\(\int x \text {sech}^{-1}(a+b x)^2 \, dx\) [10]

Optimal result

Integrand size = 10, antiderivative size = 149 \[ \int x \text {sech}^{-1}(a+b x)^2 \, dx=-\frac {\sqrt {\frac {1-a-b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)}{b^2}-\frac {a^2 \text {sech}^{-1}(a+b x)^2}{2 b^2}+\frac {1}{2} x^2 \text {sech}^{-1}(a+b x)^2+\frac {4 a \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}-\frac {\log (a+b x)}{b^2}-\frac {2 i a \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2}+\frac {2 i a \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )}{b^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.15 \[ \int x \text {sech}^{-1}(a+b x)^2 \, dx=\frac {-2 \sqrt {-\frac {-1+a+b x}{1+a+b x}} (1+a+b x) \text {sech}^{-1}(a+b x)-2 a (a+b x) \text {sech}^{-1}(a+b x)^2+(a+b x)^2 \text {sech}^{-1}(a+b x)^2-4 i a \text {sech}^{-1}(a+b x) \left (\log \left (1-i e^{-\text {sech}^{-1}(a+b x)}\right )-\log \left (1+i e^{-\text {sech}^{-1}(a+b x)}\right )\right )+2 \log \left (\frac {1}{a+b x}\right )-4 i a \left (\operatorname {PolyLog}\left (2,-i e^{-\text {sech}^{-1}(a+b x)}\right )-\operatorname {PolyLog}\left (2,i e^{-\text {sech}^{-1}(a+b x)}\right )\right )}{2 b^2} \] Input:

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

Output:

(-2*Sqrt[-((-1 + a + b*x)/(1 + a + b*x))]*(1 + a + b*x)*ArcSech[a + b*x] - 
 2*a*(a + b*x)*ArcSech[a + b*x]^2 + (a + b*x)^2*ArcSech[a + b*x]^2 - (4*I) 
*a*ArcSech[a + b*x]*(Log[1 - I/E^ArcSech[a + b*x]] - Log[1 + I/E^ArcSech[a 
 + b*x]]) + 2*Log[(a + b*x)^(-1)] - (4*I)*a*(PolyLog[2, (-I)/E^ArcSech[a + 
 b*x]] - PolyLog[2, I/E^ArcSech[a + b*x]]))/(2*b^2)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6875, 25, 5991, 3042, 4678, 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.

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

\(\Big \downarrow \) 6875

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 5991

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4678

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {1}{2} a^2 \text {sech}^{-1}(a+b x)^2-4 a \text {sech}^{-1}(a+b x) \arctan \left (e^{\text {sech}^{-1}(a+b x)}\right )-\frac {1}{2} b^2 x^2 \text {sech}^{-1}(a+b x)^2+2 i a \operatorname {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a+b x)}\right )-2 i a \operatorname {PolyLog}\left (2,i e^{\text {sech}^{-1}(a+b x)}\right )-\log \left (\frac {1}{a+b x}\right )+\sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)}{b^2}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

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 4678

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 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 = 0.42 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.22

method	result	size
derivativedivides	\(\frac {-\frac {\operatorname {arcsech}\left (b x +a \right ) \left (2 \sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}\, \left (b x +a \right )+2 \,\operatorname {arcsech}\left (b x +a \right ) a \left (b x +a \right )-\operatorname {arcsech}\left (b x +a \right ) \left (b x +a \right )^{2}-2\right )}{2}-2 \ln \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\ln \left (1+\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )-2 i a \,\operatorname {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i a \,\operatorname {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 i a \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i a \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b^{2}}\)	\(331\)
default	\(\frac {-\frac {\operatorname {arcsech}\left (b x +a \right ) \left (2 \sqrt {-\frac {b x +a -1}{b x +a}}\, \sqrt {\frac {b x +a +1}{b x +a}}\, \left (b x +a \right )+2 \,\operatorname {arcsech}\left (b x +a \right ) a \left (b x +a \right )-\operatorname {arcsech}\left (b x +a \right ) \left (b x +a \right )^{2}-2\right )}{2}-2 \ln \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\ln \left (1+\left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )^{2}\right )-2 i a \,\operatorname {arcsech}\left (b x +a \right ) \ln \left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i a \,\operatorname {arcsech}\left (b x +a \right ) \ln \left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )-2 i a \operatorname {dilog}\left (1+i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )+2 i a \operatorname {dilog}\left (1-i \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )\right )}{b^{2}}\)	\(331\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/2*x^2*log(sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*s 
qrt(-b*x - a + 1)*a + b*x + a)^2 - integrate(-(4*(b^3*x^4 + 3*a*b^2*x^3 + 
(3*a^2*b - b)*x^2 + (a^3 - a)*x)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*log( 
b*x + a)^2 + 4*(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*l 
og(b*x + a)^2 - (b^3*x^4 + 2*a*b^2*x^3 + (a^2*b - b)*x^2 + 4*(b^3*x^4 + 3* 
a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*log(b*x + a) + (2*(b^3*x^4 + 
3*a*b^2*x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*sqrt(b*x + a + 1)*log(b*x + 
 a) + (2*b^3*x^4 + 4*a*b^2*x^3 + (2*a^2*b - b)*x^2 + 2*(b^3*x^4 + 3*a*b^2* 
x^3 + (3*a^2*b - b)*x^2 + (a^3 - a)*x)*log(b*x + a))*sqrt(b*x + a + 1))*sq 
rt(-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)*sqrt(b*x + a + 1)*sqrt( 
-b*x - a + 1) + (3*a^2*b - b)*x - a), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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