\(\int \text {sech}^{-1}(a+b x)^2 \, dx\) [11]

Optimal result

Integrand size = 8, antiderivative size = 80 \[ \int \text {sech}^{-1}(a+b x)^2 \, dx=\frac {(a+b x) \text {sech}^{-1}(a+b x)^2}{b}-\frac {4 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b}+\frac {2 i \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b}-\frac {2 i \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.31 \[ \int \text {sech}^{-1}(a+b x)^2 \, dx=\frac {i \left (\text {sech}^{-1}(a+b x) \left (-i (a+b x) \text {sech}^{-1}(a+b x)+2 \log \left (1-i e^{-\text {sech}^{-1}(a+b x)}\right )-2 \log \left (1+i e^{-\text {sech}^{-1}(a+b x)}\right )\right )+2 \operatorname {PolyLog}\left (2,-i e^{-\text {sech}^{-1}(a+b x)}\right )-2 \operatorname {PolyLog}\left (2,i e^{-\text {sech}^{-1}(a+b x)}\right )\right )}{b} \] Input:

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

Output:

(I*(ArcSech[a + b*x]*((-I)*(a + b*x)*ArcSech[a + b*x] + 2*Log[1 - I/E^ArcS 
ech[a + b*x]] - 2*Log[1 + I/E^ArcSech[a + b*x]]) + 2*PolyLog[2, (-I)/E^Arc 
Sech[a + b*x]] - 2*PolyLog[2, I/E^ArcSech[a + b*x]]))/b
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {6869, 6833, 5941, 3042, 4668, 2715, 2838}

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.

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

\(\Big \downarrow \) 6869

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

\(\Big \downarrow \) 6833

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

\(\Big \downarrow \) 5941

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4668

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

\(\Big \downarrow \) 2715

\(\displaystyle -\frac {-(a+b x) \text {sech}^{-1}(a+b x)^2+2 \left (-i \int e^{-\text {sech}^{-1}(a+b x)} \log \left (1-i e^{\text {sech}^{-1}(a+b x)}\right )de^{\text {sech}^{-1}(a+b x)}+i \int e^{-\text {sech}^{-1}(a+b x)} \log \left (1+i e^{\text {sech}^{-1}(a+b x)}\right )de^{\text {sech}^{-1}(a+b x)}+2 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )\right )}{b}\)

\(\Big \downarrow \) 2838

\(\displaystyle -\frac {-(a+b x) \text {sech}^{-1}(a+b x)^2+2 \left (2 \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )-i \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )+i \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )\right )}{b}\)

Input:

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

Output:

-((-((a + b*x)*ArcSech[a + b*x]^2) + 2*(2*ArcSech[a + b*x]*ArcTan[E^ArcSec 
h[a + b*x]] - I*PolyLog[2, (-I)*E^ArcSech[a + b*x]] + I*PolyLog[2, I*E^Arc 
Sech[a + b*x]]))/b)
 

Defintions of rubi rules used

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[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 2838

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 3042

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

rule 4668

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] + (-Simp[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] + Simp[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 5941

Int[(x_)^(m_.)*Sech[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tanh[(a_.) + (b_.)*(x_) 
^(n_.)]^(q_.), x_Symbol] :> Simp[(-x^(m - n + 1))*(Sech[a + b*x^n]^p/(b*n*p 
)), x] + Simp[(m - n + 1)/(b*n*p)   Int[x^(m - n)*Sech[a + b*x^n]^p, x], x] 
 /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && 
EqQ[q, 1]
 

rule 6833

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[-c^(-1)   S 
ubst[Int[(a + b*x)^n*Sech[x]*Tanh[x], x], x, ArcSech[c*x]], x] /; FreeQ[{a, 
 b, c, n}, x] && IGtQ[n, 0]
 

rule 6869

Int[((a_.) + ArcSech[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[1/d 
   Subst[Int[(a + b*ArcSech[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d 
}, x] && IGtQ[p, 0]
 

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.40

method	result	size
derivativedivides	\(\frac {\operatorname {arcsech}\left (b x +a \right )^{2} \left (b x +a \right )+2 i \operatorname {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 i \operatorname {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b}\)	\(192\)
default	\(\frac {\operatorname {arcsech}\left (b x +a \right )^{2} \left (b x +a \right )+2 i \operatorname {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 i \operatorname {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 i \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b}\)	\(192\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

x*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 - integrate(-2*(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)*sqrt(-b*x - a + 1)*log(b*x + a)^2 
 + 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 - 
(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)*sq 
rt(-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)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1) + (3*a^2*b - b)*x - a), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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