\(\int \frac {e^{\text {csch}^{-1}(a x^2)}}{x^4} \, dx\) [33]

Optimal result

Integrand size = 12, antiderivative size = 181 \[ \int \frac {e^{\text {csch}^{-1}\left (a x^2\right )}}{x^4} \, dx=-\frac {1}{5 a x^5}-\frac {\sqrt {1+\frac {1}{a^2 x^4}}}{5 x^3}-\frac {2 a^2 \sqrt {1+\frac {1}{a^2 x^4}}}{5 \left (a+\frac {1}{x^2}\right ) x}+\frac {2 \sqrt {a} \sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) E\left (2 \cot ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{5 \sqrt {1+\frac {1}{a^2 x^4}}}-\frac {\sqrt {a} \sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\sqrt {a} x\right ),\frac {1}{2}\right )}{5 \sqrt {1+\frac {1}{a^2 x^4}}} \] Output:

-1/5/a/x^5-1/5*(1+1/a^2/x^4)^(1/2)/x^3-2/5*a^2*(1+1/a^2/x^4)^(1/2)/(a+1/x^ 
2)/x+2/5*a^(1/2)*((a^2+1/x^4)/(a+1/x^2)^2)^(1/2)*(a+1/x^2)*EllipticE(sin(2 
*arccot(a^(1/2)*x)),1/2*2^(1/2))/(1+1/a^2/x^4)^(1/2)-1/5*a^(1/2)*((a^2+1/x 
^4)/(a+1/x^2)^2)^(1/2)*(a+1/x^2)*InverseJacobiAM(2*arccot(a^(1/2)*x),1/2*2 
^(1/2))/(1+1/a^2/x^4)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.63 \[ \int \frac {e^{\text {csch}^{-1}\left (a x^2\right )}}{x^4} \, dx=\frac {\left (a x^2\right )^{3/2} \left (3 \left (1-e^{2 \text {csch}^{-1}\left (a x^2\right )}\right )^{3/2}+4 e^{2 \text {csch}^{-1}\left (a x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},e^{2 \text {csch}^{-1}\left (a x^2\right )}\right )\right )}{6 \sqrt {2-2 e^{2 \text {csch}^{-1}\left (a x^2\right )}} \sqrt {\frac {e^{\text {csch}^{-1}\left (a x^2\right )}}{-1+e^{2 \text {csch}^{-1}\left (a x^2\right )}}} x^3} \] Input:

Integrate[E^ArcCsch[a*x^2]/x^4,x]
 

Output:

((a*x^2)^(3/2)*(3*(1 - E^(2*ArcCsch[a*x^2]))^(3/2) + 4*E^(2*ArcCsch[a*x^2] 
)*Hypergeometric2F1[-1/2, 3/4, 7/4, E^(2*ArcCsch[a*x^2])]))/(6*Sqrt[2 - 2* 
E^(2*ArcCsch[a*x^2])]*Sqrt[E^ArcCsch[a*x^2]/(-1 + E^(2*ArcCsch[a*x^2]))]*x 
^3)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6890, 15, 858, 811, 834, 27, 761, 1510}

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.

\(\displaystyle \int \frac {e^{\text {csch}^{-1}\left (a x^2\right )}}{x^4} \, dx\)

\(\Big \downarrow \) 6890

\(\displaystyle \int \frac {\sqrt {1+\frac {1}{x^4 a^2}}}{x^4}dx+\frac {\int \frac {1}{x^6}dx}{a}\)

\(\Big \downarrow \) 15

\(\displaystyle \int \frac {\sqrt {1+\frac {1}{x^4 a^2}}}{x^4}dx-\frac {1}{5 a x^5}\)

\(\Big \downarrow \) 858

\(\displaystyle -\int \frac {\sqrt {1+\frac {1}{x^4 a^2}}}{x^2}d\frac {1}{x}-\frac {1}{5 a x^5}\)

\(\Big \downarrow \) 811

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

\(\displaystyle -\frac {2}{5} \left (\frac {\sqrt {a} \sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{\sqrt {a} x}\right ),\frac {1}{2}\right )}{2 \sqrt {\frac {1}{a^2 x^4}+1}}-\int \frac {a-\frac {1}{x^2}}{\sqrt {1+\frac {1}{x^4 a^2}}}d\frac {1}{x}\right )-\frac {\sqrt {\frac {1}{a^2 x^4}+1}}{5 x^3}-\frac {1}{5 a x^5}\)

\(\Big \downarrow \) 1510

\(\displaystyle -\frac {2}{5} \left (\frac {\sqrt {a} \sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {1}{\sqrt {a} x}\right ),\frac {1}{2}\right )}{2 \sqrt {\frac {1}{a^2 x^4}+1}}-\frac {\sqrt {a} \sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) E\left (2 \arctan \left (\frac {1}{\sqrt {a} x}\right )|\frac {1}{2}\right )}{\sqrt {\frac {1}{a^2 x^4}+1}}+\frac {a^2 \sqrt {\frac {1}{a^2 x^4}+1}}{x \left (a+\frac {1}{x^2}\right )}\right )-\frac {\sqrt {\frac {1}{a^2 x^4}+1}}{5 x^3}-\frac {1}{5 a x^5}\)

Input:

Int[E^ArcCsch[a*x^2]/x^4,x]
 

Output:

-1/5*1/(a*x^5) - Sqrt[1 + 1/(a^2*x^4)]/(5*x^3) - (2*((a^2*Sqrt[1 + 1/(a^2* 
x^4)])/((a + x^(-2))*x) - (Sqrt[a]*Sqrt[(a^2 + x^(-4))/(a + x^(-2))^2]*(a 
+ x^(-2))*EllipticE[2*ArcTan[1/(Sqrt[a]*x)], 1/2])/Sqrt[1 + 1/(a^2*x^4)] + 
 (Sqrt[a]*Sqrt[(a^2 + x^(-4))/(a + x^(-2))^2]*(a + x^(-2))*EllipticF[2*Arc 
Tan[1/(Sqrt[a]*x)], 1/2])/(2*Sqrt[1 + 1/(a^2*x^4)])))/5
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

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 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 811

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 
))   Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I 
GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m 
, p, x]
 

rule 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

rule 1510

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

rule 6890

Int[E^ArcCsch[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Simp[1/a   Int[x^( 
m - p), x], x] + Int[x^m*Sqrt[1 + 1/(a^2*x^(2*p))], x] /; FreeQ[{a, m, p}, 
x]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.74 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\sqrt {\frac {a^{2} x^{4}+1}{a^{2} x^{4}}}\, \left (-2 \sqrt {i a}\, a^{4} x^{8}+2 i a^{3} \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, x^{5} \operatorname {EllipticF}\left (x \sqrt {i a}, i\right )-2 i a^{3} \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, x^{5} \operatorname {EllipticE}\left (x \sqrt {i a}, i\right )-3 \sqrt {i a}\, a^{2} x^{4}-\sqrt {i a}\right )}{5 x^{3} \left (a^{2} x^{4}+1\right ) \sqrt {i a}}-\frac {1}{5 a \,x^{5}}\)	\(171\)

Input:

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

Output:

1/5*((a^2*x^4+1)/a^2/x^4)^(1/2)*(-2*(I*a)^(1/2)*a^4*x^8+2*I*a^3*(1-I*a*x^2 
)^(1/2)*(1+I*a*x^2)^(1/2)*x^5*EllipticF(x*(I*a)^(1/2),I)-2*I*a^3*(1-I*a*x^ 
2)^(1/2)*(1+I*a*x^2)^(1/2)*x^5*EllipticE(x*(I*a)^(1/2),I)-3*(I*a)^(1/2)*a^ 
2*x^4-(I*a)^(1/2))/x^3/(a^2*x^4+1)/(I*a)^(1/2)-1/5/a/x^5
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.54 \[ \int \frac {e^{\text {csch}^{-1}\left (a x^2\right )}}{x^4} \, dx=-\frac {2 \, \left (-a^{2}\right )^{\frac {3}{4}} a^{2} x^{5} E(\arcsin \left (\left (-a^{2}\right )^{\frac {1}{4}} x\right )\,|\,-1) - 2 \, \left (-a^{2}\right )^{\frac {3}{4}} a^{2} x^{5} F(\arcsin \left (\left (-a^{2}\right )^{\frac {1}{4}} x\right )\,|\,-1) + {\left (2 \, a^{3} x^{6} + a x^{2}\right )} \sqrt {\frac {a^{2} x^{4} + 1}{a^{2} x^{4}}} + 1}{5 \, a x^{5}} \] Input:

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

Output:

-1/5*(2*(-a^2)^(3/4)*a^2*x^5*elliptic_e(arcsin((-a^2)^(1/4)*x), -1) - 2*(- 
a^2)^(3/4)*a^2*x^5*elliptic_f(arcsin((-a^2)^(1/4)*x), -1) + (2*a^3*x^6 + a 
*x^2)*sqrt((a^2*x^4 + 1)/(a^2*x^4)) + 1)/(a*x^5)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.21 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.24 \[ \int \frac {e^{\text {csch}^{-1}\left (a x^2\right )}}{x^4} \, dx=- \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{a^{2} x^{4}}} \right )}}{4 x^{3} \Gamma \left (\frac {7}{4}\right )} - \frac {1}{5 a x^{5}} \] Input:

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

Output:

-gamma(3/4)*hyper((-1/2, 3/4), (7/4,), exp_polar(I*pi)/(a**2*x**4))/(4*x** 
3*gamma(7/4)) - 1/(5*a*x**5)
 

Maxima [F]

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

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

Output:

integrate(sqrt(a^2*x^4 + 1)/x^6, x)/a - 1/5/(a*x^5)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {e^{\text {csch}^{-1}\left (a x^2\right )}}{x^4} \, dx=\frac {-5 \sqrt {a^{2} x^{4}+1}-10 \left (\int \frac {\sqrt {a^{2} x^{4}+1}}{a^{2} x^{10}+x^{6}}d x \right ) x^{5}-3}{15 a \,x^{5}} \] Input:

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

Output:

( - 5*sqrt(a**2*x**4 + 1) - 10*int(sqrt(a**2*x**4 + 1)/(a**2*x**10 + x**6) 
,x)*x**5 - 3)/(15*a*x**5)