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]

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

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Rubi [A]  time = 0.03, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.07, size = 74, normalized size = 0.96 \[ -\frac {a}{3 x^3}+b \left (\frac {2 c^3}{9}+\frac {2 c^2}{9 x}+\frac {c}{9 x^2}+\frac {1}{9 x^3}\right ) \sqrt {\frac {1-c x}{c x+1}}-\frac {b \text {sech}^{-1}(c x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSech[c*x])/x^4,x]

[Out]

-1/3*a/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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fricas [A]  time = 0.58, size = 79, normalized size = 1.03 \[ -\frac {3 \, b \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (2 \, b c^{3} x^{3} + b c x\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 3 \, a}{9 \, x^{3}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsech}\left (c x\right ) + a}{x^{4}}\,{d x} \]

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)

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maple [A]  time = 0.06, size = 77, normalized size = 1.00 \[ c^{3} \left (-\frac {a}{3 c^{3} x^{3}}+b \left (-\frac {\mathrm {arcsech}\left (c x \right )}{3 c^{3} x^{3}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (2 c^{2} x^{2}+1\right )}{9 c^{2} x^{2}}\right )\right ) \]

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/c^3/x^3*a+b*(-1/3*arcsech(c*x)/c^3/x^3+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 = 0.30, size = 56, normalized size = 0.73 \[ \frac {1}{9} \, b {\left (\frac {c^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} + 3 \, c^{4} \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c} - \frac {3 \, \operatorname {arsech}\left (c x\right )}{x^{3}}\right )} - \frac {a}{3 \, x^{3}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))/x^4,x)

[Out]

int((a + b*acosh(1/(c*x)))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asech}{\left (c x \right )}}{x^{4}}\, dx \]

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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