3.40 \(\int \frac{(a+b \text{sech}^{-1}(c x))^2}{x^4} \, dx\)

Optimal. Leaf size=122 \[ \frac{4 b c^2 \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{9 x}+\frac{2 b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{9 x^3}-\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{3 x^3}-\frac{4 b^2 c^2}{9 x}-\frac{2 b^2}{27 x^3} \]

[Out]

(-2*b^2)/(27*x^3) - (4*b^2*c^2)/(9*x) + (2*b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(9*x^3)
+ (4*b*c^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(9*x) - (a + b*ArcSech[c*x])^2/(3*x^3)

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Rubi [A]  time = 0.101873, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {6285, 5447, 3310, 3296, 2638} \[ \frac{4 b c^2 \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{9 x}+\frac{2 b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (a+b \text{sech}^{-1}(c x)\right )}{9 x^3}-\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{3 x^3}-\frac{4 b^2 c^2}{9 x}-\frac{2 b^2}{27 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])^2/x^4,x]

[Out]

(-2*b^2)/(27*x^3) - (4*b^2*c^2)/(9*x) + (2*b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(9*x^3)
+ (4*b*c^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(9*x) - (a + b*ArcSech[c*x])^2/(3*x^3)

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]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{x^4} \, dx &=-\left (c^3 \operatorname{Subst}\left (\int (a+b x)^2 \cosh ^2(x) \sinh (x) \, dx,x,\text{sech}^{-1}(c x)\right )\right )\\ &=-\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{3 x^3}+\frac{1}{3} \left (2 b c^3\right ) \operatorname{Subst}\left (\int (a+b x) \cosh ^3(x) \, dx,x,\text{sech}^{-1}(c x)\right )\\ &=-\frac{2 b^2}{27 x^3}+\frac{2 b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{9 x^3}-\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{3 x^3}+\frac{1}{9} \left (4 b c^3\right ) \operatorname{Subst}\left (\int (a+b x) \cosh (x) \, dx,x,\text{sech}^{-1}(c x)\right )\\ &=-\frac{2 b^2}{27 x^3}+\frac{2 b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{9 x^3}+\frac{4 b c^2 \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{9 x}-\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{3 x^3}-\frac{1}{9} \left (4 b^2 c^3\right ) \operatorname{Subst}\left (\int \sinh (x) \, dx,x,\text{sech}^{-1}(c x)\right )\\ &=-\frac{2 b^2}{27 x^3}-\frac{4 b^2 c^2}{9 x}+\frac{2 b \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{9 x^3}+\frac{4 b c^2 \sqrt{\frac{1-c x}{1+c x}} (1+c x) \left (a+b \text{sech}^{-1}(c x)\right )}{9 x}-\frac{\left (a+b \text{sech}^{-1}(c x)\right )^2}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.249804, size = 134, normalized size = 1.1 \[ \frac{-9 a^2+6 a b \sqrt{\frac{1-c x}{c x+1}} \left (2 c^3 x^3+2 c^2 x^2+c x+1\right )+6 b \text{sech}^{-1}(c x) \left (b \sqrt{\frac{1-c x}{c x+1}} \left (2 c^3 x^3+2 c^2 x^2+c x+1\right )-3 a\right )-2 b^2 \left (6 c^2 x^2+1\right )-9 b^2 \text{sech}^{-1}(c x)^2}{27 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSech[c*x])^2/x^4,x]

[Out]

(-9*a^2 - 2*b^2*(1 + 6*c^2*x^2) + 6*a*b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x + 2*c^2*x^2 + 2*c^3*x^3) + 6*b*(-3*
a + b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x + 2*c^2*x^2 + 2*c^3*x^3))*ArcSech[c*x] - 9*b^2*ArcSech[c*x]^2)/(27*x^
3)

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Maple [B]  time = 0.225, size = 226, normalized size = 1.9 \begin{align*}{c}^{3} \left ( -{\frac{{a}^{2}}{3\,{c}^{3}{x}^{3}}}+{b}^{2} \left ({\frac{ \left ({\rm arcsech} \left (cx\right ) \right ) ^{2} \left ( cx-1 \right ) \left ( cx+1 \right ) }{3\,{c}^{3}{x}^{3}}}-{\frac{ \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}}{3\,cx}}+{\frac{2\,{\rm arcsech} \left (cx\right )}{9\,{c}^{2}{x}^{2}}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}+{\frac{4\,{\rm arcsech} \left (cx\right )}{9}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}+{\frac{ \left ( 2\,cx-2 \right ) \left ( cx+1 \right ) }{27\,{c}^{3}{x}^{3}}}-{\frac{14}{27\,cx}} \right ) +2\,ab \left ( -1/3\,{\frac{{\rm arcsech} \left (cx\right )}{{c}^{3}{x}^{3}}}+1/9\,{\frac{2\,{c}^{2}{x}^{2}+1}{{c}^{2}{x}^{2}}\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^4,x)

[Out]

c^3*(-1/3*a^2/c^3/x^3+b^2*(1/3*arcsech(c*x)^2/c^3/x^3*(c*x-1)*(c*x+1)-1/3*arcsech(c*x)^2/c/x+2/9*arcsech(c*x)/
c^2/x^2*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)+4/9*arcsech(c*x)*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)+2/2
7*(c*x-1)/c^3/x^3*(c*x+1)-14/27/c/x)+2*a*b*(-1/3/c^3/x^3*arcsech(c*x)+1/9*(-(c*x-1)/c/x)^(1/2)/c^2/x^2*((c*x+1
)/c/x)^(1/2)*(2*c^2*x^2+1)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2}{9} \, a b{\left (\frac{c^{4}{\left (\frac{1}{c^{2} x^{2}} - 1\right )}^{\frac{3}{2}} + 3 \, c^{4} \sqrt{\frac{1}{c^{2} x^{2}} - 1}}{c} - \frac{3 \, \operatorname{arsech}\left (c x\right )}{x^{3}}\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^{4}}\,{d x} - \frac{a^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^2/x^4,x, algorithm="maxima")

[Out]

2/9*a*b*((c^4*(1/(c^2*x^2) - 1)^(3/2) + 3*c^4*sqrt(1/(c^2*x^2) - 1))/c - 3*arcsech(c*x)/x^3) + b^2*integrate(l
og(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))^2/x^4, x) - 1/3*a^2/x^3

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Fricas [A]  time = 1.65368, size = 392, normalized size = 3.21 \begin{align*} -\frac{12 \, b^{2} c^{2} x^{2} + 9 \, b^{2} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} + 9 \, a^{2} + 2 \, b^{2} + 6 \,{\left (3 \, a b -{\left (2 \, b^{2} c^{3} x^{3} + 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 ) - 6 \,{\left (2 \, a b c^{3} x^{3} + a b c x\right )} \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}}}{27 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^2/x^4,x, algorithm="fricas")

[Out]

-1/27*(12*b^2*c^2*x^2 + 9*b^2*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x))^2 + 9*a^2 + 2*b^2 + 6*(3*a*b
- (2*b^2*c^3*x^3 + 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
)) - 6*(2*a*b*c^3*x^3 + a*b*c*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/x^3

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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^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asech(c*x))**2/x**4,x)

[Out]

Integral((a + b*asech(c*x))**2/x**4, 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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^2/x^4,x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)^2/x^4, x)