Integrand size = 24, antiderivative size = 85 \[ \int e^{-\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=-\frac {\left (1-a^2 x^2\right )^{\frac {1}{2}+p}}{a^4 (1+2 p)}+\frac {\left (1-a^2 x^2\right )^{\frac {3}{2}+p}}{a^4 (3+2 p)}-\frac {1}{5} a x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-p,\frac {7}{2},a^2 x^2\right ) \]
-(-a^2*x^2+1)^(1/2+p)/a^4/(1+2*p)+(-a^2*x^2+1)^(3/2+p)/a^4/(3+2*p)-1/5*a*x ^5*hypergeom([5/2, 1/2-p],[7/2],a^2*x^2)
Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91 \[ \int e^{-\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=-\frac {\left (1-a^2 x^2\right )^{\frac {1}{2}+p} \left (2+a^2 (1+2 p) x^2\right )}{a^4 \left (3+8 p+4 p^2\right )}-\frac {1}{5} a x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-p,\frac {7}{2},a^2 x^2\right ) \]
-(((1 - a^2*x^2)^(1/2 + p)*(2 + a^2*(1 + 2*p)*x^2))/(a^4*(3 + 8*p + 4*p^2) )) - (a*x^5*Hypergeometric2F1[5/2, 1/2 - p, 7/2, a^2*x^2])/5
Time = 0.33 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6699, 542, 243, 53, 278, 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.
\(\displaystyle \int x^3 e^{-\text {arctanh}(a x)} \left (1-a^2 x^2\right )^p \, dx\) |
\(\Big \downarrow \) 6699 |
\(\displaystyle \int x^3 (1-a x) \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx\) |
\(\Big \downarrow \) 542 |
\(\displaystyle \int x^3 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx-a \int x^4 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int x^2 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx^2-a \int x^4 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{2} \int \left (\frac {\left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{a^2}-\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^2}\right )dx^2-a \int x^4 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {1}{2} \int \left (\frac {\left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{a^2}-\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^2}\right )dx^2-\frac {1}{5} a x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-p,\frac {7}{2},a^2 x^2\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \left (1-a^2 x^2\right )^{p+\frac {3}{2}}}{a^4 (2 p+3)}-\frac {2 \left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^4 (2 p+1)}\right )-\frac {1}{5} a x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-p,\frac {7}{2},a^2 x^2\right )\) |
((-2*(1 - a^2*x^2)^(1/2 + p))/(a^4*(1 + 2*p)) + (2*(1 - a^2*x^2)^(3/2 + p) )/(a^4*(3 + 2*p)))/2 - (a*x^5*Hypergeometric2F1[5/2, 1/2 - p, 7/2, a^2*x^2 ])/5
3.13.16.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c Int[x^m*(a + b*x^2)^p, x], x] + Simp[d Int[x^(m + 1)*(a + b*x^2 )^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && IntegerQ[m] && !IntegerQ[2*p]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[c^p Int[x^m*((1 - a^2*x^2)^(p + n/2)/(1 - a*x)^n), x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c , 0]) && ILtQ[(n - 1)/2, 0] && !IntegerQ[p - n/2]
\[\int \frac {x^{3} \left (-a^{2} x^{2}+1\right )^{p} \sqrt {-a^{2} x^{2}+1}}{a x +1}d x\]
\[ \int e^{-\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} x^{2} + 1\right )}^{p} x^{3}}{a x + 1} \,d x } \]
\[ \int e^{-\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=\int \frac {x^{3} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{p}}{a x + 1}\, dx \]
\[ \int e^{-\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} x^{2} + 1\right )}^{p} x^{3}}{a x + 1} \,d x } \]
\[ \int e^{-\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} x^{2} + 1\right )}^{p} x^{3}}{a x + 1} \,d x } \]
Timed out. \[ \int e^{-\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=\int \frac {x^3\,{\left (1-a^2\,x^2\right )}^p\,\sqrt {1-a^2\,x^2}}{a\,x+1} \,d x \]