\(\int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx\) [662]

Optimal result

Integrand size = 27, antiderivative size = 99 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\frac {\sqrt {c-a^2 c x^2}}{3 x^3}+\frac {a \sqrt {c-a^2 c x^2}}{x^2}+\frac {5 a^2 \sqrt {c-a^2 c x^2}}{3 x}+a^3 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right ) \] Output:

1/3*(-a^2*c*x^2+c)^(1/2)/x^3+a*(-a^2*c*x^2+c)^(1/2)/x^2+5/3*a^2*(-a^2*c*x^ 
2+c)^(1/2)/x+a^3*c^(1/2)*arctanh((-a^2*c*x^2+c)^(1/2)/c^(1/2))
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.83 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\frac {\left (1+3 a x+5 a^2 x^2\right ) \sqrt {c-a^2 c x^2}}{3 x^3}-a^3 \sqrt {c} \log (x)+a^3 \sqrt {c} \log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right ) \] Input:

Integrate[(E^(2*ArcCoth[a*x])*Sqrt[c - a^2*c*x^2])/x^4,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.15, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {6717, 6701, 540, 25, 27, 539, 27, 534, 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.

\(\displaystyle \int \frac {\sqrt {c-a^2 c x^2} e^{2 \coth ^{-1}(a x)}}{x^4} \, dx\)

\(\Big \downarrow \) 6717

\(\displaystyle -\int \frac {e^{2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^4}dx\)

\(\Big \downarrow \) 6701

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

\(\Big \downarrow \) 540

\(\displaystyle -c \left (-\frac {\int -\frac {a c (5 a x+6)}{x^3 \sqrt {c-a^2 c x^2}}dx}{3 c}-\frac {\sqrt {c-a^2 c x^2}}{3 c x^3}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle -c \left (\frac {\int \frac {a c (5 a x+6)}{x^3 \sqrt {c-a^2 c x^2}}dx}{3 c}-\frac {\sqrt {c-a^2 c x^2}}{3 c x^3}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle -c \left (\frac {1}{3} a \int \frac {5 a x+6}{x^3 \sqrt {c-a^2 c x^2}}dx-\frac {\sqrt {c-a^2 c x^2}}{3 c x^3}\right )\)

\(\Big \downarrow \) 539

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 534

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[(E^(2*ArcCoth[a*x])*Sqrt[c - a^2*c*x^2])/x^4,x]
 

Output:

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

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 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 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 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 539

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[c*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*(a*d*(m + 1) - b*c*(m + 2*p + 3)*x), x], x] 
/; FreeQ[{a, b, c, d, p}, x] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

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]
 

rule 6717

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Simp[(-1)^(n/2)   Int[ 
u*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[a, x] && IntegerQ[n/2]
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.87

method	result	size
risch	\(-\frac {\left (5 a^{4} x^{4}+3 a^{3} x^{3}-4 a^{2} x^{2}-3 a x -1\right ) c}{3 x^{3} \sqrt {-c \left (a^{2} x^{2}-1\right )}}+a^{3} \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )\)	\(86\)
default	\(\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{3 c \,x^{3}}-2 a \left (-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{2 c \,x^{2}}-\frac {a^{2} \left (\sqrt {-a^{2} c \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )\right )}{2}\right )-2 a^{2} \left (-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{c x}-2 a^{2} \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )\right )-2 a^{3} \left (\sqrt {-a^{2} c \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )\right )+2 a^{3} \left (\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}-\frac {a c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )}{\sqrt {a^{2} c}}\right )\)	\(317\)

Input:

int(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^(1/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/3*(5*a^4*x^4+3*a^3*x^3-4*a^2*x^2-3*a*x-1)/x^3/(-c*(a^2*x^2-1))^(1/2)*c+ 
a^3*c^(1/2)*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^(1/2))/x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.56 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\left [\frac {3 \, a^{3} \sqrt {c} x^{3} \log \left (-\frac {a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) + 2 \, \sqrt {-a^{2} c x^{2} + c} {\left (5 \, a^{2} x^{2} + 3 \, a x + 1\right )}}{6 \, x^{3}}, -\frac {3 \, a^{3} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{c}\right ) - \sqrt {-a^{2} c x^{2} + c} {\left (5 \, a^{2} x^{2} + 3 \, a x + 1\right )}}{3 \, x^{3}}\right ] \] Input:

integrate(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^(1/2)/x^4,x, algorithm="fricas" 
)
 

Output:

[1/6*(3*a^3*sqrt(c)*x^3*log(-(a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*sqrt(c) - 
 2*c)/x^2) + 2*sqrt(-a^2*c*x^2 + c)*(5*a^2*x^2 + 3*a*x + 1))/x^3, -1/3*(3* 
a^3*sqrt(-c)*x^3*arctan(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/c) - sqrt(-a^2*c*x^2 
 + c)*(5*a^2*x^2 + 3*a*x + 1))/x^3]
 

Sympy [F]

\[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\int \frac {\sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )} \left (a x + 1\right )}{x^{4} \left (a x - 1\right )}\, dx \] Input:

integrate(1/(a*x-1)*(a*x+1)*(-a**2*c*x**2+c)**(1/2)/x**4,x)
 

Output:

Integral(sqrt(-c*(a*x - 1)*(a*x + 1))*(a*x + 1)/(x**4*(a*x - 1)), x)
 

Maxima [F]

\[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} {\left (a x + 1\right )}}{{\left (a x - 1\right )} x^{4}} \,d x } \] Input:

integrate(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^(1/2)/x^4,x, algorithm="maxima" 
)
 

Output:

integrate(sqrt(-a^2*c*x^2 + c)*(a*x + 1)/((a*x - 1)*x^4), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (83) = 166\).

Time = 0.14 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.53 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=-\frac {2 \, a^{3} c \arctan \left (-\frac {\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {2 \, {\left (3 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{5} a^{3} c - 3 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{4} a^{2} \sqrt {-c} c {\left | a \right |} + 12 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} a^{2} \sqrt {-c} c^{2} {\left | a \right |} - 3 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )} a^{3} c^{3} - 5 \, a^{2} \sqrt {-c} c^{3} {\left | a \right |}\right )}}{3 \, {\left ({\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} - c\right )}^{3}} \] Input:

integrate(1/(a*x-1)*(a*x+1)*(-a^2*c*x^2+c)^(1/2)/x^4,x, algorithm="giac")
 

Output:

-2*a^3*c*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c 
) + 2/3*(3*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^5*a^3*c - 3*(sqrt(-a^2* 
c)*x - sqrt(-a^2*c*x^2 + c))^4*a^2*sqrt(-c)*c*abs(a) + 12*(sqrt(-a^2*c)*x 
- sqrt(-a^2*c*x^2 + c))^2*a^2*sqrt(-c)*c^2*abs(a) - 3*(sqrt(-a^2*c)*x - sq 
rt(-a^2*c*x^2 + c))*a^3*c^3 - 5*a^2*sqrt(-c)*c^3*abs(a))/((sqrt(-a^2*c)*x 
- sqrt(-a^2*c*x^2 + c))^2 - c)^3
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\int \frac {\sqrt {c-a^2\,c\,x^2}\,\left (a\,x+1\right )}{x^4\,\left (a\,x-1\right )} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.70 \[ \int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^4} \, dx=\frac {\sqrt {c}\, \left (5 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+3 \sqrt {-a^{2} x^{2}+1}\, a x +\sqrt {-a^{2} x^{2}+1}-3 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{3} x^{3}\right )}{3 x^{3}} \] Input:

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

Output:

(sqrt(c)*(5*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 3*sqrt( - a**2*x**2 + 1)*a* 
x + sqrt( - a**2*x**2 + 1) - 3*log(tan(asin(a*x)/2))*a**3*x**3))/(3*x**3)