Integrand size = 12, antiderivative size = 224 \[ \int \frac {\text {sech}^{-1}(a+b x)^2}{x^2} \, dx=-\frac {b \text {sech}^{-1}(a+b x)^2}{a}-\frac {\text {sech}^{-1}(a+b x)^2}{x}+\frac {2 b \text {sech}^{-1}(a+b x) \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {2 b \text {sech}^{-1}(a+b x) \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {2 b \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {2 b \operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}} \] Output:
-b*arcsech(b*x+a)^2/a-arcsech(b*x+a)^2/x+2*b*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/(-a^2+1 )^(1/2)-2*b*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/(-a^2+1)^(1/2)+2*b*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)-2*b*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)
Result contains complex when optimal does not.
Time = 1.72 (sec) , antiderivative size = 678, normalized size of antiderivative = 3.03 \[ \int \frac {\text {sech}^{-1}(a+b x)^2}{x^2} \, dx =\text {Too large to display} \] Input:
Integrate[ArcSech[a + b*x]^2/x^2,x]
Output:
(-(((a + b*x)*ArcSech[a + b*x]^2)/x) + (2*b*(2*ArcSech[a + b*x]*ArcTan[((- 1 + a)*Coth[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]] - (2*I)*ArcCos[a^(-1)]*Ar cTan[((1 + a)*Tanh[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]] + (ArcCos[a^(-1)] + 2*(ArcTan[((-1 + a)*Coth[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]] + ArcTan[( (1 + a)*Tanh[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]]))*Log[Sqrt[-1 + a^2]/(Sq rt[2]*Sqrt[a]*E^(ArcSech[a + b*x]/2)*Sqrt[-((b*x)/(a + b*x))])] + (ArcCos[ a^(-1)] - 2*(ArcTan[((-1 + a)*Coth[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]] + ArcTan[((1 + a)*Tanh[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]]))*Log[(Sqrt[-1 + a^2]*E^(ArcSech[a + b*x]/2))/(Sqrt[2]*Sqrt[a]*Sqrt[-((b*x)/(a + b*x))])] - (ArcCos[a^(-1)] + 2*ArcTan[((1 + a)*Tanh[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]])*Log[-(((-1 + a)*(1 + a - I*Sqrt[-1 + a^2])*(-1 + Tanh[ArcSech[a + b *x]/2]))/(a*(-1 + a + I*Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/2])))] - (Arc Cos[a^(-1)] - 2*ArcTan[((1 + a)*Tanh[ArcSech[a + b*x]/2])/Sqrt[-1 + a^2]]) *Log[((-1 + a)*(1 + a + I*Sqrt[-1 + a^2])*(1 + Tanh[ArcSech[a + b*x]/2]))/ (a*(-1 + a + I*Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/2]))] + I*(PolyLog[2, ((-1 - I*Sqrt[-1 + a^2])*(-1 + a - I*Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/ 2]))/(a*(-1 + a + I*Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/2]))] - PolyLog[2 , ((I + Sqrt[-1 + a^2])*(-1 + a - I*Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/2 ]))/(a*((-I)*(-1 + a) + Sqrt[-1 + a^2]*Tanh[ArcSech[a + b*x]/2]))])))/Sqrt [-1 + a^2])/a
Time = 0.74 (sec) , antiderivative size = 227, 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.
\(\displaystyle \int \frac {\text {sech}^{-1}(a+b x)^2}{x^2} \, dx\) |
\(\Big \downarrow \) 6875 |
\(\displaystyle -b \int \frac {(a+b x) \sqrt {\frac {-a-b x+1}{a+b x+1}} (a+b x+1) \text {sech}^{-1}(a+b x)^2}{b^2 x^2}d\text {sech}^{-1}(a+b x)\) |
\(\Big \downarrow \) 5991 |
\(\displaystyle -b \left (2 \int -\frac {\text {sech}^{-1}(a+b x)}{b x}d\text {sech}^{-1}(a+b x)+\frac {\text {sech}^{-1}(a+b x)^2}{b x}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -b \left (\frac {\text {sech}^{-1}(a+b x)^2}{b x}+2 \int \frac {\text {sech}^{-1}(a+b x)}{a-\csc \left (i \text {sech}^{-1}(a+b x)+\frac {\pi }{2}\right )}d\text {sech}^{-1}(a+b x)\right )\) |
\(\Big \downarrow \) 4679 |
\(\displaystyle -b \left (2 \int \left (\frac {\text {sech}^{-1}(a+b x)}{a}+\frac {\text {sech}^{-1}(a+b x)}{a \left (\frac {a}{a+b x}-1\right )}\right )d\text {sech}^{-1}(a+b x)+\frac {\text {sech}^{-1}(a+b x)^2}{b x}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -b \left (2 \left (-\frac {\operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {\operatorname {PolyLog}\left (2,\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}-\frac {\text {sech}^{-1}(a+b x) \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {\text {sech}^{-1}(a+b x) \log \left (1-\frac {a e^{\text {sech}^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}+\frac {\text {sech}^{-1}(a+b x)^2}{2 a}\right )+\frac {\text {sech}^{-1}(a+b x)^2}{b x}\right )\) |
Input:
Int[ArcSech[a + b*x]^2/x^2,x]
Output:
-(b*(ArcSech[a + b*x]^2/(b*x) + 2*(ArcSech[a + b*x]^2/(2*a) - (ArcSech[a + b*x]*Log[1 - (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2 ]) + (ArcSech[a + b*x]*Log[1 - (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])] )/(a*Sqrt[1 - a^2]) - PolyLog[2, (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2] )]/(a*Sqrt[1 - a^2]) + PolyLog[2, (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2 ])]/(a*Sqrt[1 - a^2]))))
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]
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]
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]
Time = 0.49 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.62
method | result | size |
derivativedivides | \(b \left (-\frac {\left (b x +a \right ) \operatorname {arcsech}\left (b x +a \right )^{2}}{a b x}+\frac {2 \sqrt {-a^{2}+1}\, \operatorname {arcsech}\left (b x +a \right ) \ln \left (\frac {-a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}-\frac {2 \sqrt {-a^{2}+1}\, \operatorname {arcsech}\left (b x +a \right ) \ln \left (\frac {a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}+\frac {2 \sqrt {-a^{2}+1}\, \operatorname {dilog}\left (\frac {-a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}-\frac {2 \sqrt {-a^{2}+1}\, \operatorname {dilog}\left (\frac {a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}\right )\) | \(362\) |
default | \(b \left (-\frac {\left (b x +a \right ) \operatorname {arcsech}\left (b x +a \right )^{2}}{a b x}+\frac {2 \sqrt {-a^{2}+1}\, \operatorname {arcsech}\left (b x +a \right ) \ln \left (\frac {-a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}-\frac {2 \sqrt {-a^{2}+1}\, \operatorname {arcsech}\left (b x +a \right ) \ln \left (\frac {a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}+\frac {2 \sqrt {-a^{2}+1}\, \operatorname {dilog}\left (\frac {-a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}+1}{1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}-\frac {2 \sqrt {-a^{2}+1}\, \operatorname {dilog}\left (\frac {a \left (\frac {1}{b x +a}+\sqrt {\frac {1}{b x +a}-1}\, \sqrt {\frac {1}{b x +a}+1}\right )+\sqrt {-a^{2}+1}-1}{-1+\sqrt {-a^{2}+1}}\right )}{a \left (a^{2}-1\right )}\right )\) | \(362\) |
Input:
int(arcsech(b*x+a)^2/x^2,x,method=_RETURNVERBOSE)
Output:
b*(-(b*x+a)*arcsech(b*x+a)^2/a/b/x+2*(-a^2+1)^(1/2)/a/(a^2-1)*arcsech(b*x+ a)*ln((-a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/ 2)+1)/(1+(-a^2+1)^(1/2)))-2*(-a^2+1)^(1/2)/a/(a^2-1)*arcsech(b*x+a)*ln((a* (1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)-1)/(-1+ (-a^2+1)^(1/2)))+2*(-a^2+1)^(1/2)/a/(a^2-1)*dilog((-a*(1/(b*x+a)+(1/(b*x+a )-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)+1)/(1+(-a^2+1)^(1/2)))-2*(- a^2+1)^(1/2)/a/(a^2-1)*dilog((a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+ 1)^(1/2))+(-a^2+1)^(1/2)-1)/(-1+(-a^2+1)^(1/2))))
\[ \int \frac {\text {sech}^{-1}(a+b x)^2}{x^2} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )^{2}}{x^{2}} \,d x } \] Input:
integrate(arcsech(b*x+a)^2/x^2,x, algorithm="fricas")
Output:
integral(arcsech(b*x + a)^2/x^2, x)
\[ \int \frac {\text {sech}^{-1}(a+b x)^2}{x^2} \, dx=\int \frac {\operatorname {asech}^{2}{\left (a + b x \right )}}{x^{2}}\, dx \] Input:
integrate(asech(b*x+a)**2/x**2,x)
Output:
Integral(asech(a + b*x)**2/x**2, x)
\[ \int \frac {\text {sech}^{-1}(a+b x)^2}{x^2} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )^{2}}{x^{2}} \,d x } \] Input:
integrate(arcsech(b*x+a)^2/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)^2/x - 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)*lo g(b*x + a))*sqrt(b*x + a + 1))*sqrt(-b*x - a + 1))*log(sqrt(b*x + a + 1)*s qrt(-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)
\[ \int \frac {\text {sech}^{-1}(a+b x)^2}{x^2} \, dx=\int { \frac {\operatorname {arsech}\left (b x + a\right )^{2}}{x^{2}} \,d x } \] Input:
integrate(arcsech(b*x+a)^2/x^2,x, algorithm="giac")
Output:
integrate(arcsech(b*x + a)^2/x^2, x)
Timed out. \[ \int \frac {\text {sech}^{-1}(a+b x)^2}{x^2} \, dx=\int \frac {{\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}^2}{x^2} \,d x \] Input:
int(acosh(1/(a + b*x))^2/x^2,x)
Output:
int(acosh(1/(a + b*x))^2/x^2, x)
\[ \int \frac {\text {sech}^{-1}(a+b x)^2}{x^2} \, dx=\int \frac {\mathit {asech} \left (b x +a \right )^{2}}{x^{2}}d x \] Input:
int(asech(b*x+a)^2/x^2,x)
Output:
int(asech(a + b*x)**2/x**2,x)