3.16 \(\int \text{sech}^{-1}(a+b x)^3 \, dx\)

Optimal. Leaf size=136 \[ \frac{6 i \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{PolyLog}\left (3,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{6 i \text{PolyLog}\left (3,i e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{6 \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b} \]

[Out]

((a + b*x)*ArcSech[a + b*x]^3)/b - (6*ArcSech[a + b*x]^2*ArcTan[E^ArcSech[a + b*x]])/b + ((6*I)*ArcSech[a + b*
x]*PolyLog[2, (-I)*E^ArcSech[a + b*x]])/b - ((6*I)*ArcSech[a + b*x]*PolyLog[2, I*E^ArcSech[a + b*x]])/b - ((6*
I)*PolyLog[3, (-I)*E^ArcSech[a + b*x]])/b + ((6*I)*PolyLog[3, I*E^ArcSech[a + b*x]])/b

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Rubi [A]  time = 0.104783, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.875, Rules used = {6315, 6279, 5418, 4180, 2531, 2282, 6589} \[ \frac{6 i \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{PolyLog}\left (3,-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{6 i \text{PolyLog}\left (3,i e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{6 \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSech[a + b*x]^3,x]

[Out]

((a + b*x)*ArcSech[a + b*x]^3)/b - (6*ArcSech[a + b*x]^2*ArcTan[E^ArcSech[a + b*x]])/b + ((6*I)*ArcSech[a + b*
x]*PolyLog[2, (-I)*E^ArcSech[a + b*x]])/b - ((6*I)*ArcSech[a + b*x]*PolyLog[2, I*E^ArcSech[a + b*x]])/b - ((6*
I)*PolyLog[3, (-I)*E^ArcSech[a + b*x]])/b + ((6*I)*PolyLog[3, I*E^ArcSech[a + b*x]])/b

Rule 6315

Int[((a_.) + ArcSech[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSech[x])^p, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]

Rule 6279

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

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 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 2531

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 2282

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 6589

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 \text{sech}^{-1}(a+b x)^3 \, dx &=\frac{\operatorname{Subst}\left (\int \text{sech}^{-1}(x)^3 \, dx,x,a+b x\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int x^3 \text{sech}(x) \tanh (x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{3 \operatorname{Subst}\left (\int x^2 \text{sech}(x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{6 \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{(6 i) \operatorname{Subst}\left (\int x \log \left (1-i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}-\frac{(6 i) \operatorname{Subst}\left (\int x \log \left (1+i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{6 \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{6 i \text{sech}^{-1}(a+b x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{sech}^{-1}(a+b x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{(6 i) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}+\frac{(6 i) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{6 \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{6 i \text{sech}^{-1}(a+b x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{sech}^{-1}(a+b x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )}{b}\\ &=\frac{(a+b x) \text{sech}^{-1}(a+b x)^3}{b}-\frac{6 \text{sech}^{-1}(a+b x)^2 \tan ^{-1}\left (e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{6 i \text{sech}^{-1}(a+b x) \text{Li}_2\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{sech}^{-1}(a+b x) \text{Li}_2\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b}-\frac{6 i \text{Li}_3\left (-i e^{\text{sech}^{-1}(a+b x)}\right )}{b}+\frac{6 i \text{Li}_3\left (i e^{\text{sech}^{-1}(a+b x)}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.140201, size = 153, normalized size = 1.12 \[ -\frac{-(a+b x) \text{sech}^{-1}(a+b x)^3+3 i \left (-2 \text{sech}^{-1}(a+b x) \left (\text{PolyLog}\left (2,-i e^{-\text{sech}^{-1}(a+b x)}\right )-\text{PolyLog}\left (2,i e^{-\text{sech}^{-1}(a+b x)}\right )\right )-2 \left (\text{PolyLog}\left (3,-i e^{-\text{sech}^{-1}(a+b x)}\right )-\text{PolyLog}\left (3,i e^{-\text{sech}^{-1}(a+b x)}\right )\right )+\text{sech}^{-1}(a+b x)^2 \left (-\left (\log \left (1-i e^{-\text{sech}^{-1}(a+b x)}\right )-\log \left (1+i e^{-\text{sech}^{-1}(a+b x)}\right )\right )\right )\right )}{b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSech[a + b*x]^3,x]

[Out]

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

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Maple [F]  time = 0.353, size = 0, normalized size = 0. \begin{align*} \int \left ({\rm arcsech} \left (bx+a\right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsech(b*x+a)^3,x)

[Out]

int(arcsech(b*x+a)^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} x \log \left (\sqrt{b x + a + 1} \sqrt{-b x - a + 1} b x + \sqrt{b x + a + 1} \sqrt{-b x - a + 1} a + b x + a\right )^{3} - \int \frac{8 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt{b x + a + 1} \sqrt{-b x - a + 1} \log \left (b x + a\right )^{3} + 8 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right )^{3} + 3 \,{\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} +{\left (a^{2} b - b\right )} x + 2 \,{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right ) +{\left ({\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt{b x + a + 1} \log \left (b x + a\right ) +{\left (2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} +{\left (2 \, a^{2} b - b\right )} x +{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right )\right )} \sqrt{b x + a + 1}\right )} \sqrt{-b x - a + 1}\right )} \log \left (\sqrt{b x + a + 1} \sqrt{-b x - a + 1} b x + \sqrt{b x + a + 1} \sqrt{-b x - a + 1} a + b x + a\right )^{2} - 12 \,{\left ({\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt{b x + a + 1} \sqrt{-b x - a + 1} \log \left (b x + a\right )^{2} +{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right )^{2}\right )} \log \left (\sqrt{b x + a + 1} \sqrt{-b x - a + 1} b x + \sqrt{b x + a + 1} \sqrt{-b x - a + 1} a + b x + a\right )}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt{b x + a + 1} \sqrt{-b x - a + 1} +{\left (3 \, a^{2} b - b\right )} x - a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*a + b*x + a)^3 - integra
te((8*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*log(b*x + a)^3
+ 8*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*log(b*x + a)^3 + 3*(b^3*x^3 + 2*a*b^2*x^2 + (a^2*b - b
)*x + 2*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*log(b*x + a) + ((b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*
a^2*b - b)*x - a)*sqrt(b*x + a + 1)*log(b*x + a) + (2*b^3*x^3 + 4*a*b^2*x^2 + (2*a^2*b - b)*x + (b^3*x^3 + 3*a
*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*log(b*x + a))*sqrt(b*x + a + 1))*sqrt(-b*x - a + 1))*log(sqrt(b*x + a +
1)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*a + b*x + a)^2 - 12*((b^3*x^3 + 3*a*b^2*x^2 +
a^3 + (3*a^2*b - b)*x - a)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*log(b*x + a)^2 + (b^3*x^3 + 3*a*b^2*x^2 + a^3
+ (3*a^2*b - b)*x - a)*log(b*x + a)^2)*log(sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sqrt(
-b*x - a + 1)*a + b*x + a))/(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)
*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1) + (3*a^2*b - b)*x - a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\operatorname{arsech}\left (b x + a\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsech(b*x + a)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(a + b*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(b*x + a)^3, x)