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

Optimal. Leaf size=50 \[ 2 b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )-\frac{\left (a+b \sec ^{-1}(c x)\right )^2}{x}+\frac{2 b^2}{x} \]

[Out]

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

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Rubi [A]  time = 0.0586854, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {5222, 3296, 2638} \[ 2 b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )-\frac{\left (a+b \sec ^{-1}(c x)\right )^2}{x}+\frac{2 b^2}{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])^2/x^2,x]

[Out]

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

Rule 5222

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
ec[x]^(m + 1)*Tan[x], x], x, ArcSec[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (GtQ[n,
0] || LtQ[m, -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 \sec ^{-1}(c x)\right )^2}{x^2} \, dx &=c \operatorname{Subst}\left (\int (a+b x)^2 \sin (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \sec ^{-1}(c x)\right )^2}{x}+(2 b c) \operatorname{Subst}\left (\int (a+b x) \cos (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=2 b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )-\frac{\left (a+b \sec ^{-1}(c x)\right )^2}{x}-\left (2 b^2 c\right ) \operatorname{Subst}\left (\int \sin (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{2 b^2}{x}+2 b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )-\frac{\left (a+b \sec ^{-1}(c x)\right )^2}{x}\\ \end{align*}

Mathematica [A]  time = 0.122778, size = 75, normalized size = 1.5 \[ \frac{-a^2+2 a b c x \sqrt{1-\frac{1}{c^2 x^2}}+2 b \sec ^{-1}(c x) \left (b c x \sqrt{1-\frac{1}{c^2 x^2}}-a\right )-b^2 \sec ^{-1}(c x)^2+2 b^2}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.24, size = 117, normalized size = 2.3 \begin{align*} c \left ( -{\frac{{a}^{2}}{cx}}+{b}^{2} \left ( -{\frac{ \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}}{cx}}+2\,{\frac{1}{cx}}+2\,{\rm arcsec} \left (cx\right )\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}} \right ) +2\,ab \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))^2/x^2,x)

[Out]

c*(-a^2/c/x+b^2*(-1/c/x*arcsec(c*x)^2+2/c/x+2*arcsec(c*x)*((c^2*x^2-1)/c^2/x^2)^(1/2))+2*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 = 1.0221, size = 105, normalized size = 2.1 \begin{align*} 2 \,{\left (c \sqrt{-\frac{1}{c^{2} x^{2}} + 1} - \frac{\operatorname{arcsec}\left (c x\right )}{x}\right )} a b + 2 \,{\left (c \sqrt{-\frac{1}{c^{2} x^{2}} + 1} \operatorname{arcsec}\left (c x\right ) + \frac{1}{x}\right )} b^{2} - \frac{b^{2} \operatorname{arcsec}\left (c x\right )^{2}}{x} - \frac{a^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.66604, size = 140, normalized size = 2.8 \begin{align*} -\frac{b^{2} \operatorname{arcsec}\left (c x\right )^{2} + 2 \, a b \operatorname{arcsec}\left (c x\right ) + a^{2} - 2 \, b^{2} - 2 \, \sqrt{c^{2} x^{2} - 1}{\left (b^{2} \operatorname{arcsec}\left (c x\right ) + a b\right )}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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