\(\int \text {sech}^{-1}(a+b x)^3 \, dx\) [16]

Optimal result

Integrand size = 8, antiderivative size = 136 \[ \int \text {sech}^{-1}(a+b x)^3 \, dx=\frac {(a+b x) \text {sech}^{-1}(a+b x)^3}{b}-\frac {6 \text {sech}^{-1}(a+b x)^2 \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {6 i \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b}-\frac {6 i \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b}-\frac {6 i \operatorname {PolyLog}\left (3,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {6 i \operatorname {PolyLog}\left (3,i e^{\text {sech}^{-1}(a+b x)}\right )}{b} \]

[Out]

(b*x+a)*arcsech(b*x+a)^3/b-6*arcsech(b*x+a)^2*arctan(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/b+6*I*
arcsech(b*x+a)*polylog(2,-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))/b-6*I*arcsech(b*x+a)*polylog(
2,I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))/b-6*I*polylog(3,-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/
(b*x+a)+1)^(1/2)))/b+6*I*polylog(3,I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))/b

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {6450, 6414, 5526, 4265, 2611, 2320, 6724} \[ \int \text {sech}^{-1}(a+b x)^3 \, dx=-\frac {6 \text {sech}^{-1}(a+b x)^2 \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {6 i \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b}-\frac {6 i \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b}-\frac {6 i \operatorname {PolyLog}\left (3,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {6 i \operatorname {PolyLog}\left (3,i e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {(a+b x) \text {sech}^{-1}(a+b x)^3}{b} \]

[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 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 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 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 6414

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 6450

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12 \[ \int \text {sech}^{-1}(a+b x)^3 \, dx=\frac {(a+b x) \text {sech}^{-1}(a+b x)^3}{b}-\frac {3 i \left (-\text {sech}^{-1}(a+b x)^2 \left (\log \left (1-i e^{-\text {sech}^{-1}(a+b x)}\right )-\log \left (1+i e^{-\text {sech}^{-1}(a+b x)}\right )\right )-2 \text {sech}^{-1}(a+b x) \left (\operatorname {PolyLog}\left (2,-i e^{-\text {sech}^{-1}(a+b x)}\right )-\operatorname {PolyLog}\left (2,i e^{-\text {sech}^{-1}(a+b x)}\right )\right )-2 \left (\operatorname {PolyLog}\left (3,-i e^{-\text {sech}^{-1}(a+b x)}\right )-\operatorname {PolyLog}\left (3,i e^{-\text {sech}^{-1}(a+b x)}\right )\right )\right )}{b} \]

[In]

Integrate[ArcSech[a + b*x]^3,x]

[Out]

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

Maple [F]

\[\int \operatorname {arcsech}\left (b x +a \right )^{3}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \text {sech}^{-1}(a+b x)^3 \, dx=\int { \operatorname {arsech}\left (b x + a\right )^{3} \,d x } \]

[In]

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

[Out]

integral(arcsech(b*x + a)^3, x)

Sympy [F]

\[ \int \text {sech}^{-1}(a+b x)^3 \, dx=\int \operatorname {asech}^{3}{\left (a + b x \right )}\, dx \]

[In]

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

[Out]

Integral(asech(a + b*x)**3, x)

Maxima [F]

\[ \int \text {sech}^{-1}(a+b x)^3 \, dx=\int { \operatorname {arsech}\left (b x + a\right )^{3} \,d x } \]

[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)

Giac [F]

\[ \int \text {sech}^{-1}(a+b x)^3 \, dx=\int { \operatorname {arsech}\left (b x + a\right )^{3} \,d x } \]

[In]

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

[Out]

integrate(arcsech(b*x + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \text {sech}^{-1}(a+b x)^3 \, dx=\int {\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}^3 \,d x \]

[In]

int(acosh(1/(a + b*x))^3,x)

[Out]

int(acosh(1/(a + b*x))^3, x)