\(\int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx\) [1281]

Optimal result

Integrand size = 27, antiderivative size = 130 \[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=-\frac {\sqrt {c-a^2 c x^2}}{4 x^4}+\frac {2 a \sqrt {c-a^2 c x^2}}{3 x^3}-\frac {7 a^2 \sqrt {c-a^2 c x^2}}{8 x^2}+\frac {4 a^3 \sqrt {c-a^2 c x^2}}{3 x}-\frac {7}{8} a^4 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right ) \] Output:

-1/4*(-a^2*c*x^2+c)^(1/2)/x^4+2/3*a*(-a^2*c*x^2+c)^(1/2)/x^3-7/8*a^2*(-a^2 
*c*x^2+c)^(1/2)/x^2+4/3*a^3*(-a^2*c*x^2+c)^(1/2)/x-7/8*a^4*c^(1/2)*arctanh 
((-a^2*c*x^2+c)^(1/2)/c^(1/2))
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (-6+16 a x-21 a^2 x^2+32 a^3 x^3\right )}{24 x^4}+\frac {7}{8} a^4 \sqrt {c} \log (x)-\frac {7}{8} a^4 \sqrt {c} \log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right ) \] Input:

Integrate[Sqrt[c - a^2*c*x^2]/(E^(2*ArcTanh[a*x])*x^5),x]
 

Output:

(Sqrt[c - a^2*c*x^2]*(-6 + 16*a*x - 21*a^2*x^2 + 32*a^3*x^3))/(24*x^4) + ( 
7*a^4*Sqrt[c]*Log[x])/8 - (7*a^4*Sqrt[c]*Log[c + Sqrt[c]*Sqrt[c - a^2*c*x^ 
2]])/8
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 149, 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 = {6702, 540, 27, 539, 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.

\(\displaystyle \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx\)

\(\Big \downarrow \) 6702

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

\(\Big \downarrow \) 540

\(\displaystyle c \left (-\frac {\int \frac {a c (8-7 a x)}{x^4 \sqrt {c-a^2 c x^2}}dx}{4 c}-\frac {\sqrt {c-a^2 c x^2}}{4 c x^4}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (-\frac {1}{4} a \int \frac {8-7 a x}{x^4 \sqrt {c-a^2 c x^2}}dx-\frac {\sqrt {c-a^2 c x^2}}{4 c x^4}\right )\)

\(\Big \downarrow \) 539

\(\displaystyle c \left (-\frac {1}{4} a \left (-\frac {\int \frac {a c (21-16 a x)}{x^3 \sqrt {c-a^2 c x^2}}dx}{3 c}-\frac {8 \sqrt {c-a^2 c x^2}}{3 c x^3}\right )-\frac {\sqrt {c-a^2 c x^2}}{4 c x^4}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (-\frac {1}{4} a \left (-\frac {1}{3} a \int \frac {21-16 a x}{x^3 \sqrt {c-a^2 c x^2}}dx-\frac {8 \sqrt {c-a^2 c x^2}}{3 c x^3}\right )-\frac {\sqrt {c-a^2 c x^2}}{4 c x^4}\right )\)

\(\Big \downarrow \) 539

\(\displaystyle c \left (-\frac {1}{4} a \left (-\frac {1}{3} a \left (-\frac {\int \frac {a c (32-21 a x)}{x^2 \sqrt {c-a^2 c x^2}}dx}{2 c}-\frac {21 \sqrt {c-a^2 c x^2}}{2 c x^2}\right )-\frac {8 \sqrt {c-a^2 c x^2}}{3 c x^3}\right )-\frac {\sqrt {c-a^2 c x^2}}{4 c x^4}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (-\frac {1}{4} a \left (-\frac {1}{3} a \left (-\frac {1}{2} a \int \frac {32-21 a x}{x^2 \sqrt {c-a^2 c x^2}}dx-\frac {21 \sqrt {c-a^2 c x^2}}{2 c x^2}\right )-\frac {8 \sqrt {c-a^2 c x^2}}{3 c x^3}\right )-\frac {\sqrt {c-a^2 c x^2}}{4 c x^4}\right )\)

\(\Big \downarrow \) 534

\(\displaystyle c \left (-\frac {1}{4} a \left (-\frac {1}{3} a \left (-\frac {1}{2} a \left (-21 a \int \frac {1}{x \sqrt {c-a^2 c x^2}}dx-\frac {32 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {21 \sqrt {c-a^2 c x^2}}{2 c x^2}\right )-\frac {8 \sqrt {c-a^2 c x^2}}{3 c x^3}\right )-\frac {\sqrt {c-a^2 c x^2}}{4 c x^4}\right )\)

\(\Big \downarrow \) 243

\(\displaystyle c \left (-\frac {1}{4} a \left (-\frac {1}{3} a \left (-\frac {1}{2} a \left (-\frac {21}{2} a \int \frac {1}{x^2 \sqrt {c-a^2 c x^2}}dx^2-\frac {32 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {21 \sqrt {c-a^2 c x^2}}{2 c x^2}\right )-\frac {8 \sqrt {c-a^2 c x^2}}{3 c x^3}\right )-\frac {\sqrt {c-a^2 c x^2}}{4 c x^4}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle c \left (-\frac {1}{4} a \left (-\frac {1}{3} a \left (-\frac {1}{2} a \left (\frac {21 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2 c}}d\sqrt {c-a^2 c x^2}}{a c}-\frac {32 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {21 \sqrt {c-a^2 c x^2}}{2 c x^2}\right )-\frac {8 \sqrt {c-a^2 c x^2}}{3 c x^3}\right )-\frac {\sqrt {c-a^2 c x^2}}{4 c x^4}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle c \left (-\frac {1}{4} a \left (-\frac {1}{3} a \left (-\frac {1}{2} a \left (\frac {21 a \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {32 \sqrt {c-a^2 c x^2}}{c x}\right )-\frac {21 \sqrt {c-a^2 c x^2}}{2 c x^2}\right )-\frac {8 \sqrt {c-a^2 c x^2}}{3 c x^3}\right )-\frac {\sqrt {c-a^2 c x^2}}{4 c x^4}\right )\)

Input:

Int[Sqrt[c - a^2*c*x^2]/(E^(2*ArcTanh[a*x])*x^5),x]
 

Output:

c*(-1/4*Sqrt[c - a^2*c*x^2]/(c*x^4) - (a*((-8*Sqrt[c - a^2*c*x^2])/(3*c*x^ 
3) - (a*((-21*Sqrt[c - a^2*c*x^2])/(2*c*x^2) - (a*((-32*Sqrt[c - a^2*c*x^2 
])/(c*x) + (21*a*ArcTanh[Sqrt[c - a^2*c*x^2]/Sqrt[c]])/Sqrt[c]))/2))/3))/4 
)
 

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 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 6702

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Simp[1/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]) && ILtQ[n/2, 0]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.73

method	result	size
risch	\(-\frac {\left (32 a^{5} x^{5}-21 a^{4} x^{4}-16 a^{3} x^{3}+15 a^{2} x^{2}-16 a x +6\right ) c}{24 x^{4} \sqrt {-c \left (a^{2} x^{2}-1\right )}}-\frac {7 a^{4} \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{8}\)	\(95\)
default	\(-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{4 c \,x^{4}}+\frac {9 a^{2} \left (-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{2 c \,x^{2}}-\frac {a^{2} \left (\sqrt {-a^{2} c \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )\right )}{2}\right )}{4}+\frac {2 a \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{3 c \,x^{3}}-2 a^{3} \left (-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{c x}-2 a^{2} \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )\right )+2 a^{4} \left (\sqrt {-a^{2} c \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )\right )-2 a^{4} \left (\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 \left (x +\frac {1}{a}\right ) a c}+\frac {a c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 \left (x +\frac {1}{a}\right ) a c}}\right )}{\sqrt {a^{2} c}}\right )\)	\(332\)

Input:

int((-a^2*c*x^2+c)^(1/2)/(a*x+1)^2*(-a^2*x^2+1)/x^5,x,method=_RETURNVERBOS 
E)
 

Output:

-1/24*(32*a^5*x^5-21*a^4*x^4-16*a^3*x^3+15*a^2*x^2-16*a*x+6)/x^4/(-c*(a^2* 
x^2-1))^(1/2)*c-7/8*a^4*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 = 0.08 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.30 \[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=\left [\frac {21 \, a^{4} \sqrt {c} x^{4} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) + 2 \, {\left (32 \, a^{3} x^{3} - 21 \, a^{2} x^{2} + 16 \, a x - 6\right )} \sqrt {-a^{2} c x^{2} + c}}{48 \, x^{4}}, \frac {21 \, a^{4} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{c}\right ) + {\left (32 \, a^{3} x^{3} - 21 \, a^{2} x^{2} + 16 \, a x - 6\right )} \sqrt {-a^{2} c x^{2} + c}}{24 \, x^{4}}\right ] \] Input:

integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^2*(-a^2*x^2+1)/x^5,x, algorithm="fr 
icas")
 

Output:

[1/48*(21*a^4*sqrt(c)*x^4*log(-(a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*sqrt(c) 
 - 2*c)/x^2) + 2*(32*a^3*x^3 - 21*a^2*x^2 + 16*a*x - 6)*sqrt(-a^2*c*x^2 + 
c))/x^4, 1/24*(21*a^4*sqrt(-c)*x^4*arctan(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/c) 
 + (32*a^3*x^3 - 21*a^2*x^2 + 16*a*x - 6)*sqrt(-a^2*c*x^2 + c))/x^4]
 

Sympy [F]

\[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=- \int \left (- \frac {\sqrt {- a^{2} c x^{2} + c}}{a x^{6} + x^{5}}\right )\, dx - \int \frac {a x \sqrt {- a^{2} c x^{2} + c}}{a x^{6} + x^{5}}\, dx \] Input:

integrate((-a**2*c*x**2+c)**(1/2)/(a*x+1)**2*(-a**2*x**2+1)/x**5,x)
 

Output:

-Integral(-sqrt(-a**2*c*x**2 + c)/(a*x**6 + x**5), x) - Integral(a*x*sqrt( 
-a**2*c*x**2 + c)/(a*x**6 + x**5), x)
 

Maxima [F]

\[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=\int { -\frac {\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} - 1\right )}}{{\left (a x + 1\right )}^{2} x^{5}} \,d x } \] Input:

integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^2*(-a^2*x^2+1)/x^5,x, algorithm="ma 
xima")
 

Output:

-integrate(sqrt(-a^2*c*x^2 + c)*(a^2*x^2 - 1)/((a*x + 1)^2*x^5), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (106) = 212\).

Time = 0.14 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.98 \[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=\frac {1}{192} \, {\left (\frac {336 \, a^{3} c \arctan \left (\frac {\sqrt {-c + \frac {2 \, c}{a x + 1}}}{\sqrt {-c}}\right ) \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right )}{\sqrt {-c}} - \frac {4 \, {\left (21 \, \pi a^{3} c - 64 \, a^{3} c\right )} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right )}{\sqrt {-c}} + \frac {75 \, a^{3} {\left (c - \frac {2 \, c}{a x + 1}\right )}^{3} c \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right ) + 83 \, a^{3} {\left (c - \frac {2 \, c}{a x + 1}\right )}^{2} c^{2} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right ) + 21 \, a^{3} c^{4} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right ) - 77 \, a^{3} c^{3} {\left (-c + \frac {2 \, c}{a x + 1}\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right )}{{\left (c - \frac {c}{a x + 1}\right )}^{4}}\right )} {\left | a \right |} \] Input:

integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^2*(-a^2*x^2+1)/x^5,x, algorithm="gi 
ac")
 

Output:

1/192*(336*a^3*c*arctan(sqrt(-c + 2*c/(a*x + 1))/sqrt(-c))*sgn(1/(a*x + 1) 
)*sgn(a)/sqrt(-c) - 4*(21*pi*a^3*c - 64*a^3*c)*sgn(1/(a*x + 1))*sgn(a)/sqr 
t(-c) + (75*a^3*(c - 2*c/(a*x + 1))^3*c*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a* 
x + 1))*sgn(a) + 83*a^3*(c - 2*c/(a*x + 1))^2*c^2*sqrt(-c + 2*c/(a*x + 1)) 
*sgn(1/(a*x + 1))*sgn(a) + 21*a^3*c^4*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a*x 
+ 1))*sgn(a) - 77*a^3*c^3*(-c + 2*c/(a*x + 1))^(3/2)*sgn(1/(a*x + 1))*sgn( 
a))/(c - c/(a*x + 1))^4)*abs(a)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=-\int \frac {\sqrt {c-a^2\,c\,x^2}\,\left (a^2\,x^2-1\right )}{x^5\,{\left (a\,x+1\right )}^2} \,d x \] Input:

int(-((c - a^2*c*x^2)^(1/2)*(a^2*x^2 - 1))/(x^5*(a*x + 1)^2),x)
 

Output:

-int(((c - a^2*c*x^2)^(1/2)*(a^2*x^2 - 1))/(x^5*(a*x + 1)^2), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.69 \[ \int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x^5} \, dx=\frac {\sqrt {c}\, \left (32 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-21 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+16 \sqrt {-a^{2} x^{2}+1}\, a x -6 \sqrt {-a^{2} x^{2}+1}+21 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{4} x^{4}\right )}{24 x^{4}} \] Input:

int((-a^2*c*x^2+c)^(1/2)/(a*x+1)^2*(-a^2*x^2+1)/x^5,x)
 

Output:

(sqrt(c)*(32*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 21*sqrt( - a**2*x**2 + 1)* 
a**2*x**2 + 16*sqrt( - a**2*x**2 + 1)*a*x - 6*sqrt( - a**2*x**2 + 1) + 21* 
log(tan(asin(a*x)/2))*a**4*x**4))/(24*x**4)