\(\int \frac {e^{n \text {arctanh}(a x)} x^3}{(c-a^2 c x^2)^2} \, dx\) [1357]

Optimal result

Integrand size = 25, antiderivative size = 197 \[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {(1-a x)^{-1-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-2+n)}}{a^4 c^2 (2+n)}+\frac {\left (4+n-n^2\right ) (1-a x)^{1-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-2+n)}}{a^4 c^2 n \left (4-n^2\right )}-\frac {(4+3 n) (1-a x)^{-n/2} (1+a x)^{\frac {1}{2} (-2+n)}}{a^4 c^2 n (2+n)}+\frac {2^{1-\frac {n}{2}} (1+a x)^{n/2} \operatorname {Hypergeometric2F1}\left (\frac {n}{2},\frac {n}{2},\frac {2+n}{2},\frac {1}{2} (1+a x)\right )}{a^4 c^2 n} \] Output:

(-a*x+1)^(-1-1/2*n)*(a*x+1)^(-1+1/2*n)/a^4/c^2/(2+n)+(-n^2+n+4)*(-a*x+1)^( 
1-1/2*n)*(a*x+1)^(-1+1/2*n)/a^4/c^2/n/(-n^2+4)-(4+3*n)*(a*x+1)^(-1+1/2*n)/ 
a^4/c^2/n/(2+n)/((-a*x+1)^(1/2*n))+2^(1-1/2*n)*(a*x+1)^(1/2*n)*hypergeom([ 
1/2*n, 1/2*n],[1+1/2*n],1/2*a*x+1/2)/a^4/c^2/n
 

Mathematica [A] (verified)

Time = 6.62 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.74 \[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^2} \, dx=-\frac {e^{n \text {arctanh}(a x)} \left (n \left (-1+a n x-a^2 x^2\right )+e^{2 \text {arctanh}(a x)} (-2+n) n \left (-1+a^2 x^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+\frac {n}{2},2+\frac {n}{2},-e^{2 \text {arctanh}(a x)}\right )-\left (-4+n^2\right ) \left (-1+a^2 x^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {n}{2},1+\frac {n}{2},-e^{2 \text {arctanh}(a x)}\right )\right )}{a^4 c^2 (-2+n) n (2+n) \left (-1+a^2 x^2\right )} \] Input:

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

Output:

-((E^(n*ArcTanh[a*x])*(n*(-1 + a*n*x - a^2*x^2) + E^(2*ArcTanh[a*x])*(-2 + 
 n)*n*(-1 + a^2*x^2)*Hypergeometric2F1[1, 1 + n/2, 2 + n/2, -E^(2*ArcTanh[ 
a*x])] - (-4 + n^2)*(-1 + a^2*x^2)*Hypergeometric2F1[1, n/2, 1 + n/2, -E^( 
2*ArcTanh[a*x])]))/(a^4*c^2*(-2 + n)*n*(2 + n)*(-1 + a^2*x^2)))
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.50, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6700, 137, 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.

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

\(\Big \downarrow \) 6700

\(\displaystyle \frac {\int x^3 (1-a x)^{-\frac {n}{2}-2} (a x+1)^{\frac {n-4}{2}}dx}{c^2}\)

\(\Big \downarrow \) 137

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {2^{\frac {n}{2}+2} (1-a x)^{-\frac {n}{2}-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-n-2),-\frac {n}{2}-1,-\frac {n}{2},\frac {1}{2} (1-a x)\right )}{a^4 (n+2)}+\frac {2 (a x+1)^{\frac {n-2}{2}} (1-a x)^{1-\frac {n}{2}}}{a^4 n \left (4-n^2\right )}-\frac {(a x+1)^{\frac {n-2}{2}} (1-a x)^{-\frac {n}{2}-1}}{a^4 (n+2)}+\frac {3 (a x+1)^{n/2} (1-a x)^{-\frac {n}{2}-1}}{a^4 (n+2)}-\frac {3 (a x+1)^{\frac {n+2}{2}} (1-a x)^{-\frac {n}{2}-1}}{a^4 (n+2)}-\frac {2 (a x+1)^{\frac {n-2}{2}} (1-a x)^{-n/2}}{a^4 n (n+2)}+\frac {3 (a x+1)^{n/2} (1-a x)^{-n/2}}{a^4 n (n+2)}}{c^2}\)

Input:

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

Output:

(-(((1 - a*x)^(-1 - n/2)*(1 + a*x)^((-2 + n)/2))/(a^4*(2 + n))) + (2*(1 - 
a*x)^(1 - n/2)*(1 + a*x)^((-2 + n)/2))/(a^4*n*(4 - n^2)) - (2*(1 + a*x)^(( 
-2 + n)/2))/(a^4*n*(2 + n)*(1 - a*x)^(n/2)) + (3*(1 - a*x)^(-1 - n/2)*(1 + 
 a*x)^(n/2))/(a^4*(2 + n)) + (3*(1 + a*x)^(n/2))/(a^4*n*(2 + n)*(1 - a*x)^ 
(n/2)) - (3*(1 - a*x)^(-1 - n/2)*(1 + a*x)^((2 + n)/2))/(a^4*(2 + n)) + (2 
^(2 + n/2)*(1 - a*x)^(-1 - n/2)*Hypergeometric2F1[(-2 - n)/2, -1 - n/2, -1 
/2*n, (1 - a*x)/2])/(a^4*(2 + n)))/c^2
 

Defintions of rubi rules used

rule 137

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (IGtQ[m, 0] || (ILtQ[m, 0] && 
ILtQ[n, 0]))
 

rule 2009

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

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])
 

Maple [F]

\[\int \frac {{\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )} x^{3}}{\left (-a^{2} c \,x^{2}+c \right )^{2}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(x^3*(-(a*x + 1)/(a*x - 1))^(1/2*n)/(a^4*c^2*x^4 - 2*a^2*c^2*x^2 + 
 c^2), x)
 

Sympy [F]

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

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

Output:

Integral(x**3*exp(n*atanh(a*x))/(a**4*x**4 - 2*a**2*x**2 + 1), x)/c**2
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((e**(atanh(a*x)*n)*x**3)/(a**4*x**4 - 2*a**2*x**2 + 1),x)/c**2