\(\int e^{-2 \text {arctanh}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx\) [1274]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 27, antiderivative size = 172 \[ \int e^{-2 \text {arctanh}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=-\frac {x^{1+m} \sqrt {c-a^2 c x^2}}{2+m}+\frac {c (3+2 m) x^{1+m} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{(1+m) (2+m) \sqrt {c-a^2 c x^2}}-\frac {2 a c x^{2+m} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{(2+m) \sqrt {c-a^2 c x^2}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.13 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.65 \[ \int e^{-2 \text {arctanh}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\frac {x^{1+m} \left (\frac {2 \sqrt {c-a c x} \operatorname {AppellF1}\left (1+m,\frac {1}{2},-\frac {1}{2},2+m,-a x,a x\right )}{\sqrt {1-a x}}-\frac {\sqrt {c-a^2 c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{\sqrt {1-a^2 x^2}}\right )}{1+m} \] Input:

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

Output:

(x^(1 + m)*((2*Sqrt[c - a*c*x]*AppellF1[1 + m, 1/2, -1/2, 2 + m, -(a*x), a 
*x])/Sqrt[1 - a*x] - (Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[-1/2, (1 + m)/ 
2, (3 + m)/2, a^2*x^2])/Sqrt[1 - a^2*x^2]))/(1 + m)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6702, 559, 25, 27, 557, 279, 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.

\(\displaystyle \int x^m e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2} \, dx\)

\(\Big \downarrow \) 6702

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

\(\Big \downarrow \) 559

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 557

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

c*(-((x^(1 + m)*Sqrt[c - a^2*c*x^2])/(c*(2 + m))) + (((3 + 2*m)*x^(1 + m)* 
Sqrt[1 - a^2*x^2]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, a^2*x^2])/( 
(1 + m)*Sqrt[c - a^2*c*x^2]) - (2*a*x^(2 + m)*Sqrt[1 - a^2*x^2]*Hypergeome 
tric2F1[1/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/Sqrt[c - a^2*c*x^2])/(2 + m))
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 279
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 557
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym 
bol] :> Simp[c   Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e   Int[(e*x)^( 
m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
 

rule 559
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[d^n*(e*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*e^(n - 1)*( 
m + n + 2*p + 1))), x] + Simp[1/(b*(m + n + 2*p + 1))   Int[(e*x)^m*(a + b* 
x^2)^p*ExpandToSum[b*(m + n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1 
)*x^n - a*d^n*(m + n - 1)*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, e, m, 
 p}, x] && IGtQ[n, 1] &&  !IntegerQ[m] && NeQ[m + n + 2*p + 1, 0]
 

rule 6702
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Simp[1/c^(n/2)   Int[x^m*((c + d*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/2, 0]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int e^{-2 \text {arctanh}(a x)} x^m \sqrt {c-a^2 c x^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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