

Integrate[(E^(n*ArcTanh[a*x])*x^3)/(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]*(6 - 6* 
a*n*x + 3*a^2*(-3 + n^2)*x^2 - a^3*n*(-7 + n^2)*x^3))/(a^4*c^2*(9 - 10*n^2 
 + n^4)*Sqrt[c - a^2*c*x^2]))
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 275, normalized size of antiderivative = 0.70, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {6703, 6700, 105, 101, 25, 88, 55, 48}

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.

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

\(\Big \downarrow \) 6703

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^2 \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 101

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 88

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 48

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

Input:

Int[(E^(n*ArcTanh[a*x])*x^3)/(c - a^2*c*x^2)^(5/2),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 48

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

rule 55

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S 
implify[m + n + 2]/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^Simplify[m + 1]*( 
c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 
 2], 0] && NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ 
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] ||  !SumSimp 
lerQ[n, 1])
 

rule 88

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], 
 x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimpl 
erQ[p, 1]
 

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

rule 6700

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

rule 6703

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.24

method	result	size
gosper	\(-\frac {\left (a x -1\right ) \left (a x +1\right ) \left (a^{3} n^{3} x^{3}-7 a^{3} x^{3} n -3 a^{2} n^{2} x^{2}+9 a^{2} x^{2}+6 n a x -6\right ) {\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )}}{a^{4} \left (n^{4}-10 n^{2}+9\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\)	\(93\)
orering	\(-\frac {\left (a x -1\right ) \left (a x +1\right ) \left (a^{3} n^{3} x^{3}-7 a^{3} x^{3} n -3 a^{2} n^{2} x^{2}+9 a^{2} x^{2}+6 n a x -6\right ) {\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )}}{a^{4} \left (n^{4}-10 n^{2}+9\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\)	\(93\)

Input:

int(exp(n*arctanh(a*x))*x^3/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.45 \[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {-a^{2} c x^{2} + c} {\left ({\left (a^{3} n^{3} - 7 \, a^{3} n\right )} x^{3} + 6 \, a n x - 3 \, {\left (a^{2} n^{2} - 3 \, a^{2}\right )} x^{2} - 6\right )} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{4} c^{3} n^{4} - 10 \, a^{4} c^{3} n^{2} + 9 \, a^{4} c^{3} + {\left (a^{8} c^{3} n^{4} - 10 \, a^{8} c^{3} n^{2} + 9 \, a^{8} c^{3}\right )} x^{4} - 2 \, {\left (a^{6} c^{3} n^{4} - 10 \, a^{6} c^{3} n^{2} + 9 \, a^{6} c^{3}\right )} x^{2}} \] Input:

integrate(exp(n*arctanh(a*x))*x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="frica 
s")
 

Output:

sqrt(-a^2*c*x^2 + c)*((a^3*n^3 - 7*a^3*n)*x^3 + 6*a*n*x - 3*(a^2*n^2 - 3*a 
^2)*x^2 - 6)*(-(a*x + 1)/(a*x - 1))^(1/2*n)/(a^4*c^3*n^4 - 10*a^4*c^3*n^2 
+ 9*a^4*c^3 + (a^8*c^3*n^4 - 10*a^8*c^3*n^2 + 9*a^8*c^3)*x^4 - 2*(a^6*c^3* 
n^4 - 10*a^6*c^3*n^2 + 9*a^6*c^3)*x^2)
 

Sympy [F]

\[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^{3} e^{n \operatorname {atanh}{\left (a x \right )}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(exp(n*atanh(a*x))*x**3/(-a**2*c*x**2+c)**(5/2),x)
 

Output:

Integral(x**3*exp(n*atanh(a*x))/(-c*(a*x - 1)*(a*x + 1))**(5/2), x)
 

Maxima [F]

\[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {x^{3} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(exp(n*arctanh(a*x))*x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxim 
a")
 

Output:

integrate(x^3*(-(a*x + 1)/(a*x - 1))^(1/2*n)/(-a^2*c*x^2 + c)^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(exp(n*arctanh(a*x))*x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac" 
)
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 26.62 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.41 \[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\left (a\,x+1\right )}^{n/2}\,\left (\frac {6}{a^6\,c^2\,\left (n^4-10\,n^2+9\right )}-\frac {6\,n\,x}{a^5\,c^2\,\left (n^4-10\,n^2+9\right )}+\frac {x^2\,\left (3\,n^2-9\right )}{a^4\,c^2\,\left (n^4-10\,n^2+9\right )}-\frac {n\,x^3\,\left (n^2-7\right )}{a^3\,c^2\,\left (n^4-10\,n^2+9\right )}\right )}{{\left (1-a\,x\right )}^{n/2}\,\left (\frac {\sqrt {c-a^2\,c\,x^2}}{a^2}-x^2\,\sqrt {c-a^2\,c\,x^2}\right )} \] Input:

int((x^3*exp(n*atanh(a*x)))/(c - a^2*c*x^2)^(5/2),x)
 

Output:

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

Reduce [F]

\[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\int \frac {e^{\mathit {atanh} \left (a x \right ) n} x^{3}}{\sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}-2 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+\sqrt {-a^{2} x^{2}+1}}d x}{\sqrt {c}\, c^{2}} \] Input:

int(exp(n*atanh(a*x))*x^3/(-a^2*c*x^2+c)^(5/2),x)
 

Output:

int((e**(atanh(a*x)*n)*x**3)/(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)