\(\int e^{\text {sech}^{-1}(a x^2)} x^6 \, dx\) [46]

Optimal result

Integrand size = 12, antiderivative size = 115 \[ \int e^{\text {sech}^{-1}\left (a x^2\right )} x^6 \, dx=\frac {2 x^5}{35 a}+\frac {1}{7} e^{\text {sech}^{-1}\left (a x^2\right )} x^7-\frac {2 x \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} \sqrt {1-a^2 x^4}}{21 a^3}+\frac {2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {a} x\right ),-1\right )}{21 a^{7/2}} \]

[Out]

2/35*x^5/a+1/7*(1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^7+2/21*EllipticF(x*a^(1/2),I)*(1/(a*x^2+1))^(1/
2)*(a*x^2+1)^(1/2)/a^(7/2)-2/21*x*(1/(a*x^2+1))^(1/2)*(a*x^2+1)^(1/2)*(-a^2*x^4+1)^(1/2)/a^3

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6470, 30, 265, 327, 227} \[ \int e^{\text {sech}^{-1}\left (a x^2\right )} x^6 \, dx=\frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {a} x\right ),-1\right )}{21 a^{7/2}}-\frac {2 x \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \sqrt {1-a^2 x^4}}{21 a^3}+\frac {2 x^5}{35 a}+\frac {1}{7} x^7 e^{\text {sech}^{-1}\left (a x^2\right )} \]

[In]

Int[E^ArcSech[a*x^2]*x^6,x]

[Out]

(2*x^5)/(35*a) + (E^ArcSech[a*x^2]*x^7)/7 - (2*x*Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + a*x^2]*Sqrt[1 - a^2*x^4])/(21
*a^3) + (2*Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + a*x^2]*EllipticF[ArcSin[Sqrt[a]*x], -1])/(21*a^(7/2))

Rule 30

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

Rule 227

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

Rule 265

Int[((c_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[(c*x)
^m*(a1*a2 + b1*b2*x^(2*n))^p, x] /; FreeQ[{a1, b1, a2, b2, c, m, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (Intege
rQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rule 327

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

Rule 6470

Int[E^ArcSech[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(E^ArcSech[a*x^p]/(m + 1)), x] + (Dist
[p/(a*(m + 1)), Int[x^(m - p), x], x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{7} e^{\text {sech}^{-1}\left (a x^2\right )} x^7+\frac {2 \int x^4 \, dx}{7 a}+\frac {\left (2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \int \frac {x^4}{\sqrt {1-a x^2} \sqrt {1+a x^2}} \, dx}{7 a} \\ & = \frac {2 x^5}{35 a}+\frac {1}{7} e^{\text {sech}^{-1}\left (a x^2\right )} x^7+\frac {\left (2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \int \frac {x^4}{\sqrt {1-a^2 x^4}} \, dx}{7 a} \\ & = \frac {2 x^5}{35 a}+\frac {1}{7} e^{\text {sech}^{-1}\left (a x^2\right )} x^7-\frac {2 x \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} \sqrt {1-a^2 x^4}}{21 a^3}+\frac {\left (2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \int \frac {1}{\sqrt {1-a^2 x^4}} \, dx}{21 a^3} \\ & = \frac {2 x^5}{35 a}+\frac {1}{7} e^{\text {sech}^{-1}\left (a x^2\right )} x^7-\frac {2 x \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} \sqrt {1-a^2 x^4}}{21 a^3}+\frac {2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {a} x\right ),-1\right )}{21 a^{7/2}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.35 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.24 \[ \int e^{\text {sech}^{-1}\left (a x^2\right )} x^6 \, dx=-\frac {2 \sqrt {2} \sqrt {\frac {e^{\text {sech}^{-1}\left (a x^2\right )}}{1+e^{2 \text {sech}^{-1}\left (a x^2\right )}}} x^5 \left (-5-17 e^{2 \text {sech}^{-1}\left (a x^2\right )}-67 e^{4 \text {sech}^{-1}\left (a x^2\right )}+5 e^{6 \text {sech}^{-1}\left (a x^2\right )}+5 \left (1+e^{2 \text {sech}^{-1}\left (a x^2\right )}\right )^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 \text {sech}^{-1}\left (a x^2\right )}\right )\right )}{105 a \left (1+e^{2 \text {sech}^{-1}\left (a x^2\right )}\right )^3 \left (a x^2\right )^{5/2}} \]

[In]

Integrate[E^ArcSech[a*x^2]*x^6,x]

[Out]

(-2*Sqrt[2]*Sqrt[E^ArcSech[a*x^2]/(1 + E^(2*ArcSech[a*x^2]))]*x^5*(-5 - 17*E^(2*ArcSech[a*x^2]) - 67*E^(4*ArcS
ech[a*x^2]) + 5*E^(6*ArcSech[a*x^2]) + 5*(1 + E^(2*ArcSech[a*x^2]))^(7/2)*Hypergeometric2F1[1/4, 1/2, 5/4, -E^
(2*ArcSech[a*x^2])]))/(105*a*(1 + E^(2*ArcSech[a*x^2]))^3*(a*x^2)^(5/2))

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {\sqrt {-\frac {a \,x^{2}-1}{a \,x^{2}}}\, x^{2} \sqrt {\frac {a \,x^{2}+1}{a \,x^{2}}}\, \left (3 a^{\frac {9}{2}} x^{9}-5 a^{\frac {5}{2}} x^{5}-2 \operatorname {EllipticF}\left (x \sqrt {a}, i\right ) \sqrt {-a \,x^{2}+1}\, \sqrt {a \,x^{2}+1}+2 x \sqrt {a}\right )}{21 a^{\frac {5}{2}} \left (x^{4} a^{2}-1\right )}+\frac {x^{5}}{5 a}\)	\(114\)

[In]

int((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^6,x,method=_RETURNVERBOSE)

[Out]

1/21*(-(a*x^2-1)/a/x^2)^(1/2)*x^2*((a*x^2+1)/a/x^2)^(1/2)*(3*a^(9/2)*x^9-5*a^(5/2)*x^5-2*EllipticF(x*a^(1/2),I
)*(-a*x^2+1)^(1/2)*(a*x^2+1)^(1/2)+2*x*a^(1/2))/a^(5/2)/(a^2*x^4-1)+1/5/a*x^5

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69 \[ \int e^{\text {sech}^{-1}\left (a x^2\right )} x^6 \, dx=\frac {21 \, a^{2} x^{5} + 5 \, {\left (3 \, a^{3} x^{7} - 2 \, a x^{3}\right )} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} + \frac {10 i \, F(\arcsin \left (\frac {1}{\sqrt {a} x}\right )\,|\,-1)}{\sqrt {a}}}{105 \, a^{3}} \]

[In]

integrate((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^6,x, algorithm="fricas")

[Out]

1/105*(21*a^2*x^5 + 5*(3*a^3*x^7 - 2*a*x^3)*sqrt((a*x^2 + 1)/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2)) + 10*I*ellipt
ic_f(arcsin(1/(sqrt(a)*x)), -1)/sqrt(a))/a^3

Sympy [F]

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

[In]

integrate((1/a/x**2+(1/a/x**2-1)**(1/2)*(1/a/x**2+1)**(1/2))*x**6,x)

[Out]

(Integral(x**4, x) + Integral(a*x**6*sqrt(-1 + 1/(a*x**2))*sqrt(1 + 1/(a*x**2)), x))/a

Maxima [F]

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

[In]

integrate((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^6,x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int e^{\text {sech}^{-1}\left (a x^2\right )} x^6 \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^6,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{1,[0,4,2,1,1,1]%%%}+%%%{1,[0,4,0,0,0,2]%%%} / %%%{1,[0,0
,0,0,0,3]%%

Mupad [F(-1)]

Timed out. \[ \int e^{\text {sech}^{-1}\left (a x^2\right )} x^6 \, dx=\int x^6\,\left (\sqrt {\frac {1}{a\,x^2}-1}\,\sqrt {\frac {1}{a\,x^2}+1}+\frac {1}{a\,x^2}\right ) \,d x \]

[In]

int(x^6*((1/(a*x^2) - 1)^(1/2)*(1/(a*x^2) + 1)^(1/2) + 1/(a*x^2)),x)

[Out]

int(x^6*((1/(a*x^2) - 1)^(1/2)*(1/(a*x^2) + 1)^(1/2) + 1/(a*x^2)), x)