\(\int e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x^m \, dx\) [653]

Optimal result

Integrand size = 27, antiderivative size = 133 \[ \int e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x^m \, dx=\frac {2 \sqrt {c-\frac {c}{a x}} x^{1+m} \sqrt {1+a x}}{(1+2 m) \sqrt {1-a x}}-\frac {2 (3+4 m) \sqrt {c-\frac {c}{a x}} x^{1+m} (-a x)^{-\frac {1}{2}-m} \sqrt {1+a x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {1}{2}-m,\frac {3}{2},1+a x\right )}{(1+2 m) \sqrt {1-a x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.65 \[ \int e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x^m \, dx=-\frac {2 \sqrt {c-\frac {c}{a x}} x^{1+m} \left (-\left ((3+2 m) \sqrt {1+a x}\right )+a (3+4 m) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{2}+m,\frac {5}{2}+m,-a x\right )\right )}{\left (3+8 m+4 m^2\right ) \sqrt {1-a x}} \] Input:

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

Output:

(-2*Sqrt[c - c/(a*x)]*x^(1 + m)*(-((3 + 2*m)*Sqrt[1 + a*x]) + a*(3 + 4*m)* 
x*Hypergeometric2F1[1/2, 3/2 + m, 5/2 + m, -(a*x)]))/((3 + 8*m + 4*m^2)*Sq 
rt[1 - a*x])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.78, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6684, 6678, 516, 88, 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.

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

\(\Big \downarrow \) 6684

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

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 516

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

\(\Big \downarrow \) 88

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

\(\Big \downarrow \) 74

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

Input:

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

Output:

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

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 88

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

rule 516

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

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]
 

rule 6684

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

Maple [F]

\[\int \frac {\sqrt {c -\frac {c}{a x}}\, x^{m} \sqrt {-a^{2} x^{2}+1}}{a x +1}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(c)*(int((x**m*sqrt(a*x + 1)*i*x)/(sqrt(x)*a*x + sqrt(x)),x)*a - int( 
(x**m*sqrt(a*x + 1)*i)/(sqrt(x)*a*x + sqrt(x)),x)))/sqrt(a)