Integrand size = 10, antiderivative size = 90 \[ \int \frac {\text {sech}^{-1}(a x)^2}{x^3} \, dx=-\frac {(1-a x) (1+a x)}{4 x^2}+\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{2 x^2}-\frac {1}{4} a^2 \text {sech}^{-1}(a x)^2-\frac {(1-a x) (1+a x) \text {sech}^{-1}(a x)^2}{2 x^2} \]
-1/4*(-a*x+1)*(a*x+1)/x^2-1/4*a^2*arcsech(a*x)^2-1/2*(-a*x+1)*(a*x+1)*arcs ech(a*x)^2/x^2+1/2*(a*x+1)*arcsech(a*x)*((-a*x+1)/(a*x+1))^(1/2)/x^2
Time = 0.06 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.60 \[ \int \frac {\text {sech}^{-1}(a x)^2}{x^3} \, dx=\frac {-1+2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)+\left (-2+a^2 x^2\right ) \text {sech}^{-1}(a x)^2}{4 x^2} \]
(-1 + 2*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x] + (-2 + a^2*x^2)* ArcSech[a*x]^2)/(4*x^2)
Time = 0.31 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6839, 5895, 3042, 25, 3791, 15}
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.
\(\displaystyle \int \frac {\text {sech}^{-1}(a x)^2}{x^3} \, dx\) |
\(\Big \downarrow \) 6839 |
\(\displaystyle -a^2 \int \frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)^2}{a^2 x^2}d\text {sech}^{-1}(a x)\) |
\(\Big \downarrow \) 5895 |
\(\displaystyle -a^2 \left (\frac {(1-a x) (a x+1) \text {sech}^{-1}(a x)^2}{2 a^2 x^2}-\int \frac {(1-a x) (a x+1) \text {sech}^{-1}(a x)}{a^2 x^2}d\text {sech}^{-1}(a x)\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a^2 \left (\frac {(1-a x) (a x+1) \text {sech}^{-1}(a x)^2}{2 a^2 x^2}-\int -\text {sech}^{-1}(a x) \sin \left (i \text {sech}^{-1}(a x)\right )^2d\text {sech}^{-1}(a x)\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -a^2 \left (\frac {(1-a x) (a x+1) \text {sech}^{-1}(a x)^2}{2 a^2 x^2}+\int \text {sech}^{-1}(a x) \sin \left (i \text {sech}^{-1}(a x)\right )^2d\text {sech}^{-1}(a x)\right )\) |
\(\Big \downarrow \) 3791 |
\(\displaystyle -a^2 \left (\frac {1}{2} \int \text {sech}^{-1}(a x)d\text {sech}^{-1}(a x)+\frac {(1-a x) (a x+1)}{4 a^2 x^2}+\frac {(1-a x) (a x+1) \text {sech}^{-1}(a x)^2}{2 a^2 x^2}-\frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)}{2 a^2 x^2}\right )\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -a^2 \left (\frac {(1-a x) (a x+1)}{4 a^2 x^2}+\frac {(1-a x) (a x+1) \text {sech}^{-1}(a x)^2}{2 a^2 x^2}-\frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)}{2 a^2 x^2}+\frac {1}{4} \text {sech}^{-1}(a x)^2\right )\) |
-(a^2*(((1 - a*x)*(1 + a*x))/(4*a^2*x^2) - (Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x])/(2*a^2*x^2) + ArcSech[a*x]^2/4 + ((1 - a*x)*(1 + a*x)* ArcSech[a*x]^2)/(2*a^2*x^2)))
3.1.8.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Simp[b*(c + d*x)*Cos[e + f*x ]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^2*((n - 1)/n) Int[(c + d* x)*(b*Sin[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]
Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.) ]^(p_.), x_Symbol] :> Simp[x^(m - n + 1)*(Sinh[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] - Simp[(m - n + 1)/(b*n*(p + 1)) Int[x^(m - n)*Sinh[a + b*x^n]^ (p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]
Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[ -(c^(m + 1))^(-1) Subst[Int[(a + b*x)^n*Sech[x]^(m + 1)*Tanh[x], x], x, A rcSech[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (G tQ[n, 0] || LtQ[m, -1])
Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(a^{2} \left (-\frac {\operatorname {arcsech}\left (a x \right )^{2}}{2 a^{2} x^{2}}+\frac {\sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, \operatorname {arcsech}\left (a x \right )}{2 a x}+\frac {\operatorname {arcsech}\left (a x \right )^{2}}{4}-\frac {1}{4 a^{2} x^{2}}\right )\) | \(77\) |
default | \(a^{2} \left (-\frac {\operatorname {arcsech}\left (a x \right )^{2}}{2 a^{2} x^{2}}+\frac {\sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, \operatorname {arcsech}\left (a x \right )}{2 a x}+\frac {\operatorname {arcsech}\left (a x \right )^{2}}{4}-\frac {1}{4 a^{2} x^{2}}\right )\) | \(77\) |
a^2*(-1/2/a^2/x^2*arcsech(a*x)^2+1/2*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1 /2)/a/x*arcsech(a*x)+1/4*arcsech(a*x)^2-1/4/a^2/x^2)
Time = 0.25 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.18 \[ \int \frac {\text {sech}^{-1}(a x)^2}{x^3} \, dx=\frac {2 \, a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} \log \left (\frac {a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right ) + {\left (a^{2} x^{2} - 2\right )} \log \left (\frac {a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )^{2} - 1}{4 \, x^{2}} \]
1/4*(2*a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2))*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^ 2*x^2)) + 1)/(a*x)) + (a^2*x^2 - 2)*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2) ) + 1)/(a*x))^2 - 1)/x^2
\[ \int \frac {\text {sech}^{-1}(a x)^2}{x^3} \, dx=\int \frac {\operatorname {asech}^{2}{\left (a x \right )}}{x^{3}}\, dx \]
\[ \int \frac {\text {sech}^{-1}(a x)^2}{x^3} \, dx=\int { \frac {\operatorname {arsech}\left (a x\right )^{2}}{x^{3}} \,d x } \]
\[ \int \frac {\text {sech}^{-1}(a x)^2}{x^3} \, dx=\int { \frac {\operatorname {arsech}\left (a x\right )^{2}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\text {sech}^{-1}(a x)^2}{x^3} \, dx=\int \frac {{\mathrm {acosh}\left (\frac {1}{a\,x}\right )}^2}{x^3} \,d x \]