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

Optimal. Leaf size=77 \[ -\frac{a+b \text{sech}^{-1}(c x)}{3 x^3}+\frac{2 b c^2 \sqrt{1-c x}}{9 x \sqrt{\frac{1}{c x+1}}}+\frac{b \sqrt{1-c x}}{9 x^3 \sqrt{\frac{1}{c x+1}}} \]

[Out]

(b*Sqrt[1 - c*x])/(9*x^3*Sqrt[(1 + c*x)^(-1)]) + (2*b*c^2*Sqrt[1 - c*x])/(9*x*Sqrt[(1 + c*x)^(-1)]) - (a + b*A
rcSech[c*x])/(3*x^3)

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Rubi [A]  time = 0.0332651, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6283, 103, 12, 95} \[ -\frac{a+b \text{sech}^{-1}(c x)}{3 x^3}+\frac{2 b c^2 \sqrt{1-c x}}{9 x \sqrt{\frac{1}{c x+1}}}+\frac{b \sqrt{1-c x}}{9 x^3 \sqrt{\frac{1}{c x+1}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])/x^4,x]

[Out]

(b*Sqrt[1 - c*x])/(9*x^3*Sqrt[(1 + c*x)^(-1)]) + (2*b*c^2*Sqrt[1 - c*x])/(9*x*Sqrt[(1 + c*x)^(-1)]) - (a + b*A
rcSech[c*x])/(3*x^3)

Rule 6283

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

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 95

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] /; FreeQ[{a, b, c, d,
e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && EqQ[a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1), 0
] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0628587, size = 74, normalized size = 0.96 \[ -\frac{a}{3 x^3}+b \left (\frac{2 c^2}{9 x}+\frac{2 c^3}{9}+\frac{c}{9 x^2}+\frac{1}{9 x^3}\right ) \sqrt{\frac{1-c x}{c x+1}}-\frac{b \text{sech}^{-1}(c x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSech[c*x])/x^4,x]

[Out]

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

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Maple [A]  time = 0.181, size = 77, normalized size = 1. \begin{align*}{c}^{3} \left ( -{\frac{a}{3\,{c}^{3}{x}^{3}}}+b \left ( -{\frac{{\rm arcsech} \left (cx\right )}{3\,{c}^{3}{x}^{3}}}+{\frac{2\,{c}^{2}{x}^{2}+1}{9\,{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))/x^4,x)

[Out]

c^3*(-1/3*a/c^3/x^3+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 [A]  time = 0.979279, size = 76, normalized size = 0.99 \begin{align*} \frac{1}{9} \, 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 )} - \frac{a}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*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) - 1/3*a/x^3

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Fricas [A]  time = 1.89021, size = 174, normalized size = 2.26 \begin{align*} -\frac{3 \, b \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) -{\left (2 \, b c^{3} x^{3} + b c x\right )} \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 3 \, a}{9 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{asech}{\left (c x \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asech(c*x))/x**4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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