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

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

[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

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Rubi [A]  time = 0.0793479, antiderivative size = 94, 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, 4404, 3310} \[ \frac{b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{2 x}+\frac{1}{2} \left (c^2-\frac{1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )^2-\frac{1}{2} a b c^2 \sec ^{-1}(c x)-\frac{1}{4} b^2 c^2 \sec ^{-1}(c x)^2+\frac{b^2}{4 x^2} \]

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 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 4404

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 3310

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]

Rubi steps

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

Mathematica [A]  time = 0.113702, size = 102, normalized size = 1.09 \[ \frac{-2 a^2+2 a b c x \sqrt{1-\frac{1}{c^2 x^2}}-2 a b c^2 x^2 \sin ^{-1}\left (\frac{1}{c x}\right )+2 b \sec ^{-1}(c x) \left (b c x \sqrt{1-\frac{1}{c^2 x^2}}-2 a\right )+b^2 \left (c^2 x^2-2\right ) \sec ^{-1}(c x)^2+b^2}{4 x^2} \]

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]  time = 0.244, size = 199, normalized size = 2.1 \begin{align*} -{\frac{{a}^{2}}{2\,{x}^{2}}}-{\frac{{b}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}}{2\,{x}^{2}}}+{\frac{{b}^{2}{c}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}}{4}}+{\frac{c{b}^{2}{\rm arcsec} \left (cx\right )}{2\,x}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}-{\frac{{b}^{2}{c}^{2}}{4}}+{\frac{{b}^{2}}{4\,{x}^{2}}}-{\frac{ab{\rm arcsec} \left (cx\right )}{{x}^{2}}}-{\frac{acb}{2\,x}\sqrt{{c}^{2}{x}^{2}-1}\arctan \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{acb}{2\,x}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{ab}{2\,c{x}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

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 = 2.38939, size = 196, normalized size = 2.09 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asec}{\left (c x \right )}\right )^{2}}{x^{3}}\, dx \end{align*}

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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )}^{2}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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