3.1.12 \(\int x^2 \text {sech}^{-1}(a x)^3 \, dx\) [12]

Optimal. Leaf size=198 \[ -\frac {x \text {sech}^{-1}(a x)}{a^2}-\frac {x \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{2 a^2}+\frac {1}{3} x^3 \text {sech}^{-1}(a x)^3-\frac {\text {sech}^{-1}(a x)^2 \text {ArcTan}\left (e^{\text {sech}^{-1}(a x)}\right )}{a^3}+\frac {\text {ArcTan}\left (\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)}{a x}\right )}{a^3}+\frac {i \text {sech}^{-1}(a x) \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a x)}\right )}{a^3}-\frac {i \text {sech}^{-1}(a x) \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(a x)}\right )}{a^3}-\frac {i \text {PolyLog}\left (3,-i e^{\text {sech}^{-1}(a x)}\right )}{a^3}+\frac {i \text {PolyLog}\left (3,i e^{\text {sech}^{-1}(a x)}\right )}{a^3} \]

[Out]

-x*arcsech(a*x)/a^2+1/3*x^3*arcsech(a*x)^3-arcsech(a*x)^2*arctan(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))/a^3+ar
ctan((a*x+1)*((-a*x+1)/(a*x+1))^(1/2)/a/x)/a^3+I*arcsech(a*x)*polylog(2,-I*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1
/2)))/a^3-I*arcsech(a*x)*polylog(2,I*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)))/a^3-I*polylog(3,-I*(1/a/x+(1/a/x
-1)^(1/2)*(1+1/a/x)^(1/2)))/a^3+I*polylog(3,I*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)))/a^3-1/2*x*(a*x+1)*arcse
ch(a*x)^2*((-a*x+1)/(a*x+1))^(1/2)/a^2

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Rubi [A]
time = 0.10, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {6420, 5526, 4271, 3855, 4265, 2611, 2320, 6724} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1)}{a x}\right )}{a^3}-\frac {\text {sech}^{-1}(a x)^2 \text {ArcTan}\left (e^{\text {sech}^{-1}(a x)}\right )}{a^3}+\frac {i \text {sech}^{-1}(a x) \text {Li}_2\left (-i e^{\text {sech}^{-1}(a x)}\right )}{a^3}-\frac {i \text {sech}^{-1}(a x) \text {Li}_2\left (i e^{\text {sech}^{-1}(a x)}\right )}{a^3}-\frac {i \text {Li}_3\left (-i e^{\text {sech}^{-1}(a x)}\right )}{a^3}+\frac {i \text {Li}_3\left (i e^{\text {sech}^{-1}(a x)}\right )}{a^3}-\frac {x \sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)^2}{2 a^2}-\frac {x \text {sech}^{-1}(a x)}{a^2}+\frac {1}{3} x^3 \text {sech}^{-1}(a x)^3 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3855

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

Rule 4265

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 4271

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

Rule 5526

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 6420

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 6724

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

Rubi steps

\begin {align*} \int x^2 \text {sech}^{-1}(a x)^3 \, dx &=-\frac {\text {Subst}\left (\int x^3 \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)^3-\frac {\text {Subst}\left (\int x^2 \text {sech}^3(x) \, dx,x,\text {sech}^{-1}(a x)\right )}{a^3}\\ &=-\frac {x \text {sech}^{-1}(a x)}{a^2}-\frac {x \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{2 a^2}+\frac {1}{3} x^3 \text {sech}^{-1}(a x)^3-\frac {\text {Subst}\left (\int x^2 \text {sech}(x) \, dx,x,\text {sech}^{-1}(a x)\right )}{2 a^3}+\frac {\text {Subst}\left (\int \text {sech}(x) \, dx,x,\text {sech}^{-1}(a x)\right )}{a^3}\\ &=-\frac {x \text {sech}^{-1}(a x)}{a^2}-\frac {x \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{2 a^2}+\frac {1}{3} x^3 \text {sech}^{-1}(a x)^3-\frac {\text {sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(a x)}\right )}{a^3}+\frac {\tan ^{-1}\left (\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)}{a x}\right )}{a^3}+\frac {i \text {Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\text {sech}^{-1}(a x)\right )}{a^3}-\frac {i \text {Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\text {sech}^{-1}(a x)\right )}{a^3}\\ &=-\frac {x \text {sech}^{-1}(a x)}{a^2}-\frac {x \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{2 a^2}+\frac {1}{3} x^3 \text {sech}^{-1}(a x)^3-\frac {\text {sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(a x)}\right )}{a^3}+\frac {\tan ^{-1}\left (\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)}{a x}\right )}{a^3}+\frac {i \text {sech}^{-1}(a x) \text {Li}_2\left (-i e^{\text {sech}^{-1}(a x)}\right )}{a^3}-\frac {i \text {sech}^{-1}(a x) \text {Li}_2\left (i e^{\text {sech}^{-1}(a x)}\right )}{a^3}-\frac {i \text {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\text {sech}^{-1}(a x)\right )}{a^3}+\frac {i \text {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\text {sech}^{-1}(a x)\right )}{a^3}\\ &=-\frac {x \text {sech}^{-1}(a x)}{a^2}-\frac {x \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{2 a^2}+\frac {1}{3} x^3 \text {sech}^{-1}(a x)^3-\frac {\text {sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(a x)}\right )}{a^3}+\frac {\tan ^{-1}\left (\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)}{a x}\right )}{a^3}+\frac {i \text {sech}^{-1}(a x) \text {Li}_2\left (-i e^{\text {sech}^{-1}(a x)}\right )}{a^3}-\frac {i \text {sech}^{-1}(a x) \text {Li}_2\left (i e^{\text {sech}^{-1}(a x)}\right )}{a^3}-\frac {i \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a x)}\right )}{a^3}+\frac {i \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a x)}\right )}{a^3}\\ &=-\frac {x \text {sech}^{-1}(a x)}{a^2}-\frac {x \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{2 a^2}+\frac {1}{3} x^3 \text {sech}^{-1}(a x)^3-\frac {\text {sech}^{-1}(a x)^2 \tan ^{-1}\left (e^{\text {sech}^{-1}(a x)}\right )}{a^3}+\frac {\tan ^{-1}\left (\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)}{a x}\right )}{a^3}+\frac {i \text {sech}^{-1}(a x) \text {Li}_2\left (-i e^{\text {sech}^{-1}(a x)}\right )}{a^3}-\frac {i \text {sech}^{-1}(a x) \text {Li}_2\left (i e^{\text {sech}^{-1}(a x)}\right )}{a^3}-\frac {i \text {Li}_3\left (-i e^{\text {sech}^{-1}(a x)}\right )}{a^3}+\frac {i \text {Li}_3\left (i e^{\text {sech}^{-1}(a x)}\right )}{a^3}\\ \end {align*}

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Mathematica [A]
time = 0.33, size = 199, normalized size = 1.01 \begin {gather*} \frac {-6 a x \text {sech}^{-1}(a x)-3 a x \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2+2 a^3 x^3 \text {sech}^{-1}(a x)^3+3 i \left (-4 i \text {ArcTan}\left (\tanh \left (\frac {1}{2} \text {sech}^{-1}(a x)\right )\right )+\text {sech}^{-1}(a x)^2 \log \left (1-i e^{-\text {sech}^{-1}(a x)}\right )-\text {sech}^{-1}(a x)^2 \log \left (1+i e^{-\text {sech}^{-1}(a x)}\right )+2 \text {sech}^{-1}(a x) \text {PolyLog}\left (2,-i e^{-\text {sech}^{-1}(a x)}\right )-2 \text {sech}^{-1}(a x) \text {PolyLog}\left (2,i e^{-\text {sech}^{-1}(a x)}\right )+2 \text {PolyLog}\left (3,-i e^{-\text {sech}^{-1}(a x)}\right )-2 \text {PolyLog}\left (3,i e^{-\text {sech}^{-1}(a x)}\right )\right )}{6 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*x*ArcSech[a*x] - 3*a*x*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x]^2 + 2*a^3*x^3*ArcSech[a*x]^3 + (
3*I)*((-4*I)*ArcTan[Tanh[ArcSech[a*x]/2]] + ArcSech[a*x]^2*Log[1 - I/E^ArcSech[a*x]] - ArcSech[a*x]^2*Log[1 +
I/E^ArcSech[a*x]] + 2*ArcSech[a*x]*PolyLog[2, (-I)/E^ArcSech[a*x]] - 2*ArcSech[a*x]*PolyLog[2, I/E^ArcSech[a*x
]] + 2*PolyLog[3, (-I)/E^ArcSech[a*x]] - 2*PolyLog[3, I/E^ArcSech[a*x]]))/(6*a^3)

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Maple [F]
time = 0.38, size = 0, normalized size = 0.00 \[\int x^{2} \mathrm {arcsech}\left (a x \right )^{3}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arcsech(a*x)^3, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*arcsech(a*x)^3, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \operatorname {asech}^{3}{\left (a x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*asech(a*x)**3, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arcsech(a*x)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\mathrm {acosh}\left (\frac {1}{a\,x}\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*acosh(1/(a*x))^3, x)

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