\(\int e^{2 \text {arctanh}(a x)} (e x)^m \sqrt {c-a c x} \, dx\) [469]

Optimal result

Integrand size = 25, antiderivative size = 87 \[ \int e^{2 \text {arctanh}(a x)} (e x)^m \sqrt {c-a c x} \, dx=-\frac {2 (e x)^{1+m} \sqrt {c-a c x}}{e (3+2 m)}-\frac {2 (5+4 m) (a x)^{-m} (e x)^m \sqrt {c-a c x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1-a x\right )}{a (3+2 m)} \] Output:

-2*(e*x)^(1+m)*(-a*c*x+c)^(1/2)/e/(3+2*m)-2*(5+4*m)*(e*x)^m*(-a*c*x+c)^(1/ 
2)*hypergeom([1/2, -m],[3/2],-a*x+1)/a/(3+2*m)/((a*x)^m)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.86 \[ \int e^{2 \text {arctanh}(a x)} (e x)^m \sqrt {c-a c x} \, dx=\frac {x (e x)^m \sqrt {c-a c x} \left ((2+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+m,2+m,a x\right )+a (1+m) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},2+m,3+m,a x\right )\right )}{(1+m) (2+m) \sqrt {1-a x}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6680, 35, 90, 77, 75}

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.

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

\(\Big \downarrow \) 6680

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

\(\Big \downarrow \) 35

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

\(\Big \downarrow \) 90

\(\displaystyle c \left (\frac {(4 m+5) \int \frac {(e x)^m}{\sqrt {c-a c x}}dx}{2 m+3}-\frac {2 \sqrt {c-a c x} (e x)^{m+1}}{c e (2 m+3)}\right )\)

\(\Big \downarrow \) 77

\(\displaystyle c \left (\frac {(4 m+5) (a x)^{-m} (e x)^m \int \frac {(a x)^m}{\sqrt {c-a c x}}dx}{2 m+3}-\frac {2 \sqrt {c-a c x} (e x)^{m+1}}{c e (2 m+3)}\right )\)

\(\Big \downarrow \) 75

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 35

Int[(u_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} 
, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] &&  !(IntegerQ[n] && SimplerQ[a + 
b*x, c + d*x])
 

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 77

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

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 6680

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int e^{2 \text {arctanh}(a x)} (e x)^m \sqrt {c-a c x} \, dx=\frac {2 e^{m} \sqrt {c}\, \left (-2 x^{m} \sqrt {-a x +1}\, a m x -x^{m} \sqrt {-a x +1}\, a x -4 x^{m} \sqrt {-a x +1}\, m -5 x^{m} \sqrt {-a x +1}-16 \left (\int \frac {x^{m} \sqrt {-a x +1}}{4 a \,m^{2} x^{2}+8 a m \,x^{2}+3 a \,x^{2}-4 m^{2} x -8 m x -3 x}d x \right ) m^{4}-52 \left (\int \frac {x^{m} \sqrt {-a x +1}}{4 a \,m^{2} x^{2}+8 a m \,x^{2}+3 a \,x^{2}-4 m^{2} x -8 m x -3 x}d x \right ) m^{3}-52 \left (\int \frac {x^{m} \sqrt {-a x +1}}{4 a \,m^{2} x^{2}+8 a m \,x^{2}+3 a \,x^{2}-4 m^{2} x -8 m x -3 x}d x \right ) m^{2}-15 \left (\int \frac {x^{m} \sqrt {-a x +1}}{4 a \,m^{2} x^{2}+8 a m \,x^{2}+3 a \,x^{2}-4 m^{2} x -8 m x -3 x}d x \right ) m \right )}{a \left (4 m^{2}+8 m +3\right )} \] Input:

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

Output:

(2*e**m*sqrt(c)*( - 2*x**m*sqrt( - a*x + 1)*a*m*x - x**m*sqrt( - a*x + 1)* 
a*x - 4*x**m*sqrt( - a*x + 1)*m - 5*x**m*sqrt( - a*x + 1) - 16*int((x**m*s 
qrt( - a*x + 1))/(4*a*m**2*x**2 + 8*a*m*x**2 + 3*a*x**2 - 4*m**2*x - 8*m*x 
 - 3*x),x)*m**4 - 52*int((x**m*sqrt( - a*x + 1))/(4*a*m**2*x**2 + 8*a*m*x* 
*2 + 3*a*x**2 - 4*m**2*x - 8*m*x - 3*x),x)*m**3 - 52*int((x**m*sqrt( - a*x 
 + 1))/(4*a*m**2*x**2 + 8*a*m*x**2 + 3*a*x**2 - 4*m**2*x - 8*m*x - 3*x),x) 
*m**2 - 15*int((x**m*sqrt( - a*x + 1))/(4*a*m**2*x**2 + 8*a*m*x**2 + 3*a*x 
**2 - 4*m**2*x - 8*m*x - 3*x),x)*m))/(a*(4*m**2 + 8*m + 3))