\(\int e^{\text {csch}^{-1}(a x^2)} x^4 \, dx\) [25]

Optimal result

Integrand size = 12, antiderivative size = 202 \[ \int e^{\text {csch}^{-1}\left (a x^2\right )} x^4 \, dx=-\frac {2 \sqrt {1+\frac {1}{a^2 x^4}}}{5 a^2 \left (a+\frac {1}{x^2}\right ) x}+\frac {2 \sqrt {1+\frac {1}{a^2 x^4}} x}{5 a^2}+\frac {x^3}{3 a}+\frac {1}{5} \sqrt {1+\frac {1}{a^2 x^4}} x^5+\frac {2 \sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) E\left (2 \cot ^{-1}\left (\sqrt {a} x\right )|\frac {1}{2}\right )}{5 a^{7/2} \sqrt {1+\frac {1}{a^2 x^4}}}-\frac {\sqrt {\frac {a^2+\frac {1}{x^4}}{\left (a+\frac {1}{x^2}\right )^2}} \left (a+\frac {1}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\sqrt {a} x\right ),\frac {1}{2}\right )}{5 a^{7/2} \sqrt {1+\frac {1}{a^2 x^4}}} \] Output:

-2/5*(1+1/a^2/x^4)^(1/2)/a^2/(a+1/x^2)/x+2/5*(1+1/a^2/x^4)^(1/2)*x/a^2+1/3 
*x^3/a+1/5*(1+1/a^2/x^4)^(1/2)*x^5+2/5*((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))/a^(7/2)/(1+1/a^2/x^ 
4)^(1/2)-1/5*((a^2+1/x^4)/(a+1/x^2)^2)^(1/2)*(a+1/x^2)*InverseJacobiAM(2*a 
rccot(a^(1/2)*x),1/2*2^(1/2))/a^(7/2)/(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 = 0.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.55 \[ \int e^{\text {csch}^{-1}\left (a x^2\right )} x^4 \, dx=\frac {4 \sqrt {2} e^{-\text {csch}^{-1}\left (a x^2\right )} \left (\frac {e^{\text {csch}^{-1}\left (a x^2\right )}}{-1+e^{2 \text {csch}^{-1}\left (a x^2\right )}}\right )^{5/2} x^5 \left (-4+7 e^{2 \text {csch}^{-1}\left (a x^2\right )}+4 \left (1-e^{2 \text {csch}^{-1}\left (a x^2\right )}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{2},\frac {7}{4},e^{2 \text {csch}^{-1}\left (a x^2\right )}\right )\right )}{21 \left (a x^2\right )^{5/2}} \] Input:

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

Output:

(4*Sqrt[2]*(E^ArcCsch[a*x^2]/(-1 + E^(2*ArcCsch[a*x^2])))^(5/2)*x^5*(-4 + 
7*E^(2*ArcCsch[a*x^2]) + 4*(1 - E^(2*ArcCsch[a*x^2]))^(5/2)*Hypergeometric 
2F1[3/4, 7/2, 7/4, E^(2*ArcCsch[a*x^2])]))/(21*E^ArcCsch[a*x^2]*(a*x^2)^(5 
/2))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6890, 15, 858, 809, 847, 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.

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

\(\Big \downarrow \) 6890

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 858

\(\displaystyle \frac {x^3}{3 a}-\int \sqrt {1+\frac {1}{x^4 a^2}} x^6d\frac {1}{x}\)

\(\Big \downarrow \) 809

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

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

Input:

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

Output:

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

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 809

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1))), x] - Simp[b*n*(p/(c^n*(m + 1)))   I 
nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ 
[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntB 
inomialQ[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 847

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

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 = 0.54 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {\sqrt {\frac {a^{2} x^{4}+1}{a^{2} x^{4}}}\, x^{2} \left (\sqrt {i a}\, a^{3} x^{7}+x^{3} a \sqrt {i a}+2 i \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i a}, i\right )-2 i \sqrt {-i a \,x^{2}+1}\, \sqrt {i a \,x^{2}+1}\, \operatorname {EllipticE}\left (x \sqrt {i a}, i\right )\right )}{5 \left (a^{2} x^{4}+1\right ) a \sqrt {i a}}+\frac {x^{3}}{3 a}\)	\(150\)

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)*x^2*((I*a)^(1/2)*a^3*x^7+x^3*a*(I*a)^(1/2) 
+2*I*(1-I*a*x^2)^(1/2)*(1+I*a*x^2)^(1/2)*EllipticF(x*(I*a)^(1/2),I)-2*I*(1 
-I*a*x^2)^(1/2)*(1+I*a*x^2)^(1/2)*EllipticE(x*(I*a)^(1/2),I))/(a^2*x^4+1)/ 
a/(I*a)^(1/2)+1/3*x^3/a
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.44 \[ \int e^{\text {csch}^{-1}\left (a x^2\right )} x^4 \, dx=\frac {5 \, a x^{3} + 6 \, \left (-\frac {1}{a^{2}}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {1}{a^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 6 \, \left (-\frac {1}{a^{2}}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {1}{a^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 3 \, {\left (a^{2} x^{5} + 2 \, x\right )} \sqrt {\frac {a^{2} x^{4} + 1}{a^{2} x^{4}}}}{15 \, a^{2}} \] Input:

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

Output:

1/15*(5*a*x^3 + 6*(-1/a^2)^(3/4)*elliptic_e(arcsin((-1/a^2)^(1/4)/x), -1) 
- 6*(-1/a^2)^(3/4)*elliptic_f(arcsin((-1/a^2)^(1/4)/x), -1) + 3*(a^2*x^5 + 
 2*x)*sqrt((a^2*x^4 + 1)/(a^2*x^4)))/a^2
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.24 \[ \int e^{\text {csch}^{-1}\left (a x^2\right )} x^4 \, dx=- \frac {x^{5} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{a^{2} x^{4}}} \right )}}{4 \Gamma \left (- \frac {1}{4}\right )} + \frac {x^{3}}{3 a} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

1/3*x^3/a + integrate(sqrt(a^2*x^4 + 1)*x^2, x)/a
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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