\(\int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx\) [649]

Optimal result

Integrand size = 27, antiderivative size = 166 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {158 a^2 \sqrt {c-\frac {c}{a x}}}{15 \sqrt {1-a x} \sqrt {1+a x}}-\frac {2 \sqrt {c-\frac {c}{a x}}}{5 x^2 \sqrt {1-a x} \sqrt {1+a x}}+\frac {32 a \sqrt {c-\frac {c}{a x}}}{15 x \sqrt {1-a x} \sqrt {1+a x}}-\frac {316 a^2 \sqrt {c-\frac {c}{a x}} \sqrt {1+a x}}{15 \sqrt {1-a x}} \] Output:

158/15*a^2*(c-c/a/x)^(1/2)/(-a*x+1)^(1/2)/(a*x+1)^(1/2)-2/5*(c-c/a/x)^(1/2 
)/x^2/(-a*x+1)^(1/2)/(a*x+1)^(1/2)+32/15*a*(c-c/a/x)^(1/2)/x/(-a*x+1)^(1/2 
)/(a*x+1)^(1/2)-316/15*a^2*(c-c/a/x)^(1/2)*(a*x+1)^(1/2)/(-a*x+1)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.35 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=-\frac {2 \sqrt {c-\frac {c}{a x}} \left (3-16 a x+79 a^2 x^2+158 a^3 x^3\right )}{15 x^2 \sqrt {1-a^2 x^2}} \] Input:

Integrate[Sqrt[c - c/(a*x)]/(E^(3*ArcTanh[a*x])*x^3),x]
 

Output:

(-2*Sqrt[c - c/(a*x)]*(3 - 16*a*x + 79*a^2*x^2 + 158*a^3*x^3))/(15*x^2*Sqr 
t[1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.67, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6684, 6679, 100, 27, 87, 55, 48}

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.

\(\displaystyle \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx\)

\(\Big \downarrow \) 6684

\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {1-a x}}{x^{7/2}}dx}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 6679

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

\(\Big \downarrow \) 100

\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (\frac {2}{5} \int -\frac {a (16-5 a x)}{2 x^{5/2} (a x+1)^{3/2}}dx-\frac {2}{5 x^{5/2} \sqrt {a x+1}}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (-\frac {1}{5} a \int \frac {16-5 a x}{x^{5/2} (a x+1)^{3/2}}dx-\frac {2}{5 x^{5/2} \sqrt {a x+1}}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (-\frac {1}{5} a \left (-\frac {79}{3} a \int \frac {1}{x^{3/2} (a x+1)^{3/2}}dx-\frac {32}{3 x^{3/2} \sqrt {a x+1}}\right )-\frac {2}{5 x^{5/2} \sqrt {a x+1}}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (-\frac {1}{5} a \left (-\frac {79}{3} a \left (2 \int \frac {1}{x^{3/2} \sqrt {a x+1}}dx+\frac {2}{\sqrt {x} \sqrt {a x+1}}\right )-\frac {32}{3 x^{3/2} \sqrt {a x+1}}\right )-\frac {2}{5 x^{5/2} \sqrt {a x+1}}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {\sqrt {x} \left (-\frac {1}{5} a \left (-\frac {32}{3 x^{3/2} \sqrt {a x+1}}-\frac {79}{3} a \left (\frac {2}{\sqrt {x} \sqrt {a x+1}}-\frac {4 \sqrt {a x+1}}{\sqrt {x}}\right )\right )-\frac {2}{5 x^{5/2} \sqrt {a x+1}}\right ) \sqrt {c-\frac {c}{a x}}}{\sqrt {1-a x}}\)

Input:

Int[Sqrt[c - c/(a*x)]/(E^(3*ArcTanh[a*x])*x^3),x]
 

Output:

(Sqrt[c - c/(a*x)]*Sqrt[x]*(-2/(5*x^(5/2)*Sqrt[1 + a*x]) - (a*(-32/(3*x^(3 
/2)*Sqrt[1 + a*x]) - (79*a*(2/(Sqrt[x]*Sqrt[1 + a*x]) - (4*Sqrt[1 + a*x])/ 
Sqrt[x]))/3))/5))/Sqrt[1 - a*x]
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 48

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

rule 55

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S 
implify[m + n + 2]/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^Simplify[m + 1]*( 
c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 
 2], 0] && NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ 
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] ||  !SumSimp 
lerQ[n, 1])
 

rule 87

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)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

rule 6679

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

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 = 0.21 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.40

method	result	size
orering	\(-\frac {2 \left (158 a^{3} x^{3}+79 a^{2} x^{2}-16 a x +3\right ) \sqrt {c -\frac {c}{a x}}\, \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{15 \left (a x +1\right )^{2} x^{2} \left (a x -1\right )^{2}}\)	\(67\)
gosper	\(-\frac {2 \left (158 a^{3} x^{3}+79 a^{2} x^{2}-16 a x +3\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{15 x^{2} \left (a x +1\right )^{2} \left (a x -1\right )^{2}}\)	\(69\)
default	\(\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (158 a^{3} x^{3}+79 a^{2} x^{2}-16 a x +3\right ) \sqrt {-a^{2} x^{2}+1}}{15 x^{2} \left (a x +1\right ) \left (a x -1\right )}\)	\(69\)
risch	\(-\frac {2 \left (98 a^{3} x^{3}+79 a^{2} x^{2}-16 a x +3\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}}{15 x^{2} \sqrt {-\left (a x +1\right ) a c x}\, \sqrt {-a^{2} x^{2}+1}}-\frac {8 a^{3} x \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}}{\sqrt {-\left (a x +1\right ) a c x}\, \sqrt {-a^{2} x^{2}+1}}\)	\(159\)

Input:

int((c-c/a/x)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x,method=_RETURNVERBO 
SE)
 

Output:

-2/15*(158*a^3*x^3+79*a^2*x^2-16*a*x+3)/(a*x+1)^2/x^2/(a*x-1)^2*(c-c/a/x)^ 
(1/2)*(-a^2*x^2+1)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.41 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2 \, {\left (158 \, a^{3} x^{3} + 79 \, a^{2} x^{2} - 16 \, a x + 3\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{15 \, {\left (a^{2} x^{4} - x^{2}\right )}} \] Input:

integrate((c-c/a/x)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x, algorithm="f 
ricas")
 

Output:

2/15*(158*a^3*x^3 + 79*a^2*x^2 - 16*a*x + 3)*sqrt(-a^2*x^2 + 1)*sqrt((a*c* 
x - c)/(a*x))/(a^2*x^4 - x^2)
 

Sympy [F]

\[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\int \frac {\sqrt {- c \left (-1 + \frac {1}{a x}\right )} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{x^{3} \left (a x + 1\right )^{3}}\, dx \] Input:

integrate((c-c/a/x)**(1/2)/(a*x+1)**3*(-a**2*x**2+1)**(3/2)/x**3,x)
 

Output:

Integral(sqrt(-c*(-1 + 1/(a*x)))*(-(a*x - 1)*(a*x + 1))**(3/2)/(x**3*(a*x 
+ 1)**3), x)
 

Maxima [F]

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

integrate((c-c/a/x)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x, algorithm="m 
axima")
 

Output:

integrate((-a^2*x^2 + 1)^(3/2)*sqrt(c - c/(a*x))/((a*x + 1)^3*x^3), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((c-c/a/x)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x, algorithm="g 
iac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 14.51 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.60 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {\sqrt {c-\frac {c}{a\,x}}\,\left (\frac {2\,\sqrt {1-a^2\,x^2}}{5\,a^2}+\frac {158\,x^2\,\sqrt {1-a^2\,x^2}}{15}-\frac {32\,x\,\sqrt {1-a^2\,x^2}}{15\,a}+\frac {316\,a\,x^3\,\sqrt {1-a^2\,x^2}}{15}\right )}{x^4-\frac {x^2}{a^2}} \] Input:

int(((c - c/(a*x))^(1/2)*(1 - a^2*x^2)^(3/2))/(x^3*(a*x + 1)^3),x)
 

Output:

((c - c/(a*x))^(1/2)*((2*(1 - a^2*x^2)^(1/2))/(5*a^2) + (158*x^2*(1 - a^2* 
x^2)^(1/2))/15 - (32*x*(1 - a^2*x^2)^(1/2))/(15*a) + (316*a*x^3*(1 - a^2*x 
^2)^(1/2))/15))/(x^4 - x^2/a^2)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.43 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2 \sqrt {c}\, i \left (-158 \sqrt {a x +1}\, a^{3} x^{3}+158 \sqrt {x}\, \sqrt {a}\, a^{3} x^{3}+79 \sqrt {x}\, \sqrt {a}\, a^{2} x^{2}-16 \sqrt {x}\, \sqrt {a}\, a x +3 \sqrt {x}\, \sqrt {a}\right )}{15 \sqrt {a x +1}\, a \,x^{3}} \] Input:

int((c-c/a/x)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x)
 

Output:

(2*sqrt(c)*i*( - 158*sqrt(a*x + 1)*a**3*x**3 + 158*sqrt(x)*sqrt(a)*a**3*x* 
*3 + 79*sqrt(x)*sqrt(a)*a**2*x**2 - 16*sqrt(x)*sqrt(a)*a*x + 3*sqrt(x)*sqr 
t(a)))/(15*sqrt(a*x + 1)*a*x**3)