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

Optimal. Leaf size=84 \[ \frac {\sqrt {\frac {1}{a x+1}} \sqrt {a x+1} \sin ^{-1}(a x)}{8 a^4}-\frac {x \sqrt {1-a x}}{8 a^3 \sqrt {\frac {1}{a x+1}}}+\frac {1}{4} x^4 e^{\text {sech}^{-1}(a x)}+\frac {x^3}{12 a} \]

[Out]

1/12*x^3/a+1/4*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))*x^4-1/8*x*(-a*x+1)^(1/2)/a^3/(1/(a*x+1))^(1/2)+1/8*arcs
in(a*x)*(1/(a*x+1))^(1/2)*(a*x+1)^(1/2)/a^4

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Rubi [A]  time = 0.03, antiderivative size = 84, normalized size of antiderivative = 1.00, 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, 90, 41, 216} \[ -\frac {x \sqrt {1-a x}}{8 a^3 \sqrt {\frac {1}{a x+1}}}+\frac {\sqrt {\frac {1}{a x+1}} \sqrt {a x+1} \sin ^{-1}(a x)}{8 a^4}+\frac {x^3}{12 a}+\frac {1}{4} x^4 e^{\text {sech}^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^3/(12*a) + (E^ArcSech[a*x]*x^4)/4 - (x*Sqrt[1 - a*x])/(8*a^3*Sqrt[(1 + a*x)^(-1)]) + (Sqrt[(1 + a*x)^(-1)]*S
qrt[1 + a*x]*ArcSin[a*x])/(8*a^4)

Rule 30

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

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

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 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^3 \, dx &=\frac {1}{4} e^{\text {sech}^{-1}(a x)} x^4+\frac {\int x^2 \, dx}{4 a}+\frac {\left (\sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int \frac {x^2}{\sqrt {1-a x} \sqrt {1+a x}} \, dx}{4 a}\\ &=\frac {x^3}{12 a}+\frac {1}{4} e^{\text {sech}^{-1}(a x)} x^4-\frac {x \sqrt {1-a x}}{8 a^3 \sqrt {\frac {1}{1+a x}}}+\frac {\left (\sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int \frac {1}{\sqrt {1-a x} \sqrt {1+a x}} \, dx}{8 a^3}\\ &=\frac {x^3}{12 a}+\frac {1}{4} e^{\text {sech}^{-1}(a x)} x^4-\frac {x \sqrt {1-a x}}{8 a^3 \sqrt {\frac {1}{1+a x}}}+\frac {\left (\sqrt {\frac {1}{1+a x}} \sqrt {1+a x}\right ) \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{8 a^3}\\ &=\frac {x^3}{12 a}+\frac {1}{4} e^{\text {sech}^{-1}(a x)} x^4-\frac {x \sqrt {1-a x}}{8 a^3 \sqrt {\frac {1}{1+a x}}}+\frac {\sqrt {\frac {1}{1+a x}} \sqrt {1+a x} \sin ^{-1}(a x)}{8 a^4}\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 97, normalized size = 1.15 \[ \frac {8 a^3 x^3-3 a \sqrt {\frac {1-a x}{a x+1}} \left (-2 a^3 x^4-2 a^2 x^3+a x^2+x\right )+3 i \log \left (2 \sqrt {\frac {1-a x}{a x+1}} (a x+1)-2 i a x\right )}{24 a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(8*a^3*x^3 - 3*a*Sqrt[(1 - a*x)/(1 + a*x)]*(x + a*x^2 - 2*a^2*x^3 - 2*a^3*x^4) + (3*I)*Log[(-2*I)*a*x + 2*Sqrt
[(1 - a*x)/(1 + a*x)]*(1 + a*x)])/(24*a^4)

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fricas [A]  time = 0.51, size = 95, normalized size = 1.13 \[ \frac {8 \, a^{3} x^{3} + 3 \, {\left (2 \, a^{4} x^{4} - a^{2} x^{2}\right )} \sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}} - 3 \, \arctan \left (\sqrt {\frac {a x + 1}{a x}} \sqrt {-\frac {a x - 1}{a x}}\right )}{24 \, a^{4}} \]

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^3,x, algorithm="fricas")

[Out]

1/24*(8*a^3*x^3 + 3*(2*a^4*x^4 - a^2*x^2)*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) - 3*arctan(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^3,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,[
2,2,2,0,0]%%%}+%%%{1,[1,0,1,1,1]%%%} / %%%{1,[0,2,3,0,0]%%%} Error: Bad Argument Value

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maple [C]  time = 0.06, size = 118, normalized size = 1.40 \[ \frac {\sqrt {-\frac {a x -1}{a x}}\, x \sqrt {\frac {a x +1}{a x}}\, \left (2 \,\mathrm {csgn}\relax (a ) x^{3} a^{3} \sqrt {-a^{2} x^{2}+1}-x \sqrt {-a^{2} x^{2}+1}\, \mathrm {csgn}\relax (a ) a +\arctan \left (\frac {\mathrm {csgn}\relax (a ) a x}{\sqrt {-a^{2} x^{2}+1}}\right )\right ) \mathrm {csgn}\relax (a )}{8 \sqrt {-a^{2} x^{2}+1}\, a^{3}}+\frac {x^{3}}{3 a} \]

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^3,x)

[Out]

1/8*(-(a*x-1)/a/x)^(1/2)*x*((a*x+1)/a/x)^(1/2)*(2*csgn(a)*x^3*a^3*(-a^2*x^2+1)^(1/2)-x*(-a^2*x^2+1)^(1/2)*csgn
(a)*a+arctan(csgn(a)*a*x/(-a^2*x^2+1)^(1/2)))*csgn(a)/(-a^2*x^2+1)^(1/2)/a^3+1/3*x^3/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {x^{3}}{3 \, a} + \frac {-\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{4 \, a^{2}} + \frac {\sqrt {-a^{2} x^{2} + 1} x}{8 \, a^{2}} + \frac {\arcsin \left (a x\right )}{8 \, a^{3}}}{a} \]

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^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 11.93, size = 521, normalized size = 6.20 \[ \frac {\ln \left (\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+1\right )\,1{}\mathrm {i}}{8\,a^4}-\frac {\frac {1{}\mathrm {i}}{1024\,a^4}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{128\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4\,11{}\mathrm {i}}{512\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6\,7{}\mathrm {i}}{256\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^8\,239{}\mathrm {i}}{1024\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^8}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{10}\,1{}\mathrm {i}}{256\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{10}}}{\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4}+\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^6}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^6}+\frac {6\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^8}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^8}+\frac {4\,{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{10}}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{10}}+\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^{12}}{{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^{12}}}-\frac {\ln \left (\frac {\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{a\,x}+1}-1}\right )\,1{}\mathrm {i}}{8\,a^4}+\frac {x^3}{3\,a}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^2\,1{}\mathrm {i}}{256\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^2}-\frac {{\left (\sqrt {\frac {1}{a\,x}-1}-\mathrm {i}\right )}^4\,1{}\mathrm {i}}{1024\,a^4\,{\left (\sqrt {\frac {1}{a\,x}+1}-1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(((1/(a*x) - 1)^(1/2) - 1i)^2/((1/(a*x) + 1)^(1/2) - 1)^2 + 1)*1i)/(8*a^4) - (1i/(1024*a^4) + (((1/(a*x) -
1)^(1/2) - 1i)^2*1i)/(128*a^4*((1/(a*x) + 1)^(1/2) - 1)^2) + (((1/(a*x) - 1)^(1/2) - 1i)^4*11i)/(512*a^4*((1/
(a*x) + 1)^(1/2) - 1)^4) + (((1/(a*x) - 1)^(1/2) - 1i)^6*7i)/(256*a^4*((1/(a*x) + 1)^(1/2) - 1)^6) - (((1/(a*x
) - 1)^(1/2) - 1i)^8*239i)/(1024*a^4*((1/(a*x) + 1)^(1/2) - 1)^8) + (((1/(a*x) - 1)^(1/2) - 1i)^10*1i)/(256*a^
4*((1/(a*x) + 1)^(1/2) - 1)^10))/(((1/(a*x) - 1)^(1/2) - 1i)^4/((1/(a*x) + 1)^(1/2) - 1)^4 + (4*((1/(a*x) - 1)
^(1/2) - 1i)^6)/((1/(a*x) + 1)^(1/2) - 1)^6 + (6*((1/(a*x) - 1)^(1/2) - 1i)^8)/((1/(a*x) + 1)^(1/2) - 1)^8 + (
4*((1/(a*x) - 1)^(1/2) - 1i)^10)/((1/(a*x) + 1)^(1/2) - 1)^10 + ((1/(a*x) - 1)^(1/2) - 1i)^12/((1/(a*x) + 1)^(
1/2) - 1)^12) - (log(((1/(a*x) - 1)^(1/2) - 1i)/((1/(a*x) + 1)^(1/2) - 1))*1i)/(8*a^4) + x^3/(3*a) - (((1/(a*x
) - 1)^(1/2) - 1i)^2*1i)/(256*a^4*((1/(a*x) + 1)^(1/2) - 1)^2) - (((1/(a*x) - 1)^(1/2) - 1i)^4*1i)/(1024*a^4*(
(1/(a*x) + 1)^(1/2) - 1)^4)

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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**3,x)

[Out]

Timed out

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