Integrand size = 27, antiderivative size = 102 \[ \int \frac {e^{n \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {e^{n \text {arctanh}(a x)} (n-3 a x)}{a^3 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}+\frac {e^{n \text {arctanh}(a x)} \left (3-n^2\right ) (n-a x)}{a^3 c^2 \left (9-10 n^2+n^4\right ) \sqrt {c-a^2 c x^2}} \] Output:
-exp(n*arctanh(a*x))*(-3*a*x+n)/a^3/c/(-n^2+9)/(-a^2*c*x^2+c)^(3/2)+exp(n* arctanh(a*x))*(-n^2+3)*(-a*x+n)/a^3/c^2/(n^4-10*n^2+9)/(-a^2*c*x^2+c)^(1/2 )
Time = 0.20 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.23 \[ \int \frac {e^{n \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {(1-a x)^{\frac {1}{2} (-3-n)} (1+a x)^{\frac {1}{2} (-3+n)} \sqrt {1-a^2 x^2} \left (-a^2 n^3 x^2-3 a^3 x^3+a n^2 x \left (2+a^2 x^2\right )+n \left (-2+3 a^2 x^2\right )\right )}{a^3 c^2 \left (9-10 n^2+n^4\right ) \sqrt {c-a^2 c x^2}} \] Input:
Integrate[(E^(n*ArcTanh[a*x])*x^2)/(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]*(-(a^2* n^3*x^2) - 3*a^3*x^3 + a*n^2*x*(2 + a^2*x^2) + n*(-2 + 3*a^2*x^2)))/(a^3*c ^2*(9 - 10*n^2 + n^4)*Sqrt[c - a^2*c*x^2]))
Time = 0.75 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {6697, 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.
| \(\displaystyle \int \frac {x^2 e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6697 |
| \(\displaystyle -\frac {\left (3-n^2\right ) \int \frac {e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}}dx}{a^2 c \left (9-n^2\right )}-\frac {(n-3 a x) e^{n \text {arctanh}(a x)}}{a^3 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 6685 |
| \(\displaystyle \frac {\left (3-n^2\right ) (n-a x) e^{n \text {arctanh}(a x)}}{a^3 c^2 \left (1-n^2\right ) \left (9-n^2\right ) \sqrt {c-a^2 c x^2}}-\frac {(n-3 a x) e^{n \text {arctanh}(a x)}}{a^3 c \left (9-n^2\right ) \left (c-a^2 c x^2\right )^{3/2}}\) |
Input:
Int[(E^(n*ArcTanh[a*x])*x^2)/(c - a^2*c*x^2)^(5/2),x]
Output:
-((E^(n*ArcTanh[a*x])*(n - 3*a*x))/(a^3*c*(9 - n^2)*(c - a^2*c*x^2)^(3/2)) ) + (E^(n*ArcTanh[a*x])*(3 - n^2)*(n - a*x))/(a^3*c^2*(1 - n^2)*(9 - n^2)* Sqrt[c - a^2*c*x^2])
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]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^2*((c_) + (d_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[(-(n + 2*(p + 1)*a*x))*(c + d*x^2)^(p + 1)*(E^(n*ArcTanh[a*x])/( a*d*(n^2 - 4*(p + 1)^2))), x] + Simp[(n^2 + 2*(p + 1))/(d*(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]
Time = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.94
| method | result | size |
| gosper | \(\frac {\left (a x -1\right ) \left (a x +1\right ) \left (a^{3} n^{2} x^{3}-a^{2} n^{3} x^{2}-3 a^{3} x^{3}+3 n \,x^{2} a^{2}+2 n^{2} x a -2 n \right ) {\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )}}{\left (n^{4}-10 n^{2}+9\right ) a^{3} \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\) | \(96\) |
| orering | \(\frac {\left (a x -1\right ) \left (a x +1\right ) \left (a^{3} n^{2} x^{3}-a^{2} n^{3} x^{2}-3 a^{3} x^{3}+3 n \,x^{2} a^{2}+2 n^{2} x a -2 n \right ) {\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )}}{\left (n^{4}-10 n^{2}+9\right ) a^{3} \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\) | \(96\) |
Input:
int(exp(n*arctanh(a*x))*x^2/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
(a*x-1)*(a*x+1)*(a^3*n^2*x^3-a^2*n^3*x^2-3*a^3*x^3+3*a^2*n*x^2+2*a*n^2*x-2 *n)*exp(n*arctanh(a*x))/(n^4-10*n^2+9)/a^3/(-a^2*c*x^2+c)^(5/2)
Time = 0.09 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.77 \[ \int \frac {e^{n \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {\sqrt {-a^{2} c x^{2} + c} {\left (2 \, a n^{2} x + {\left (a^{3} n^{2} - 3 \, a^{3}\right )} x^{3} - {\left (a^{2} n^{3} - 3 \, a^{2} n\right )} x^{2} - 2 \, n\right )} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{3} c^{3} n^{4} - 10 \, a^{3} c^{3} n^{2} + 9 \, a^{3} c^{3} + {\left (a^{7} c^{3} n^{4} - 10 \, a^{7} c^{3} n^{2} + 9 \, a^{7} c^{3}\right )} x^{4} - 2 \, {\left (a^{5} c^{3} n^{4} - 10 \, a^{5} c^{3} n^{2} + 9 \, a^{5} c^{3}\right )} x^{2}} \] Input:
integrate(exp(n*arctanh(a*x))*x^2/(-a^2*c*x^2+c)^(5/2),x, algorithm="frica s")
Output:
-sqrt(-a^2*c*x^2 + c)*(2*a*n^2*x + (a^3*n^2 - 3*a^3)*x^3 - (a^2*n^3 - 3*a^ 2*n)*x^2 - 2*n)*(-(a*x + 1)/(a*x - 1))^(1/2*n)/(a^3*c^3*n^4 - 10*a^3*c^3*n ^2 + 9*a^3*c^3 + (a^7*c^3*n^4 - 10*a^7*c^3*n^2 + 9*a^7*c^3)*x^4 - 2*(a^5*c ^3*n^4 - 10*a^5*c^3*n^2 + 9*a^5*c^3)*x^2)
\[ \int \frac {e^{n \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^{2} e^{n \operatorname {atanh}{\left (a x \right )}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(exp(n*atanh(a*x))*x**2/(-a**2*c*x**2+c)**(5/2),x)
Output:
Integral(x**2*exp(n*atanh(a*x))/(-c*(a*x - 1)*(a*x + 1))**(5/2), x)
\[ \int \frac {e^{n \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {x^{2} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(exp(n*arctanh(a*x))*x^2/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxim a")
Output:
integrate(x^2*(-(a*x + 1)/(a*x - 1))^(1/2*n)/(-a^2*c*x^2 + c)^(5/2), x)
\[ \int \frac {e^{n \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {x^{2} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(exp(n*arctanh(a*x))*x^2/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac" )
Output:
integrate(x^2*(-(a*x + 1)/(a*x - 1))^(1/2*n)/(-a^2*c*x^2 + c)^(5/2), x)
Time = 13.97 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.59 \[ \int \frac {e^{n \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {{\mathrm {e}}^{\frac {n\,\ln \left (a\,x+1\right )}{2}-\frac {n\,\ln \left (1-a\,x\right )}{2}}\,\left (\frac {2\,n}{a^5\,c^2\,\left (n^4-10\,n^2+9\right )}-\frac {x^3\,\left (n^2-3\right )}{a^2\,c^2\,\left (n^4-10\,n^2+9\right )}-\frac {2\,n^2\,x}{a^4\,c^2\,\left (n^4-10\,n^2+9\right )}+\frac {n\,x^2\,\left (n^2-3\right )}{a^3\,c^2\,\left (n^4-10\,n^2+9\right )}\right )}{\frac {\sqrt {c-a^2\,c\,x^2}}{a^2}-x^2\,\sqrt {c-a^2\,c\,x^2}} \] Input:
int((x^2*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)*((2*n)/(a^5*c^2*(n^4 - 10*n^ 2 + 9)) - (x^3*(n^2 - 3))/(a^2*c^2*(n^4 - 10*n^2 + 9)) - (2*n^2*x)/(a^4*c^ 2*(n^4 - 10*n^2 + 9)) + (n*x^2*(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))
\[ \int \frac {e^{n \text {arctanh}(a x)} x^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\int \frac {e^{\mathit {atanh} \left (a x \right ) n} x^{2}}{\sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}-2 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+\sqrt {-a^{2} x^{2}+1}}d x}{\sqrt {c}\, c^{2}} \] Input:
int(exp(n*atanh(a*x))*x^2/(-a^2*c*x^2+c)^(5/2),x)
Output:
int((e**(atanh(a*x)*n)*x**2)/(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)