3.1.20 \(\int \frac {(a+b \sec ^{-1}(c x))^2}{x^2} \, dx\) [20]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 = {5330, 3377, 2718} \begin {gather*} 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} \end {gather*}

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 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

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 5330

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])

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 75, normalized size = 1.50 \begin {gather*} \frac {-a^2+2 b^2+2 a b c \sqrt {1-\frac {1}{c^2 x^2}} x+2 b \left (-a+b c \sqrt {1-\frac {1}{c^2 x^2}} x\right ) \sec ^{-1}(c x)-b^2 \sec ^{-1}(c x)^2}{x} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(116\) vs. \(2(48)=96\).
time = 0.16, size = 117, normalized size = 2.34

method	result	size
derivativedivides	\(c \left (-\frac {a^{2}}{c x}+b^{2} \left (-\frac {\mathrm {arcsec}\left (c x \right )^{2}}{c x}+\frac {2}{c x}+2 \,\mathrm {arcsec}\left (c x \right ) \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\right )+2 a b \left (-\frac {\mathrm {arcsec}\left (c x \right )}{c x}+\frac {c^{2} x^{2}-1}{\sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{2} x^{2}}\right )\right )\)	\(117\)
default	\(c \left (-\frac {a^{2}}{c x}+b^{2} \left (-\frac {\mathrm {arcsec}\left (c x \right )^{2}}{c x}+\frac {2}{c x}+2 \,\mathrm {arcsec}\left (c x \right ) \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\right )+2 a b \left (-\frac {\mathrm {arcsec}\left (c x \right )}{c x}+\frac {c^{2} x^{2}-1}{\sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{2} x^{2}}\right )\right )\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(-a^2/c/x+b^2*(-arcsec(c*x)^2/c/x+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 = 0.27, size = 78, normalized size = 1.56 \begin {gather*} 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 {gather*}

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 = 1.94, size = 57, normalized size = 1.14 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {asec}{\left (c x \right )}\right )^{2}}{x^{2}}\, dx \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (48) = 96\).
time = 0.40, size = 105, normalized size = 2.10 \begin {gather*} {\left (2 \, b^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} \arccos \left (\frac {1}{c x}\right ) + 2 \, a b \sqrt {-\frac {1}{c^{2} x^{2}} + 1} - \frac {b^{2} \arccos \left (\frac {1}{c x}\right )^{2}}{c x} - \frac {2 \, a b \arccos \left (\frac {1}{c x}\right )}{c x} - \frac {a^{2}}{c x} + \frac {2 \, b^{2}}{c x}\right )} c \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.82, size = 89, normalized size = 1.78 \begin {gather*} 2\,b^2\,c\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\,\sqrt {1-\frac {1}{c^2\,x^2}}-\frac {b^2\,\left ({\mathrm {acos}\left (\frac {1}{c\,x}\right )}^2-2\right )}{x}-\frac {a^2}{x}+2\,a\,b\,c\,\left (\sqrt {1-\frac {1}{c^2\,x^2}}-\frac {\mathrm {acos}\left (\frac {1}{c\,x}\right )}{c\,x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(1/(c*x)))^2/x^2,x)

[Out]

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

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