\(\int e^{\text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx\) [472]

Optimal result

Integrand size = 22, antiderivative size = 105 \[ \int e^{\text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=-\frac {2 c (e x)^{1+m} \sqrt {1-a^2 x^2}}{e (3+2 m) \sqrt {c+a c x}}+\frac {(5+4 m) (e x)^{1+m} \sqrt {c+a c x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+m,2+m,a x\right )}{e (1+m) (3+2 m) \sqrt {1+a x}} \] 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 = 76, 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^ArcTanh[a*x]*(e*x)^m*Sqrt[c + a*c*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 = 0.52 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.87, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6680, 37, 90, 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.

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

\(\Big \downarrow \) 6680

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

\(\Big \downarrow \) 37

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 74

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 37

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

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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int e^{\text {arctanh}(a x)} (e x)^m \sqrt {c+a c x} \, dx=\frac {2 e^{m} \sqrt {c}\, \left (-2 x^{m} \sqrt {-a x +1}\, a m x -x^{m} \sqrt {-a x +1}\, a x -4 x^{m} \sqrt {-a x +1}\, m -5 x^{m} \sqrt {-a x +1}-8 \left (\int \frac {x^{m} \sqrt {-a x +1}}{4 a \,m^{2} x^{2}+8 a m \,x^{2}+3 a \,x^{2}-4 m^{2} x -8 m x -3 x}d x \right ) m^{4}-24 \left (\int \frac {x^{m} \sqrt {-a x +1}}{4 a \,m^{2} x^{2}+8 a m \,x^{2}+3 a \,x^{2}-4 m^{2} x -8 m x -3 x}d x \right ) m^{3}-22 \left (\int \frac {x^{m} \sqrt {-a x +1}}{4 a \,m^{2} x^{2}+8 a m \,x^{2}+3 a \,x^{2}-4 m^{2} x -8 m x -3 x}d x \right ) m^{2}-6 \left (\int \frac {x^{m} \sqrt {-a x +1}}{4 a \,m^{2} x^{2}+8 a m \,x^{2}+3 a \,x^{2}-4 m^{2} x -8 m x -3 x}d x \right ) m -4 \left (\int \frac {x^{m} \sqrt {-a x +1}}{2 a m \,x^{2}+a \,x^{2}-2 m x -x}d x \right ) m^{3}-8 \left (\int \frac {x^{m} \sqrt {-a x +1}}{2 a m \,x^{2}+a \,x^{2}-2 m x -x}d x \right ) m^{2}-3 \left (\int \frac {x^{m} \sqrt {-a x +1}}{2 a m \,x^{2}+a \,x^{2}-2 m x -x}d x \right ) m \right )}{a \left (4 m^{2}+8 m +3\right )} \] Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(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) - 8*int((x**m*sq 
rt( - 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 - 24*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 - 22*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 - 6*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*int((x**m*sqrt( - a*x + 1))/(2*a*m*x* 
*2 + a*x**2 - 2*m*x - x),x)*m**3 - 8*int((x**m*sqrt( - a*x + 1))/(2*a*m*x* 
*2 + a*x**2 - 2*m*x - x),x)*m**2 - 3*int((x**m*sqrt( - a*x + 1))/(2*a*m*x* 
*2 + a*x**2 - 2*m*x - x),x)*m))/(a*(4*m**2 + 8*m + 3))