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

Optimal result

Integrand size = 10, antiderivative size = 153 \[ \int x^2 \text {sech}^{-1}(a+b x) \, dx=\frac {5 a \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{6 b^3}-\frac {x \sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{6 b^2}+\frac {a^3 \text {sech}^{-1}(a+b x)}{3 b^3}+\frac {1}{3} x^3 \text {sech}^{-1}(a+b x)-\frac {\left (1+6 a^2\right ) \arctan \left (\frac {\sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{a+b x}\right )}{6 b^3} \] Output:

5/6*a*((-b*x-a+1)/(b*x+a+1))^(1/2)*(b*x+a+1)/b^3-1/6*x*((-b*x-a+1)/(b*x+a+ 
1))^(1/2)*(b*x+a+1)/b^2+1/3*a^3*arcsech(b*x+a)/b^3+1/3*x^3*arcsech(b*x+a)- 
1/6*(6*a^2+1)*arctan(((-b*x-a+1)/(b*x+a+1))^(1/2)*(b*x+a+1)/(b*x+a))/b^3
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.31 \[ \int x^2 \text {sech}^{-1}(a+b x) \, dx=\frac {\sqrt {-\frac {-1+a+b x}{1+a+b x}} \left (5 a^2-b x (1+b x)+a (5+4 b x)\right )+2 b^3 x^3 \text {sech}^{-1}(a+b x)-2 a^3 \log (a+b x)+2 a^3 \log \left (1+\sqrt {-\frac {-1+a+b x}{1+a+b x}}+a \sqrt {-\frac {-1+a+b x}{1+a+b x}}+b x \sqrt {-\frac {-1+a+b x}{1+a+b x}}\right )+i \left (1+6 a^2\right ) \log \left (-2 i (a+b x)+2 \sqrt {-\frac {-1+a+b x}{1+a+b x}} (1+a+b x)\right )}{6 b^3} \] Input:

Integrate[x^2*ArcSech[a + b*x],x]
 

Output:

(Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(5*a^2 - b*x*(1 + b*x) + a*(5 + 4*b 
*x)) + 2*b^3*x^3*ArcSech[a + b*x] - 2*a^3*Log[a + b*x] + 2*a^3*Log[1 + Sqr 
t[-((-1 + a + b*x)/(1 + a + b*x))] + a*Sqrt[-((-1 + a + b*x)/(1 + a + b*x) 
)] + b*x*Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]] + I*(1 + 6*a^2)*Log[(-2*I) 
*(a + b*x) + 2*Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(1 + a + b*x)])/(6*b^ 
3)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6875, 5991, 3042, 4269, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 6875

\(\displaystyle -\frac {\int b^2 x^2 (a+b x) \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)d\text {sech}^{-1}(a+b x)}{b^3}\)

\(\Big \downarrow \) 5991

\(\displaystyle -\frac {-\frac {1}{3} \int -b^3 x^3d\text {sech}^{-1}(a+b x)-\frac {1}{3} b^3 x^3 \text {sech}^{-1}(a+b x)}{b^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {1}{3} b^3 x^3 \text {sech}^{-1}(a+b x)-\frac {1}{3} \int \left (a-\csc \left (i \text {sech}^{-1}(a+b x)+\frac {\pi }{2}\right )\right )^3d\text {sech}^{-1}(a+b x)}{b^3}\)

\(\Big \downarrow \) 4269

\(\displaystyle -\frac {\frac {1}{3} \left (\frac {1}{2} b x \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)-\frac {1}{2} \int \left (2 a^3+5 (a+b x)^2 a-\left (6 a^2+1\right ) (a+b x)\right )d\text {sech}^{-1}(a+b x)\right )-\frac {1}{3} b^3 x^3 \text {sech}^{-1}(a+b x)}{b^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {1}{3} \left (\frac {1}{2} \left (-2 a^3 \text {sech}^{-1}(a+b x)+\left (6 a^2+1\right ) \arctan \left (\frac {\sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)}{a+b x}\right )-5 a \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)\right )+\frac {1}{2} b x \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1)\right )-\frac {1}{3} b^3 x^3 \text {sech}^{-1}(a+b x)}{b^3}\)

Input:

Int[x^2*ArcSech[a + b*x],x]
 

Output:

-((-1/3*(b^3*x^3*ArcSech[a + b*x]) + ((b*x*Sqrt[(1 - a - b*x)/(1 + a + b*x 
)]*(1 + a + b*x))/2 + (-5*a*Sqrt[(1 - a - b*x)/(1 + a + b*x)]*(1 + a + b*x 
) - 2*a^3*ArcSech[a + b*x] + (1 + 6*a^2)*ArcTan[(Sqrt[(1 - a - b*x)/(1 + a 
 + b*x)]*(1 + a + b*x))/(a + b*x)])/2)/3)/b^3)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4269

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-b^2)*C 
ot[c + d*x]*((a + b*Csc[c + d*x])^(n - 2)/(d*(n - 1))), x] + Simp[1/(n - 1) 
   Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2) + 3* 
a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] / 
; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]
 

rule 5991

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] + Simp[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 6875

Int[((a_.) + ArcSech[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( 
m_.), x_Symbol] :> Simp[-(d^(m + 1))^(-1)   Subst[Int[(a + b*x)^p*Sech[x]*T 
anh[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]
 

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.24

method	result	size
derivativedivides	\(\frac {-\frac {\operatorname {arcsech}\left (b x +a \right ) a^{3}}{3}+\operatorname {arcsech}\left (b x +a \right ) a^{2} \left (b x +a \right )-\operatorname {arcsech}\left (b x +a \right ) a \left (b x +a \right )^{2}+\frac {\operatorname {arcsech}\left (b x +a \right ) \left (b x +a \right )^{3}}{3}+\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \left (2 a^{3} \operatorname {arctanh}\left (\frac {1}{\sqrt {1-\left (b x +a \right )^{2}}}\right )+6 a^{2} \arcsin \left (b x +a \right )+6 a \sqrt {1-\left (b x +a \right )^{2}}-\left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}+\arcsin \left (b x +a \right )\right )}{6 \sqrt {1-\left (b x +a \right )^{2}}}}{b^{3}}\)	\(189\)
default	\(\frac {-\frac {\operatorname {arcsech}\left (b x +a \right ) a^{3}}{3}+\operatorname {arcsech}\left (b x +a \right ) a^{2} \left (b x +a \right )-\operatorname {arcsech}\left (b x +a \right ) a \left (b x +a \right )^{2}+\frac {\operatorname {arcsech}\left (b x +a \right ) \left (b x +a \right )^{3}}{3}+\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \left (2 a^{3} \operatorname {arctanh}\left (\frac {1}{\sqrt {1-\left (b x +a \right )^{2}}}\right )+6 a^{2} \arcsin \left (b x +a \right )+6 a \sqrt {1-\left (b x +a \right )^{2}}-\left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}+\arcsin \left (b x +a \right )\right )}{6 \sqrt {1-\left (b x +a \right )^{2}}}}{b^{3}}\)	\(189\)
parts	\(\frac {x^{3} \operatorname {arcsech}\left (b x +a \right )}{3}+\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \left (2 \,\operatorname {csgn}\left (b \right ) \operatorname {arctanh}\left (\frac {1}{\sqrt {-b^{2} x^{2}-2 b x a -a^{2}+1}}\right ) a^{3}-\sqrt {-b^{2} x^{2}-2 b x a -a^{2}+1}\, \operatorname {csgn}\left (b \right ) b x +5 \sqrt {-b^{2} x^{2}-2 b x a -a^{2}+1}\, \operatorname {csgn}\left (b \right ) a +6 \arctan \left (\frac {\operatorname {csgn}\left (b \right ) \left (b x +a \right )}{\sqrt {-b^{2} x^{2}-2 b x a -a^{2}+1}}\right ) a^{2}+\arctan \left (\frac {\operatorname {csgn}\left (b \right ) \left (b x +a \right )}{\sqrt {-b^{2} x^{2}-2 b x a -a^{2}+1}}\right )\right ) \operatorname {csgn}\left (b \right )}{6 b^{3} \sqrt {-b^{2} x^{2}-2 b x a -a^{2}+1}}\)	\(233\)

Input:

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

Output:

1/b^3*(-1/3*arcsech(b*x+a)*a^3+arcsech(b*x+a)*a^2*(b*x+a)-arcsech(b*x+a)*a 
*(b*x+a)^2+1/3*arcsech(b*x+a)*(b*x+a)^3+1/6*(-(b*x+a-1)/(b*x+a))^(1/2)*(b* 
x+a)*((b*x+a+1)/(b*x+a))^(1/2)*(2*a^3*arctanh(1/(1-(b*x+a)^2)^(1/2))+6*a^2 
*arcsin(b*x+a)+6*a*(1-(b*x+a)^2)^(1/2)-(b*x+a)*(1-(b*x+a)^2)^(1/2)+arcsin( 
b*x+a))/(1-(b*x+a)^2)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 327 vs. \(2 (131) = 262\).

Time = 0.12 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.14 \[ \int x^2 \text {sech}^{-1}(a+b x) \, dx=\frac {2 \, b^{3} x^{3} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{b x + a}\right ) + a^{3} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{x}\right ) - a^{3} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - 1}{x}\right ) - {\left (6 \, a^{2} + 1\right )} \arctan \left (\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - {\left (b^{2} x^{2} - 4 \, a b x - 5 \, a^{2}\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{6 \, b^{3}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

integrate(x^2*arcsech(b*x + a), x)
 

Mupad [F(-1)]

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

int(x^2*acosh(1/(a + b*x)),x)
 

Output:

int(x^2*acosh(1/(a + b*x)), x)
 

Reduce [F]

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

int(x^2*asech(b*x+a),x)
 

Output:

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