\(\int e^{-\text {arctanh}(a x)} (e x)^m \sqrt {c-a c x} \, dx\) [481]

Optimal result

Integrand size = 25, antiderivative size = 109 \[ \int e^{-\text {arctanh}(a x)} (e x)^m \sqrt {c-a c x} \, dx=-\frac {2 c (e x)^{1+m} \sqrt {1-a^2 x^2}}{e (3+2 m) \sqrt {c-a c x}}+\frac {(5+4 m) (e x)^{1+m} \sqrt {c-a c x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+m,2+m,-a x\right )}{e (1+m) (3+2 m) \sqrt {1-a x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.72 \[ \int e^{-\text {arctanh}(a x)} (e x)^m \sqrt {c-a c x} \, dx=-\frac {c x (e x)^m \sqrt {1-a x} \left (2 (1+m) \sqrt {1+a x}-(5+4 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+m,2+m,-a x\right )\right )}{(1+m) (3+2 m) \sqrt {c-a c x}} \] Input:

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

Output:

-((c*x*(e*x)^m*Sqrt[1 - a*x]*(2*(1 + m)*Sqrt[1 + a*x] - (5 + 4*m)*Hypergeo 
metric2F1[1/2, 1 + m, 2 + m, -(a*x)]))/((1 + m)*(3 + 2*m)*Sqrt[c - a*c*x]) 
)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6678, 574, 585, 74}

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.

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

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 574

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

\(\Big \downarrow \) 585

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

\(\Big \downarrow \) 74

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 74

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

rule 574

Int[((e_.)*(x_))^(n_)*((c_) + (d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[d^2*(e*x)^(n + 1)*(c + d*x)^(m - 2)*((a + b*x^2)^(p + 1)/ 
(b*e*(n + p + 2))), x] + Simp[c*((2*n + p + 3)/(n + p + 2))   Int[(e*x)^n*( 
c + d*x)^(m - 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x 
] && EqQ[b*c^2 + a*d^2, 0] && EqQ[m + p - 1, 0] &&  !LtQ[n, -1] && IntegerQ 
[2*p]
 

rule 585

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_) 
, x_Symbol] :> Simp[a^p*c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^F 
racPart[n])   Int[(e*x)^m*(1 - d*(x/c))^p*(1 + d*(x/c))^(n + p), x], x] /; 
FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[a, 0]
 

rule 6678

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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