3.49 \(\int e^{\text {sech}^{-1}(a x^2)} x^3 \, dx\)

Optimal. Leaf size=63 \[ \frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \sin ^{-1}\left (a x^2\right )}{4 a^2}+\frac {x^2}{4 a}+\frac {1}{4} x^4 e^{\text {sech}^{-1}\left (a x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 63, 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, 275, 216} \[ \frac {\sqrt {\frac {1}{a x^2+1}} \sqrt {a x^2+1} \sin ^{-1}\left (a x^2\right )}{4 a^2}+\frac {x^2}{4 a}+\frac {1}{4} x^4 e^{\text {sech}^{-1}\left (a x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSech[a*x^2]*x^3,x]

[Out]

x^2/(4*a) + (E^ArcSech[a*x^2]*x^4)/4 + (Sqrt[(1 + a*x^2)^(-1)]*Sqrt[1 + a*x^2]*ArcSin[a*x^2])/(4*a^2)

Rule 30

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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^3 \, dx &=\frac {1}{4} e^{\text {sech}^{-1}\left (a x^2\right )} x^4+\frac {\int x \, dx}{2 a}+\frac {\left (\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \int \frac {x}{\sqrt {1-a x^2} \sqrt {1+a x^2}} \, dx}{2 a}\\ &=\frac {x^2}{4 a}+\frac {1}{4} e^{\text {sech}^{-1}\left (a x^2\right )} x^4+\frac {\left (\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \int \frac {x}{\sqrt {1-a^2 x^4}} \, dx}{2 a}\\ &=\frac {x^2}{4 a}+\frac {1}{4} e^{\text {sech}^{-1}\left (a x^2\right )} x^4+\frac {\left (\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx,x,x^2\right )}{4 a}\\ &=\frac {x^2}{4 a}+\frac {1}{4} e^{\text {sech}^{-1}\left (a x^2\right )} x^4+\frac {\sqrt {\frac {1}{1+a x^2}} \sqrt {1+a x^2} \sin ^{-1}\left (a x^2\right )}{4 a^2}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 92, normalized size = 1.46 \[ \frac {2 a x^2+i \log \left (2 \sqrt {\frac {1-a x^2}{a x^2+1}} \left (a x^2+1\right )-2 i a x^2\right )+a \sqrt {\frac {1-a x^2}{a x^2+1}} \left (a x^4+x^2\right )}{4 a^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSech[a*x^2]*x^3,x]

[Out]

(2*a*x^2 + a*Sqrt[(1 - a*x^2)/(1 + a*x^2)]*(x^2 + a*x^4) + I*Log[(-2*I)*a*x^2 + 2*Sqrt[(1 - a*x^2)/(1 + a*x^2)
]*(1 + a*x^2)])/(4*a^2)

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fricas [A]  time = 0.76, size = 102, normalized size = 1.62 \[ \frac {a^{2} x^{4} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} + 2 \, a x^{2} - 2 \, \arctan \left (\frac {a x^{2} \sqrt {\frac {a x^{2} + 1}{a x^{2}}} \sqrt {-\frac {a x^{2} - 1}{a x^{2}}} - 1}{a x^{2}}\right )}{4 \, a^{2}} \]

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^3,x, algorithm="fricas")

[Out]

1/4*(a^2*x^4*sqrt((a*x^2 + 1)/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2)) + 2*a*x^2 - 2*arctan((a*x^2*sqrt((a*x^2 + 1)
/(a*x^2))*sqrt(-(a*x^2 - 1)/(a*x^2)) - 1)/(a*x^2)))/a^2

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giac [B]  time = 0.17, size = 132, normalized size = 2.10 \[ \frac {2 \, a^{2} x^{2} + 4 \, a \arcsin \left (\frac {\sqrt {2} \sqrt {a^{2} x^{2} + a}}{2 \, \sqrt {a}}\right ) + 2 \, \sqrt {a^{2} x^{2} + a} \sqrt {-a^{2} x^{2} + a} + 2 \, a - \frac {2 \, a^{2} \arcsin \left (\frac {\sqrt {2} \sqrt {a^{2} x^{2} + a}}{2 \, \sqrt {a}}\right ) - \sqrt {a^{2} x^{2} + a} {\left (a^{2} x^{2} - 2 \, a\right )} \sqrt {-a^{2} x^{2} + a}}{a}}{4 \, a^{3}} \]

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^3,x, algorithm="giac")

[Out]

1/4*(2*a^2*x^2 + 4*a*arcsin(1/2*sqrt(2)*sqrt(a^2*x^2 + a)/sqrt(a)) + 2*sqrt(a^2*x^2 + a)*sqrt(-a^2*x^2 + a) +
2*a - (2*a^2*arcsin(1/2*sqrt(2)*sqrt(a^2*x^2 + a)/sqrt(a)) - sqrt(a^2*x^2 + a)*(a^2*x^2 - 2*a)*sqrt(-a^2*x^2 +
a))/a)/a^3

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maple [A]  time = 0.15, size = 112, normalized size = 1.78 \[ \frac {\sqrt {-\frac {a \,x^{2}-1}{a \,x^{2}}}\, x^{2} \sqrt {\frac {a \,x^{2}+1}{a \,x^{2}}}\, \left (x^{2} \sqrt {-\frac {a^{2} x^{4}-1}{a^{2}}}\, a^{2}+\arctan \left (\frac {x^{2}}{\sqrt {-\frac {a^{2} x^{4}-1}{a^{2}}}}\right )\right )}{4 \sqrt {-\frac {a^{2} x^{4}-1}{a^{2}}}\, a^{2}}+\frac {x^{2}}{2 a} \]

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^3,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {x^{2}}{2 \, a} + \frac {\frac {1}{4} \, \sqrt {-a^{2} x^{4} + 1} x^{2} + \frac {\arcsin \left (a x^{2}\right )}{4 \, a}}{a} \]

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^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 7.14, size = 306, normalized size = 4.86 \[ \frac {\ln \left (\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^2}+1\right )\,1{}\mathrm {i}}{4\,a^2}-\frac {\ln \left (\frac {\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x^2}+1}-1}\right )\,1{}\mathrm {i}}{4\,a^2}+\frac {\frac {1{}\mathrm {i}}{64\,a^2}+\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{32\,a^2\,{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^2}-\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^4\,15{}\mathrm {i}}{64\,a^2\,{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^4}}{\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^2}+\frac {2\,{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^4}+\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^6}}+\frac {x^2}{2\,a}+\frac {{\left (\sqrt {\frac {1}{a\,x^2}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{64\,a^2\,{\left (\sqrt {\frac {1}{a\,x^2}+1}-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(((1/(a*x^2) - 1)^(1/2) - 1i)^2/((1/(a*x^2) + 1)^(1/2) - 1)^2 + 1)*1i)/(4*a^2) - (log(((1/(a*x^2) - 1)^(1/
2) - 1i)/((1/(a*x^2) + 1)^(1/2) - 1))*1i)/(4*a^2) + (1i/(64*a^2) + (((1/(a*x^2) - 1)^(1/2) - 1i)^2*1i)/(32*a^2
*((1/(a*x^2) + 1)^(1/2) - 1)^2) - (((1/(a*x^2) - 1)^(1/2) - 1i)^4*15i)/(64*a^2*((1/(a*x^2) + 1)^(1/2) - 1)^4))
/(((1/(a*x^2) - 1)^(1/2) - 1i)^2/((1/(a*x^2) + 1)^(1/2) - 1)^2 + (2*((1/(a*x^2) - 1)^(1/2) - 1i)^4)/((1/(a*x^2
) + 1)^(1/2) - 1)^4 + ((1/(a*x^2) - 1)^(1/2) - 1i)^6/((1/(a*x^2) + 1)^(1/2) - 1)^6) + x^2/(2*a) + (((1/(a*x^2)
- 1)^(1/2) - 1i)^2*1i)/(64*a^2*((1/(a*x^2) + 1)^(1/2) - 1)^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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**3,x)

[Out]

Timed out

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