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

Optimal. Leaf size=537 \[ -\frac {2 b^2 \text {Li}_2\left (\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a^2 \sqrt {1-a^2}}+\frac {b^2 \text {Li}_2\left (\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a^2 \left (1-a^2\right )^{3/2}}+\frac {2 b^2 \text {Li}_2\left (\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a^2 \sqrt {1-a^2}}-\frac {b^2 \text {Li}_2\left (\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a^2 \left (1-a^2\right )^{3/2}}+\frac {b^2 \log \left (\frac {x}{a+b x}\right )}{a^2 \left (1-a^2\right )}+\frac {b^2 \text {sech}^{-1}(a+b x)^2}{2 a^2}+\frac {b^2 \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)}{a \left (1-a^2\right ) (a+b x) \left (1-\frac {a}{a+b x}\right )}-\frac {2 b^2 \text {sech}^{-1}(a+b x) \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a^2 \sqrt {1-a^2}}+\frac {b^2 \text {sech}^{-1}(a+b x) \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a^2 \left (1-a^2\right )^{3/2}}+\frac {2 b^2 \text {sech}^{-1}(a+b x) \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a^2 \sqrt {1-a^2}}-\frac {b^2 \text {sech}^{-1}(a+b x) \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a^2 \left (1-a^2\right )^{3/2}}-\frac {\text {sech}^{-1}(a+b x)^2}{2 x^2} \]

[Out]

1/2*b^2*arcsech(b*x+a)^2/a^2-1/2*arcsech(b*x+a)^2/x^2+b^2*ln(x/(b*x+a))/a^2/(-a^2+1)+b^2*arcsech(b*x+a)*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^2/(-a^2+1)^(3/2)-b^2*arcsech(b*x+a)
*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^2/(-a^2+1)^(3/2)+b^2*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^2/(-a^2+1)^(3/2)-b^2*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^2/(-a^2+1)^(3/2)-2*b^2*arcsech(b*x
+a)*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^2/(-a^2+1)^(1/2)+2*b^2*ar
csech(b*x+a)*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^2/(-a^2+1)^(1/2)
-2*b^2*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^2/(-a^2+1)^(1/2)+
2*b^2*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^2/(-a^2+1)^(1/2)+b
^2*(b*x+a+1)*arcsech(b*x+a)*((-b*x-a+1)/(b*x+a+1))^(1/2)/a/(-a^2+1)/(b*x+a)/(1-a/(b*x+a))

________________________________________________________________________________________

Rubi [A]  time = 0.75, antiderivative size = 537, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 11, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {6321, 5468, 4191, 3324, 3320, 2264, 2190, 2279, 2391, 2668, 31} \[ -\frac {2 b^2 \text {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a^2 \sqrt {1-a^2}}+\frac {b^2 \text {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a^2 \left (1-a^2\right )^{3/2}}+\frac {2 b^2 \text {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a^2 \sqrt {1-a^2}}-\frac {b^2 \text {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a^2 \left (1-a^2\right )^{3/2}}+\frac {b^2 \log \left (\frac {x}{a+b x}\right )}{a^2 \left (1-a^2\right )}+\frac {b^2 \text {sech}^{-1}(a+b x)^2}{2 a^2}+\frac {b^2 \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)}{a \left (1-a^2\right ) (a+b x) \left (1-\frac {a}{a+b x}\right )}-\frac {2 b^2 \text {sech}^{-1}(a+b x) \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a^2 \sqrt {1-a^2}}+\frac {b^2 \text {sech}^{-1}(a+b x) \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a^2 \left (1-a^2\right )^{3/2}}+\frac {2 b^2 \text {sech}^{-1}(a+b x) \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a^2 \sqrt {1-a^2}}-\frac {b^2 \text {sech}^{-1}(a+b x) \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a^2 \left (1-a^2\right )^{3/2}}-\frac {\text {sech}^{-1}(a+b x)^2}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcSech[a + b*x]^2/x^3,x]

[Out]

(b^2*Sqrt[(1 - a - b*x)/(1 + a + b*x)]*(1 + a + b*x)*ArcSech[a + b*x])/(a*(1 - a^2)*(a + b*x)*(1 - a/(a + b*x)
)) + (b^2*ArcSech[a + b*x]^2)/(2*a^2) - ArcSech[a + b*x]^2/(2*x^2) + (b^2*ArcSech[a + b*x]*Log[1 - (a*E^ArcSec
h[a + b*x])/(1 - Sqrt[1 - a^2])])/(a^2*(1 - a^2)^(3/2)) - (2*b^2*ArcSech[a + b*x]*Log[1 - (a*E^ArcSech[a + b*x
])/(1 - Sqrt[1 - a^2])])/(a^2*Sqrt[1 - a^2]) - (b^2*ArcSech[a + b*x]*Log[1 - (a*E^ArcSech[a + b*x])/(1 + Sqrt[
1 - a^2])])/(a^2*(1 - a^2)^(3/2)) + (2*b^2*ArcSech[a + b*x]*Log[1 - (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])
])/(a^2*Sqrt[1 - a^2]) + (b^2*Log[x/(a + b*x)])/(a^2*(1 - a^2)) + (b^2*PolyLog[2, (a*E^ArcSech[a + b*x])/(1 -
Sqrt[1 - a^2])])/(a^2*(1 - a^2)^(3/2)) - (2*b^2*PolyLog[2, (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])])/(a^2*S
qrt[1 - a^2]) - (b^2*PolyLog[2, (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])])/(a^2*(1 - a^2)^(3/2)) + (2*b^2*Po
lyLog[2, (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])])/(a^2*Sqrt[1 - a^2])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2264

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 2279

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 2391

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 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 3320

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))/(E^(I*Pi*(k - 1/2))*(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))), x], x] /; FreeQ[{a, b, c, d, e, f, fz}, x] && IntegerQ[
2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3324

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(c + d*x)^m*Cos[
e + f*x])/(f*(a^2 - b^2)*(a + b*Sin[e + f*x])), x] + (Dist[a/(a^2 - b^2), Int[(c + d*x)^m/(a + b*Sin[e + f*x])
, x], x] - Dist[(b*d*m)/(f*(a^2 - b^2)), Int[((c + d*x)^(m - 1)*Cos[e + f*x])/(a + b*Sin[e + f*x]), x], x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 4191

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 5468

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 6321

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]

Rubi steps

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

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Mathematica [C]  time = 7.58, size = 1439, normalized size = 2.68 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSech[a + b*x]^2/x^3,x]

[Out]

-1/2*((a + b*x)^2*ArcSech[a + b*x]^2)/(a^2*x^2) + (b*ArcSech[a + b*x]*(-(a*Sqrt[-((-1 + a + b*x)/(1 + a + b*x)
)]*(1 + a + b*x)) + (-1 + a^2)*(a + b*x)*ArcSech[a + b*x]))/((-1 + a)*a^2*(1 + a)*x) + (b^2*Log[(b*x)/(a + b*x
)])/(a^2 - a^4) - (2*b^2*(2*ArcSech[a + b*x]*ArcTan[((-1 + a)*Coth[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]] - (2*I
)*ArcCos[a^(-1)]*ArcTan[((1 + a)*Tanh[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]] + (ArcCos[a^(-1)] + 2*(ArcTan[((-1
+ a)*Coth[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]] + ArcTan[((1 + a)*Tanh[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]]))*L
og[Sqrt[-1 + a^2]/(Sqrt[2]*Sqrt[a]*E^(ArcSech[a + b*x]/2)*Sqrt[-((b*x)/(a + b*x))])] + (ArcCos[a^(-1)] - 2*(Ar
cTan[((-1 + a)*Coth[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]] + ArcTan[((1 + a)*Tanh[ArcSech[a + b*x]/2])/Sqrt[-1 +
a^2]]))*Log[(Sqrt[-1 + a^2]*E^(ArcSech[a + b*x]/2))/(Sqrt[2]*Sqrt[a]*Sqrt[-((b*x)/(a + b*x))])] - (ArcCos[a^(
-1)] + 2*ArcTan[((1 + a)*Tanh[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]])*Log[-(((-1 + a)*(1 + a - I*Sqrt[-1 + a^2])
*(-1 + Tanh[ArcSech[a + b*x]/2]))/(a*(-1 + a + I*Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/2])))] - (ArcCos[a^(-1)]
- 2*ArcTan[((1 + a)*Tanh[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]])*Log[((-1 + a)*(1 + a + I*Sqrt[-1 + a^2])*(1 +
Tanh[ArcSech[a + b*x]/2]))/(a*(-1 + a + I*Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/2]))] + I*(PolyLog[2, ((-1 - I*
Sqrt[-1 + a^2])*(-1 + a - I*Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/2]))/(a*(-1 + a + I*Sqrt[-1 + a^2]*Tanh[ArcSe
ch[a + b*x]/2]))] - PolyLog[2, ((I + Sqrt[-1 + a^2])*(-1 + a - I*Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/2]))/(a*
((-I)*(-1 + a) + Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/2]))])))/(-1 + a^2)^(3/2) + (b^2*(2*ArcSech[a + b*x]*Arc
Tan[((-1 + a)*Coth[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]] - (2*I)*ArcCos[a^(-1)]*ArcTan[((1 + a)*Tanh[ArcSech[a
+ b*x]/2])/Sqrt[-1 + a^2]] + (ArcCos[a^(-1)] + 2*(ArcTan[((-1 + a)*Coth[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]] +
ArcTan[((1 + a)*Tanh[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]]))*Log[Sqrt[-1 + a^2]/(Sqrt[2]*Sqrt[a]*E^(ArcSech[a
+ b*x]/2)*Sqrt[-((b*x)/(a + b*x))])] + (ArcCos[a^(-1)] - 2*(ArcTan[((-1 + a)*Coth[ArcSech[a + b*x]/2])/Sqrt[-1
+ a^2]] + ArcTan[((1 + a)*Tanh[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]]))*Log[(Sqrt[-1 + a^2]*E^(ArcSech[a + b*x]
/2))/(Sqrt[2]*Sqrt[a]*Sqrt[-((b*x)/(a + b*x))])] - (ArcCos[a^(-1)] + 2*ArcTan[((1 + a)*Tanh[ArcSech[a + b*x]/2
])/Sqrt[-1 + a^2]])*Log[-(((-1 + a)*(1 + a - I*Sqrt[-1 + a^2])*(-1 + Tanh[ArcSech[a + b*x]/2]))/(a*(-1 + a + I
*Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/2])))] - (ArcCos[a^(-1)] - 2*ArcTan[((1 + a)*Tanh[ArcSech[a + b*x]/2])/S
qrt[-1 + a^2]])*Log[((-1 + a)*(1 + a + I*Sqrt[-1 + a^2])*(1 + Tanh[ArcSech[a + b*x]/2]))/(a*(-1 + a + I*Sqrt[-
1 + a^2]*Tanh[ArcSech[a + b*x]/2]))] + I*(PolyLog[2, ((-1 - I*Sqrt[-1 + a^2])*(-1 + a - I*Sqrt[-1 + a^2]*Tanh[
ArcSech[a + b*x]/2]))/(a*(-1 + a + I*Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/2]))] - PolyLog[2, ((I + Sqrt[-1 + a
^2])*(-1 + a - I*Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/2]))/(a*((-I)*(-1 + a) + Sqrt[-1 + a^2]*Tanh[ArcSech[a +
b*x]/2]))])))/(a^2*(-1 + a^2)^(3/2))

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arsech}\left (b x + a\right )^{2}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsech(b*x + a)^2/x^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsech}\left (b x + a\right )^{2}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(b*x + a)^2/x^3, x)

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maple [A]  time = 1.29, size = 1026, normalized size = 1.91 \[ \frac {b^{2} \mathrm {arcsech}\left (b x +a \right )^{2}}{2 a^{2}-2}-\frac {b^{2} \mathrm {arcsech}\left (b x +a \right ) \sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}}{a \left (a^{2}-1\right )}-\frac {b \,\mathrm {arcsech}\left (b x +a \right ) \sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}}{\left (a^{2}-1\right ) x}-\frac {b^{2} \mathrm {arcsech}\left (b x +a \right )^{2}}{2 a^{2} \left (a^{2}-1\right )}-\frac {\mathrm {arcsech}\left (b x +a \right )^{2} a^{2}}{2 \left (a^{2}-1\right ) x^{2}}-\frac {b^{2} \mathrm {arcsech}\left (b x +a \right )}{a^{2} \left (a^{2}-1\right )}+\frac {\mathrm {arcsech}\left (b x +a \right )^{2}}{2 \left (a^{2}-1\right ) x^{2}}+\frac {2 b^{2} \ln \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )}{a^{2} \left (a^{2}-1\right )}-\frac {b^{2} \ln \left (a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}+a -\frac {2}{b x +a}-2 \sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )}{a^{2} \left (a^{2}-1\right )}+\frac {b^{2} \sqrt {-a^{2}+1}\, \mathrm {arcsech}\left (b x +a \right ) \ln \left (\frac {-a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{a^{2} \left (a^{2}-1\right )^{2}}-\frac {b^{2} \sqrt {-a^{2}+1}\, \mathrm {arcsech}\left (b x +a \right ) \ln \left (\frac {a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{a^{2} \left (a^{2}-1\right )^{2}}+\frac {b^{2} \sqrt {-a^{2}+1}\, \dilog \left (\frac {-a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{a^{2} \left (a^{2}-1\right )^{2}}-\frac {b^{2} \sqrt {-a^{2}+1}\, \dilog \left (\frac {a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{a^{2} \left (a^{2}-1\right )^{2}}-\frac {2 b^{2} \sqrt {-a^{2}+1}\, \mathrm {arcsech}\left (b x +a \right ) \ln \left (\frac {-a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{\left (a^{2}-1\right )^{2}}+\frac {2 b^{2} \sqrt {-a^{2}+1}\, \mathrm {arcsech}\left (b x +a \right ) \ln \left (\frac {a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{\left (a^{2}-1\right )^{2}}-\frac {2 b^{2} \sqrt {-a^{2}+1}\, \dilog \left (\frac {-a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{\left (a^{2}-1\right )^{2}}+\frac {2 b^{2} \sqrt {-a^{2}+1}\, \dilog \left (\frac {a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{\left (a^{2}-1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*arcsech(b*x+a)^2/(a^2-1)-b^2*arcsech(b*x+a)/a/(a^2-1)*(-(b*x+a-1)/(b*x+a))^(1/2)*((b*x+a+1)/(b*x+a))^(
1/2)-b*arcsech(b*x+a)/(a^2-1)/x*(-(b*x+a-1)/(b*x+a))^(1/2)*((b*x+a+1)/(b*x+a))^(1/2)-1/2*b^2*arcsech(b*x+a)^2/
a^2/(a^2-1)-1/2*arcsech(b*x+a)^2*a^2/(a^2-1)/x^2-b^2*arcsech(b*x+a)/a^2/(a^2-1)+1/2*arcsech(b*x+a)^2/(a^2-1)/x
^2+2*b^2/a^2/(a^2-1)*ln(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))-b^2/a^2/(a^2-1)*ln(a*(1/(b*x+a)+(1/
(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))^2+a-2/(b*x+a)-2*(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+b^2*(-a^2+1)^(1
/2)/a^2/(a^2-1)^2*arcsech(b*x+a)*ln((-a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)+1)/
(1+(-a^2+1)^(1/2)))-b^2*(-a^2+1)^(1/2)/a^2/(a^2-1)^2*arcsech(b*x+a)*ln((a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b
*x+a)+1)^(1/2))+(-a^2+1)^(1/2)-1)/(-1+(-a^2+1)^(1/2)))+b^2*(-a^2+1)^(1/2)/a^2/(a^2-1)^2*dilog((-a*(1/(b*x+a)+(
1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)+1)/(1+(-a^2+1)^(1/2)))-b^2*(-a^2+1)^(1/2)/a^2/(a^2-1)^2
*dilog((a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)-1)/(-1+(-a^2+1)^(1/2)))-2*b^2*(-a
^2+1)^(1/2)/(a^2-1)^2*arcsech(b*x+a)*ln((-a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)
+1)/(1+(-a^2+1)^(1/2)))+2*b^2*(-a^2+1)^(1/2)/(a^2-1)^2*arcsech(b*x+a)*ln((a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/
(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)-1)/(-1+(-a^2+1)^(1/2)))-2*b^2*(-a^2+1)^(1/2)/(a^2-1)^2*dilog((-a*(1/(b*x+a)+(
1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)+1)/(1+(-a^2+1)^(1/2)))+2*b^2*(-a^2+1)^(1/2)/(a^2-1)^2*d
ilog((a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)-1)/(-1+(-a^2+1)^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {\log \left (\sqrt {b x + a + 1} \sqrt {-b x - a + 1} b x + \sqrt {b x + a + 1} \sqrt {-b x - a + 1} a + b x + a\right )^{2}}{2 \, x^{2}} - \int -\frac {4 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt {b x + a + 1} \sqrt {-b x - a + 1} \log \left (b x + a\right )^{2} + 4 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right )^{2} + {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + {\left (a^{2} b - b\right )} x - 4 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right ) - {\left (2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b - b\right )} x - a\right )} \sqrt {b x + a + 1} \log \left (b x + a\right ) - {\left (2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} + {\left (2 \, a^{2} b - b\right )} x - 2 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} + {\left (3 \, a^{2} b - b\right )} x - a\right )} \log \left (b x + a\right )\right )} \sqrt {b x + a + 1}\right )} \sqrt {-b x - a + 1}\right )} \log \left (\sqrt {b x + a + 1} \sqrt {-b x - a + 1} b x + \sqrt {b x + a + 1} \sqrt {-b x - a + 1} a + b x + a\right )}{b^{3} x^{6} + 3 \, a b^{2} x^{5} + {\left (3 \, a^{2} b - b\right )} x^{4} + {\left (a^{3} - a\right )} x^{3} + {\left (b^{3} x^{6} + 3 \, a b^{2} x^{5} + {\left (3 \, a^{2} b - b\right )} x^{4} + {\left (a^{3} - a\right )} x^{3}\right )} \sqrt {b x + a + 1} \sqrt {-b x - a + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*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)^2/x^2 -
integrate(-(4*(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 + 4*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*log(b*x + a)^2 + (b^3*x^3 + 2*a*b^2*x^2 + (a^2
*b - b)*x - 4*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*log(b*x + a) - (2*(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 - 2*(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))/(b^3*x^6 + 3*a*b^2*x^
5 + (3*a^2*b - b)*x^4 + (a^3 - a)*x^3 + (b^3*x^6 + 3*a*b^2*x^5 + (3*a^2*b - b)*x^4 + (a^3 - a)*x^3)*sqrt(b*x +
a + 1)*sqrt(-b*x - a + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}^2}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(1/(a + b*x))^2/x^3,x)

[Out]

int(acosh(1/(a + b*x))^2/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asech}^{2}{\left (a + b x \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(a + b*x)**2/x**3, x)

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