Integrand size = 20, antiderivative size = 58 \[ \int e^{\text {arctanh}(a x)} x \left (1-a^2 x^2\right )^p \, dx=-\frac {\left (1-a^2 x^2\right )^{\frac {1}{2}+p}}{a^2 (1+2 p)}+\frac {1}{3} a x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-p,\frac {5}{2},a^2 x^2\right ) \]
Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03 \[ \int e^{\text {arctanh}(a x)} x \left (1-a^2 x^2\right )^p \, dx=-\frac {\left (1-a^2 x^2\right )^{\frac {1}{2}+p}}{2 a^2 \left (\frac {1}{2}+p\right )}+\frac {1}{3} a x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-p,\frac {5}{2},a^2 x^2\right ) \]
-1/2*(1 - a^2*x^2)^(1/2 + p)/(a^2*(1/2 + p)) + (a*x^3*Hypergeometric2F1[3/ 2, 1/2 - p, 5/2, a^2*x^2])/3
Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6698, 542, 241, 278}
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.
\(\displaystyle \int x e^{\text {arctanh}(a x)} \left (1-a^2 x^2\right )^p \, dx\) |
\(\Big \downarrow \) 6698 |
\(\displaystyle \int x (a x+1) \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx\) |
\(\Big \downarrow \) 542 |
\(\displaystyle \int x \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx+a \int x^2 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx\) |
\(\Big \downarrow \) 241 |
\(\displaystyle a \int x^2 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}dx-\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^2 (2 p+1)}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {1}{3} a x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2}-p,\frac {5}{2},a^2 x^2\right )-\frac {\left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^2 (2 p+1)}\) |
-((1 - a^2*x^2)^(1/2 + p)/(a^2*(1 + 2*p))) + (a*x^3*Hypergeometric2F1[3/2, 1/2 - p, 5/2, a^2*x^2])/3
3.11.7.3.1 Defintions of rubi rules used
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
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 = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.81
method | result | size |
meijerg | \(\frac {a \,x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {1}{2}-p \right ], \left [\frac {5}{2}\right ], a^{2} x^{2}\right )}{3}+\frac {x^{2} \operatorname {hypergeom}\left (\left [1, \frac {1}{2}-p \right ], \left [2\right ], a^{2} x^{2}\right )}{2}\) | \(47\) |
\[ \int e^{\text {arctanh}(a x)} x \left (1-a^2 x^2\right )^p \, dx=\int { \frac {{\left (a x + 1\right )} {\left (-a^{2} x^{2} + 1\right )}^{p} x}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \]
Result contains complex when optimal does not.
Time = 5.85 (sec) , antiderivative size = 243, normalized size of antiderivative = 4.19 \[ \int e^{\text {arctanh}(a x)} x \left (1-a^2 x^2\right )^p \, dx=- \frac {a a^{2 p} x^{2 p + 3} e^{i \pi p} \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} 1, - p, - p - \frac {3}{2} \\ \frac {1}{2}, - p - \frac {1}{2} \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } \Gamma \left (- p - \frac {1}{2}\right ) \Gamma \left (p + 1\right )} + \begin {cases} \frac {x^{2}}{2} & \text {for}\: a^{2} = 0 \\\frac {\begin {cases} \frac {\sqrt {- a^{2} x^{2} + 1} \left (- a^{2} x^{2} + 1\right )^{p}}{- 2 p - 1} & \text {for}\: p \neq - \frac {1}{2} \\\sqrt {- a^{2} x^{2} + 1} \left (- a^{2} x^{2} + 1\right )^{p} \log {\left (\frac {1}{\sqrt {- a^{2} x^{2} + 1}} \right )} & \text {otherwise} \end {cases}}{a^{2}} & \text {otherwise} \end {cases} - \frac {a^{2 p + 3} x^{2 p + 3} e^{i \pi p} \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, p + \frac {3}{2} \\ p + 1, p + \frac {5}{2} \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } a^{2} \Gamma \left (- p - \frac {1}{2}\right ) \Gamma \left (p + 1\right )} \]
-a*a**(2*p)*x**(2*p + 3)*exp(I*pi*p)*gamma(-p - 3/2)*gamma(p + 1/2)*hyper( (1, -p, -p - 3/2), (1/2, -p - 1/2), 1/(a**2*x**2))/(2*sqrt(pi)*gamma(-p - 1/2)*gamma(p + 1)) + Piecewise((x**2/2, Eq(a**2, 0)), (Piecewise((sqrt(-a* *2*x**2 + 1)*(-a**2*x**2 + 1)**p/(-2*p - 1), Ne(p, -1/2)), (sqrt(-a**2*x** 2 + 1)*(-a**2*x**2 + 1)**p*log(1/sqrt(-a**2*x**2 + 1)), True))/a**2, True) ) - a**(2*p + 3)*x**(2*p + 3)*exp(I*pi*p)*gamma(-p - 3/2)*gamma(p + 1/2)*h yper((1/2, 1, p + 3/2), (p + 1, p + 5/2), a**2*x**2*exp_polar(2*I*pi))/(2* sqrt(pi)*a**2*gamma(-p - 1/2)*gamma(p + 1))
\[ \int e^{\text {arctanh}(a x)} x \left (1-a^2 x^2\right )^p \, dx=\int { \frac {{\left (a x + 1\right )} {\left (-a^{2} x^{2} + 1\right )}^{p} x}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \]
a*integrate(x^2*e^(p*log(a*x + 1) + p*log(-a*x + 1))/(sqrt(a*x + 1)*sqrt(- a*x + 1)), x) - (-a^2*x^2 + 1)^(p + 1/2)/(a^2*(2*p + 1))
\[ \int e^{\text {arctanh}(a x)} x \left (1-a^2 x^2\right )^p \, dx=\int { \frac {{\left (a x + 1\right )} {\left (-a^{2} x^{2} + 1\right )}^{p} x}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \]
Timed out. \[ \int e^{\text {arctanh}(a x)} x \left (1-a^2 x^2\right )^p \, dx=\int \frac {x\,{\left (1-a^2\,x^2\right )}^p\,\left (a\,x+1\right )}{\sqrt {1-a^2\,x^2}} \,d x \]