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

Optimal. Leaf size=51 \[ -\frac{a+b \sec ^{-1}(c x)}{2 x^2}+\frac{b c \sqrt{1-\frac{1}{c^2 x^2}}}{4 x}-\frac{1}{4} b c^2 \csc ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.0332673, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5220, 335, 321, 216} \[ -\frac{a+b \sec ^{-1}(c x)}{2 x^2}+\frac{b c \sqrt{1-\frac{1}{c^2 x^2}}}{4 x}-\frac{1}{4} b c^2 \csc ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])/x^3,x]

[Out]

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

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 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 321

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

Mathematica [A]  time = 0.0325762, size = 66, normalized size = 1.29 \[ -\frac{a}{2 x^2}+\frac{b c \sqrt{\frac{c^2 x^2-1}{c^2 x^2}}}{4 x}-\frac{1}{4} b c^2 \sin ^{-1}\left (\frac{1}{c x}\right )-\frac{b \sec ^{-1}(c x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSec[c*x])/x^3,x]

[Out]

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

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Maple [B]  time = 0.163, size = 118, normalized size = 2.3 \begin{align*} -{\frac{a}{2\,{x}^{2}}}-{\frac{b{\rm arcsec} \left (cx\right )}{2\,{x}^{2}}}-{\frac{cb}{4\,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{cb}{4\,x}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{b}{4\,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))/x^3,x)

[Out]

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

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Maxima [A]  time = 1.44524, size = 112, normalized size = 2.2 \begin{align*} -\frac{1}{4} \, b{\left (\frac{\frac{c^{4} x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c^{2} x^{2}{\left (\frac{1}{c^{2} x^{2}} - 1\right )} - 1} - c^{3} \arctan \left (c x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}\right )}{c} + \frac{2 \, \operatorname{arcsec}\left (c x\right )}{x^{2}}\right )} - \frac{a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*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) - 1/2*a/x^2

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Fricas [A]  time = 2.8523, size = 93, normalized size = 1.82 \begin{align*} \frac{{\left (b c^{2} x^{2} - 2 \, b\right )} \operatorname{arcsec}\left (c x\right ) + \sqrt{c^{2} x^{2} - 1} b - 2 \, a}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asec(c*x))/x**3, x)

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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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsec(c*x) + a)/x^3, x)