3.1.18 \(\int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx\) [18]

3.1.18.1 Optimal result

Integrand size = 12, antiderivative size = 330 \[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=-\frac {b \text {sech}^{-1}(a+b x)^3}{a}-\frac {\text {sech}^{-1}(a+b x)^3}{x}+\frac {3 b \text {sech}^{-1}(a+b x)^2 \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {3 b \text {sech}^{-1}(a+b x)^2 \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {6 b \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {6 b \text {sech}^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {6 b \operatorname {PolyLog}\left (3,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {6 b \operatorname {PolyLog}\left (3,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}} \]

output

-b*arcsech(b*x+a)^3/a-arcsech(b*x+a)^3/x+3*b*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/(-a^2 
+1)^(1/2)-3*b*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/(-a^2+1)^(1/2)+6*b*arcsech(b*x+a)*po 
lylog(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/(-a^2+1)^(1/2)-6*b*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/(-a^2+1)^(1/2)-6*b 
*polylog(3,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/(-a^2+1)^(1/2)+6*b*polylog(3,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/(-a^2+1)^(1/2)
 

3.1.18.2 Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 69.42 (sec) , antiderivative size = 8527, normalized size of antiderivative = 25.84 \[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=\text {Result too large to show} \]

input

Integrate[ArcSech[a + b*x]^3/x^2,x]
 

output

Result too large to show
 

3.1.18.3 Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6875, 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.

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

\(\Big \downarrow \) 6875

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

\(\Big \downarrow \) 5991

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

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

input

Int[ArcSech[a + b*x]^3/x^2,x]
 

output

-(b*(ArcSech[a + b*x]^3/(b*x) + 3*(ArcSech[a + b*x]^3/(3*a) - (ArcSech[a + 
 b*x]^2*Log[1 - (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])])/(a*Sqrt[1 - a 
^2]) + (ArcSech[a + b*x]^2*Log[1 - (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^ 
2])])/(a*Sqrt[1 - a^2]) - (2*ArcSech[a + b*x]*PolyLog[2, (a*E^ArcSech[a + 
b*x])/(1 - Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2]) + (2*ArcSech[a + b*x]*PolyLo 
g[2, (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2]) + (2*P 
olyLog[3, (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2]) - 
 (2*PolyLog[3, (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])])/(a*Sqrt[1 - a^ 
2]))))
 

3.1.18.3.1 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 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]
 

3.1.18.4 Maple [F]

\[\int \frac {\operatorname {arcsech}\left (b x +a \right )^{3}}{x^{2}}d x\]

input

int(arcsech(b*x+a)^3/x^2,x)
 

output

int(arcsech(b*x+a)^3/x^2,x)
 

3.1.18.5 Fricas [F]

\[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )^{3}}{x^{2}} \,d x } \]

input

integrate(arcsech(b*x+a)^3/x^2,x, algorithm="fricas")
 

output

integral(arcsech(b*x + a)^3/x^2, x)
 

3.1.18.6 Sympy [F]

\[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=\int \frac {\operatorname {asech}^{3}{\left (a + b x \right )}}{x^{2}}\, dx \]

input

integrate(asech(b*x+a)**3/x**2,x)
 

output

Integral(asech(a + b*x)**3/x**2, x)
 

3.1.18.7 Maxima [F]

\[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )^{3}}{x^{2}} \,d x } \]

input

integrate(arcsech(b*x+a)^3/x^2,x, algorithm="maxima")
 

output

-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 - integrate((8*(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 + 
 8*(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 - 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)^2 
 - 12*((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)*sqrt(-b*x - a + 1)* 
b*x + sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)*a + b*x + a))/(b^3*x^5 + 3*a*b^ 
2*x^4 + (3*a^2*b - b)*x^3 + (a^3 - a)*x^2 + (b^3*x^5 + 3*a*b^2*x^4 + (3*a^ 
2*b - b)*x^3 + (a^3 - a)*x^2)*sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)), x)
 

3.1.18.8 Giac [F]

\[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )^{3}}{x^{2}} \,d x } \]

input

integrate(arcsech(b*x+a)^3/x^2,x, algorithm="giac")
 

output

integrate(arcsech(b*x + a)^3/x^2, x)
 

3.1.18.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}^{-1}(a+b x)^3}{x^2} \, dx=\int \frac {{\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}^3}{x^2} \,d x \]

input

int(acosh(1/(a + b*x))^3/x^2,x)
 

output

int(acosh(1/(a + b*x))^3/x^2, x)