3.1.21 \(\int \frac {(a+b \sec ^{-1}(c x))^2}{x^3} \, dx\) [21]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 94, 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, 4489, 3391} \begin {gather*} \frac {b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{2 x}+\frac {1}{2} \left (c^2-\frac {1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )^2-\frac {1}{2} a b c^2 \sec ^{-1}(c x)-\frac {1}{4} b^2 c^2 \sec ^{-1}(c x)^2+\frac {b^2}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])^2/x^3,x]

[Out]

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

Rule 3391

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^
2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[b*(c + d*x)*Cos[e + f*x
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 4489

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[(c + d
*x)^m*(Sin[a + b*x]^(n + 1)/(b*(n + 1))), x] - Dist[d*(m/(b*(n + 1))), Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n +
1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

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

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Mathematica [A]
time = 0.07, size = 102, normalized size = 1.09 \begin {gather*} \frac {-2 a^2+b^2+2 a b c \sqrt {1-\frac {1}{c^2 x^2}} x+2 b \left (-2 a+b c \sqrt {1-\frac {1}{c^2 x^2}} x\right ) \sec ^{-1}(c x)+b^2 \left (-2+c^2 x^2\right ) \sec ^{-1}(c x)^2-2 a b c^2 x^2 \text {ArcSin}\left (\frac {1}{c x}\right )}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSec[c*x])^2/x^3,x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(192\) vs. \(2(82)=164\).
time = 0.16, size = 193, normalized size = 2.05

method	result	size
derivativedivides	\(c^{2} \left (-\frac {a^{2}}{2 c^{2} x^{2}}+b^{2} \left (-\frac {\mathrm {arcsec}\left (c x \right )^{2}}{2 c^{2} x^{2}}+\frac {\mathrm {arcsec}\left (c x \right ) \left (\mathrm {arcsec}\left (c x \right ) c x +\sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\right )}{2 c x}-\frac {\mathrm {arcsec}\left (c x \right )^{2}}{4}-\frac {1}{4}+\frac {1}{4 c^{2} x^{2}}\right )-\frac {a b \,\mathrm {arcsec}\left (c x \right )}{c^{2} x^{2}}-\frac {a b \sqrt {c^{2} x^{2}-1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{2 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}+\frac {a b \left (c^{2} x^{2}-1\right )}{2 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{3} x^{3}}\right )\)	\(193\)
default	\(c^{2} \left (-\frac {a^{2}}{2 c^{2} x^{2}}+b^{2} \left (-\frac {\mathrm {arcsec}\left (c x \right )^{2}}{2 c^{2} x^{2}}+\frac {\mathrm {arcsec}\left (c x \right ) \left (\mathrm {arcsec}\left (c x \right ) c x +\sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\right )}{2 c x}-\frac {\mathrm {arcsec}\left (c x \right )^{2}}{4}-\frac {1}{4}+\frac {1}{4 c^{2} x^{2}}\right )-\frac {a b \,\mathrm {arcsec}\left (c x \right )}{c^{2} x^{2}}-\frac {a b \sqrt {c^{2} x^{2}-1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{2 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}+\frac {a b \left (c^{2} x^{2}-1\right )}{2 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{3} x^{3}}\right )\)	\(193\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsec(c*x))^2/x^3,x,method=_RETURNVERBOSE)

[Out]

c^2*(-1/2*a^2/c^2/x^2+b^2*(-1/2*arcsec(c*x)^2/c^2/x^2+1/2*arcsec(c*x)*(arcsec(c*x)*c*x+((c^2*x^2-1)/c^2/x^2)^(
1/2))/c/x-1/4*arcsec(c*x)^2-1/4+1/4/c^2/x^2)-a*b/c^2/x^2*arcsec(c*x)-1/2*a*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^
2/x^2)^(1/2)/c/x*arctan(1/(c^2*x^2-1)^(1/2))+1/2*a*b*(c^2*x^2-1)/((c^2*x^2-1)/c^2/x^2)^(1/2)/c^3/x^3)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))^2/x^3,x, algorithm="maxima")

[Out]

-1/2*a*b*((c^4*x*sqrt(-1/(c^2*x^2) + 1)/(c^2*x^2*(1/(c^2*x^2) - 1) - 1) - c^3*arctan(c*x*sqrt(-1/(c^2*x^2) + 1
)))/c + 2*arcsec(c*x)/x^2) - 1/8*(4*(c^2*(log(c*x + 1) + log(c*x - 1) - 2*log(x))*log(c)^2 - 4*c^2*integrate(1
/2*x^2*log(c^2*x^2)/(c^2*x^5 - x^3), x)*log(c) + 8*c^2*integrate(1/2*x^2*log(x)/(c^2*x^5 - x^3), x)*log(c) - 4
*c^2*integrate(1/2*x^2*log(c^2*x^2)*log(x)/(c^2*x^5 - x^3), x) + 4*c^2*integrate(1/2*x^2*log(x)^2/(c^2*x^5 - x
^3), x) + 2*c^2*integrate(1/2*x^2*log(c^2*x^2)/(c^2*x^5 - x^3), x) - (c^2*log(c*x + 1) + c^2*log(c*x - 1) - 2*
c^2*log(x) + 1/x^2)*log(c)^2 + 4*integrate(1/2*log(c^2*x^2)/(c^2*x^5 - x^3), x)*log(c) - 8*integrate(1/2*log(x
)/(c^2*x^5 - x^3), x)*log(c) - 4*integrate(1/2*sqrt(c*x + 1)*sqrt(c*x - 1)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))
/(c^2*x^5 - x^3), x) + 4*integrate(1/2*log(c^2*x^2)*log(x)/(c^2*x^5 - x^3), x) - 4*integrate(1/2*log(x)^2/(c^2
*x^5 - x^3), x) - 2*integrate(1/2*log(c^2*x^2)/(c^2*x^5 - x^3), x))*x^2 + 4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)
)^2 - log(c^2*x^2)^2)*b^2/x^2 - 1/2*a^2/x^2

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Fricas [A]
time = 2.49, size = 82, normalized size = 0.87 \begin {gather*} \frac {{\left (b^{2} c^{2} x^{2} - 2 \, b^{2}\right )} \operatorname {arcsec}\left (c x\right )^{2} - 2 \, a^{2} + b^{2} + 2 \, {\left (a b c^{2} x^{2} - 2 \, a b\right )} \operatorname {arcsec}\left (c x\right ) + 2 \, \sqrt {c^{2} x^{2} - 1} {\left (b^{2} \operatorname {arcsec}\left (c x\right ) + a b\right )}}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))^2/x^3,x, algorithm="fricas")

[Out]

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

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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^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asec(c*x))**2/x**3,x)

[Out]

Integral((a + b*asec(c*x))**2/x**3, x)

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Giac [A]
time = 0.43, size = 147, normalized size = 1.56 \begin {gather*} \frac {1}{8} \, {\left (2 \, b^{2} c \arccos \left (\frac {1}{c x}\right )^{2} + 4 \, a b c \arccos \left (\frac {1}{c x}\right ) - b^{2} c + \frac {4 \, b^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} \arccos \left (\frac {1}{c x}\right )}{x} + \frac {4 \, a b \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x} - \frac {4 \, b^{2} \arccos \left (\frac {1}{c x}\right )^{2}}{c x^{2}} - \frac {8 \, a b \arccos \left (\frac {1}{c x}\right )}{c x^{2}} - \frac {4 \, a^{2}}{c x^{2}} + \frac {2 \, b^{2}}{c x^{2}}\right )} c \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))^2/x^3,x, algorithm="giac")

[Out]

1/8*(2*b^2*c*arccos(1/(c*x))^2 + 4*a*b*c*arccos(1/(c*x)) - b^2*c + 4*b^2*sqrt(-1/(c^2*x^2) + 1)*arccos(1/(c*x)
)/x + 4*a*b*sqrt(-1/(c^2*x^2) + 1)/x - 4*b^2*arccos(1/(c*x))^2/(c*x^2) - 8*a*b*arccos(1/(c*x))/(c*x^2) - 4*a^2
/(c*x^2) + 2*b^2/(c*x^2))*c

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^2}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(1/(c*x)))^2/x^3,x)

[Out]

int((a + b*acos(1/(c*x)))^2/x^3, x)

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