\(\int x^2 \text {sech}^{-1}(a+b x)^2 \, dx\) [9]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 12, antiderivative size = 279 \[ \int x^2 \text {sech}^{-1}(a+b x)^2 \, dx=-\frac {x}{3 b^2}+\frac {2 a \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{3 b^3}+\frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {4 a^2 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {2 a \log (a+b x)}{b^3}+\frac {i \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}+\frac {2 i a^2 \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {i \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {2 i a^2 \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6456, 5576, 4275, 4265, 2317, 2438, 4269, 3556, 4270} \[ \int x^2 \text {sech}^{-1}(a+b x)^2 \, dx=\frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}-\frac {4 a^2 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {2 i a^2 \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {2 i a^2 \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {2 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}+\frac {i \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {i \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}+\frac {2 a \log (a+b x)}{b^3}+\frac {2 a \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {x}{3 b^2} \]

[In]

Int[x^2*ArcSech[a + b*x]^2,x]

[Out]

-1/3*x/b^2 + (2*a*Sqrt[(1 - a - b*x)/(1 + a + b*x)]*(1 + a + b*x)*ArcSech[a + b*x])/b^3 - ((a + b*x)*Sqrt[(1 -
 a - b*x)/(1 + a + b*x)]*(1 + a + b*x)*ArcSech[a + b*x])/(3*b^3) + (a^3*ArcSech[a + b*x]^2)/(3*b^3) + (x^3*Arc
Sech[a + b*x]^2)/3 - (2*ArcSech[a + b*x]*ArcTan[E^ArcSech[a + b*x]])/(3*b^3) - (4*a^2*ArcSech[a + b*x]*ArcTan[
E^ArcSech[a + b*x]])/b^3 + (2*a*Log[a + b*x])/b^3 + ((I/3)*PolyLog[2, (-I)*E^ArcSech[a + b*x]])/b^3 + ((2*I)*a
^2*PolyLog[2, (-I)*E^ArcSech[a + b*x]])/b^3 - ((I/3)*PolyLog[2, I*E^ArcSech[a + b*x]])/b^3 - ((2*I)*a^2*PolyLo
g[2, I*E^ArcSech[a + b*x]])/b^3

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 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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 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 4270

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(-b^2)*(c + d*x)*Cot[e + f*x]
*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), x] + (Dist[b^2*((n - 2)/(n - 1)), Int[(c + d*x)*(b*Csc[e + f*x])^(n -
 2), x], x] - Simp[b^2*d*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x]) /; FreeQ[{b, c, d, e, f}, x] &&
 GtQ[n, 1] && NeQ[n, 2]

Rule 4275

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int x^2 \text {sech}(x) (-a+\text {sech}(x))^2 \tanh (x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^3} \\ & = \frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {Subst}\left (\int x (-a+\text {sech}(x))^3 \, dx,x,\text {sech}^{-1}(a+b x)\right )}{3 b^3} \\ & = \frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {Subst}\left (\int \left (-a^3 x+3 a^2 x \text {sech}(x)-3 a x \text {sech}^2(x)+x \text {sech}^3(x)\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{3 b^3} \\ & = \frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {Subst}\left (\int x \text {sech}^3(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{3 b^3}+\frac {(2 a) \text {Subst}\left (\int x \text {sech}^2(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^3}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int x \text {sech}(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^3} \\ & = -\frac {x}{3 b^2}+\frac {2 a \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{3 b^3}+\frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {4 a^2 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {\text {Subst}\left (\int x \text {sech}(x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{3 b^3}-\frac {(2 a) \text {Subst}\left (\int \tanh (x) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^3}+\frac {\left (2 i a^2\right ) \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^3}-\frac {\left (2 i a^2\right ) \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{b^3} \\ & = -\frac {x}{3 b^2}+\frac {2 a \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{3 b^3}+\frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {4 a^2 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {2 a \log (a+b x)}{b^3}+\frac {i \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{3 b^3}-\frac {i \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text {sech}^{-1}(a+b x)\right )}{3 b^3}+\frac {\left (2 i a^2\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {\left (2 i a^2\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a+b x)}\right )}{b^3} \\ & = -\frac {x}{3 b^2}+\frac {2 a \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{3 b^3}+\frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {4 a^2 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {2 a \log (a+b x)}{b^3}+\frac {2 i a^2 \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {2 i a^2 \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {i \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {i \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3} \\ & = -\frac {x}{3 b^2}+\frac {2 a \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{b^3}-\frac {(a+b x) \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{3 b^3}+\frac {a^3 \text {sech}^{-1}(a+b x)^2}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)^2-\frac {2 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {4 a^2 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}+\frac {2 a \log (a+b x)}{b^3}+\frac {i \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}+\frac {2 i a^2 \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3}-\frac {i \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{3 b^3}-\frac {2 i a^2 \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.28 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.09 \[ \int x^2 \text {sech}^{-1}(a+b x)^2 \, dx=-\frac {2 (a+b x) \sqrt {-\frac {-1+a+b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)+6 a (a+b x)^2 \text {sech}^{-1}(a+b x)^2-2 (a+b x)^3 \text {sech}^{-1}(a+b x)^2+2 \left (a+b x-6 a \sqrt {-\frac {-1+a+b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)-3 a^2 (a+b x) \text {sech}^{-1}(a+b x)^2\right )+12 a \log \left (\frac {1}{a+b x}\right )-\left (1+6 a^2\right ) \left (\pi \log \left (1-i e^{\text {sech}^{-1}(a+b x)}\right )-2 i \text {sech}^{-1}(a+b x) \log \left (1-i e^{\text {sech}^{-1}(a+b x)}\right )-\pi \log \left (1+i e^{\text {sech}^{-1}(a+b x)}\right )+2 i \text {sech}^{-1}(a+b x) \log \left (1+i e^{\text {sech}^{-1}(a+b x)}\right )-\pi \log \left (\cot \left (\frac {1}{4} \left (\pi +2 i \text {sech}^{-1}(a+b x)\right )\right )\right )+2 i \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )-2 i \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )\right )}{6 b^3} \]

[In]

Integrate[x^2*ArcSech[a + b*x]^2,x]

[Out]

-1/6*(2*(a + b*x)*Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(1 + a + b*x)*ArcSech[a + b*x] + 6*a*(a + b*x)^2*ArcSe
ch[a + b*x]^2 - 2*(a + b*x)^3*ArcSech[a + b*x]^2 + 2*(a + b*x - 6*a*Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(1 +
 a + b*x)*ArcSech[a + b*x] - 3*a^2*(a + b*x)*ArcSech[a + b*x]^2) + 12*a*Log[(a + b*x)^(-1)] - (1 + 6*a^2)*(Pi*
Log[1 - I*E^ArcSech[a + b*x]] - (2*I)*ArcSech[a + b*x]*Log[1 - I*E^ArcSech[a + b*x]] - Pi*Log[1 + I*E^ArcSech[
a + b*x]] + (2*I)*ArcSech[a + b*x]*Log[1 + I*E^ArcSech[a + b*x]] - Pi*Log[Cot[(Pi + (2*I)*ArcSech[a + b*x])/4]
] + (2*I)*PolyLog[2, (-I)*E^ArcSech[a + b*x]] - (2*I)*PolyLog[2, I*E^ArcSech[a + b*x]]))/b^3

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 599, normalized size of antiderivative = 2.15

method result size
derivativedivides \(\frac {\operatorname {arcsech}\left (b x +a \right )^{2} a^{2} \left (b x +a \right )-\operatorname {arcsech}\left (b x +a \right )^{2} a \left (b x +a \right )^{2}+\frac {\operatorname {arcsech}\left (b x +a \right )^{2} \left (b x +a \right )^{3}}{3}+2 \,\operatorname {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 (b x +a \right )-\frac {\operatorname {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 (b x +a \right )^{2}}{3}-2 a \,\operatorname {arcsech}\left (b x +a \right )-\frac {b x}{3}-\frac {a}{3}+\frac {i \operatorname {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}-\frac {i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}-2 i a^{2} \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i a^{2} \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 \ln \left (1+\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right ) a +4 a \ln \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )-\frac {i \operatorname {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}-2 i a^{2} \operatorname {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i a^{2} \operatorname {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+\frac {i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}}{b^{3}}\) \(599\)
default \(\frac {\operatorname {arcsech}\left (b x +a \right )^{2} a^{2} \left (b x +a \right )-\operatorname {arcsech}\left (b x +a \right )^{2} a \left (b x +a \right )^{2}+\frac {\operatorname {arcsech}\left (b x +a \right )^{2} \left (b x +a \right )^{3}}{3}+2 \,\operatorname {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 (b x +a \right )-\frac {\operatorname {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 (b x +a \right )^{2}}{3}-2 a \,\operatorname {arcsech}\left (b x +a \right )-\frac {b x}{3}-\frac {a}{3}+\frac {i \operatorname {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}-\frac {i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}-2 i a^{2} \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i a^{2} \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 \ln \left (1+\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right ) a +4 a \ln \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )-\frac {i \operatorname {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}-2 i a^{2} \operatorname {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i a^{2} \operatorname {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+\frac {i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{3}}{b^{3}}\) \(599\)

[In]

int(x^2*arcsech(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b^3*(arcsech(b*x+a)^2*a^2*(b*x+a)-arcsech(b*x+a)^2*a*(b*x+a)^2+1/3*arcsech(b*x+a)^2*(b*x+a)^3+2*arcsech(b*x+
a)*(-(b*x+a-1)/(b*x+a))^(1/2)*((b*x+a+1)/(b*x+a))^(1/2)*a*(b*x+a)-1/3*arcsech(b*x+a)*(-(b*x+a-1)/(b*x+a))^(1/2
)*((b*x+a+1)/(b*x+a))^(1/2)*(b*x+a)^2-2*a*arcsech(b*x+a)-1/3*b*x-1/3*a+1/3*I*arcsech(b*x+a)*ln(1+I*(1/(b*x+a)+
(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))-1/3*I*dilog(1-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))
-2*I*a^2*dilog(1-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))+2*I*a^2*dilog(1+I*(1/(b*x+a)+(1/(b*x+a
)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))-2*ln(1+(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))^2)*a+4*a*ln(1/(b*x
+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))-1/3*I*arcsech(b*x+a)*ln(1-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*
x+a)+1)^(1/2)))-2*I*a^2*arcsech(b*x+a)*ln(1-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))+2*I*a^2*arc
sech(b*x+a)*ln(1+I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))+1/3*I*dilog(1+I*(1/(b*x+a)+(1/(b*x+a)-
1)^(1/2)*(1/(b*x+a)+1)^(1/2))))

Fricas [F]

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

[In]

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

[Out]

integral(x^2*arcsech(b*x + a)^2, x)

Sympy [F]

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

[In]

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

[Out]

Integral(x**2*asech(a + b*x)**2, x)

Maxima [F]

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

[In]

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

[Out]

1/3*x^3*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 - i
ntegrate(-2/3*(6*(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)*log(b*x + a)^2 + 6*(b^3*x^5 + 3*a*b^2*x^4 + (3*a^2*b - b)*x^3 + (a^3 - a)*x^2)*log(b*x + a)^2 - (b^3*x^5 +
 2*a*b^2*x^4 + (a^2*b - b)*x^3 + 6*(b^3*x^5 + 3*a*b^2*x^4 + (3*a^2*b - b)*x^3 + (a^3 - a)*x^2)*log(b*x + a) +
(3*(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)*log(b*x + a) + (2*b^3*x^5 + 4
*a*b^2*x^4 + (2*a^2*b - b)*x^3 + 3*(b^3*x^5 + 3*a*b^2*x^4 + (3*a^2*b - b)*x^3 + (a^3 - a)*x^2)*log(b*x + a))*s
qrt(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^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)*s
qrt(b*x + a + 1)*sqrt(-b*x - a + 1) + (3*a^2*b - b)*x - a), x)

Giac [F]

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

[In]

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

[Out]

integrate(x^2*arcsech(b*x + a)^2, x)

Mupad [F(-1)]

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

[In]

int(x^2*acosh(1/(a + b*x))^2,x)

[Out]

int(x^2*acosh(1/(a + b*x))^2, x)