\(\int \frac {e^{\text {arctanh}(a x)} (1-a^2 x^2)^p}{x^3} \, dx\) [1033]

Optimal result

Integrand size = 22, antiderivative size = 78 \[ \int \frac {e^{\text {arctanh}(a x)} \left (1-a^2 x^2\right )^p}{x^3} \, dx=-\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2}-p,\frac {1}{2},a^2 x^2\right )}{x}-\frac {a^2 \left (1-a^2 x^2\right )^{\frac {1}{2}+p} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2}+p,\frac {3}{2}+p,1-a^2 x^2\right )}{1+2 p} \] Output:

-a*hypergeom([-1/2, 1/2-p],[1/2],a^2*x^2)/x-a^2*(-a^2*x^2+1)^(1/2+p)*hyper 
geom([2, 1/2+p],[3/2+p],-a^2*x^2+1)/(1+2*p)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\text {arctanh}(a x)} \left (1-a^2 x^2\right )^p}{x^3} \, dx=-\frac {a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2}-p,\frac {1}{2},a^2 x^2\right )}{x}-\frac {a^2 \left (1-a^2 x^2\right )^{\frac {1}{2}+p} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2}+p,\frac {3}{2}+p,1-a^2 x^2\right )}{2 \left (\frac {1}{2}+p\right )} \] Input:

Integrate[(E^ArcTanh[a*x]*(1 - a^2*x^2)^p)/x^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6698, 542, 243, 75, 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.

\(\displaystyle \int \frac {e^{\text {arctanh}(a x)} \left (1-a^2 x^2\right )^p}{x^3} \, dx\)

\(\Big \downarrow \) 6698

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

\(\Big \downarrow \) 542

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 75

\(\displaystyle a \int \frac {\left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{x^2}dx-\frac {a^2 \left (1-a^2 x^2\right )^{p+\frac {1}{2}} \operatorname {Hypergeometric2F1}\left (2,p+\frac {1}{2},p+\frac {3}{2},1-a^2 x^2\right )}{2 p+1}\)

\(\Big \downarrow \) 278

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

Input:

Int[(E^ArcTanh[a*x]*(1 - a^2*x^2)^p)/x^3,x]
 

Output:

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

Defintions of rubi rules used

rule 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 243

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]
 

rule 278

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])
 

rule 542

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]
 

rule 6698

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]
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.44

method	result	size
meijerg	\(-\frac {a \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {1}{2}-p \right ], \left [\frac {1}{2}\right ], a^{2} x^{2}\right )}{x}-\frac {a^{2} \left (\frac {\Gamma \left (\frac {1}{2}-p \right )}{x^{2} a^{2}}-\left (\Psi \left (\frac {3}{2}-p \right )+\gamma -1+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \Gamma \left (\frac {3}{2}-p \right )-\frac {\Gamma \left (\frac {5}{2}-p \right ) a^{2} x^{2} \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{2}-p \right ], \left [2, 3\right ], a^{2} x^{2}\right )}{2}\right )}{2 \Gamma \left (\frac {1}{2}-p \right )}\)	\(112\)

Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*x^2+1)^p/x^3,x,method=_RETURNVERBOSE)
 

Output:

-a*hypergeom([-1/2,1/2-p],[1/2],a^2*x^2)/x-1/2*a^2*(GAMMA(1/2-p)/x^2/a^2-( 
Psi(3/2-p)+gamma-1+2*ln(x)+ln(-a^2))*GAMMA(3/2-p)-1/2*GAMMA(5/2-p)*a^2*x^2 
*hypergeom([1,1,5/2-p],[2,3],a^2*x^2))/GAMMA(1/2-p)
 

Fricas [F]

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

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*x^2+1)^p/x^3,x, algorithm="fric 
as")
 

Output:

integral(-sqrt(-a^2*x^2 + 1)*(-a^2*x^2 + 1)^p/(a*x^4 - x^3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 11.35 (sec) , antiderivative size = 292, normalized size of antiderivative = 3.74 \[ \int \frac {e^{\text {arctanh}(a x)} \left (1-a^2 x^2\right )^p}{x^3} \, dx=- \frac {a^{2} a^{2 p - 2} x^{2 p - 2} e^{i \pi p} \Gamma \left (1 - p\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, p - 1 \\ p, p + 1 \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } \Gamma \left (2 - p\right ) \Gamma \left (p + 1\right )} - \frac {a^{2} a^{2 p - 1} x^{2 p - 1} e^{i \pi p} \Gamma \left (\frac {1}{2} - p\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, p - \frac {1}{2} \\ p + \frac {1}{2}, p + 1 \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } \Gamma \left (\frac {3}{2} - p\right ) \Gamma \left (p + 1\right )} - \frac {a a^{2 p} x^{2 p - 1} e^{i \pi p} \Gamma \left (\frac {1}{2} - p\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} 1, - p, \frac {1}{2} - p \\ \frac {1}{2}, \frac {3}{2} - p \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } \Gamma \left (\frac {3}{2} - p\right ) \Gamma \left (p + 1\right )} - \frac {a^{2 p} x^{2 p - 2} e^{i \pi p} \Gamma \left (1 - p\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} 1, - p, 1 - p \\ \frac {1}{2}, 2 - p \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } \Gamma \left (2 - p\right ) \Gamma \left (p + 1\right )} \] Input:

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a**2*x**2+1)**p/x**3,x)
 

Output:

-a**2*a**(2*p - 2)*x**(2*p - 2)*exp(I*pi*p)*gamma(1 - p)*gamma(p + 1/2)*hy 
per((1/2, 1, p - 1), (p, p + 1), a**2*x**2*exp_polar(2*I*pi))/(2*sqrt(pi)* 
gamma(2 - p)*gamma(p + 1)) - a**2*a**(2*p - 1)*x**(2*p - 1)*exp(I*pi*p)*ga 
mma(1/2 - p)*gamma(p + 1/2)*hyper((1/2, 1, p - 1/2), (p + 1/2, p + 1), a** 
2*x**2*exp_polar(2*I*pi))/(2*sqrt(pi)*gamma(3/2 - p)*gamma(p + 1)) - a*a** 
(2*p)*x**(2*p - 1)*exp(I*pi*p)*gamma(1/2 - p)*gamma(p + 1/2)*hyper((1, -p, 
 1/2 - p), (1/2, 3/2 - p), 1/(a**2*x**2))/(2*sqrt(pi)*gamma(3/2 - p)*gamma 
(p + 1)) - a**(2*p)*x**(2*p - 2)*exp(I*pi*p)*gamma(1 - p)*gamma(p + 1/2)*h 
yper((1, -p, 1 - p), (1/2, 2 - p), 1/(a**2*x**2))/(2*sqrt(pi)*gamma(2 - p) 
*gamma(p + 1))
 

Maxima [F]

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

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*x^2+1)^p/x^3,x, algorithm="maxi 
ma")
 

Output:

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

Giac [F]

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

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*x^2+1)^p/x^3,x, algorithm="giac 
")
 

Output:

integrate((a*x + 1)*(-a^2*x^2 + 1)^p/(sqrt(-a^2*x^2 + 1)*x^3), x)
 

Mupad [F(-1)]

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

int(((1 - a^2*x^2)^p*(a*x + 1))/(x^3*(1 - a^2*x^2)^(1/2)),x)
 

Output:

int(((1 - a^2*x^2)^p*(a*x + 1))/(x^3*(1 - a^2*x^2)^(1/2)), x)
 

Reduce [F]

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

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a^2*x^2+1)^p/x^3,x)
 

Output:

int(( - a**2*x**2 + 1)**p/(sqrt( - a**2*x**2 + 1)*x**3),x) + int(( - a**2* 
x**2 + 1)**p/(sqrt( - a**2*x**2 + 1)*x**2),x)*a