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

Optimal. Leaf size=118 \[ -\frac{1}{2} a b c^2 \text{sech}^{-1}(c x)+\frac{b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{2 x^2}-\frac{(1-c x) (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 x^2}-\frac{1}{4} b^2 c^2 \text{sech}^{-1}(c x)^2-\frac{b^2 (1-c x) (c x+1)}{4 x^2} \]

[Out]

-(b^2*(1 - c*x)*(1 + c*x))/(4*x^2) - (a*b*c^2*ArcSech[c*x])/2 - (b^2*c^2*ArcSech[c*x]^2)/4 + (b*Sqrt[(1 - c*x)
/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(2*x^2) - ((1 - c*x)*(1 + c*x)*(a + b*ArcSech[c*x])^2)/(2*x^2)

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Rubi [A]  time = 0.0847385, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {6285, 5446, 3310} \[ -\frac{1}{2} a b c^2 \text{sech}^{-1}(c x)+\frac{b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{2 x^2}-\frac{(1-c x) (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 x^2}-\frac{1}{4} b^2 c^2 \text{sech}^{-1}(c x)^2-\frac{b^2 (1-c x) (c x+1)}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])^2/x^3,x]

[Out]

-(b^2*(1 - c*x)*(1 + c*x))/(4*x^2) - (a*b*c^2*ArcSech[c*x])/2 - (b^2*c^2*ArcSech[c*x]^2)/4 + (b*Sqrt[(1 - c*x)
/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(2*x^2) - ((1 - c*x)*(1 + c*x)*(a + b*ArcSech[c*x])^2)/(2*x^2)

Rule 6285

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b
*x)^n*Sech[x]^(m + 1)*Tanh[x], x], x, ArcSech[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] &
& (GtQ[n, 0] || LtQ[m, -1])

Rule 5446

Int[Cosh[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[((c
+ d*x)^m*Sinh[a + b*x]^(n + 1))/(b*(n + 1)), x] - Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Sinh[a + b*x]^
(n + 1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

Rule 3310

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] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rubi steps

\begin{align*} \int \frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{x^3} \, dx &=-\left (c^2 \operatorname{Subst}\left (\int (a+b x)^2 \cosh (x) \sinh (x) \, dx,x,\text{sech}^{-1}(c x)\right )\right )\\ &=-\frac{(1-c x) (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 x^2}+\left (b c^2\right ) \operatorname{Subst}\left (\int (a+b x) \sinh ^2(x) \, dx,x,\text{sech}^{-1}(c x)\right )\\ &=-\frac{b^2 (1-c x) (1+c x)}{4 x^2}+\frac{b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{2 x^2}-\frac{(1-c x) (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 x^2}-\frac{1}{2} \left (b c^2\right ) \operatorname{Subst}\left (\int (a+b x) \, dx,x,\text{sech}^{-1}(c x)\right )\\ &=-\frac{b^2 (1-c x) (1+c x)}{4 x^2}-\frac{1}{2} a b c^2 \text{sech}^{-1}(c x)-\frac{1}{4} b^2 c^2 \text{sech}^{-1}(c x)^2+\frac{b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{2 x^2}-\frac{(1-c x) (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )^2}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.163408, size = 183, normalized size = 1.55 \[ \frac{-2 a^2-2 a b c^2 x^2 \log (x)+2 a b c^2 x^2 \log \left (c x \sqrt{\frac{1-c x}{c x+1}}+\sqrt{\frac{1-c x}{c x+1}}+1\right )+2 a b \sqrt{\frac{1-c x}{c x+1}}+2 a b c x \sqrt{\frac{1-c x}{c x+1}}+2 b \text{sech}^{-1}(c x) \left (b \sqrt{\frac{1-c x}{c x+1}} (c x+1)-2 a\right )+b^2 \left (c^2 x^2-2\right ) \text{sech}^{-1}(c x)^2-b^2}{4 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2 - b^2 + 2*a*b*Sqrt[(1 - c*x)/(1 + c*x)] + 2*a*b*c*x*Sqrt[(1 - c*x)/(1 + c*x)] + 2*b*(-2*a + b*Sqrt[(1
- c*x)/(1 + c*x)]*(1 + c*x))*ArcSech[c*x] + b^2*(-2 + c^2*x^2)*ArcSech[c*x]^2 - 2*a*b*c^2*x^2*Log[x] + 2*a*b*c
^2*x^2*Log[1 + Sqrt[(1 - c*x)/(1 + c*x)] + c*x*Sqrt[(1 - c*x)/(1 + c*x)]])/(4*x^2)

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Maple [A]  time = 0.236, size = 192, normalized size = 1.6 \begin{align*}{c}^{2} \left ( -{\frac{{a}^{2}}{2\,{c}^{2}{x}^{2}}}+{b}^{2} \left ( -{\frac{ \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}}{2\,{c}^{2}{x}^{2}}}+{\frac{{\rm arcsech} \left (cx\right )}{2\,cx}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}+{\frac{ \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}}{4}}-{\frac{1}{4\,{c}^{2}{x}^{2}}} \right ) +2\,ab \left ( -1/2\,{\frac{{\rm arcsech} \left (cx\right )}{{c}^{2}{x}^{2}}}+1/4\,{\frac{{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}} \right ){c}^{2}{x}^{2}+\sqrt{-{c}^{2}{x}^{2}+1}}{cx\sqrt{-{c}^{2}{x}^{2}+1}}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*(-1/2*a^2/c^2/x^2+b^2*(-1/2/c^2/x^2*arcsech(c*x)^2+1/2*arcsech(c*x)/c/x*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)
^(1/2)+1/4*arcsech(c*x)^2-1/4/c^2/x^2)+2*a*b*(-1/2/c^2/x^2*arcsech(c*x)+1/4*(-(c*x-1)/c/x)^(1/2)/c/x*((c*x+1)/
c/x)^(1/2)*(arctanh(1/(-c^2*x^2+1)^(1/2))*c^2*x^2+(-c^2*x^2+1)^(1/2))/(-c^2*x^2+1)^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{4} \, a b{\left (\frac{\frac{2 \, c^{4} x \sqrt{\frac{1}{c^{2} x^{2}} - 1}}{c^{2} x^{2}{\left (\frac{1}{c^{2} x^{2}} - 1\right )} - 1} - c^{3} \log \left (c x \sqrt{\frac{1}{c^{2} x^{2}} - 1} + 1\right ) + c^{3} \log \left (c x \sqrt{\frac{1}{c^{2} x^{2}} - 1} - 1\right )}{c} + \frac{4 \, \operatorname{arsech}\left (c x\right )}{x^{2}}\right )} + b^{2} \int \frac{\log \left (\sqrt{\frac{1}{c x} + 1} \sqrt{\frac{1}{c x} - 1} + \frac{1}{c x}\right )^{2}}{x^{3}}\,{d x} - \frac{a^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a*b*((2*c^4*x*sqrt(1/(c^2*x^2) - 1)/(c^2*x^2*(1/(c^2*x^2) - 1) - 1) - c^3*log(c*x*sqrt(1/(c^2*x^2) - 1) +
1) + c^3*log(c*x*sqrt(1/(c^2*x^2) - 1) - 1))/c + 4*arcsech(c*x)/x^2) + b^2*integrate(log(sqrt(1/(c*x) + 1)*sq
rt(1/(c*x) - 1) + 1/(c*x))^2/x^3, x) - 1/2*a^2/x^2

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Fricas [A]  time = 1.61809, size = 355, normalized size = 3.01 \begin{align*} \frac{2 \, a b c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} +{\left (b^{2} c^{2} x^{2} - 2 \, b^{2}\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} - 2 \, a^{2} - b^{2} + 2 \,{\left (a b c^{2} x^{2} + b^{2} c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 2 \, a b\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*b*c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + (b^2*c^2*x^2 - 2*b^2)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))
+ 1)/(c*x))^2 - 2*a^2 - b^2 + 2*(a*b*c^2*x^2 + b^2*c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 2*a*b)*log((c*x*sqrt(-
(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)))/x^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asech}{\left (c x \right )}\right )^{2}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsech}\left (c x\right ) + a\right )}^{2}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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