\(\int e^{\text {sech}^{-1}(a x^2)} x^4 \, dx\) [56]

Optimal result

Integrand size = 12, antiderivative size = 193 \[ \int e^{\text {sech}^{-1}\left (a x^2\right )} x^4 \, dx=\frac {x^3}{3 a}+\frac {x^3 \sqrt {1+a x^2} \sqrt {1-a^2 x^4} \sqrt {-1+\frac {2}{1+a x^2}}}{5 a \sqrt {1-a x^2}}+\frac {2 \sqrt {1+a x^2} \sqrt {-1+\frac {2}{1+a x^2}} E\left (\left .\arcsin \left (\sqrt {a} x\right )\right |-1\right )}{5 a^{5/2} \sqrt {1-a x^2}}-\frac {2 \sqrt {1+a x^2} \sqrt {-1+\frac {2}{1+a x^2}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {a} x\right ),-1\right )}{5 a^{5/2} \sqrt {1-a x^2}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.51 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.73 \[ \int e^{\text {sech}^{-1}\left (a x^2\right )} x^4 \, dx=\frac {1}{15} \left (\frac {5 x^3}{a}+\frac {3 \sqrt {\frac {1-a x^2}{1+a x^2}} \left (x^3+a x^5\right )}{a}+\frac {6 i \sqrt {\frac {1-a x^2}{1+a x^2}} \sqrt {1-a^2 x^4} \left (E\left (\left .i \text {arcsinh}\left (\sqrt {-a} x\right )\right |-1\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-a} x\right ),-1\right )\right )}{(-a)^{5/2} \left (-1+a x^2\right )}\right ) \] Input:

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

Output:

((5*x^3)/a + (3*Sqrt[(1 - a*x^2)/(1 + a*x^2)]*(x^3 + a*x^5))/a + ((6*I)*Sq 
rt[(1 - a*x^2)/(1 + a*x^2)]*Sqrt[1 - a^2*x^4]*(EllipticE[I*ArcSinh[Sqrt[-a 
]*x], -1] - EllipticF[I*ArcSinh[Sqrt[-a]*x], -1]))/((-a)^(5/2)*(-1 + a*x^2 
)))/15
 

Rubi [A] (warning: unable to verify)

Time = 0.52 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.47, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6889, 15, 335, 836, 762, 1388, 327}

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.

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

\(\Big \downarrow \) 6889

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

\(\Big \downarrow \) 15

\(\displaystyle \frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \int \frac {x^2}{\sqrt {1-a x^2} \sqrt {a x^2+1}}dx}{5 a}+\frac {2 x^3}{15 a}+\frac {1}{5} x^5 e^{\text {sech}^{-1}\left (a x^2\right )}\)

\(\Big \downarrow \) 335

\(\displaystyle \frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \int \frac {x^2}{\sqrt {1-a^2 x^4}}dx}{5 a}+\frac {2 x^3}{15 a}+\frac {1}{5} x^5 e^{\text {sech}^{-1}\left (a x^2\right )}\)

\(\Big \downarrow \) 836

\(\displaystyle \frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \left (\frac {\int \frac {a x^2+1}{\sqrt {1-a^2 x^4}}dx}{a}-\frac {\int \frac {1}{\sqrt {1-a^2 x^4}}dx}{a}\right )}{5 a}+\frac {2 x^3}{15 a}+\frac {1}{5} x^5 e^{\text {sech}^{-1}\left (a x^2\right )}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \left (\frac {\int \frac {a x^2+1}{\sqrt {1-a^2 x^4}}dx}{a}-\frac {\operatorname {EllipticF}\left (\arcsin \left (\sqrt {a} x\right ),-1\right )}{a^{3/2}}\right )}{5 a}+\frac {2 x^3}{15 a}+\frac {1}{5} x^5 e^{\text {sech}^{-1}\left (a x^2\right )}\)

\(\Big \downarrow \) 1388

\(\displaystyle \frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \left (\frac {\int \frac {\sqrt {a x^2+1}}{\sqrt {1-a x^2}}dx}{a}-\frac {\operatorname {EllipticF}\left (\arcsin \left (\sqrt {a} x\right ),-1\right )}{a^{3/2}}\right )}{5 a}+\frac {2 x^3}{15 a}+\frac {1}{5} x^5 e^{\text {sech}^{-1}\left (a x^2\right )}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \left (\frac {E\left (\left .\arcsin \left (\sqrt {a} x\right )\right |-1\right )}{a^{3/2}}-\frac {\operatorname {EllipticF}\left (\arcsin \left (\sqrt {a} x\right ),-1\right )}{a^{3/2}}\right )}{5 a}+\frac {2 x^3}{15 a}+\frac {1}{5} x^5 e^{\text {sech}^{-1}\left (a x^2\right )}\)

Input:

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

Output:

(2*x^3)/(15*a) + (E^ArcSech[a*x^2]*x^5)/5 + (2*Sqrt[(1 + a*x^2)^(-1)]*Sqrt 
[1 + a*x^2]*(EllipticE[ArcSin[Sqrt[a]*x], -1]/a^(3/2) - EllipticF[ArcSin[S 
qrt[a]*x], -1]/a^(3/2)))/(5*a)
 

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 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 335

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(p 
_.), x_Symbol] :> Int[(e*x)^m*(a*c + b*d*x^4)^p, x] /; FreeQ[{a, b, c, d, e 
, m, p}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] 
))
 

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 836

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

rule 1388

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), 
x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, 
 c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer 
Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
 

rule 6889

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

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.70

method	result	size
default	\(\frac {\sqrt {-\frac {a \,x^{2}-1}{a \,x^{2}}}\, x^{2} \sqrt {\frac {a \,x^{2}+1}{a \,x^{2}}}\, \left (a^{\frac {7}{2}} x^{7}-x^{3} a^{\frac {3}{2}}+2 \operatorname {EllipticF}\left (\sqrt {a}\, x , i\right ) \sqrt {-a \,x^{2}+1}\, \sqrt {a \,x^{2}+1}-2 \sqrt {-a \,x^{2}+1}\, \sqrt {a \,x^{2}+1}\, \operatorname {EllipticE}\left (\sqrt {a}\, x , i\right )\right )}{5 \left (a^{2} x^{4}-1\right ) a^{\frac {3}{2}}}+\frac {x^{3}}{3 a}\)	\(136\)

Input:

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

Output:

1/5*(-(a*x^2-1)/a/x^2)^(1/2)*x^2*((a*x^2+1)/a/x^2)^(1/2)*(a^(7/2)*x^7-x^3* 
a^(3/2)+2*EllipticF(a^(1/2)*x,I)*(-a*x^2+1)^(1/2)*(a*x^2+1)^(1/2)-2*(-a*x^ 
2+1)^(1/2)*(a*x^2+1)^(1/2)*EllipticE(a^(1/2)*x,I))/(a^2*x^4-1)/a^(3/2)+1/3 
*x^3/a
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.47 \[ \int e^{\text {sech}^{-1}\left (a x^2\right )} x^4 \, dx=\frac {5 \, a^{2} x^{3} + 3 \, {\left (a^{3} x^{5} - 2 \, a x\right )} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} - \frac {6 i \, E(\arcsin \left (\frac {1}{\sqrt {a} x}\right )\,|\,-1)}{\sqrt {a}} + \frac {6 i \, F(\arcsin \left (\frac {1}{\sqrt {a} x}\right )\,|\,-1)}{\sqrt {a}}}{15 \, a^{3}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

(Integral(x**2, x) + Integral(a*x**4*sqrt(-1 + 1/(a*x**2))*sqrt(1 + 1/(a*x 
**2)), x))/a
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int e^{\text {sech}^{-1}\left (a x^2\right )} x^4 \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{1,[0,2,2,1,1,1]%%%}+%%%{1,[0,2,0,0,0,2]%%%} / %%%{1,[0,0,0 
,0,0,3]%%
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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