\(\int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^2} \, dx\) [1119]

Optimal result

Integrand size = 27, antiderivative size = 82 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^2} \, dx=-\frac {\sqrt {c-a^2 c x^2}}{x}+a \sqrt {c} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )-2 a \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.29 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^2} \, dx=-\frac {\sqrt {c-a^2 c x^2}}{x}-a \sqrt {c} \arctan \left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (-1+a^2 x^2\right )}\right )+2 a \sqrt {c} \log (x)-2 a \sqrt {c} \log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right ) \] Input:

Integrate[(E^(2*ArcTanh[a*x])*Sqrt[c - a^2*c*x^2])/x^2,x]
 

Output:

-(Sqrt[c - a^2*c*x^2]/x) - a*Sqrt[c]*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqr 
t[c]*(-1 + a^2*x^2))] + 2*a*Sqrt[c]*Log[x] - 2*a*Sqrt[c]*Log[c + Sqrt[c]*S 
qrt[c - a^2*c*x^2]]
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {6701, 540, 25, 27, 538, 224, 216, 243, 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.

\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^2} \, dx\)

\(\Big \downarrow \) 6701

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

\(\Big \downarrow \) 540

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 538

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[(E^(2*ArcTanh[a*x])*Sqrt[c - a^2*c*x^2])/x^2,x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 216

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

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 224

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 538

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

rule 540

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain 
der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) 
, x] + Simp[1/(a*(m + 1))   Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 
1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG 
tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 6701

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.22

method	result	size
risch	\(\frac {\left (a^{2} x^{2}-1\right ) c}{x \sqrt {-c \left (a^{2} x^{2}-1\right )}}+\left (\frac {a^{2} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{\sqrt {a^{2} c}}-\frac {2 a \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{\sqrt {c}}\right ) c\)	\(100\)
default	\(-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{c x}-2 a^{2} \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )+2 a \left (\sqrt {-a^{2} c \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )\right )-2 a \left (\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}-\frac {a c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )}{\sqrt {a^{2} c}}\right )\)	\(210\)

Input:

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

Output:

(a^2*x^2-1)/x/(-c*(a^2*x^2-1))^(1/2)*c+(a^2/(a^2*c)^(1/2)*arctan((a^2*c)^( 
1/2)*x/(-a^2*c*x^2+c)^(1/2))-2*a/c^(1/2)*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^ 
(1/2))/x))*c
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 200, normalized size of antiderivative = 2.44 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^2} \, dx=\left [-\frac {a \sqrt {c} x \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) - a \sqrt {c} x \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) + \sqrt {-a^{2} c x^{2} + c}}{x}, \frac {4 \, a \sqrt {-c} x \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{c}\right ) + a \sqrt {-c} x \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) - 2 \, \sqrt {-a^{2} c x^{2} + c}}{2 \, x}\right ] \] Input:

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

Output:

[-(a*sqrt(c)*x*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) - 
a*sqrt(c)*x*log(-(a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) + 
 sqrt(-a^2*c*x^2 + c))/x, 1/2*(4*a*sqrt(-c)*x*arctan(sqrt(-a^2*c*x^2 + c)* 
sqrt(-c)/c) + a*sqrt(-c)*x*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt 
(-c)*x - c) - 2*sqrt(-a^2*c*x^2 + c))/x]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.62 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^2} \, dx=\frac {4 \, a c \arctan \left (-\frac {\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {a^{2} \sqrt {-c} \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{{\left | a \right |}} + \frac {2 \, a^{2} \sqrt {-c} c}{{\left ({\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} - c\right )} {\left | a \right |}} \] Input:

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

Output:

4*a*c*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c) + 
 a^2*sqrt(-c)*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/abs(a) + 2* 
a^2*sqrt(-c)*c/(((sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2 - c)*abs(a))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.48 \[ \int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^2} \, dx=\frac {\sqrt {c}\, \left (\mathit {asin} \left (a x \right ) a x -\sqrt {-a^{2} x^{2}+1}+2 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a x \right )}{x} \] Input:

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

Output:

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