\(\int e^{-2 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx\) [477]

Optimal result

Integrand size = 24, antiderivative size = 92 \[ \int e^{-2 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=-\frac {2 (e x)^{1+m} \sqrt {c+a c x}}{e (3+2 m)}+\frac {c (5+4 m) (e x)^{1+m} \sqrt {1+a x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+m,2+m,-a x\right )}{e (1+m) (3+2 m) \sqrt {c+a c x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.84 \[ \int e^{-2 \text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=-\frac {x (e x)^m \sqrt {c+a c x} \left (-\left ((2+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+m,2+m,-a x\right )\right )+a (1+m) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},2+m,3+m,-a x\right )\right )}{(1+m) (2+m) \sqrt {1+a x}} \] Input:

Integrate[((e*x)^m*Sqrt[c + a*c*x])/E^(2*ArcTanh[a*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 = 0.56 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, 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.

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

\(\Big \downarrow \) 6680

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

\(\Big \downarrow \) 35

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 77

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

\(\Big \downarrow \) 75

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

Input:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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