3.32 \(\int e^{\text {sech}^{-1}(a x)} x^4 \, dx\)

Optimal. Leaf size=64 \[ -\frac {2 x e^{\text {sech}^{-1}(a x)}}{15 a^4}+\frac {x^2}{15 a^3}-\frac {x^3 e^{\text {sech}^{-1}(a x)}}{15 a^2}+\frac {1}{5} x^5 e^{\text {sech}^{-1}(a x)}+\frac {x^4}{20 a} \]

[Out]

-2/15*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))*x/a^4+1/15*x^2/a^3-1/15*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))*
x^3/a^2+1/20*x^4/a+1/5*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))*x^5

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Rubi [A]  time = 0.04, antiderivative size = 83, normalized size of antiderivative = 1.30, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6335, 30, 100, 12, 74} \[ -\frac {x^2 \sqrt {1-a x}}{15 a^3 \sqrt {\frac {1}{a x+1}}}-\frac {2 \sqrt {1-a x}}{15 a^5 \sqrt {\frac {1}{a x+1}}}+\frac {x^4}{20 a}+\frac {1}{5} x^5 e^{\text {sech}^{-1}(a x)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

x^4/(20*a) + (E^ArcSech[a*x]*x^5)/5 - (2*Sqrt[1 - a*x])/(15*a^5*Sqrt[(1 + a*x)^(-1)]) - (x^2*Sqrt[1 - a*x])/(1
5*a^3*Sqrt[(1 + a*x)^(-1)])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

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

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 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}(a x)} x^4 \, dx &=\frac {1}{5} e^{\text {sech}^{-1}(a x)} x^5+\frac {\int x^3 \, dx}{5 a}+\frac {\left (\sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int \frac {x^3}{\sqrt {1-a x} \sqrt {1+a x}} \, dx}{5 a}\\ &=\frac {x^4}{20 a}+\frac {1}{5} e^{\text {sech}^{-1}(a x)} x^5-\frac {x^2 \sqrt {1-a x}}{15 a^3 \sqrt {\frac {1}{1+a x}}}-\frac {\left (\sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int -\frac {2 x}{\sqrt {1-a x} \sqrt {1+a x}} \, dx}{15 a^3}\\ &=\frac {x^4}{20 a}+\frac {1}{5} e^{\text {sech}^{-1}(a x)} x^5-\frac {x^2 \sqrt {1-a x}}{15 a^3 \sqrt {\frac {1}{1+a x}}}+\frac {\left (2 \sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int \frac {x}{\sqrt {1-a x} \sqrt {1+a x}} \, dx}{15 a^3}\\ &=\frac {x^4}{20 a}+\frac {1}{5} e^{\text {sech}^{-1}(a x)} x^5-\frac {2 \sqrt {1-a x}}{15 a^5 \sqrt {\frac {1}{1+a x}}}-\frac {x^2 \sqrt {1-a x}}{15 a^3 \sqrt {\frac {1}{1+a x}}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 65, normalized size = 1.02 \[ \frac {15 a^4 x^4+4 \sqrt {\frac {1-a x}{a x+1}} (a x+1)^2 \left (3 a^3 x^3-3 a^2 x^2+2 a x-2\right )}{60 a^5} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(15*a^4*x^4 + 4*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)^2*(-2 + 2*a*x - 3*a^2*x^2 + 3*a^3*x^3))/(60*a^5)

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fricas [A]  time = 0.55, size = 65, normalized size = 1.02 \[ \frac {15 \, a^{3} x^{4} + 4 \, {\left (3 \, a^{4} x^{5} - a^{2} x^{3} - 2 \, x\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}}{60 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*a^3*x^4 + 4*(3*a^4*x^5 - a^2*x^3 - 2*x)*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)))/a^4

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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+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))*x^4,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,[1,1]%%%},0,%%%{4,[4,4]%%%}] at parameters values [86,-97]Warning,
choosing root of [1,0,%%%{4,[1,1]%%%},0,%%%{4,[4,4]%%%}] at parameters values [-82,7]Warning, choosing root of
[1,0,%%%{-4,[1,1]%%%},0,%%%{4,[4,4]%%%}] at parameters values [-27,26]Warning, choosing root of [1,0,%%%{4,[1
,1]%%%},0,%%%{4,[4,4]%%%}] at parameters values [-89,63]Unable to divide, perhaps due to rounding error%%%{1,[
3,2,2,0,0]%%%}+%%%{1,[2,0,1,1,1]%%%} / %%%{1,[0,2,3,0,0]%%%} Error: Bad Argument Value

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maple [A]  time = 0.06, size = 64, normalized size = 1.00 \[ \frac {\sqrt {-\frac {a x -1}{a x}}\, x \sqrt {\frac {a x +1}{a x}}\, \left (a^{2} x^{2}-1\right ) \left (3 a^{2} x^{2}+2\right )}{15 a^{4}}+\frac {x^{4}}{4 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(-(a*x-1)/a/x)^(1/2)*x*((a*x+1)/a/x)^(1/2)*(a^2*x^2-1)*(3*a^2*x^2+2)/a^4+1/4*x^4/a

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maxima [A]  time = 0.38, size = 47, normalized size = 0.73 \[ \frac {x^{4}}{4 \, a} + \frac {{\left (3 \, a^{4} x^{4} - a^{2} x^{2} - 2\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}{15 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4/a + 1/15*(3*a^4*x^4 - a^2*x^2 - 2)*sqrt(a*x + 1)*sqrt(-a*x + 1)/a^5

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mupad [B]  time = 1.47, size = 75, normalized size = 1.17 \[ \frac {x^4}{4\,a}-\sqrt {\frac {1}{a\,x}-1}\,\left (\frac {2\,x\,\sqrt {\frac {1}{a\,x}+1}}{15\,a^4}-\frac {x^5\,\sqrt {\frac {1}{a\,x}+1}}{5}+\frac {x^3\,\sqrt {\frac {1}{a\,x}+1}}{15\,a^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^4/(4*a) - (1/(a*x) - 1)^(1/2)*((2*x*(1/(a*x) + 1)^(1/2))/(15*a^4) - (x^5*(1/(a*x) + 1)^(1/2))/5 + (x^3*(1/(a
*x) + 1)^(1/2))/(15*a^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+(1/a/x-1)**(1/2)*(1+1/a/x)**(1/2))*x**4,x)

[Out]

Timed out

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