\(\int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx\) [453]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 151 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx=\frac {47 a^2 \sqrt {c-a c x}}{4 \sqrt {1-a x} \sqrt {1+a x}}-\frac {\sqrt {c-a c x}}{2 x^2 \sqrt {1-a x} \sqrt {1+a x}}+\frac {13 a \sqrt {c-a c x}}{4 x \sqrt {1-a x} \sqrt {1+a x}}-\frac {47 a^2 \sqrt {c-a c x} \text {arctanh}\left (\sqrt {1+a x}\right )}{4 \sqrt {1-a x}} \] Output:

47/4*a^2*(-a*c*x+c)^(1/2)/(-a*x+1)^(1/2)/(a*x+1)^(1/2)-1/2*(-a*c*x+c)^(1/2 
)/x^2/(-a*x+1)^(1/2)/(a*x+1)^(1/2)+13/4*a*(-a*c*x+c)^(1/2)/x/(-a*x+1)^(1/2 
)/(a*x+1)^(1/2)-47/4*a^2*(-a*c*x+c)^(1/2)*arctanh((a*x+1)^(1/2))/(-a*x+1)^ 
(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.43 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx=\frac {c \sqrt {1-a x} \left (-2+13 a x+47 a^2 x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+a x\right )\right )}{4 x^2 \sqrt {1+a x} \sqrt {c-a c x}} \] Input:

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

Output:

(c*Sqrt[1 - a*x]*(-2 + 13*a*x + 47*a^2*x^2*Hypergeometric2F1[-1/2, 1, 1/2, 
 1 + a*x]))/(4*x^2*Sqrt[1 + a*x]*Sqrt[c - a*c*x])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.58, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {6680, 37, 100, 27, 87, 61, 73, 221}

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.

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

\(\Big \downarrow \) 6680

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

\(\Big \downarrow \) 37

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

\(\Big \downarrow \) 100

\(\displaystyle \frac {\sqrt {c-a c x} \left (\frac {1}{2} \int -\frac {a (13-4 a x)}{2 x^2 (a x+1)^{3/2}}dx-\frac {1}{2 x^2 \sqrt {a x+1}}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {c-a c x} \left (-\frac {1}{4} a \int \frac {13-4 a x}{x^2 (a x+1)^{3/2}}dx-\frac {1}{2 x^2 \sqrt {a x+1}}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {\sqrt {c-a c x} \left (-\frac {1}{4} a \left (-\frac {47}{2} a \int \frac {1}{x (a x+1)^{3/2}}dx-\frac {13}{x \sqrt {a x+1}}\right )-\frac {1}{2 x^2 \sqrt {a x+1}}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {\sqrt {c-a c x} \left (-\frac {1}{4} a \left (-\frac {47}{2} a \left (\int \frac {1}{x \sqrt {a x+1}}dx+\frac {2}{\sqrt {a x+1}}\right )-\frac {13}{x \sqrt {a x+1}}\right )-\frac {1}{2 x^2 \sqrt {a x+1}}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\sqrt {c-a c x} \left (-\frac {1}{4} a \left (-\frac {47}{2} a \left (\frac {2 \int \frac {1}{\frac {a x+1}{a}-\frac {1}{a}}d\sqrt {a x+1}}{a}+\frac {2}{\sqrt {a x+1}}\right )-\frac {13}{x \sqrt {a x+1}}\right )-\frac {1}{2 x^2 \sqrt {a x+1}}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\left (-\frac {1}{4} a \left (-\frac {47}{2} a \left (\frac {2}{\sqrt {a x+1}}-2 \text {arctanh}\left (\sqrt {a x+1}\right )\right )-\frac {13}{x \sqrt {a x+1}}\right )-\frac {1}{2 x^2 \sqrt {a x+1}}\right ) \sqrt {c-a c x}}{\sqrt {1-a x}}\)

Input:

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

Output:

(Sqrt[c - a*c*x]*(-1/2*1/(x^2*Sqrt[1 + a*x]) - (a*(-13/(x*Sqrt[1 + a*x]) - 
 (47*a*(2/Sqrt[1 + a*x] - 2*ArcTanh[Sqrt[1 + a*x]]))/2))/4))/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 37
Int[(u_.)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> S 
imp[(a + b*x)^m/(c + d*x)^m   Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, 
b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !SimplerQ[a + b*x, c + d*x]
 

rule 61
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*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

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 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 6680
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol 
] :> Int[u*(c + d*x)^p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; FreeQ[{a, c 
, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] &&  !(IntegerQ[p] || GtQ[c, 0])
 
Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.66

method result size
default \(\frac {\sqrt {-c \left (a x -1\right )}\, \sqrt {-a^{2} x^{2}+1}\, \left (47 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (a x +1\right )}}{\sqrt {c}}\right ) a^{2} x^{2} \sqrt {c \left (a x +1\right )}-47 \sqrt {c}\, a^{2} x^{2}-13 \sqrt {c}\, a x +2 \sqrt {c}\right )}{4 \sqrt {c}\, \left (a x -1\right ) \left (a x +1\right ) x^{2}}\) \(100\)
risch \(-\frac {\left (15 a^{2} x^{2}+13 a x -2\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right ) c}{4 x^{2} \sqrt {c \left (a x +1\right )}\, \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}-\frac {a^{2} \left (\frac {64}{\sqrt {a c x +c}}-\frac {94 \,\operatorname {arctanh}\left (\frac {\sqrt {a c x +c}}{\sqrt {c}}\right )}{\sqrt {c}}\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right ) c}{8 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}\) \(164\)

Input:

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

Output:

1/4*(-c*(a*x-1))^(1/2)*(-a^2*x^2+1)^(1/2)*(47*arctanh((c*(a*x+1))^(1/2)/c^ 
(1/2))*a^2*x^2*(c*(a*x+1))^(1/2)-47*c^(1/2)*a^2*x^2-13*c^(1/2)*a*x+2*c^(1/ 
2))/c^(1/2)/(a*x-1)/(a*x+1)/x^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.64 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx=\left [\frac {47 \, {\left (a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + a c x + 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{a x^{2} - x}\right ) - 2 \, {\left (47 \, a^{2} x^{2} + 13 \, a x - 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{8 \, {\left (a^{2} x^{4} - x^{2}\right )}}, -\frac {47 \, {\left (a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{a c x - c}\right ) + {\left (47 \, a^{2} x^{2} + 13 \, a x - 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{4 \, {\left (a^{2} x^{4} - x^{2}\right )}}\right ] \] Input:

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

Output:

[1/8*(47*(a^4*x^4 - a^2*x^2)*sqrt(c)*log(-(a^2*c*x^2 + a*c*x + 2*sqrt(-a^2 
*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(c) - 2*c)/(a*x^2 - x)) - 2*(47*a^2*x^2 + 1 
3*a*x - 2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^2*x^4 - x^2), -1/4*(47* 
(a^4*x^4 - a^2*x^2)*sqrt(-c)*arctan(sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sq 
rt(-c)/(a*c*x - c)) + (47*a^2*x^2 + 13*a*x - 2)*sqrt(-a^2*x^2 + 1)*sqrt(-a 
*c*x + c))/(a^2*x^4 - x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.50 \[ \int \frac {e^{-3 \text {arctanh}(a x)} \sqrt {c-a c x}}{x^3} \, dx=\frac {\sqrt {c}\, \left (47 \sqrt {a x +1}\, \mathrm {log}\left (\sqrt {a x +1}-1\right ) a^{2} x^{2}-47 \sqrt {a x +1}\, \mathrm {log}\left (\sqrt {a x +1}+1\right ) a^{2} x^{2}+94 a^{2} x^{2}+26 a x -4\right )}{8 \sqrt {a x +1}\, x^{2}} \] Input:

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

Output:

(sqrt(c)*(47*sqrt(a*x + 1)*log(sqrt(a*x + 1) - 1)*a**2*x**2 - 47*sqrt(a*x 
+ 1)*log(sqrt(a*x + 1) + 1)*a**2*x**2 + 94*a**2*x**2 + 26*a*x - 4))/(8*sqr 
t(a*x + 1)*x**2)