\(\int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx\) [18]

Optimal result

Integrand size = 12, antiderivative size = 330 \[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=-\frac {b \text {sech}^{-1}(a+b x)^3}{a}-\frac {\text {sech}^{-1}(a+b x)^3}{x}+\frac {3 b \text {sech}^{-1}(a+b x)^2 \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {3 b \text {sech}^{-1}(a+b x)^2 \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {6 b \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {6 b \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {6 b \operatorname {PolyLog}\left (3,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {6 b \operatorname {PolyLog}\left (3,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6456, 5576, 4276, 3401, 2296, 2221, 2611, 2320, 6724} \[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=\frac {6 b \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {6 b \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}-\frac {6 b \operatorname {PolyLog}\left (3,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {6 b \operatorname {PolyLog}\left (3,\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}+\frac {3 b \text {sech}^{-1}(a+b x)^2 \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {3 b \text {sech}^{-1}(a+b x)^2 \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}-\frac {b \text {sech}^{-1}(a+b x)^3}{a}-\frac {\text {sech}^{-1}(a+b x)^3}{x} \]

[In]

Int[ArcSech[a + b*x]^3/x^2,x]

[Out]

-((b*ArcSech[a + b*x]^3)/a) - ArcSech[a + b*x]^3/x + (3*b*ArcSech[a + b*x]^2*Log[1 - (a*E^ArcSech[a + b*x])/(1
- Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2]) - (3*b*ArcSech[a + b*x]^2*Log[1 - (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a
^2])])/(a*Sqrt[1 - a^2]) + (6*b*ArcSech[a + b*x]*PolyLog[2, (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])])/(a*Sq
rt[1 - a^2]) - (6*b*ArcSech[a + b*x]*PolyLog[2, (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2])
- (6*b*PolyLog[3, (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2]) + (6*b*PolyLog[3, (a*E^ArcSe
ch[a + b*x])/(1 + Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2])

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 2296

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[2*(c/q), Int[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Dist[2*(c/q), Int[(f + g
*x)^m*(F^u/(b + q + 2*c*F^u)), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

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 3401

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

Rule 4276

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& IGtQ[m, 0]

Rule 5576

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

Rule 6456

Int[((a_.) + ArcSech[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[-(d^(m + 1)
)^(-1), Subst[Int[(a + b*x)^p*Sech[x]*Tanh[x]*(d*e - c*f + f*Sech[x])^m, x], x, ArcSech[c + d*x]], x] /; FreeQ
[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 69.42 (sec) , antiderivative size = 8527, normalized size of antiderivative = 25.84 \[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=\text {Result too large to show} \]

[In]

Integrate[ArcSech[a + b*x]^3/x^2,x]

[Out]

Result too large to show

Maple [F]

\[\int \frac {\operatorname {arcsech}\left (b x +a \right )^{3}}{x^{2}}d x\]

[In]

int(arcsech(b*x+a)^3/x^2,x)

[Out]

int(arcsech(b*x+a)^3/x^2,x)

Fricas [F]

\[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )^{3}}{x^{2}} \,d x } \]

[In]

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

[Out]

integral(arcsech(b*x + a)^3/x^2, x)

Sympy [F]

\[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=\int \frac {\operatorname {asech}^{3}{\left (a + b x \right )}}{x^{2}}\, dx \]

[In]

integrate(asech(b*x+a)**3/x**2,x)

[Out]

Integral(asech(a + b*x)**3/x**2, x)

Maxima [F]

\[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )^{3}}{x^{2}} \,d x } \]

[In]

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

[Out]

-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/x - integr
ate((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^5 + 3*a*b^2*x^4 + (3*a^2*b - b)*x^3 + (a^3 - a)*x^2 + (b^3*x^5 + 3*a*b^2*x
^4 + (3*a^2*b - b)*x^3 + (a^3 - a)*x^2)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)), x)

Giac [F]

\[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )^{3}}{x^{2}} \,d x } \]

[In]

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

[Out]

integrate(arcsech(b*x + a)^3/x^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=\int \frac {{\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}^3}{x^2} \,d x \]

[In]

int(acosh(1/(a + b*x))^3/x^2,x)

[Out]

int(acosh(1/(a + b*x))^3/x^2, x)