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

Optimal. Leaf size=117 \[ \frac {i \text {Li}_2\left (-i e^{\text {sech}^{-1}(a x)}\right )}{3 a^3}-\frac {i \text {Li}_2\left (i e^{\text {sech}^{-1}(a x)}\right )}{3 a^3}-\frac {2 \text {sech}^{-1}(a x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a x)}\right )}{3 a^3}-\frac {x}{3 a^2}-\frac {x \sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)}{3 a^2}+\frac {1}{3} x^3 \text {sech}^{-1}(a x)^2 \]

[Out]

-1/3*x/a^2+1/3*x^3*arcsech(a*x)^2-2/3*arcsech(a*x)*arctan(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))/a^3+1/3*I*pol
ylog(2,-I*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)))/a^3-1/3*I*polylog(2,I*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2
)))/a^3-1/3*x*(a*x+1)*arcsech(a*x)*((-a*x+1)/(a*x+1))^(1/2)/a^2

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Rubi [A]  time = 0.10, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6285, 5418, 4185, 4180, 2279, 2391} \[ \frac {i \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a x)}\right )}{3 a^3}-\frac {i \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(a x)}\right )}{3 a^3}-\frac {x}{3 a^2}-\frac {x \sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)}{3 a^2}-\frac {2 \text {sech}^{-1}(a x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a x)}\right )}{3 a^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a x)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(3*a^2) - (x*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x])/(3*a^2) + (x^3*ArcSech[a*x]^2)/3 - (2*ArcSec
h[a*x]*ArcTan[E^ArcSech[a*x]])/(3*a^3) + ((I/3)*PolyLog[2, (-I)*E^ArcSech[a*x]])/a^3 - ((I/3)*PolyLog[2, I*E^A
rcSech[a*x]])/a^3

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

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 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 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])

Rubi steps

\begin {align*} \int x^2 \text {sech}^{-1}(a x)^2 \, dx &=-\frac {\operatorname {Subst}\left (\int x^2 \text {sech}^3(x) \tanh (x) \, dx,x,\text {sech}^{-1}(a x)\right )}{a^3}\\ &=\frac {1}{3} x^3 \text {sech}^{-1}(a x)^2-\frac {2 \operatorname {Subst}\left (\int x \text {sech}^3(x) \, dx,x,\text {sech}^{-1}(a x)\right )}{3 a^3}\\ &=-\frac {x}{3 a^2}-\frac {x \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{3 a^2}+\frac {1}{3} x^3 \text {sech}^{-1}(a x)^2-\frac {\operatorname {Subst}\left (\int x \text {sech}(x) \, dx,x,\text {sech}^{-1}(a x)\right )}{3 a^3}\\ &=-\frac {x}{3 a^2}-\frac {x \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{3 a^2}+\frac {1}{3} x^3 \text {sech}^{-1}(a x)^2-\frac {2 \text {sech}^{-1}(a x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a x)}\right )}{3 a^3}+\frac {i \operatorname {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text {sech}^{-1}(a x)\right )}{3 a^3}-\frac {i \operatorname {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text {sech}^{-1}(a x)\right )}{3 a^3}\\ &=-\frac {x}{3 a^2}-\frac {x \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{3 a^2}+\frac {1}{3} x^3 \text {sech}^{-1}(a x)^2-\frac {2 \text {sech}^{-1}(a x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a x)}\right )}{3 a^3}+\frac {i \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a x)}\right )}{3 a^3}-\frac {i \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a x)}\right )}{3 a^3}\\ &=-\frac {x}{3 a^2}-\frac {x \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{3 a^2}+\frac {1}{3} x^3 \text {sech}^{-1}(a x)^2-\frac {2 \text {sech}^{-1}(a x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a x)}\right )}{3 a^3}+\frac {i \text {Li}_2\left (-i e^{\text {sech}^{-1}(a x)}\right )}{3 a^3}-\frac {i \text {Li}_2\left (i e^{\text {sech}^{-1}(a x)}\right )}{3 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 138, normalized size = 1.18 \[ \frac {a^3 x^3 \text {sech}^{-1}(a x)^2+i \text {Li}_2\left (-i e^{-\text {sech}^{-1}(a x)}\right )-i \text {Li}_2\left (i e^{-\text {sech}^{-1}(a x)}\right )-a x-a x \sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)+i \text {sech}^{-1}(a x) \log \left (1-i e^{-\text {sech}^{-1}(a x)}\right )-i \text {sech}^{-1}(a x) \log \left (1+i e^{-\text {sech}^{-1}(a x)}\right )}{3 a^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(a*x) - a*x*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x] + a^3*x^3*ArcSech[a*x]^2 + I*ArcSech[a*x]*Log[1
- I/E^ArcSech[a*x]] - I*ArcSech[a*x]*Log[1 + I/E^ArcSech[a*x]] + I*PolyLog[2, (-I)/E^ArcSech[a*x]] - I*PolyLo
g[2, I/E^ArcSech[a*x]])/(3*a^3)

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \operatorname {arsech}\left (a x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*arcsech(a*x)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {arsech}\left (a x\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arcsech(a*x)^2, x)

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maple [A]  time = 0.72, size = 240, normalized size = 2.05 \[ \frac {x^{3} \mathrm {arcsech}\left (a x \right )^{2}}{3}-\frac {\mathrm {arcsech}\left (a x \right ) \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, x^{2}}{3 a}-\frac {x}{3 a^{2}}+\frac {i \mathrm {arcsech}\left (a x \right ) \ln \left (1+i \left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )\right )}{3 a^{3}}-\frac {i \mathrm {arcsech}\left (a x \right ) \ln \left (1-i \left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )\right )}{3 a^{3}}+\frac {i \dilog \left (1+i \left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )\right )}{3 a^{3}}-\frac {i \dilog \left (1-i \left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )\right )}{3 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arcsech(a*x)^2-1/3/a*arcsech(a*x)*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)*x^2-1/3*x/a^2+1/3*I/a^3*arc
sech(a*x)*ln(1+I*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)))-1/3*I/a^3*arcsech(a*x)*ln(1-I*(1/a/x+(1/a/x-1)^(1/2)
*(1+1/a/x)^(1/2)))+1/3*I/a^3*dilog(1+I*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)))-1/3*I/a^3*dilog(1-I*(1/a/x+(1/
a/x-1)^(1/2)*(1+1/a/x)^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {arsech}\left (a x\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arcsech(a*x)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\mathrm {acosh}\left (\frac {1}{a\,x}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acosh(1/(a*x))^2,x)

[Out]

int(x^2*acosh(1/(a*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {asech}^{2}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*asech(a*x)**2, x)

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