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

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 8, antiderivative size = 107 \[ \int x \text {sech}^{-1}(a+b x) \, dx=-\frac {\sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{2 b^2}-\frac {a^2 \text {sech}^{-1}(a+b x)}{2 b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)+\frac {a \arctan \left (\frac {\sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x)}{a+b x}\right )}{b^2} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.64 \[ \int x \text {sech}^{-1}(a+b x) \, dx=\frac {-\sqrt {-\frac {-1+a+b x}{1+a+b x}} (1+a+b x)+b^2 x^2 \text {sech}^{-1}(a+b x)+a^2 \log (a+b x)-a^2 \log \left (1+\sqrt {-\frac {-1+a+b x}{1+a+b x}}+a \sqrt {-\frac {-1+a+b x}{1+a+b x}}+b x \sqrt {-\frac {-1+a+b x}{1+a+b x}}\right )-2 i a \log \left (-2 i (a+b x)+2 \sqrt {-\frac {-1+a+b x}{1+a+b x}} (1+a+b x)\right )}{2 b^2} \] Input:

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

Output:

(-(Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(1 + a + b*x)) + b^2*x^2*ArcSech[ 
a + b*x] + a^2*Log[a + b*x] - a^2*Log[1 + Sqrt[-((-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))]] - (2*I)*a*Log[(-2*I)*(a + b*x) + 2*Sqrt[-((-1 + a + b*x 
)/(1 + a + b*x))]*(1 + a + b*x)])/(2*b^2)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {6875, 25, 5991, 3042, 4260, 3042, 4254, 24, 4257}

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.

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

\(\Big \downarrow \) 6875

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 5991

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4260

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4257

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4254
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4260
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[a^2*x, x] + 
 (Simp[2*a*b   Int[Csc[c + d*x], x], x] + Simp[b^2   Int[Csc[c + d*x]^2, x] 
, x]) /; FreeQ[{a, b, c, d}, x]
 

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 = 0.82 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.04

method result size
derivativedivides \(\frac {\frac {\operatorname {arcsech}\left (b x +a \right ) \left (b x +a \right )^{2}}{2}-\operatorname {arcsech}\left (b x +a \right ) a \left (b x +a \right )-\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \left (2 a \arcsin \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )}{2 \sqrt {1-\left (b x +a \right )^{2}}}}{b^{2}}\) \(111\)
default \(\frac {\frac {\operatorname {arcsech}\left (b x +a \right ) \left (b x +a \right )^{2}}{2}-\operatorname {arcsech}\left (b x +a \right ) a \left (b x +a \right )-\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \left (2 a \arcsin \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )}{2 \sqrt {1-\left (b x +a \right )^{2}}}}{b^{2}}\) \(111\)
parts \(\frac {x^{2} \operatorname {arcsech}\left (b x +a \right )}{2}-\frac {\sqrt {-\frac {b x +a -1}{b x +a}}\, \left (b x +a \right ) \sqrt {\frac {b x +a +1}{b x +a}}\, \left (\operatorname {csgn}\left (b \right ) \operatorname {arctanh}\left (\frac {1}{\sqrt {-b^{2} x^{2}-2 b x a -a^{2}+1}}\right ) a^{2}+\sqrt {-b^{2} x^{2}-2 b x a -a^{2}+1}\, \operatorname {csgn}\left (b \right )+2 \arctan \left (\frac {\operatorname {csgn}\left (b \right ) \left (b x +a \right )}{\sqrt {-\left (b x +a +1\right ) \left (b x +a -1\right )}}\right ) a \right ) \operatorname {csgn}\left (b \right )}{2 b^{2} \sqrt {-b^{2} x^{2}-2 b x a -a^{2}+1}}\) \(163\)

Input:

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

Output:

1/b^2*(1/2*arcsech(b*x+a)*(b*x+a)^2-arcsech(b*x+a)*a*(b*x+a)-1/2*(-(b*x+a- 
1)/(b*x+a))^(1/2)*(b*x+a)*((b*x+a+1)/(b*x+a))^(1/2)*(2*a*arcsin(b*x+a)+(1- 
(b*x+a)^2)^(1/2))/(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. 308 vs. \(2 (93) = 186\).

Time = 0.16 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.88 \[ \int x \text {sech}^{-1}(a+b x) \, dx=\frac {2 \, b^{2} x^{2} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{b x + a}\right ) - a^{2} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{x}\right ) + a^{2} \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - 1}{x}\right ) + 4 \, a \arctan \left (\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - 2 \, {\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{4 \, b^{2}} \] Input:

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

Output:

1/4*(2*b^2*x^2*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^2*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^2*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) + 4 
*a*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)) - 2*(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)))/b^2
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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