3.1.11 \(\int x^3 \text {sech}^{-1}(a x)^3 \, dx\) [11]

Optimal. Leaf size=184 \[ \frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)}{4 a^4}-\frac {x^2 \text {sech}^{-1}(a x)}{4 a^2}-\frac {\text {sech}^{-1}(a x)^2}{2 a^4}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{2 a^4}-\frac {x^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{4 a^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a x)^3+\frac {\text {sech}^{-1}(a x) \log \left (1+e^{2 \text {sech}^{-1}(a x)}\right )}{a^4}+\frac {\text {PolyLog}\left (2,-e^{2 \text {sech}^{-1}(a x)}\right )}{2 a^4} \]

[Out]

-1/4*x^2*arcsech(a*x)/a^2-1/2*arcsech(a*x)^2/a^4+1/4*x^4*arcsech(a*x)^3+arcsech(a*x)*ln(1+(1/a/x+(1/a/x-1)^(1/
2)*(1+1/a/x)^(1/2))^2)/a^4+1/2*polylog(2,-(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))^2)/a^4+1/4*(a*x+1)*((-a*x+1)
/(a*x+1))^(1/2)/a^4-1/2*(a*x+1)*arcsech(a*x)^2*((-a*x+1)/(a*x+1))^(1/2)/a^4-1/4*x^2*(a*x+1)*arcsech(a*x)^2*((-
a*x+1)/(a*x+1))^(1/2)/a^2

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Rubi [A]
time = 0.12, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6420, 5526, 4271, 3852, 8, 4269, 3799, 2221, 2317, 2438} \begin {gather*} \frac {\text {Li}_2\left (-e^{2 \text {sech}^{-1}(a x)}\right )}{2 a^4}+\frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1)}{4 a^4}-\frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)^2}{2 a^4}-\frac {\text {sech}^{-1}(a x)^2}{2 a^4}+\frac {\text {sech}^{-1}(a x) \log \left (e^{2 \text {sech}^{-1}(a x)}+1\right )}{a^4}-\frac {x^2 \sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)^2}{4 a^2}-\frac {x^2 \text {sech}^{-1}(a x)}{4 a^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a x)^3 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2221

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

Rule 2317

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 2438

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 3799

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

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 4269

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 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])

Rubi steps

\begin {align*} \int x^3 \text {sech}^{-1}(a x)^3 \, dx &=-\frac {\text {Subst}\left (\int x^3 \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)^3-\frac {3 \text {Subst}\left (\int x^2 \text {sech}^4(x) \, dx,x,\text {sech}^{-1}(a x)\right )}{4 a^4}\\ &=-\frac {x^2 \text {sech}^{-1}(a x)}{4 a^2}-\frac {x^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{4 a^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a x)^3+\frac {\text {Subst}\left (\int \text {sech}^2(x) \, dx,x,\text {sech}^{-1}(a x)\right )}{4 a^4}-\frac {\text {Subst}\left (\int x^2 \text {sech}^2(x) \, dx,x,\text {sech}^{-1}(a x)\right )}{2 a^4}\\ &=-\frac {x^2 \text {sech}^{-1}(a x)}{4 a^2}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{2 a^4}-\frac {x^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{4 a^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a x)^3+\frac {i \text {Subst}\left (\int 1 \, dx,x,-i \sqrt {\frac {1-a x}{1+a x}} (1+a x)\right )}{4 a^4}+\frac {\text {Subst}\left (\int x \tanh (x) \, dx,x,\text {sech}^{-1}(a x)\right )}{a^4}\\ &=\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)}{4 a^4}-\frac {x^2 \text {sech}^{-1}(a x)}{4 a^2}-\frac {\text {sech}^{-1}(a x)^2}{2 a^4}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{2 a^4}-\frac {x^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{4 a^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a x)^3+\frac {2 \text {Subst}\left (\int \frac {e^{2 x} x}{1+e^{2 x}} \, dx,x,\text {sech}^{-1}(a x)\right )}{a^4}\\ &=\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)}{4 a^4}-\frac {x^2 \text {sech}^{-1}(a x)}{4 a^2}-\frac {\text {sech}^{-1}(a x)^2}{2 a^4}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{2 a^4}-\frac {x^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{4 a^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a x)^3+\frac {\text {sech}^{-1}(a x) \log \left (1+e^{2 \text {sech}^{-1}(a x)}\right )}{a^4}-\frac {\text {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\text {sech}^{-1}(a x)\right )}{a^4}\\ &=\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)}{4 a^4}-\frac {x^2 \text {sech}^{-1}(a x)}{4 a^2}-\frac {\text {sech}^{-1}(a x)^2}{2 a^4}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{2 a^4}-\frac {x^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{4 a^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a x)^3+\frac {\text {sech}^{-1}(a x) \log \left (1+e^{2 \text {sech}^{-1}(a x)}\right )}{a^4}-\frac {\text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \text {sech}^{-1}(a x)}\right )}{2 a^4}\\ &=\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)}{4 a^4}-\frac {x^2 \text {sech}^{-1}(a x)}{4 a^2}-\frac {\text {sech}^{-1}(a x)^2}{2 a^4}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{2 a^4}-\frac {x^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)^2}{4 a^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a x)^3+\frac {\text {sech}^{-1}(a x) \log \left (1+e^{2 \text {sech}^{-1}(a x)}\right )}{a^4}+\frac {\text {Li}_2\left (-e^{2 \text {sech}^{-1}(a x)}\right )}{2 a^4}\\ \end {align*}

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Mathematica [A]
time = 0.40, size = 188, normalized size = 1.02 \begin {gather*} \frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x)-\left (-2+2 \sqrt {\frac {1-a x}{1+a x}}+2 a x \sqrt {\frac {1-a x}{1+a x}}+a^2 x^2 \sqrt {\frac {1-a x}{1+a x}}+a^3 x^3 \sqrt {\frac {1-a x}{1+a x}}\right ) \text {sech}^{-1}(a x)^2+a^4 x^4 \text {sech}^{-1}(a x)^3+\text {sech}^{-1}(a x) \left (-a^2 x^2+4 \log \left (1+e^{-2 \text {sech}^{-1}(a x)}\right )\right )-2 \text {PolyLog}\left (2,-e^{-2 \text {sech}^{-1}(a x)}\right )}{4 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.40, size = 236, normalized size = 1.28

method	result	size
derivativedivides	\(\frac {\frac {a^{4} x^{4} \mathrm {arcsech}\left (a x \right )^{3}}{4}-\frac {\mathrm {arcsech}\left (a x \right )^{2} \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, a^{3} x^{3}}{4}-\frac {\mathrm {arcsech}\left (a x \right )^{2} \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, a x}{2}-\frac {\mathrm {arcsech}\left (a x \right ) a^{2} x^{2}}{4}+\frac {\sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, a x}{4}-\frac {\mathrm {arcsech}\left (a x \right )^{2}}{2}-\frac {1}{4}+\mathrm {arcsech}\left (a x \right ) \ln \left (1+\left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )^{2}\right )+\frac {\polylog \left (2, -\left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )^{2}\right )}{2}}{a^{4}}\)	\(236\)
default	\(\frac {\frac {a^{4} x^{4} \mathrm {arcsech}\left (a x \right )^{3}}{4}-\frac {\mathrm {arcsech}\left (a x \right )^{2} \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, a^{3} x^{3}}{4}-\frac {\mathrm {arcsech}\left (a x \right )^{2} \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, a x}{2}-\frac {\mathrm {arcsech}\left (a x \right ) a^{2} x^{2}}{4}+\frac {\sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, a x}{4}-\frac {\mathrm {arcsech}\left (a x \right )^{2}}{2}-\frac {1}{4}+\mathrm {arcsech}\left (a x \right ) \ln \left (1+\left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )^{2}\right )+\frac {\polylog \left (2, -\left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )^{2}\right )}{2}}{a^{4}}\)	\(236\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsech(a*x)^3,x,method=_RETURNVERBOSE)

[Out]

1/a^4*(1/4*a^4*x^4*arcsech(a*x)^3-1/4*arcsech(a*x)^2*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)*a^3*x^3-1/2*arcs
ech(a*x)^2*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)*a*x-1/4*arcsech(a*x)*a^2*x^2+1/4*(-(a*x-1)/a/x)^(1/2)*((a*
x+1)/a/x)^(1/2)*a*x-1/2*arcsech(a*x)^2-1/4+arcsech(a*x)*ln(1+(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))^2)+1/2*po
lylog(2,-(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))^2))

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

[Out]

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

[Out]

integral(x^3*arcsech(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(x^3*arcsech(a*x)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,{\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^3*acosh(1/(a*x))^3,x)

[Out]

int(x^3*acosh(1/(a*x))^3, x)

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