3.1.48 \(\int e^{\text {sech}^{-1}(a x^2)} x^4 \, dx\) [48]

Optimal. Leaf size=112 \[ \frac {2 x^3}{15 a}+\frac {1}{5} e^{\text {sech}^{-1}\left (a x^2\right )} x^5+\frac {2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} E\left (\left .\text {ArcSin}\left (\sqrt {a} x\right )\right |-1\right )}{5 a^{5/2}}-\frac {2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} F\left (\left .\text {ArcSin}\left (\sqrt {a} x\right )\right |-1\right )}{5 a^{5/2}} \]

[Out]

2/15*x^3/a+1/5*(1/a/x^2+(1/a/x^2-1)^(1/2)*(1/a/x^2+1)^(1/2))*x^5+2/5*EllipticE(x*a^(1/2),I)*(1/(a*x^2+1))^(1/2
)*(a*x^2+1)^(1/2)/a^(5/2)-2/5*EllipticF(x*a^(1/2),I)*(1/(a*x^2+1))^(1/2)*(a*x^2+1)^(1/2)/a^(5/2)

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Rubi [A]
time = 0.04, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6470, 30, 265, 313, 227, 1213, 435} \begin {gather*} -\frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} F\left (\left .\text {ArcSin}\left (\sqrt {a} x\right )\right |-1\right )}{5 a^{5/2}}+\frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} E\left (\left .\text {ArcSin}\left (\sqrt {a} x\right )\right |-1\right )}{5 a^{5/2}}+\frac {2 x^3}{15 a}+\frac {1}{5} x^5 e^{\text {sech}^{-1}\left (a x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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])/(5*a^(5/2)) - (2*Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + a*x^2]*EllipticF[ArcSin[Sqrt[a]*x], -1])/(5*a^(5/2)
)

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 313

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

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[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 1213

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + e*(x^2/d)]/Sqrt
[1 - e*(x^2/d)], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

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*} \int e^{\text {sech}^{-1}\left (a x^2\right )} x^4 \, dx &=\frac {1}{5} e^{\text {sech}^{-1}\left (a x^2\right )} x^5+\frac {2 \int x^2 \, dx}{5 a}+\frac {\left (2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \int \frac {x^2}{\sqrt {1-a x^2} \sqrt {1+a x^2}} \, dx}{5 a}\\ &=\frac {2 x^3}{15 a}+\frac {1}{5} e^{\text {sech}^{-1}\left (a x^2\right )} x^5+\frac {\left (2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \int \frac {x^2}{\sqrt {1-a^2 x^4}} \, dx}{5 a}\\ &=\frac {2 x^3}{15 a}+\frac {1}{5} e^{\text {sech}^{-1}\left (a x^2\right )} x^5-\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}{5 a^2}+\frac {\left (2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \int \frac {1+a x^2}{\sqrt {1-a^2 x^4}} \, dx}{5 a^2}\\ &=\frac {2 x^3}{15 a}+\frac {1}{5} e^{\text {sech}^{-1}\left (a x^2\right )} x^5-\frac {2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} F\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right )}{5 a^{5/2}}+\frac {\left (2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \int \frac {\sqrt {1+a x^2}}{\sqrt {1-a x^2}} \, dx}{5 a^2}\\ &=\frac {2 x^3}{15 a}+\frac {1}{5} e^{\text {sech}^{-1}\left (a x^2\right )} x^5+\frac {2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} E\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right )}{5 a^{5/2}}-\frac {2 \sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} F\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right )}{5 a^{5/2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.30, size = 140, normalized size = 1.25 \begin {gather*} \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 \sinh ^{-1}\left (\sqrt {-a} x\right )\right |-1\right )-F\left (\left .i \sinh ^{-1}\left (\sqrt {-a} x\right )\right |-1\right )\right )}{(-a)^{5/2} \left (-1+a x^2\right )}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((5*x^3)/a + (3*Sqrt[(1 - a*x^2)/(1 + a*x^2)]*(x^3 + a*x^5))/a + ((6*I)*Sqrt[(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

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Maple [A]
time = 0.03, size = 136, normalized size = 1.21

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.09, size = 58, normalized size = 0.52 \begin {gather*} \frac {5 \, a x^{3} + 3 \, {\left (a^{2} x^{5} - 2 \, x\right )} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}}}{15 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(5*a*x^3 + 3*(a^2*x^5 - 2*x)*sqrt((a*x^2 + 1)/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2)))/a^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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