Integrand size = 22, antiderivative size = 85 \[ \int e^{\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=-\frac {\left (1-a^2 x^2\right )^{\frac {1}{2}+p}}{a^4 (1+2 p)}+\frac {\left (1-a^2 x^2\right )^{\frac {3}{2}+p}}{a^4 (3+2 p)}+\frac {1}{5} a x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-p,\frac {7}{2},a^2 x^2\right ) \] Output:
-(-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 \[ \int e^{\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=-\frac {\left (1-a^2 x^2\right )^{\frac {1}{2}+p} \left (2+a^2 (1+2 p) x^2\right )}{a^4 \left (3+8 p+4 p^2\right )}+\frac {1}{5} a x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-p,\frac {7}{2},a^2 x^2\right ) \] Input:
Integrate[E^ArcTanh[a*x]*x^3*(1 - a^2*x^2)^p,x]
Output:
-(((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 = 0.36 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6698, 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.
\(\displaystyle \int x^3 e^{\text {arctanh}(a x)} \left (1-a^2 x^2\right )^p \, dx\) |
\(\Big \downarrow \) 6698 |
\(\displaystyle \int x^3 (a x+1) \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx\) |
\(\Big \downarrow \) 542 |
\(\displaystyle a \int x^4 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx+\int x^3 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx\) |
\(\Big \downarrow \) 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 |
\(\displaystyle \frac {1}{5} a x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-p,\frac {7}{2},a^2 x^2\right )+\frac {1}{2} \left (\frac {2 \left (1-a^2 x^2\right )^{p+\frac {3}{2}}}{a^4 (2 p+3)}-\frac {2 \left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^4 (2 p+1)}\right )\) |
Input:
Int[E^ArcTanh[a*x]*x^3*(1 - a^2*x^2)^p,x]
Output:
((-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
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]) && IGtQ[(n + 1)/2, 0] && !IntegerQ[p - n/2]
Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.55
method | result | size |
meijerg | \(\frac {a \,x^{5} \operatorname {hypergeom}\left (\left [\frac {5}{2}, \frac {1}{2}-p \right ], \left [\frac {7}{2}\right ], a^{2} x^{2}\right )}{5}+\frac {x^{4} \operatorname {hypergeom}\left (\left [2, \frac {1}{2}-p \right ], \left [3\right ], a^{2} x^{2}\right )}{4}\) | \(47\) |
Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a^2*x^2+1)^p,x,method=_RETURNVERBOSE)
Output:
1/5*a*x^5*hypergeom([5/2,1/2-p],[7/2],a^2*x^2)+1/4*x^4*hypergeom([2,1/2-p] ,[3],a^2*x^2)
\[ \int e^{\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=\int { \frac {{\left (a x + 1\right )} {\left (-a^{2} x^{2} + 1\right )}^{p} x^{3}}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a^2*x^2+1)^p,x, algorithm="fric as")
Output:
integral(-sqrt(-a^2*x^2 + 1)*(-a^2*x^2 + 1)^p*x^3/(a*x - 1), x)
Result contains complex when optimal does not.
Time = 9.31 (sec) , antiderivative size = 258, normalized size of antiderivative = 3.04 \[ \int e^{\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=- \frac {a a^{2 p} x^{2 p + 5} e^{i \pi p} \Gamma \left (- p - \frac {5}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} 1, - p, - p - \frac {5}{2} \\ \frac {1}{2}, - p - \frac {3}{2} \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + 1\right )} - \frac {a^{2 p + 5} x^{2 p + 5} e^{i \pi p} \Gamma \left (- p - \frac {5}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, p + \frac {5}{2} \\ p + 1, p + \frac {7}{2} \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } a^{4} \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + 1\right )} - \frac {{G_{3, 3}^{2, 2}\left (\begin {matrix} - p - 1, 1 & -1 \\- p - \frac {3}{2}, - p - 1 & 0 \end {matrix} \middle | {\frac {e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac {1}{2}\right )}{2 \pi a^{4}} - \frac {{G_{3, 3}^{1, 3}\left (\begin {matrix} -1, - p - 2, 1 & \\- p - 2 & - p - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac {1}{2}\right )}{2 a^{4} \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} \] Input:
integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**3*(-a**2*x**2+1)**p,x)
Output:
-a*a**(2*p)*x**(2*p + 5)*exp(I*pi*p)*gamma(-p - 5/2)*gamma(p + 1/2)*hyper( (1, -p, -p - 5/2), (1/2, -p - 3/2), 1/(a**2*x**2))/(2*sqrt(pi)*gamma(-p - 3/2)*gamma(p + 1)) - a**(2*p + 5)*x**(2*p + 5)*exp(I*pi*p)*gamma(-p - 5/2) *gamma(p + 1/2)*hyper((1/2, 1, p + 5/2), (p + 1, p + 7/2), a**2*x**2*exp_p olar(2*I*pi))/(2*sqrt(pi)*a**4*gamma(-p - 3/2)*gamma(p + 1)) - meijerg(((- p - 1, 1), (-1,)), ((-p - 3/2, -p - 1), (0,)), exp_polar(-I*pi)/(a**2*x**2 ))*gamma(p + 1/2)/(2*pi*a**4) - meijerg(((-1, -p - 2, 1), ()), ((-p - 2,), (-p - 3/2, 0)), exp_polar(-I*pi)/(a**2*x**2))*gamma(p + 1/2)/(2*a**4*gamm a(-p)*gamma(p + 1))
\[ \int e^{\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=\int { \frac {{\left (a x + 1\right )} {\left (-a^{2} x^{2} + 1\right )}^{p} x^{3}}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a^2*x^2+1)^p,x, algorithm="maxi ma")
Output:
a*integrate(x^4*e^(p*log(a*x + 1) + p*log(-a*x + 1))/(sqrt(a*x + 1)*sqrt(- a*x + 1)), x) + (a^4*(2*p + 1)*x^4 - a^2*(2*p - 1)*x^2 - 2)*(-a^2*x^2 + 1) ^p/(sqrt(-a^2*x^2 + 1)*(4*p^2 + 8*p + 3)*a^4)
\[ \int e^{\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=\int { \frac {{\left (a x + 1\right )} {\left (-a^{2} x^{2} + 1\right )}^{p} x^{3}}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a^2*x^2+1)^p,x, algorithm="giac ")
Output:
integrate((a*x + 1)*(-a^2*x^2 + 1)^p*x^3/sqrt(-a^2*x^2 + 1), x)
Timed out. \[ \int e^{\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=\int \frac {x^3\,{\left (1-a^2\,x^2\right )}^p\,\left (a\,x+1\right )}{\sqrt {1-a^2\,x^2}} \,d x \] Input:
int((x^3*(1 - a^2*x^2)^p*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)
Output:
int((x^3*(1 - a^2*x^2)^p*(a*x + 1))/(1 - a^2*x^2)^(1/2), x)
\[ \int e^{\text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^p \, dx=\frac {-2 \left (-a^{2} x^{2}+1\right )^{p +\frac {1}{2}} a^{2} p \,x^{2}-\left (-a^{2} x^{2}+1\right )^{p +\frac {1}{2}} a^{2} x^{2}-2 \left (-a^{2} x^{2}+1\right )^{p +\frac {1}{2}}+4 \left (\int \frac {\left (-a^{2} x^{2}+1\right )^{p} x^{4}}{\sqrt {-a^{2} x^{2}+1}}d x \right ) a^{5} p^{2}+8 \left (\int \frac {\left (-a^{2} x^{2}+1\right )^{p} x^{4}}{\sqrt {-a^{2} x^{2}+1}}d x \right ) a^{5} p +3 \left (\int \frac {\left (-a^{2} x^{2}+1\right )^{p} x^{4}}{\sqrt {-a^{2} x^{2}+1}}d x \right ) a^{5}}{a^{4} \left (4 p^{2}+8 p +3\right )} \] Input:
int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3*(-a^2*x^2+1)^p,x)
Output:
( - 2*( - a**2*x**2 + 1)**((2*p + 1)/2)*a**2*p*x**2 - ( - a**2*x**2 + 1)** ((2*p + 1)/2)*a**2*x**2 - 2*( - a**2*x**2 + 1)**((2*p + 1)/2) + 4*int((( - a**2*x**2 + 1)**p*x**4)/sqrt( - a**2*x**2 + 1),x)*a**5*p**2 + 8*int((( - a**2*x**2 + 1)**p*x**4)/sqrt( - a**2*x**2 + 1),x)*a**5*p + 3*int((( - a**2 *x**2 + 1)**p*x**4)/sqrt( - a**2*x**2 + 1),x)*a**5)/(a**4*(4*p**2 + 8*p + 3))