\(\int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx\) [624]

Optimal result

Integrand size = 27, antiderivative size = 193 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=-\frac {76 a^2 \sqrt {c-\frac {c}{a x}} \sqrt {1+a x}}{15 \sqrt {1-a x}}-\frac {2 \sqrt {c-\frac {c}{a x}} \sqrt {1+a x}}{5 x^2 \sqrt {1-a x}}-\frac {22 a \sqrt {c-\frac {c}{a x}} \sqrt {1+a x}}{15 x \sqrt {1-a x}}+\frac {4 \sqrt {2} a^{5/2} \sqrt {c-\frac {c}{a x}} \sqrt {x} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {1+a x}}\right )}{\sqrt {1-a x}} \] Output:

-76/15*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 
)*(a*x+1)^(1/2)/x^2/(-a*x+1)^(1/2)-22/15*a*(c-c/a/x)^(1/2)*(a*x+1)^(1/2)/x 
/(-a*x+1)^(1/2)+4*2^(1/2)*a^(5/2)*(c-c/a/x)^(1/2)*x^(1/2)*arctanh(2^(1/2)* 
a^(1/2)*x^(1/2)/(a*x+1)^(1/2))/(-a*x+1)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.52 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2 \sqrt {c-\frac {c}{a x}} \left (-\sqrt {1+a x} \left (3+11 a x+38 a^2 x^2\right )+30 \sqrt {2} a^{5/2} x^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {1+a x}}\right )\right )}{15 x^2 \sqrt {1-a x}} \] Input:

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

Output:

(2*Sqrt[c - c/(a*x)]*(-(Sqrt[1 + a*x]*(3 + 11*a*x + 38*a^2*x^2)) + 30*Sqrt 
[2]*a^(5/2)*x^(5/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[x])/Sqrt[1 + a*x]]))/(15 
*x^2*Sqrt[1 - a*x])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.67, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6684, 6679, 107, 105, 105, 104, 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.

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

\(\Big \downarrow \) 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

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

\(\Big \downarrow \) 107

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(Sqrt[c - c/(a*x)]*Sqrt[x]*((-2*(1 + a*x)^(5/2))/(5*x^(5/2)) + a*((-2*(1 + 
 a*x)^(3/2))/(3*x^(3/2)) + 2*a*((-2*Sqrt[1 + a*x])/Sqrt[x] + 2*Sqrt[2]*Sqr 
t[a]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[x])/Sqrt[1 + a*x]]))))/Sqrt[1 - a*x]
 

Defintions of rubi rules used

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 105

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

rule 107

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

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 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 = 0.23 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {-a^{2} x^{2}+1}\, \left (38 x^{2} \sqrt {-x \left (a x +1\right )}\, a^{2} \sqrt {2}\, \sqrt {-\frac {1}{a}}+30 a^{2} \ln \left (\frac {2 \sqrt {2}\, \sqrt {-\frac {1}{a}}\, \sqrt {-x \left (a x +1\right )}\, a -3 a x -1}{a x -1}\right ) x^{3}+11 x \sqrt {-x \left (a x +1\right )}\, a \sqrt {2}\, \sqrt {-\frac {1}{a}}+3 \sqrt {-x \left (a x +1\right )}\, \sqrt {2}\, \sqrt {-\frac {1}{a}}\right ) \sqrt {2}}{15 x^{2} \left (a x -1\right ) \sqrt {-x \left (a x +1\right )}\, \sqrt {-\frac {1}{a}}}\)	\(181\)
risch	\(-\frac {2 \left (38 a^{3} x^{3}+49 a^{2} x^{2}+14 a x +3\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}}{15 x^{2} \sqrt {-\left (a x +1\right ) a c x}\, \sqrt {-a^{2} x^{2}+1}}+\frac {4 a^{2} \ln \left (\frac {-4 c -3 \left (x -\frac {1}{a}\right ) a c +2 \sqrt {-2 c}\, \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -3 \left (x -\frac {1}{a}\right ) a c -2 c}}{x -\frac {1}{a}}\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}}{\sqrt {-2 c}\, \sqrt {-a^{2} x^{2}+1}}\)	\(216\)

Input:

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

Output:

1/15*(c*(a*x-1)/a/x)^(1/2)/x^2*(-a^2*x^2+1)^(1/2)*(38*x^2*(-x*(a*x+1))^(1/ 
2)*a^2*2^(1/2)*(-1/a)^(1/2)+30*a^2*ln((2*2^(1/2)*(-1/a)^(1/2)*(-x*(a*x+1)) 
^(1/2)*a-3*a*x-1)/(a*x-1))*x^3+11*x*(-x*(a*x+1))^(1/2)*a*2^(1/2)*(-1/a)^(1 
/2)+3*(-x*(a*x+1))^(1/2)*2^(1/2)*(-1/a)^(1/2))*2^(1/2)/(a*x-1)/(-x*(a*x+1) 
)^(1/2)/(-1/a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.75 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\left [\frac {15 \, \sqrt {2} {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \sqrt {-c} \log \left (-\frac {17 \, a^{3} c x^{3} - 3 \, a^{2} c x^{2} - 13 \, a c x + 4 \, \sqrt {2} {\left (3 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) + 2 \, {\left (38 \, a^{2} x^{2} + 11 \, a x + 3\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{15 \, {\left (a x^{3} - x^{2}\right )}}, -\frac {2 \, {\left (15 \, \sqrt {2} {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{3 \, a^{2} c x^{2} - 2 \, a c x - c}\right ) - {\left (38 \, a^{2} x^{2} + 11 \, a x + 3\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}\right )}}{15 \, {\left (a x^{3} - x^{2}\right )}}\right ] \] Input:

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

Output:

[1/15*(15*sqrt(2)*(a^3*x^3 - a^2*x^2)*sqrt(-c)*log(-(17*a^3*c*x^3 - 3*a^2* 
c*x^2 - 13*a*c*x + 4*sqrt(2)*(3*a^2*x^2 + a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c) 
*sqrt((a*c*x - c)/(a*x)) - c)/(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)) + 2*(38*a 
^2*x^2 + 11*a*x + 3)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a*x^3 - 
x^2), -2/15*(15*sqrt(2)*(a^3*x^3 - a^2*x^2)*sqrt(c)*arctan(2*sqrt(2)*sqrt( 
-a^2*x^2 + 1)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x))/(3*a^2*c*x^2 - 2*a*c*x - 
 c)) - (38*a^2*x^2 + 11*a*x + 3)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x) 
))/(a*x^3 - x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate((a*x + 1)^3*sqrt(c - c/(a*x))/((-a^2*x^2 + 1)^(3/2)*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((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^(1/2)/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 [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.47 \[ \int \frac {e^{3 \text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^3} \, dx=\frac {2 \sqrt {c}\, \left (-30 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, \sqrt {2}\, i}{a x +1}\right ) a^{3} x^{3}+38 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a^{2} i \,x^{2}+11 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a i x +3 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, i \right )}{15 a \,x^{3}} \] Input:

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

Output:

(2*sqrt(c)*( - 30*sqrt(2)*atan((sqrt(x)*sqrt(a)*sqrt(a*x + 1)*sqrt(2)*i)/( 
a*x + 1))*a**3*x**3 + 38*sqrt(x)*sqrt(a)*sqrt(a*x + 1)*a**2*i*x**2 + 11*sq 
rt(x)*sqrt(a)*sqrt(a*x + 1)*a*i*x + 3*sqrt(x)*sqrt(a)*sqrt(a*x + 1)*i))/(1 
5*a*x**3)