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

Optimal. Leaf size=94 \[ -\frac {a+b \text {sech}^{-1}(c x)}{2 x^2}+\frac {1}{4} b c^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\sqrt {1-c x} \sqrt {c x+1}\right )+\frac {b \sqrt {1-c x}}{4 x^2 \sqrt {\frac {1}{c x+1}}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6283, 103, 12, 92, 208} \[ -\frac {a+b \text {sech}^{-1}(c x)}{2 x^2}+\frac {1}{4} b c^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\sqrt {1-c x} \sqrt {c x+1}\right )+\frac {b \sqrt {1-c x}}{4 x^2 \sqrt {\frac {1}{c x+1}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])/x^3,x]

[Out]

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

Rule 12

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

Rule 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

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

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Mathematica [A]  time = 0.08, size = 117, normalized size = 1.24 \[ -\frac {a}{2 x^2}-\frac {1}{4} b c^2 \log (x)+\frac {1}{4} b c^2 \log \left (c x \sqrt {\frac {1-c x}{c x+1}}+\sqrt {\frac {1-c x}{c x+1}}+1\right )+b \left (\frac {c}{4 x}+\frac {1}{4 x^2}\right ) \sqrt {\frac {1-c x}{c x+1}}-\frac {b \text {sech}^{-1}(c x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSech[c*x])/x^3,x]

[Out]

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

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fricas [A]  time = 0.43, size = 77, normalized size = 0.82 \[ \frac {b c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + {\left (b c^{2} x^{2} - 2 \, b\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - 2 \, a}{4 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsech}\left (c x\right ) + a}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 112, normalized size = 1.19 \[ c^{2} \left (-\frac {a}{2 c^{2} x^{2}}+b \left (-\frac {\mathrm {arcsech}\left (c x \right )}{2 c^{2} x^{2}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (\arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right ) c^{2} x^{2}+\sqrt {-c^{2} x^{2}+1}\right )}{4 c x \sqrt {-c^{2} x^{2}+1}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*(-1/2*a/c^2/x^2+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 [A]  time = 0.31, size = 105, normalized size = 1.12 \[ -\frac {1}{8} \, b {\left (\frac {\frac {2 \, c^{4} x \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{2} x^{2} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} - 1} - c^{3} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} - 1} + 1\right ) + c^{3} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} - 1} - 1\right )}{c} + \frac {4 \, \operatorname {arsech}\left (c x\right )}{x^{2}}\right )} - \frac {a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*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) - 1/2*a/x^2

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mupad [B]  time = 1.46, size = 61, normalized size = 0.65 \[ \frac {b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\,\left (\frac {c^2\,x}{4}-\frac {1}{2\,x}\right )}{x}-\frac {a}{2\,x^2}+\frac {b\,c\,\sqrt {\frac {1}{c\,x}-1}\,\sqrt {\frac {1}{c\,x}+1}}{4\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))/x^3,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asech}{\left (c x \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asech(c*x))/x**3, x)

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