\(\int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx\) [504]

Optimal result

Integrand size = 23, antiderivative size = 124 \[ \int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=\frac {c \sqrt {1-\frac {1}{a^2 x^2}} x}{4 a \sqrt {c-\frac {c}{a x}}}+\frac {c \sqrt {1-\frac {1}{a^2 x^2}} x^2}{2 \sqrt {c-\frac {c}{a x}}}-\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{4 a^2} \] Output:

1/4*c*(1-1/a^2/x^2)^(1/2)*x/a/(c-c/a/x)^(1/2)+1/2*c*(1-1/a^2/x^2)^(1/2)*x^ 
2/(c-c/a/x)^(1/2)-1/4*c^(1/2)*arctanh(c^(1/2)*(1-1/a^2/x^2)^(1/2)/(c-c/a/x 
)^(1/2))/a^2
 

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.19 \[ \int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=\frac {2 a^2 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}} x^2 (1+2 a x)+\sqrt {c} (-1+a x) \log (1-a x)+\sqrt {c} (1-a x) \log \left (2 a^2 \sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}} x^2+c \left (-1-a x+2 a^2 x^2\right )\right )}{8 a^2 (-1+a x)} \] Input:

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

Output:

(2*a^2*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)]*x^2*(1 + 2*a*x) + Sqrt[c]*( 
-1 + a*x)*Log[1 - a*x] + Sqrt[c]*(1 - a*x)*Log[2*a^2*Sqrt[c]*Sqrt[1 - 1/(a 
^2*x^2)]*Sqrt[c - c/(a*x)]*x^2 + c*(-1 - a*x + 2*a^2*x^2)])/(8*a^2*(-1 + a 
*x))
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6733, 575, 579, 573, 219}

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.

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

\(\Big \downarrow \) 6733

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

\(\Big \downarrow \) 575

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

\(\Big \downarrow \) 579

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

\(\Big \downarrow \) 573

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

\(\Big \downarrow \) 219

\(\displaystyle -c \left (\frac {\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{a}-\frac {c x \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}}{4 a c}-\frac {x^2 \sqrt {1-\frac {1}{a^2 x^2}}}{2 \sqrt {c-\frac {c}{a x}}}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 573

Int[Sqrt[(c_) + (d_.)*(x_)]/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> 
Simp[-2*c   Subst[Int[1/(a - c*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x]], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
 

rule 575

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

rule 579

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

rule 6733

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_S 
ymbol] :> Simp[-c^n   Subst[Int[(c + d*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^(m 
 + 2)), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] && Int 
egerQ[(n - 1)/2] && IntegerQ[m] && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (4 a^{\frac {3}{2}} x \sqrt {x \left (a x +1\right )}+2 \sqrt {x \left (a x +1\right )}\, \sqrt {a}-\ln \left (\frac {2 \sqrt {x \left (a x +1\right )}\, \sqrt {a}+2 a x +1}{2 \sqrt {a}}\right )\right )}{8 \sqrt {\frac {a x -1}{a x +1}}\, a^{\frac {3}{2}} \sqrt {x \left (a x +1\right )}}\)	\(104\)
risch	\(\frac {\left (2 a x +1\right ) x \sqrt {\frac {c \left (a x -1\right )}{a x}}}{4 a \sqrt {\frac {a x -1}{a x +1}}}-\frac {\ln \left (\frac {\frac {1}{2} a c +a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}+a c x}\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\left (a x +1\right ) a c x}}{8 a \sqrt {a^{2} c}\, \sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right )}\)	\(140\)

Input:

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

Output:

1/8/((a*x-1)/(a*x+1))^(1/2)*(c*(a*x-1)/a/x)^(1/2)*x*(4*a^(3/2)*x*(x*(a*x+1 
))^(1/2)+2*(x*(a*x+1))^(1/2)*a^(1/2)-ln(1/2*(2*(x*(a*x+1))^(1/2)*a^(1/2)+2 
*a*x+1)/a^(1/2)))/a^(3/2)/(x*(a*x+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.56 \[ \int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=\left [\frac {{\left (a x - 1\right )} \sqrt {c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt {c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{16 \, {\left (a^{3} x - a^{2}\right )}}, \frac {{\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, {\left (a^{2} x^{2} + a x\right )} \sqrt {-c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) + 2 \, {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{8 \, {\left (a^{3} x - a^{2}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.39 \[ \int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx=\frac {\sqrt {c}\, \left (2 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a x +\sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}-\mathrm {log}\left (\sqrt {a x +1}+\sqrt {x}\, \sqrt {a}\right )\right )}{4 a^{2}} \] Input:

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

Output:

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