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

Optimal. Leaf size=115 \[ \frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} F\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right )}{21 a^{7/2}}-\frac {2 x \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \sqrt {1-a^2 x^4}}{21 a^3}+\frac {2 x^5}{35 a}+\frac {1}{7} x^7 e^{\text {sech}^{-1}\left (a x^2\right )} \]

[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

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Rubi [A]  time = 0.05, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6335, 30, 259, 321, 221} \[ -\frac {2 x \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \sqrt {1-a^2 x^4}}{21 a^3}+\frac {2 \sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} F\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right )}{21 a^{7/2}}+\frac {2 x^5}{35 a}+\frac {1}{7} x^7 e^{\text {sech}^{-1}\left (a x^2\right )} \]

Antiderivative was successfully verified.

[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 221

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 259

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 321

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 6335

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]*Sqrt[1/(1 + a*x^p)])/(a*(m + 1)), 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^6 \, dx &=\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} F\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right )}{21 a^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.32, size = 139, normalized size = 1.21 \[ -\frac {2 i \sqrt {\frac {1-a x^2}{a x^2+1}} \sqrt {1-a^2 x^4} F\left (\left .i \sinh ^{-1}\left (\sqrt {-a} x\right )\right |-1\right )}{21 (-a)^{7/2} \left (a x^2-1\right )}+\frac {x \sqrt {\frac {1-a x^2}{a x^2+1}} \left (3 a^3 x^6+3 a^2 x^4-2 a x^2-2\right )}{21 a^3}+\frac {x^5}{5 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

x^5/(5*a) + (x*Sqrt[(1 - a*x^2)/(1 + a*x^2)]*(-2 - 2*a*x^2 + 3*a^2*x^4 + 3*a^3*x^6))/(21*a^3) - (((2*I)/21)*Sq
rt[(1 - a*x^2)/(1 + a*x^2)]*Sqrt[1 - a^2*x^4]*EllipticF[I*ArcSinh[Sqrt[-a]*x], -1])/((-a)^(7/2)*(-1 + a*x^2))

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a x^{6} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} + x^{4}}{a}, x\right ) \]

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^6,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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^6,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Warning, choosing root of [1,0,%%%{-4,[1,0]%%%},0,%%%{4,[4,4]%%%}] at parameters values [86,-97]Warning,
choosing root of [1,0,%%%{-4,[1,0]%%%},0,%%%{4,[4,4]%%%}] at parameters values [-82,7]Warning, choosing root o
f [1,0,%%%{-4,[1,0]%%%},0,%%%{4,[4,4]%%%}] at parameters values [-27,26]Warning, choosing root of [1,0,%%%{-4,
[1,0]%%%},0,%%%{4,[4,4]%%%}] at parameters values [-89,63]Unable to divide, perhaps due to rounding error%%%{1
,[4,2,1,1,1]%%%}+%%%{1,[4,0,0,0,2]%%%} / %%%{1,[0,0,0,0,3]%%%} Error: Bad Argument Value

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maple [A]  time = 0.06, size = 114, normalized size = 0.99 \[ \frac {\sqrt {-\frac {a \,x^{2}-1}{a \,x^{2}}}\, x^{2} \sqrt {\frac {a \,x^{2}+1}{a \,x^{2}}}\, \left (3 x^{9} a^{\frac {9}{2}}-5 x^{5} a^{\frac {5}{2}}-2 \EllipticF \left (x \sqrt {a}, i\right ) \sqrt {-a \,x^{2}+1}\, \sqrt {a \,x^{2}+1}+2 x \sqrt {a}\right )}{21 a^{\frac {5}{2}} \left (a^{2} x^{4}-1\right )}+\frac {x^{5}}{5 a} \]

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^6,x)

[Out]

1/21*(-(a*x^2-1)/a/x^2)^(1/2)*x^2*((a*x^2+1)/a/x^2)^(1/2)*(3*x^9*a^(9/2)-5*x^5*a^(5/2)-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*x^5/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {x^{5}}{5 \, a} + \frac {\int \sqrt {a x^{2} + 1} \sqrt {-a x^{2} + 1} x^{4}\,{d x}}{a} \]

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^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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^6\,\left (\sqrt {\frac {1}{a\,x^2}-1}\,\sqrt {\frac {1}{a\,x^2}+1}+\frac {1}{a\,x^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int x^{4}\, dx + \int a x^{6} \sqrt {-1 + \frac {1}{a x^{2}}} \sqrt {1 + \frac {1}{a x^{2}}}\, dx}{a} \]

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**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

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