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

Optimal. Leaf size=104 \[ -\frac{x^2}{12 a^2}-\frac{x^2 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)}{6 a^2}-\frac{\log (x)}{3 a^4}-\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)}{3 a^4}+\frac{1}{4} x^4 \text{sech}^{-1}(a x)^2 \]

[Out]

-x^2/(12*a^2) - (Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x])/(3*a^4) - (x^2*Sqrt[(1 - a*x)/(1 + a*x)]*(1
+ a*x)*ArcSech[a*x])/(6*a^2) + (x^4*ArcSech[a*x]^2)/4 - Log[x]/(3*a^4)

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Rubi [A]  time = 0.0884237, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6285, 5418, 4185, 4184, 3475} \[ -\frac{x^2}{12 a^2}-\frac{x^2 \sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)}{6 a^2}-\frac{\log (x)}{3 a^4}-\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1) \text{sech}^{-1}(a x)}{3 a^4}+\frac{1}{4} x^4 \text{sech}^{-1}(a x)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^2/(12*a^2) - (Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x])/(3*a^4) - (x^2*Sqrt[(1 - a*x)/(1 + a*x)]*(1
+ a*x)*ArcSech[a*x])/(6*a^2) + (x^4*ArcSech[a*x]^2)/4 - Log[x]/(3*a^4)

Rule 6285

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b
*x)^n*Sech[x]^(m + 1)*Tanh[x], x], x, ArcSech[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] &
& (GtQ[n, 0] || LtQ[m, -1])

Rule 5418

Int[(x_)^(m_.)*Sech[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tanh[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> -Simp[(
x^(m - n + 1)*Sech[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sech[a + b*x^n]^p, x],
x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

Rule 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x^3 \text{sech}^{-1}(a x)^2 \, dx &=-\frac{\operatorname{Subst}\left (\int x^2 \text{sech}^4(x) \tanh (x) \, dx,x,\text{sech}^{-1}(a x)\right )}{a^4}\\ &=\frac{1}{4} x^4 \text{sech}^{-1}(a x)^2-\frac{\operatorname{Subst}\left (\int x \text{sech}^4(x) \, dx,x,\text{sech}^{-1}(a x)\right )}{2 a^4}\\ &=-\frac{x^2}{12 a^2}-\frac{x^2 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)}{6 a^2}+\frac{1}{4} x^4 \text{sech}^{-1}(a x)^2-\frac{\operatorname{Subst}\left (\int x \text{sech}^2(x) \, dx,x,\text{sech}^{-1}(a x)\right )}{3 a^4}\\ &=-\frac{x^2}{12 a^2}-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)}{3 a^4}-\frac{x^2 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)}{6 a^2}+\frac{1}{4} x^4 \text{sech}^{-1}(a x)^2+\frac{\operatorname{Subst}\left (\int \tanh (x) \, dx,x,\text{sech}^{-1}(a x)\right )}{3 a^4}\\ &=-\frac{x^2}{12 a^2}-\frac{\sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)}{3 a^4}-\frac{x^2 \sqrt{\frac{1-a x}{1+a x}} (1+a x) \text{sech}^{-1}(a x)}{6 a^2}+\frac{1}{4} x^4 \text{sech}^{-1}(a x)^2-\frac{\log (x)}{3 a^4}\\ \end{align*}

Mathematica [A]  time = 0.100651, size = 77, normalized size = 0.74 \[ -\frac{a^2 x^2-3 a^4 x^4 \text{sech}^{-1}(a x)^2+2 \sqrt{\frac{1-a x}{a x+1}} \left (a^3 x^3+a^2 x^2+2 a x+2\right ) \text{sech}^{-1}(a x)+4 \log (x)}{12 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*x^2 + 2*Sqrt[(1 - a*x)/(1 + a*x)]*(2 + 2*a*x + a^2*x^2 + a^3*x^3)*ArcSech[a*x] - 3*a^4*x^4*ArcSech[a*x]^
2 + 4*Log[x])/(12*a^4)

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Maple [A]  time = 0.328, size = 151, normalized size = 1.5 \begin{align*} -{\frac{{\rm arcsech} \left (ax\right )}{3\,{a}^{4}}}+{\frac{{x}^{4} \left ({\rm arcsech} \left (ax\right ) \right ) ^{2}}{4}}-{\frac{{\rm arcsech} \left (ax\right ){x}^{3}}{6\,a}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}}}-{\frac{{\rm arcsech} \left (ax\right )x}{3\,{a}^{3}}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}}}-{\frac{{x}^{2}}{12\,{a}^{2}}}+{\frac{1}{3\,{a}^{4}}\ln \left ( 1+ \left ({\frac{1}{ax}}+\sqrt{{\frac{1}{ax}}-1}\sqrt{1+{\frac{1}{ax}}} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsech(a*x)^2,x)

[Out]

-1/3/a^4*arcsech(a*x)+1/4*x^4*arcsech(a*x)^2-1/6/a*arcsech(a*x)*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)*x^3-1
/3/a^3*arcsech(a*x)*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)*x-1/12*x^2/a^2+1/3/a^4*ln(1+(1/a/x+(1/a/x-1)^(1/2
)*(1+1/a/x)^(1/2))^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{arsech}\left (a x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsech(a*x)^2,x, algorithm="maxima")

[Out]

integrate(x^3*arcsech(a*x)^2, x)

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Fricas [A]  time = 1.69881, size = 273, normalized size = 2.62 \begin{align*} \frac{3 \, a^{4} x^{4} \log \left (\frac{a x \sqrt{-\frac{a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )^{2} - a^{2} x^{2} - 2 \,{\left (a^{3} x^{3} + 2 \, a x\right )} \sqrt{-\frac{a^{2} x^{2} - 1}{a^{2} x^{2}}} \log \left (\frac{a x \sqrt{-\frac{a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right ) - 4 \, \log \left (x\right )}{12 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsech(a*x)^2,x, algorithm="fricas")

[Out]

1/12*(3*a^4*x^4*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x))^2 - a^2*x^2 - 2*(a^3*x^3 + 2*a*x)*sqrt(-(a
^2*x^2 - 1)/(a^2*x^2))*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x)) - 4*log(x))/a^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{asech}^{2}{\left (a x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asech(a*x)**2,x)

[Out]

Integral(x**3*asech(a*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{arsech}\left (a x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsech(a*x)^2,x, algorithm="giac")

[Out]

integrate(x^3*arcsech(a*x)^2, x)