\(\int e^{2 \text {arctanh}(a x)} x^m (c-a^2 c x^2)^{5/2} \, dx\) [1168]

Optimal result

Integrand size = 27, antiderivative size = 176 \[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{5/2} \, dx=-\frac {x^{1+m} \left (c-a^2 c x^2\right )^{5/2}}{6+m}+\frac {c^2 (7+2 m) x^{1+m} \sqrt {c-a^2 c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{(1+m) (6+m) \sqrt {1-a^2 x^2}}+\frac {2 a c^2 x^{2+m} \sqrt {c-a^2 c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{(2+m) \sqrt {1-a^2 x^2}} \] Output:

-x^(1+m)*(-a^2*c*x^2+c)^(5/2)/(6+m)+c^2*(7+2*m)*x^(1+m)*(-a^2*c*x^2+c)^(1/ 
2)*hypergeom([-3/2, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/(1+m)/(6+m)/(-a^2*x^2+ 
1)^(1/2)+2*a*c^2*x^(2+m)*(-a^2*c*x^2+c)^(1/2)*hypergeom([-3/2, 1+1/2*m],[2 
+1/2*m],a^2*x^2)/(2+m)/(-a^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.02 \[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{5/2} \, dx=\frac {c^2 x^{1+m} \sqrt {c-a^2 c x^2} \left (\frac {2 a x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1+\frac {m}{2},2+\frac {m}{2},a^2 x^2\right )}{2+m}-\frac {2 a^3 x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2+\frac {m}{2},3+\frac {m}{2},a^2 x^2\right )}{4+m}+\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{1+m}-\frac {a^4 x^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {5+m}{2},\frac {7+m}{2},a^2 x^2\right )}{5+m}\right )}{\sqrt {1-a^2 x^2}} \] Input:

Integrate[E^(2*ArcTanh[a*x])*x^m*(c - a^2*c*x^2)^(5/2),x]
 

Output:

(c^2*x^(1 + m)*Sqrt[c - a^2*c*x^2]*((2*a*x*Hypergeometric2F1[-1/2, 1 + m/2 
, 2 + m/2, a^2*x^2])/(2 + m) - (2*a^3*x^3*Hypergeometric2F1[-1/2, 2 + m/2, 
 3 + m/2, a^2*x^2])/(4 + m) + Hypergeometric2F1[-1/2, (1 + m)/2, (3 + m)/2 
, a^2*x^2]/(1 + m) - (a^4*x^4*Hypergeometric2F1[-1/2, (5 + m)/2, (7 + m)/2 
, a^2*x^2])/(5 + m)))/Sqrt[1 - a^2*x^2]
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6701, 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.

\(\displaystyle \int x^m e^{2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{5/2} \, dx\)

\(\Big \downarrow \) 6701

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

\(\Big \downarrow \) 559

\(\displaystyle c \left (-\frac {\int -a^2 c x^m (2 m+2 a (m+6) x+7) \left (c-a^2 c x^2\right )^{3/2}dx}{a^2 c (m+6)}-\frac {x^{m+1} \left (c-a^2 c x^2\right )^{5/2}}{c (m+6)}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (\frac {\int a^2 c x^m (2 m+2 a (m+6) x+7) \left (c-a^2 c x^2\right )^{3/2}dx}{a^2 c (m+6)}-\frac {x^{m+1} \left (c-a^2 c x^2\right )^{5/2}}{c (m+6)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {\int x^m (2 m+2 a (m+6) x+7) \left (c-a^2 c x^2\right )^{3/2}dx}{m+6}-\frac {x^{m+1} \left (c-a^2 c x^2\right )^{5/2}}{c (m+6)}\right )\)

\(\Big \downarrow \) 557

\(\displaystyle c \left (\frac {2 a (m+6) \int x^{m+1} \left (c-a^2 c x^2\right )^{3/2}dx+(2 m+7) \int x^m \left (c-a^2 c x^2\right )^{3/2}dx}{m+6}-\frac {x^{m+1} \left (c-a^2 c x^2\right )^{5/2}}{c (m+6)}\right )\)

\(\Big \downarrow \) 279

\(\displaystyle c \left (\frac {\frac {2 a c (m+6) \sqrt {c-a^2 c x^2} \int x^{m+1} \left (1-a^2 x^2\right )^{3/2}dx}{\sqrt {1-a^2 x^2}}+\frac {c (2 m+7) \sqrt {c-a^2 c x^2} \int x^m \left (1-a^2 x^2\right )^{3/2}dx}{\sqrt {1-a^2 x^2}}}{m+6}-\frac {x^{m+1} \left (c-a^2 c x^2\right )^{5/2}}{c (m+6)}\right )\)

\(\Big \downarrow \) 278

\(\displaystyle c \left (\frac {\frac {c (2 m+7) x^{m+1} \sqrt {c-a^2 c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{(m+1) \sqrt {1-a^2 x^2}}+\frac {2 a c (m+6) x^{m+2} \sqrt {c-a^2 c x^2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{(m+2) \sqrt {1-a^2 x^2}}}{m+6}-\frac {x^{m+1} \left (c-a^2 c x^2\right )^{5/2}}{c (m+6)}\right )\)

Input:

Int[E^(2*ArcTanh[a*x])*x^m*(c - a^2*c*x^2)^(5/2),x]
 

Output:

c*(-((x^(1 + m)*(c - a^2*c*x^2)^(5/2))/(c*(6 + m))) + ((c*(7 + 2*m)*x^(1 + 
 m)*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[-3/2, (1 + m)/2, (3 + m)/2, a^2* 
x^2])/((1 + m)*Sqrt[1 - a^2*x^2]) + (2*a*c*(6 + m)*x^(2 + m)*Sqrt[c - a^2* 
c*x^2]*Hypergeometric2F1[-3/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/((2 + m)*Sq 
rt[1 - a^2*x^2]))/(6 + 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 6701

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Simp[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]) && IGtQ[n/2, 0]
 

Maple [F]

\[\int \frac {\left (a x +1\right )^{2} x^{m} \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{-a^{2} x^{2}+1}d x\]

Input:

int((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c)^(5/2),x)
 

Output:

int((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c)^(5/2),x)
 

Fricas [F]

\[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{5/2} \, dx=\int { -\frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (a x + 1\right )}^{2} x^{m}}{a^{2} x^{2} - 1} \,d x } \] Input:

integrate((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c)^(5/2),x, algorithm="fr 
icas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 11.97 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.26 \[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{5/2} \, dx=- \frac {a^{4} c^{\frac {5}{2}} x^{m + 5} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {5}{2} \\ \frac {m}{2} + \frac {7}{2} \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} - \frac {a^{3} c^{\frac {5}{2}} x^{m + 4} \Gamma \left (\frac {m}{2} + 2\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 2 \\ \frac {m}{2} + 3 \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{\Gamma \left (\frac {m}{2} + 3\right )} + \frac {a c^{\frac {5}{2}} x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{\Gamma \left (\frac {m}{2} + 2\right )} + \frac {c^{\frac {5}{2}} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} \] Input:

integrate((a*x+1)**2/(-a**2*x**2+1)*x**m*(-a**2*c*x**2+c)**(5/2),x)
 

Output:

-a**4*c**(5/2)*x**(m + 5)*gamma(m/2 + 5/2)*hyper((-1/2, m/2 + 5/2), (m/2 + 
 7/2,), a**2*x**2*exp_polar(2*I*pi))/(2*gamma(m/2 + 7/2)) - a**3*c**(5/2)* 
x**(m + 4)*gamma(m/2 + 2)*hyper((-1/2, m/2 + 2), (m/2 + 3,), a**2*x**2*exp 
_polar(2*I*pi))/gamma(m/2 + 3) + a*c**(5/2)*x**(m + 2)*gamma(m/2 + 1)*hype 
r((-1/2, m/2 + 1), (m/2 + 2,), a**2*x**2*exp_polar(2*I*pi))/gamma(m/2 + 2) 
 + c**(5/2)*x**(m + 1)*gamma(m/2 + 1/2)*hyper((-1/2, m/2 + 1/2), (m/2 + 3/ 
2,), a**2*x**2*exp_polar(2*I*pi))/(2*gamma(m/2 + 3/2))
 

Maxima [F]

\[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{5/2} \, dx=\int { -\frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} {\left (a x + 1\right )}^{2} x^{m}}{a^{2} x^{2} - 1} \,d x } \] Input:

integrate((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c)^(5/2),x, algorithm="ma 
xima")
 

Output:

-integrate((-a^2*c*x^2 + c)^(5/2)*(a*x + 1)^2*x^m/(a^2*x^2 - 1), x)
 

Giac [F(-2)]

Exception generated. \[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c)^(5/2),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. \[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{5/2} \, dx=\int -\frac {x^m\,{\left (c-a^2\,c\,x^2\right )}^{5/2}\,{\left (a\,x+1\right )}^2}{a^2\,x^2-1} \,d x \] Input:

int(-(x^m*(c - a^2*c*x^2)^(5/2)*(a*x + 1)^2)/(a^2*x^2 - 1),x)
 

Output:

int(-(x^m*(c - a^2*c*x^2)^(5/2)*(a*x + 1)^2)/(a^2*x^2 - 1), x)
 

Reduce [F]

\[ \int e^{2 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^{5/2} \, dx=\sqrt {c}\, c^{2} \left (-\left (\int x^{m} \sqrt {-a^{2} x^{2}+1}\, x^{4}d x \right ) a^{4}-2 \left (\int x^{m} \sqrt {-a^{2} x^{2}+1}\, x^{3}d x \right ) a^{3}+2 \left (\int x^{m} \sqrt {-a^{2} x^{2}+1}\, x d x \right ) a +\int x^{m} \sqrt {-a^{2} x^{2}+1}d x \right ) \] Input:

int((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c)^(5/2),x)
 

Output:

sqrt(c)*c**2*( - int(x**m*sqrt( - a**2*x**2 + 1)*x**4,x)*a**4 - 2*int(x**m 
*sqrt( - a**2*x**2 + 1)*x**3,x)*a**3 + 2*int(x**m*sqrt( - a**2*x**2 + 1)*x 
,x)*a + int(x**m*sqrt( - a**2*x**2 + 1),x))