3.9 \(\int \frac{a+b \sec ^{-1}(c x)}{x^2} \, dx\)

Optimal. Leaf size=31 \[ b c \sqrt{1-\frac{1}{c^2 x^2}}-\frac{a+b \sec ^{-1}(c x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.0202613, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {5220, 261} \[ b c \sqrt{1-\frac{1}{c^2 x^2}}-\frac{a+b \sec ^{-1}(c x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])/x^2,x]

[Out]

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

Rule 5220

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSec[c*x]
))/(d*(m + 1)), x] - Dist[(b*d)/(c*(m + 1)), Int[(d*x)^(m - 1)/Sqrt[1 - 1/(c^2*x^2)], x], x] /; FreeQ[{a, b, c
, d, m}, x] && NeQ[m, -1]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{a+b \sec ^{-1}(c x)}{x^2} \, dx &=-\frac{a+b \sec ^{-1}(c x)}{x}+\frac{b \int \frac{1}{\sqrt{1-\frac{1}{c^2 x^2}} x^3} \, dx}{c}\\ &=b c \sqrt{1-\frac{1}{c^2 x^2}}-\frac{a+b \sec ^{-1}(c x)}{x}\\ \end{align*}

Mathematica [A]  time = 0.0276364, size = 40, normalized size = 1.29 \[ -\frac{a}{x}+b c \sqrt{\frac{c^2 x^2-1}{c^2 x^2}}-\frac{b \sec ^{-1}(c x)}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSec[c*x])/x^2,x]

[Out]

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

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Maple [A]  time = 0.164, size = 62, normalized size = 2. \begin{align*} c \left ( -{\frac{a}{cx}}+b \left ( -{\frac{{\rm arcsec} \left (cx\right )}{cx}}+{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsec(c*x))/x^2,x)

[Out]

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

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Maxima [A]  time = 0.981653, size = 45, normalized size = 1.45 \begin{align*}{\left (c \sqrt{-\frac{1}{c^{2} x^{2}} + 1} - \frac{\operatorname{arcsec}\left (c x\right )}{x}\right )} b - \frac{a}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.44086, size = 62, normalized size = 2. \begin{align*} -\frac{b \operatorname{arcsec}\left (c x\right ) - \sqrt{c^{2} x^{2} - 1} b + a}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.78728, size = 36, normalized size = 1.16 \begin{align*} \begin{cases} - \frac{a}{x} + b c \sqrt{1 - \frac{1}{c^{2} x^{2}}} - \frac{b \operatorname{asec}{\left (c x \right )}}{x} & \text{for}\: c \neq 0 \\- \frac{a + \tilde{\infty } b}{x} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a/x + b*c*sqrt(1 - 1/(c**2*x**2)) - b*asec(c*x)/x, Ne(c, 0)), (-(a + zoo*b)/x, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arcsec}\left (c x\right ) + a}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsec(c*x) + a)/x^2, x)