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

Optimal. Leaf size=163 \[ -\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1)^4}{4 a^4}+\frac{\left (9 \sqrt{\frac{1-a x}{a x+1}}+4\right ) (a x+1)^3}{12 a^4}-\frac{\left (5 \sqrt{\frac{1-a x}{a x+1}}+8\right ) (a x+1)^2}{8 a^4}+\frac{\left (\sqrt{\frac{1-a x}{a x+1}}+8\right ) (a x+1)}{8 a^4}+\frac{\tan ^{-1}\left (\sqrt{\frac{1-a x}{a x+1}}\right )}{4 a^4} \]

[Out]

-(Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)^4)/(4*a^4) + ((1 + a*x)*(8 + Sqrt[(1 - a*x)/(1 + a*x)]))/(8*a^4) - ((1 +
a*x)^2*(8 + 5*Sqrt[(1 - a*x)/(1 + a*x)]))/(8*a^4) + ((1 + a*x)^3*(4 + 9*Sqrt[(1 - a*x)/(1 + a*x)]))/(12*a^4)
+ ArcTan[Sqrt[(1 - a*x)/(1 + a*x)]]/(4*a^4)

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Rubi [A]  time = 0.56946, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {6337, 1804, 1814, 639, 203} \[ -\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1)^4}{4 a^4}+\frac{\left (9 \sqrt{\frac{1-a x}{a x+1}}+4\right ) (a x+1)^3}{12 a^4}-\frac{\left (5 \sqrt{\frac{1-a x}{a x+1}}+8\right ) (a x+1)^2}{8 a^4}+\frac{\left (\sqrt{\frac{1-a x}{a x+1}}+8\right ) (a x+1)}{8 a^4}+\frac{\tan ^{-1}\left (\sqrt{\frac{1-a x}{a x+1}}\right )}{4 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)^4)/(4*a^4) + ((1 + a*x)*(8 + Sqrt[(1 - a*x)/(1 + a*x)]))/(8*a^4) - ((1 +
a*x)^2*(8 + 5*Sqrt[(1 - a*x)/(1 + a*x)]))/(8*a^4) + ((1 + a*x)^3*(4 + 9*Sqrt[(1 - a*x)/(1 + a*x)]))/(12*a^4)
+ ArcTan[Sqrt[(1 - a*x)/(1 + a*x)]]/(4*a^4)

Rule 6337

Int[E^(ArcSech[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[(1 - u)/(1 + u)] + (1*Sqrt[(1 - u)/(1 +
u)])/u)^n, x] /; FreeQ[m, x] && IntegerQ[n]

Rule 1804

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x
^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x
], x, 1]}, Simp[((c*x)^m*(a + b*x^2)^(p + 1)*(a*g - b*f*x))/(2*a*b*(p + 1)), x] + Dist[c/(2*a*b*(p + 1)), Int[
(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 639

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

Rule 203

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

Rubi steps

\begin{align*} \int e^{-\text{sech}^{-1}(a x)} x^3 \, dx &=\int \frac{x^3}{\frac{1}{a x}+\sqrt{\frac{1-a x}{1+a x}}+\frac{\sqrt{\frac{1-a x}{1+a x}}}{a x}} \, dx\\ &=-\frac{4 \operatorname{Subst}\left (\int \frac{(-1+x)^4 x (1+x)^2}{\left (1+x^2\right )^5} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{a^4}\\ &=-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^4}{4 a^4}+\frac{\operatorname{Subst}\left (\int \frac{8-8 x-48 x^2+16 x^3+16 x^4-8 x^5}{\left (1+x^2\right )^4} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{2 a^4}\\ &=-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^4}{4 a^4}+\frac{(1+a x)^3 \left (4+9 \sqrt{\frac{1-a x}{1+a x}}\right )}{12 a^4}-\frac{\operatorname{Subst}\left (\int \frac{24-144 x-96 x^2+48 x^3}{\left (1+x^2\right )^3} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{12 a^4}\\ &=-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^4}{4 a^4}-\frac{(1+a x)^2 \left (8+5 \sqrt{\frac{1-a x}{1+a x}}\right )}{8 a^4}+\frac{(1+a x)^3 \left (4+9 \sqrt{\frac{1-a x}{1+a x}}\right )}{12 a^4}+\frac{\operatorname{Subst}\left (\int \frac{24-192 x}{\left (1+x^2\right )^2} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{48 a^4}\\ &=-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^4}{4 a^4}+\frac{(1+a x) \left (8+\sqrt{\frac{1-a x}{1+a x}}\right )}{8 a^4}-\frac{(1+a x)^2 \left (8+5 \sqrt{\frac{1-a x}{1+a x}}\right )}{8 a^4}+\frac{(1+a x)^3 \left (4+9 \sqrt{\frac{1-a x}{1+a x}}\right )}{12 a^4}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{4 a^4}\\ &=-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x)^4}{4 a^4}+\frac{(1+a x) \left (8+\sqrt{\frac{1-a x}{1+a x}}\right )}{8 a^4}-\frac{(1+a x)^2 \left (8+5 \sqrt{\frac{1-a x}{1+a x}}\right )}{8 a^4}+\frac{(1+a x)^3 \left (4+9 \sqrt{\frac{1-a x}{1+a x}}\right )}{12 a^4}+\frac{\tan ^{-1}\left (\sqrt{\frac{1-a x}{1+a x}}\right )}{4 a^4}\\ \end{align*}

Mathematica [C]  time = 0.113146, size = 97, normalized size = 0.6 \[ \frac{8 a^3 x^3+3 a \sqrt{\frac{1-a x}{a x+1}} \left (-2 a^3 x^4-2 a^2 x^3+a x^2+x\right )-3 i \log \left (2 \sqrt{\frac{1-a x}{a x+1}} (a x+1)-2 i a x\right )}{24 a^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3/E^ArcSech[a*x],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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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{{x}^{3} \left ({\frac{1}{ax}}+\sqrt{{\frac{1}{ax}}-1}\sqrt{1+{\frac{1}{ax}}} \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\sqrt{\frac{1}{a x} + 1} \sqrt{\frac{1}{a x} - 1} + \frac{1}{a x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.06981, size = 203, normalized size = 1.25 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)),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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \int \frac{x^{4}}{a x \sqrt{-1 + \frac{1}{a x}} \sqrt{1 + \frac{1}{a x}} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\sqrt{\frac{1}{a x} + 1} \sqrt{\frac{1}{a x} - 1} + \frac{1}{a x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)),x, algorithm="giac")

[Out]

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