3.41 \(\int \frac{(a+b \text{sech}^{-1}(c x))^2}{x^5} \, dx\)

Optimal. Leaf size=151 \[ \frac{3 b c^2 \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{16 x^2}+\frac{3}{16} a b c^4 \text{sech}^{-1}(c x)-\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{4 x^4}+\frac{b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{8 x^4}-\frac{3 b^2 c^2}{32 x^2}+\frac{3}{32} b^2 c^4 \text{sech}^{-1}(c x)^2-\frac{b^2}{32 x^4} \]

[Out]

-b^2/(32*x^4) - (3*b^2*c^2)/(32*x^2) + (3*a*b*c^4*ArcSech[c*x])/16 + (3*b^2*c^4*ArcSech[c*x]^2)/32 + (b*Sqrt[(
1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(8*x^4) + (3*b*c^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a
+ b*ArcSech[c*x]))/(16*x^2) - (a + b*ArcSech[c*x])^2/(4*x^4)

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Rubi [A]  time = 0.119685, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {6285, 5447, 3310} \[ \frac{3 b c^2 \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{16 x^2}+\frac{3}{16} a b c^4 \text{sech}^{-1}(c x)-\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{4 x^4}+\frac{b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{8 x^4}-\frac{3 b^2 c^2}{32 x^2}+\frac{3}{32} b^2 c^4 \text{sech}^{-1}(c x)^2-\frac{b^2}{32 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])^2/x^5,x]

[Out]

-b^2/(32*x^4) - (3*b^2*c^2)/(32*x^2) + (3*a*b*c^4*ArcSech[c*x])/16 + (3*b^2*c^4*ArcSech[c*x]^2)/32 + (b*Sqrt[(
1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(8*x^4) + (3*b*c^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a
+ b*ArcSech[c*x]))/(16*x^2) - (a + b*ArcSech[c*x])^2/(4*x^4)

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 5447

Int[Cosh[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[((c
+ d*x)^m*Cosh[a + b*x]^(n + 1))/(b*(n + 1)), x] - Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Cosh[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^5} \, dx &=-\left (c^4 \operatorname{Subst}\left (\int (a+b x)^2 \cosh ^3(x) \sinh (x) \, dx,x,\text{sech}^{-1}(c x)\right )\right )\\ &=-\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{2} \left (b c^4\right ) \operatorname{Subst}\left (\int (a+b x) \cosh ^4(x) \, dx,x,\text{sech}^{-1}(c x)\right )\\ &=-\frac{b^2}{32 x^4}+\frac{b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{8 x^4}-\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{8} \left (3 b c^4\right ) \operatorname{Subst}\left (\int (a+b x) \cosh ^2(x) \, dx,x,\text{sech}^{-1}(c x)\right )\\ &=-\frac{b^2}{32 x^4}-\frac{3 b^2 c^2}{32 x^2}+\frac{b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{8 x^4}+\frac{3 b c^2 \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{16 x^2}-\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{16} \left (3 b c^4\right ) \operatorname{Subst}\left (\int (a+b x) \, dx,x,\text{sech}^{-1}(c x)\right )\\ &=-\frac{b^2}{32 x^4}-\frac{3 b^2 c^2}{32 x^2}+\frac{3}{16} a b c^4 \text{sech}^{-1}(c x)+\frac{3}{32} b^2 c^4 \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 )}{8 x^4}+\frac{3 b c^2 \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{16 x^2}-\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{4 x^4}\\ \end{align*}

Mathematica [A]  time = 0.259171, size = 268, normalized size = 1.77 \[ \frac{-8 a^2+6 a b c^3 x^3 \sqrt{\frac{1-c x}{c x+1}}+6 a b c^2 x^2 \sqrt{\frac{1-c x}{c x+1}}-6 a b c^4 x^4 \log (x)+6 a b c^4 x^4 \log \left (c x \sqrt{\frac{1-c x}{c x+1}}+\sqrt{\frac{1-c x}{c x+1}}+1\right )+2 b \text{sech}^{-1}(c x) \left (b \sqrt{\frac{1-c x}{c x+1}} \left (3 c^3 x^3+3 c^2 x^2+2 c x+2\right )-8 a\right )+4 a b c x \sqrt{\frac{1-c x}{c x+1}}+4 a b \sqrt{\frac{1-c x}{c x+1}}-3 b^2 c^2 x^2+b^2 \left (3 c^4 x^4-8\right ) \text{sech}^{-1}(c x)^2-b^2}{32 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSech[c*x])^2/x^5,x]

[Out]

(-8*a^2 - b^2 - 3*b^2*c^2*x^2 + 4*a*b*Sqrt[(1 - c*x)/(1 + c*x)] + 4*a*b*c*x*Sqrt[(1 - c*x)/(1 + c*x)] + 6*a*b*
c^2*x^2*Sqrt[(1 - c*x)/(1 + c*x)] + 6*a*b*c^3*x^3*Sqrt[(1 - c*x)/(1 + c*x)] + 2*b*(-8*a + b*Sqrt[(1 - c*x)/(1
+ c*x)]*(2 + 2*c*x + 3*c^2*x^2 + 3*c^3*x^3))*ArcSech[c*x] + b^2*(-8 + 3*c^4*x^4)*ArcSech[c*x]^2 - 6*a*b*c^4*x^
4*Log[x] + 6*a*b*c^4*x^4*Log[1 + Sqrt[(1 - c*x)/(1 + c*x)] + c*x*Sqrt[(1 - c*x)/(1 + c*x)]])/(32*x^4)

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Maple [B]  time = 0.233, size = 298, normalized size = 2. \begin{align*}{c}^{4} \left ( -{\frac{{a}^{2}}{4\,{c}^{4}{x}^{4}}}+{b}^{2} \left ({\frac{ \left ({\rm arcsech} \left (cx\right ) \right ) ^{2} \left ( cx-1 \right ) \left ( cx+1 \right ) }{4\,{c}^{4}{x}^{4}}}-{\frac{ \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}}{4\,{c}^{2}{x}^{2}}}+{\frac{{\rm arcsech} \left (cx\right )}{8\,{c}^{3}{x}^{3}}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}+{\frac{3\,{\rm arcsech} \left (cx\right )}{16\,cx}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}+{\frac{3\, \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}}{32}}+{\frac{ \left ( cx+1 \right ) \left ( cx-1 \right ) }{32\,{c}^{4}{x}^{4}}}-{\frac{1}{8\,{c}^{2}{x}^{2}}} \right ) +2\,ab \left ( -1/4\,{\frac{{\rm arcsech} \left (cx\right )}{{c}^{4}{x}^{4}}}+1/32\,{\frac{3\,{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}} \right ){c}^{4}{x}^{4}+3\,\sqrt{-{c}^{2}{x}^{2}+1}{c}^{2}{x}^{2}+2\,\sqrt{-{c}^{2}{x}^{2}+1}}{{c}^{3}{x}^{3}\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^5,x)

[Out]

c^4*(-1/4*a^2/c^4/x^4+b^2*(1/4*arcsech(c*x)^2/c^4/x^4*(c*x-1)*(c*x+1)-1/4/c^2/x^2*arcsech(c*x)^2+1/8*arcsech(c
*x)/c^3/x^3*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)+3/16*arcsech(c*x)/c/x*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^
(1/2)+3/32*arcsech(c*x)^2+1/32*(c*x-1)/c^4/x^4*(c*x+1)-1/8/c^2/x^2)+2*a*b*(-1/4/c^4/x^4*arcsech(c*x)+1/32*(-(c
*x-1)/c/x)^(1/2)/c^3/x^3*((c*x+1)/c/x)^(1/2)*(3*arctanh(1/(-c^2*x^2+1)^(1/2))*c^4*x^4+3*(-c^2*x^2+1)^(1/2)*c^2
*x^2+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}{32} \, a b{\left (\frac{3 \, c^{5} \log \left (c x \sqrt{\frac{1}{c^{2} x^{2}} - 1} + 1\right ) - 3 \, c^{5} \log \left (c x \sqrt{\frac{1}{c^{2} x^{2}} - 1} - 1\right ) - \frac{2 \,{\left (3 \, c^{8} x^{3}{\left (\frac{1}{c^{2} x^{2}} - 1\right )}^{\frac{3}{2}} - 5 \, c^{6} x \sqrt{\frac{1}{c^{2} x^{2}} - 1}\right )}}{c^{4} x^{4}{\left (\frac{1}{c^{2} x^{2}} - 1\right )}^{2} - 2 \, c^{2} x^{2}{\left (\frac{1}{c^{2} x^{2}} - 1\right )} + 1}}{c} - \frac{16 \, \operatorname{arsech}\left (c x\right )}{x^{4}}\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^{5}}\,{d x} - \frac{a^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^2/x^5,x, algorithm="maxima")

[Out]

1/32*a*b*((3*c^5*log(c*x*sqrt(1/(c^2*x^2) - 1) + 1) - 3*c^5*log(c*x*sqrt(1/(c^2*x^2) - 1) - 1) - 2*(3*c^8*x^3*
(1/(c^2*x^2) - 1)^(3/2) - 5*c^6*x*sqrt(1/(c^2*x^2) - 1))/(c^4*x^4*(1/(c^2*x^2) - 1)^2 - 2*c^2*x^2*(1/(c^2*x^2)
- 1) + 1))/c - 16*arcsech(c*x)/x^4) + b^2*integrate(log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))^2/x^5,
x) - 1/4*a^2/x^4

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Fricas [A]  time = 1.66952, size = 439, normalized size = 2.91 \begin{align*} -\frac{3 \, b^{2} c^{2} x^{2} -{\left (3 \, b^{2} c^{4} x^{4} - 8 \, b^{2}\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} + 8 \, a^{2} + b^{2} - 2 \,{\left (3 \, a b c^{4} x^{4} - 8 \, a b +{\left (3 \, b^{2} c^{3} x^{3} + 2 \, b^{2} c x\right )} \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}}\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - 2 \,{\left (3 \, a b c^{3} x^{3} + 2 \, a b c x\right )} \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}}}{32 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^2/x^5,x, algorithm="fricas")

[Out]

-1/32*(3*b^2*c^2*x^2 - (3*b^2*c^4*x^4 - 8*b^2)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x))^2 + 8*a^2 +
b^2 - 2*(3*a*b*c^4*x^4 - 8*a*b + (3*b^2*c^3*x^3 + 2*b^2*c*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*log((c*x*sqrt(-(
c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - 2*(3*a*b*c^3*x^3 + 2*a*b*c*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/x^4

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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^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asech(c*x))**2/x**5,x)

[Out]

Integral((a + b*asech(c*x))**2/x**5, 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^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^2/x^5,x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)^2/x^5, x)