\(\int e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx\) [643]

Optimal result

Integrand size = 25, antiderivative size = 135 \[ \int e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=\frac {7 \sqrt {c-\frac {c}{a x}} x \sqrt {1+a x}}{4 a \sqrt {1-a x}}-\frac {\sqrt {c-\frac {c}{a x}} x^2 \sqrt {1+a x}}{2 \sqrt {1-a x}}-\frac {7 \sqrt {c-\frac {c}{a x}} \sqrt {x} \text {arcsinh}\left (\sqrt {a} \sqrt {x}\right )}{4 a^{3/2} \sqrt {1-a x}} \] Output:

7/4*(c-c/a/x)^(1/2)*x*(a*x+1)^(1/2)/a/(-a*x+1)^(1/2)-1/2*(c-c/a/x)^(1/2)*x 
^2*(a*x+1)^(1/2)/(-a*x+1)^(1/2)-7/4*(c-c/a/x)^(1/2)*x^(1/2)*arcsinh(a^(1/2 
)*x^(1/2))/a^(3/2)/(-a*x+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.59 \[ \int e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=-\frac {\sqrt {c-\frac {c}{a x}} \sqrt {x} \left (\sqrt {a} \sqrt {x} \sqrt {1+a x} (-7+2 a x)+7 \text {arcsinh}\left (\sqrt {a} \sqrt {x}\right )\right )}{4 a^{3/2} \sqrt {1-a x}} \] Input:

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

Output:

-1/4*(Sqrt[c - c/(a*x)]*Sqrt[x]*(Sqrt[a]*Sqrt[x]*Sqrt[1 + a*x]*(-7 + 2*a*x 
) + 7*ArcSinh[Sqrt[a]*Sqrt[x]]))/(a^(3/2)*Sqrt[1 - a*x])
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.68, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {6684, 6678, 516, 90, 60, 63, 222}

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.

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

\(\Big \downarrow \) 6684

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

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 516

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

\(\Big \downarrow \) 90

\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (\frac {7}{4} \int \frac {\sqrt {x}}{\sqrt {a x+1}}dx-\frac {1}{2} x^{3/2} \sqrt {a x+1}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (\frac {7}{4} \left (\frac {\sqrt {x} \sqrt {a x+1}}{a}-\frac {\int \frac {1}{\sqrt {x} \sqrt {a x+1}}dx}{2 a}\right )-\frac {1}{2} x^{3/2} \sqrt {a x+1}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 63

\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (\frac {7}{4} \left (\frac {\sqrt {x} \sqrt {a x+1}}{a}-\frac {\int \frac {1}{\sqrt {a x+1}}d\sqrt {x}}{a}\right )-\frac {1}{2} x^{3/2} \sqrt {a x+1}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\sqrt {x} \left (\frac {7}{4} \left (\frac {\sqrt {x} \sqrt {a x+1}}{a}-\frac {\text {arcsinh}\left (\sqrt {a} \sqrt {x}\right )}{a^{3/2}}\right )-\frac {1}{2} x^{3/2} \sqrt {a x+1}\right ) \sqrt {c-\frac {c}{a x}}}{\sqrt {1-a x}}\)

Input:

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

Output:

(Sqrt[c - c/(a*x)]*Sqrt[x]*(-1/2*(x^(3/2)*Sqrt[1 + a*x]) + (7*((Sqrt[x]*Sq 
rt[1 + a*x])/a - ArcSinh[Sqrt[a]*Sqrt[x]]/a^(3/2)))/4))/Sqrt[1 - a*x]
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 63

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2/b   S 
ubst[Int[1/Sqrt[c + d*(x^2/b)], x], x, Sqrt[b*x]], x] /; FreeQ[{b, c, d}, x 
] && GtQ[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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

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 [A] (verified)

Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \sqrt {-a^{2} x^{2}+1}\, \left (4 a^{\frac {3}{2}} x \sqrt {-x \left (a x +1\right )}-14 \sqrt {a}\, \sqrt {-x \left (a x +1\right )}-7 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-x \left (a x +1\right )}}\right )\right )}{8 a^{\frac {3}{2}} \left (a x -1\right ) \sqrt {-x \left (a x +1\right )}}\)	\(107\)
risch	\(-\frac {\left (2 a x -7\right ) \left (a x +1\right ) x \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}}{4 a \sqrt {-\left (a x +1\right ) a c x}\, \sqrt {-a^{2} x^{2}+1}}-\frac {7 \arctan \left (\frac {\sqrt {a^{2} c}\, \left (x +\frac {1}{2 a}\right )}{\sqrt {-a^{2} c \,x^{2}-a c x}}\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}}{8 a \sqrt {a^{2} c}\, \sqrt {-a^{2} x^{2}+1}}\)	\(176\)

Input:

int((c-c/a/x)^(1/2)*x/(a*x+1)*(-a^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/8*(c*(a*x-1)/a/x)^(1/2)*x*(-a^2*x^2+1)^(1/2)*(4*a^(3/2)*x*(-x*(a*x+1))^( 
1/2)-14*a^(1/2)*(-x*(a*x+1))^(1/2)-7*arctan(1/2/a^(1/2)*(2*a*x+1)/(-x*(a*x 
+1))^(1/2)))/a^(3/2)/(a*x-1)/(-x*(a*x+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.04 \[ \int e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=\left [\frac {7 \, {\left (a x - 1\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, {\left (2 \, a^{2} x^{2} - 7 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{16 \, {\left (a^{3} x - a^{2}\right )}}, \frac {7 \, {\left (a x - 1\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) + 2 \, {\left (2 \, a^{2} x^{2} - 7 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{8 \, {\left (a^{3} x - a^{2}\right )}}\right ] \] Input:

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

Output:

[1/16*(7*(a*x - 1)*sqrt(-c)*log(-(8*a^3*c*x^3 - 7*a*c*x - 4*(2*a^2*x^2 + a 
*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a*c*x - c)/(a*x)) - c)/(a*x - 1)) + 
4*(2*a^2*x^2 - 7*a*x)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a^3*x - 
 a^2), 1/8*(7*(a*x - 1)*sqrt(c)*arctan(2*sqrt(-a^2*x^2 + 1)*a*sqrt(c)*x*sq 
rt((a*c*x - c)/(a*x))/(2*a^2*c*x^2 - a*c*x - c)) + 2*(2*a^2*x^2 - 7*a*x)*s 
qrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a^3*x - a^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.39 \[ \int e^{-\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=\frac {\sqrt {c}\, i \left (2 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a x -7 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}+7 \,\mathrm {log}\left (\sqrt {a x +1}\, i +\sqrt {x}\, \sqrt {a}\, i \right )\right )}{4 a^{2}} \] Input:

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

Output:

(sqrt(c)*i*(2*sqrt(x)*sqrt(a)*sqrt(a*x + 1)*a*x - 7*sqrt(x)*sqrt(a)*sqrt(a 
*x + 1) + 7*log(sqrt(a*x + 1)*i + sqrt(x)*sqrt(a)*i)))/(4*a**2)