\(\int \frac {\text {sech}^{-1}(a+b x)^3}{x^3} \, dx\) [19]

Optimal result

Integrand size = 12, antiderivative size = 965 \[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^3} \, dx =\text {Too large to display} \] Output:

-3/2*b^2*arcsech(b*x+a)^2/a^2/(-a^2+1)+3/2*b^2*((-b*x-a+1)/(b*x+a+1))^(1/2 
)*(b*x+a+1)*arcsech(b*x+a)^2/a/(-a^2+1)/(b*x+a)/(1-a/(b*x+a))+1/2*b^2*arcs 
ech(b*x+a)^3/a^2-1/2*arcsech(b*x+a)^3/x^2+3*b^2*arcsech(b*x+a)*ln(1-a*(1/( 
b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/(1-(-a^2+1)^(1/2)))/a^2/(- 
a^2+1)+3/2*b^2*arcsech(b*x+a)^2*ln(1-a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/( 
b*x+a)+1)^(1/2))/(1-(-a^2+1)^(1/2)))/a^2/(-a^2+1)^(3/2)-3*b^2*arcsech(b*x+ 
a)^2*ln(1-a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/(1-(-a^2+1 
)^(1/2)))/a^2/(-a^2+1)^(1/2)+3*b^2*arcsech(b*x+a)*ln(1-a*(1/(b*x+a)+(1/(b* 
x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/(1+(-a^2+1)^(1/2)))/a^2/(-a^2+1)-3/2*b^ 
2*arcsech(b*x+a)^2*ln(1-a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/ 
2))/(1+(-a^2+1)^(1/2)))/a^2/(-a^2+1)^(3/2)+3*b^2*arcsech(b*x+a)^2*ln(1-a*( 
1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/(1+(-a^2+1)^(1/2)))/a^2 
/(-a^2+1)^(1/2)+3*b^2*polylog(2,a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a 
)+1)^(1/2))/(1-(-a^2+1)^(1/2)))/a^2/(-a^2+1)+3*b^2*arcsech(b*x+a)*polylog( 
2,a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/(1-(-a^2+1)^(1/2)) 
)/a^2/(-a^2+1)^(3/2)-6*b^2*arcsech(b*x+a)*polylog(2,a*(1/(b*x+a)+(1/(b*x+a 
)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/(1-(-a^2+1)^(1/2)))/a^2/(-a^2+1)^(1/2)+3*b 
^2*polylog(2,a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/(1+(-a^ 
2+1)^(1/2)))/a^2/(-a^2+1)-3*b^2*arcsech(b*x+a)*polylog(2,a*(1/(b*x+a)+(1/( 
b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/(1+(-a^2+1)^(1/2)))/a^2/(-a^2+1)^(...
 

Mathematica [F]

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

Integrate[ArcSech[a + b*x]^3/x^3,x]
 

Output:

Integrate[ArcSech[a + b*x]^3/x^3, x]
 

Rubi [A] (verified)

Time = 1.78 (sec) , antiderivative size = 911, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6875, 25, 5991, 3042, 4679, 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.

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

\(\Big \downarrow \) 6875

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 5991

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

\(\displaystyle -b^2 \left (\frac {\text {sech}^{-1}(a+b x)^3}{2 b^2 x^2}-\frac {3}{2} \left (\frac {\text {sech}^{-1}(a+b x)^3}{3 a^2}-\frac {2 \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right ) \text {sech}^{-1}(a+b x)^2}{a^2 \sqrt {1-a^2}}+\frac {\log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right ) \text {sech}^{-1}(a+b x)^2}{a^2 \left (1-a^2\right )^{3/2}}+\frac {2 \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right ) \text {sech}^{-1}(a+b x)^2}{a^2 \sqrt {1-a^2}}-\frac {\log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right ) \text {sech}^{-1}(a+b x)^2}{a^2 \left (1-a^2\right )^{3/2}}-\frac {\text {sech}^{-1}(a+b x)^2}{a^2 \left (1-a^2\right )}+\frac {\sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)^2}{a \left (1-a^2\right ) (a+b x) \left (1-\frac {a}{a+b x}\right )}+\frac {2 \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right ) \text {sech}^{-1}(a+b x)}{a^2 \left (1-a^2\right )}+\frac {2 \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right ) \text {sech}^{-1}(a+b x)}{a^2 \left (1-a^2\right )}-\frac {4 \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right ) \text {sech}^{-1}(a+b x)}{a^2 \sqrt {1-a^2}}+\frac {2 \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right ) \text {sech}^{-1}(a+b x)}{a^2 \left (1-a^2\right )^{3/2}}+\frac {4 \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right ) \text {sech}^{-1}(a+b x)}{a^2 \sqrt {1-a^2}}-\frac {2 \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right ) \text {sech}^{-1}(a+b x)}{a^2 \left (1-a^2\right )^{3/2}}+\frac {2 \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a^2 \left (1-a^2\right )}+\frac {2 \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a^2 \left (1-a^2\right )}+\frac {4 \operatorname {PolyLog}\left (3,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a^2 \sqrt {1-a^2}}-\frac {2 \operatorname {PolyLog}\left (3,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a^2 \left (1-a^2\right )^{3/2}}-\frac {4 \operatorname {PolyLog}\left (3,\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a^2 \sqrt {1-a^2}}+\frac {2 \operatorname {PolyLog}\left (3,\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a^2 \left (1-a^2\right )^{3/2}}\right )\right )\)

Input:

Int[ArcSech[a + b*x]^3/x^3,x]
 

Output:

-(b^2*(ArcSech[a + b*x]^3/(2*b^2*x^2) - (3*(-(ArcSech[a + b*x]^2/(a^2*(1 - 
 a^2))) + (Sqrt[(1 - a - b*x)/(1 + a + b*x)]*(1 + a + b*x)*ArcSech[a + b*x 
]^2)/(a*(1 - a^2)*(a + b*x)*(1 - a/(a + b*x))) + ArcSech[a + b*x]^3/(3*a^2 
) + (2*ArcSech[a + b*x]*Log[1 - (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2]) 
])/(a^2*(1 - a^2)) + (ArcSech[a + b*x]^2*Log[1 - (a*E^ArcSech[a + b*x])/(1 
 - Sqrt[1 - a^2])])/(a^2*(1 - a^2)^(3/2)) - (2*ArcSech[a + b*x]^2*Log[1 - 
(a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])])/(a^2*Sqrt[1 - a^2]) + (2*ArcS 
ech[a + b*x]*Log[1 - (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])])/(a^2*(1 
- a^2)) - (ArcSech[a + b*x]^2*Log[1 - (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - 
 a^2])])/(a^2*(1 - a^2)^(3/2)) + (2*ArcSech[a + b*x]^2*Log[1 - (a*E^ArcSec 
h[a + b*x])/(1 + Sqrt[1 - a^2])])/(a^2*Sqrt[1 - a^2]) + (2*PolyLog[2, (a*E 
^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])])/(a^2*(1 - a^2)) + (2*ArcSech[a + 
b*x]*PolyLog[2, (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])])/(a^2*(1 - a^2 
)^(3/2)) - (4*ArcSech[a + b*x]*PolyLog[2, (a*E^ArcSech[a + b*x])/(1 - Sqrt 
[1 - a^2])])/(a^2*Sqrt[1 - a^2]) + (2*PolyLog[2, (a*E^ArcSech[a + b*x])/(1 
 + Sqrt[1 - a^2])])/(a^2*(1 - a^2)) - (2*ArcSech[a + b*x]*PolyLog[2, (a*E^ 
ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])])/(a^2*(1 - a^2)^(3/2)) + (4*ArcSech 
[a + b*x]*PolyLog[2, (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])])/(a^2*Sqr 
t[1 - a^2]) - (2*PolyLog[3, (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])])/( 
a^2*(1 - a^2)^(3/2)) + (4*PolyLog[3, (a*E^ArcSech[a + b*x])/(1 - Sqrt[1...
 

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 4679

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si 
n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt 
Q[m, 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 [F]

\[\int \frac {\operatorname {arcsech}\left (b x +a \right )^{3}}{x^{3}}d x\]

Input:

int(arcsech(b*x+a)^3/x^3,x)
 

Output:

int(arcsech(b*x+a)^3/x^3,x)
 

Fricas [F]

\[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^3} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )^{3}}{x^{3}} \,d x } \] Input:

integrate(arcsech(b*x+a)^3/x^3,x, algorithm="fricas")
 

Output:

integral(arcsech(b*x + a)^3/x^3, x)
 

Sympy [F]

\[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^3} \, dx=\int \frac {\operatorname {asech}^{3}{\left (a + b x \right )}}{x^{3}}\, dx \] Input:

integrate(asech(b*x+a)**3/x**3,x)
 

Output:

Integral(asech(a + b*x)**3/x**3, x)
 

Maxima [F]

\[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^3} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )^{3}}{x^{3}} \,d x } \] Input:

integrate(arcsech(b*x+a)^3/x^3,x, algorithm="maxima")
 

Output:

-1/2*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)^3/x^2 - integrate(1/2*(16*(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)^3 + 16*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2*b - b)*x - a)*log(b*x 
 + a)^3 - 3*(b^3*x^3 + 2*a*b^2*x^2 + (a^2*b - b)*x - 4*(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 + 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 - 2*(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)*sqrt(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sqrt(-b*x - a + 1 
)*a + b*x + a)^2 - 24*((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 + (b^3*x^3 + 3*a*b^2* 
x^2 + a^3 + (3*a^2*b - b)*x - a)*log(b*x + a)^2)*log(sqrt(b*x + a + 1)*sqr 
t(-b*x - a + 1)*b*x + sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*a + b*x + a))/( 
b^3*x^6 + 3*a*b^2*x^5 + (3*a^2*b - b)*x^4 + (a^3 - a)*x^3 + (b^3*x^6 + 3*a 
*b^2*x^5 + (3*a^2*b - b)*x^4 + (a^3 - a)*x^3)*sqrt(b*x + a + 1)*sqrt(-b*x 
- a + 1)), x)
 

Giac [F]

\[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^3} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )^{3}}{x^{3}} \,d x } \] Input:

integrate(arcsech(b*x+a)^3/x^3,x, algorithm="giac")
 

Output:

integrate(arcsech(b*x + a)^3/x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^3} \, dx=\int \frac {{\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}^3}{x^3} \,d x \] Input:

int(acosh(1/(a + b*x))^3/x^3,x)
 

Output:

int(acosh(1/(a + b*x))^3/x^3, x)
 

Reduce [F]

\[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^3} \, dx=\int \frac {\mathit {asech} \left (b x +a \right )^{3}}{x^{3}}d x \] Input:

int(asech(b*x+a)^3/x^3,x)
                                                                                    
                                                                                    
 

Output:

int(asech(a + b*x)**3/x**3,x)