\(\int \frac {e^{\text {arctanh}(a x)} x^m}{(c-a^2 c x^2)^{3/2}} \, dx\) [1039]

Optimal result

Integrand size = 25, antiderivative size = 134 \[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {x^{1+m} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{c (1+m) \sqrt {c-a^2 c x^2}}+\frac {a x^{2+m} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{c (2+m) \sqrt {c-a^2 c x^2}} \] Output:

x^(1+m)*(-a^2*x^2+1)^(1/2)*hypergeom([2, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/c 
/(1+m)/(-a^2*c*x^2+c)^(1/2)+a*x^(2+m)*(-a^2*x^2+1)^(1/2)*hypergeom([2, 1+1 
/2*m],[2+1/2*m],a^2*x^2)/c/(2+m)/(-a^2*c*x^2+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},1+\frac {1+m}{2},a^2 x^2\right )}{1+m}+\frac {a x^{2+m} \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},1+\frac {2+m}{2},a^2 x^2\right )}{2+m}\right )}{c \sqrt {c-a^2 c x^2}} \] Input:

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

Output:

(Sqrt[1 - a^2*x^2]*((x^(1 + m)*Hypergeometric2F1[2, (1 + m)/2, 1 + (1 + m) 
/2, a^2*x^2])/(1 + m) + (a*x^(2 + m)*Hypergeometric2F1[2, (2 + m)/2, 1 + ( 
2 + m)/2, a^2*x^2])/(2 + m)))/(c*Sqrt[c - a^2*c*x^2])
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.77, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6703, 6700, 92, 82, 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.

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

\(\Big \downarrow \) 6703

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{\text {arctanh}(a x)} x^m}{\left (1-a^2 x^2\right )^{3/2}}dx}{c \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 6700

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {x^m}{(1-a x)^2 (a x+1)}dx}{c \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 92

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (a \int \frac {x^{m+1}}{(1-a x)^2 (a x+1)^2}dx+\int \frac {x^m}{(1-a x)^2 (a x+1)^2}dx\right )}{c \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 82

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (a \int \frac {x^{m+1}}{\left (1-a^2 x^2\right )^2}dx+\int \frac {x^m}{\left (1-a^2 x^2\right )^2}dx\right )}{c \sqrt {c-a^2 c x^2}}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (2,\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \operatorname {Hypergeometric2F1}\left (2,\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{m+2}\right )}{c \sqrt {c-a^2 c x^2}}\)

Input:

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

Output:

(Sqrt[1 - a^2*x^2]*((x^(1 + m)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, 
a^2*x^2])/(1 + m) + (a*x^(2 + m)*Hypergeometric2F1[2, (2 + m)/2, (4 + m)/2 
, a^2*x^2])/(2 + m)))/(c*Sqrt[c - a^2*c*x^2])
 

Defintions of rubi rules used

rule 82

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) 
)^(p_.), x_] :> Int[(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, 
 e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && IntegerQ[m]
 

rule 92

Int[((f_.)*(x_))^(p_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), 
x_] :> Simp[a   Int[(a + b*x)^n*(c + d*x)^n*(f*x)^p, x], x] + Simp[b/f   In 
t[(a + b*x)^n*(c + d*x)^n*(f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, f, m, 
 n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n - 1, 0] &&  !RationalQ[p] && 
!IGtQ[m, 0] && NeQ[m + n + p + 2, 0]
 

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 6700

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[c^p   Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], 
 x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || 
 GtQ[c, 0])
 

rule 6703

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_ 
Symbol] :> Simp[c^IntPart[p]*((c + d*x^2)^FracPart[p]/(1 - a^2*x^2)^FracPar 
t[p])   Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, 
d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !I 
ntegerQ[n/2]
 

Maple [F]

\[\int \frac {\left (a x +1\right ) x^{m}}{\sqrt {-a^{2} x^{2}+1}\, \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^m/(-a^2*c*x^2+c)^(3/2),x, algorithm 
="fricas")
 

Output:

integral(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*x^m/(a^5*c^2*x^5 - a^4*c 
^2*x^4 - 2*a^3*c^2*x^3 + 2*a^2*c^2*x^2 + a*c^2*x - c^2), x)
 

Sympy [F]

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

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

Output:

Integral(x**m*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1 
))**(3/2)), x)
 

Maxima [F]

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

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^m/(-a^2*c*x^2+c)^(3/2),x, algorithm 
="maxima")
 

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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