3.11 \(\int \frac{a+b \sec ^{-1}(c x)}{x^4} \, dx\)

Optimal. Leaf size=60 \[ -\frac{a+b \sec ^{-1}(c x)}{3 x^3}-\frac{1}{9} b c^3 \left (1-\frac{1}{c^2 x^2}\right )^{3/2}+\frac{1}{3} b c^3 \sqrt{1-\frac{1}{c^2 x^2}} \]

[Out]

(b*c^3*Sqrt[1 - 1/(c^2*x^2)])/3 - (b*c^3*(1 - 1/(c^2*x^2))^(3/2))/9 - (a + b*ArcSec[c*x])/(3*x^3)

________________________________________________________________________________________

Rubi [A]  time = 0.0390767, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5220, 266, 43} \[ -\frac{a+b \sec ^{-1}(c x)}{3 x^3}-\frac{1}{9} b c^3 \left (1-\frac{1}{c^2 x^2}\right )^{3/2}+\frac{1}{3} b c^3 \sqrt{1-\frac{1}{c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])/x^4,x]

[Out]

(b*c^3*Sqrt[1 - 1/(c^2*x^2)])/3 - (b*c^3*(1 - 1/(c^2*x^2))^(3/2))/9 - (a + b*ArcSec[c*x])/(3*x^3)

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 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{a+b \sec ^{-1}(c x)}{x^4} \, dx &=-\frac{a+b \sec ^{-1}(c x)}{3 x^3}+\frac{b \int \frac{1}{\sqrt{1-\frac{1}{c^2 x^2}} x^5} \, dx}{3 c}\\ &=-\frac{a+b \sec ^{-1}(c x)}{3 x^3}-\frac{b \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-\frac{x}{c^2}}} \, dx,x,\frac{1}{x^2}\right )}{6 c}\\ &=-\frac{a+b \sec ^{-1}(c x)}{3 x^3}-\frac{b \operatorname{Subst}\left (\int \left (\frac{c^2}{\sqrt{1-\frac{x}{c^2}}}-c^2 \sqrt{1-\frac{x}{c^2}}\right ) \, dx,x,\frac{1}{x^2}\right )}{6 c}\\ &=\frac{1}{3} b c^3 \sqrt{1-\frac{1}{c^2 x^2}}-\frac{1}{9} b c^3 \left (1-\frac{1}{c^2 x^2}\right )^{3/2}-\frac{a+b \sec ^{-1}(c x)}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0453873, size = 59, normalized size = 0.98 \[ -\frac{a}{3 x^3}+b \left (\frac{2 c^3}{9}+\frac{c}{9 x^2}\right ) \sqrt{\frac{c^2 x^2-1}{c^2 x^2}}-\frac{b \sec ^{-1}(c x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSec[c*x])/x^4,x]

[Out]

-a/(3*x^3) + b*((2*c^3)/9 + c/(9*x^2))*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)] - (b*ArcSec[c*x])/(3*x^3)

________________________________________________________________________________________

Maple [A]  time = 0.161, size = 75, normalized size = 1.3 \begin{align*}{c}^{3} \left ( -{\frac{a}{3\,{c}^{3}{x}^{3}}}+b \left ( -{\frac{{\rm arcsec} \left (cx\right )}{3\,{c}^{3}{x}^{3}}}+{\frac{ \left ({c}^{2}{x}^{2}-1 \right ) \left ( 2\,{c}^{2}{x}^{2}+1 \right ) }{9\,{c}^{4}{x}^{4}}{\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))/x^4,x)

[Out]

c^3*(-1/3*a/c^3/x^3+b*(-1/3/c^3/x^3*arcsec(c*x)+1/9*(c^2*x^2-1)*(2*c^2*x^2+1)/((c^2*x^2-1)/c^2/x^2)^(1/2)/c^4/
x^4))

________________________________________________________________________________________

Maxima [A]  time = 0.969521, size = 78, normalized size = 1.3 \begin{align*} -\frac{1}{9} \, b{\left (\frac{c^{4}{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} - 3 \, c^{4} \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c} + \frac{3 \, \operatorname{arcsec}\left (c x\right )}{x^{3}}\right )} - \frac{a}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x^4,x, algorithm="maxima")

[Out]

-1/9*b*((c^4*(-1/(c^2*x^2) + 1)^(3/2) - 3*c^4*sqrt(-1/(c^2*x^2) + 1))/c + 3*arcsec(c*x)/x^3) - 1/3*a/x^3

________________________________________________________________________________________

Fricas [A]  time = 2.65475, size = 97, normalized size = 1.62 \begin{align*} -\frac{3 \, b \operatorname{arcsec}\left (c x\right ) -{\left (2 \, b c^{2} x^{2} + b\right )} \sqrt{c^{2} x^{2} - 1} + 3 \, a}{9 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x^4,x, algorithm="fricas")

[Out]

-1/9*(3*b*arcsec(c*x) - (2*b*c^2*x^2 + b)*sqrt(c^2*x^2 - 1) + 3*a)/x^3

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{asec}{\left (c x \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asec(c*x))/x**4,x)

[Out]

Integral((a + b*asec(c*x))/x**4, x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arcsec}\left (c x\right ) + a}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x^4,x, algorithm="giac")

[Out]

integrate((b*arcsec(c*x) + a)/x^4, x)