3.650 \(\int \frac{\sec ^{-1}(x)^4}{x^5} \, dx\)

Optimal. Leaf size=148 \[ -\frac{45}{128 x^2}-\frac{3}{128 x^4}-\frac{\sec ^{-1}(x)^4}{4 x^4}+\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac{\sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac{9 \sec ^{-1}(x)^2}{16 x^2}+\frac{3 \sec ^{-1}(x)^2}{16 x^4}-\frac{45 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)}{64 x}-\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)}{32 x^3}+\frac{3}{32} \sec ^{-1}(x)^4-\frac{45}{128} \sec ^{-1}(x)^2 \]

[Out]

-3/(128*x^4) - 45/(128*x^2) - (3*Sqrt[1 - x^(-2)]*ArcSec[x])/(32*x^3) - (45*Sqrt[1 - x^(-2)]*ArcSec[x])/(64*x)
- (45*ArcSec[x]^2)/128 + (3*ArcSec[x]^2)/(16*x^4) + (9*ArcSec[x]^2)/(16*x^2) + (Sqrt[1 - x^(-2)]*ArcSec[x]^3)
/(4*x^3) + (3*Sqrt[1 - x^(-2)]*ArcSec[x]^3)/(8*x) + (3*ArcSec[x]^4)/32 - ArcSec[x]^4/(4*x^4)

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Rubi [A]  time = 0.147337, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {5222, 3444, 3311, 30, 3310} \[ -\frac{45}{128 x^2}-\frac{3}{128 x^4}-\frac{\sec ^{-1}(x)^4}{4 x^4}+\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac{\sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac{9 \sec ^{-1}(x)^2}{16 x^2}+\frac{3 \sec ^{-1}(x)^2}{16 x^4}-\frac{45 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)}{64 x}-\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)}{32 x^3}+\frac{3}{32} \sec ^{-1}(x)^4-\frac{45}{128} \sec ^{-1}(x)^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[x]^4/x^5,x]

[Out]

-3/(128*x^4) - 45/(128*x^2) - (3*Sqrt[1 - x^(-2)]*ArcSec[x])/(32*x^3) - (45*Sqrt[1 - x^(-2)]*ArcSec[x])/(64*x)
- (45*ArcSec[x]^2)/128 + (3*ArcSec[x]^2)/(16*x^4) + (9*ArcSec[x]^2)/(16*x^2) + (Sqrt[1 - x^(-2)]*ArcSec[x]^3)
/(4*x^3) + (3*Sqrt[1 - x^(-2)]*ArcSec[x]^3)/(8*x) + (3*ArcSec[x]^4)/32 - ArcSec[x]^4/(4*x^4)

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 3444

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

Rule 3311

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -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{\sec ^{-1}(x)^4}{x^5} \, dx &=\operatorname{Subst}\left (\int x^4 \cos ^3(x) \sin (x) \, dx,x,\sec ^{-1}(x)\right )\\ &=-\frac{\sec ^{-1}(x)^4}{4 x^4}+\operatorname{Subst}\left (\int x^3 \cos ^4(x) \, dx,x,\sec ^{-1}(x)\right )\\ &=\frac{3 \sec ^{-1}(x)^2}{16 x^4}+\frac{\sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}-\frac{\sec ^{-1}(x)^4}{4 x^4}-\frac{3}{8} \operatorname{Subst}\left (\int x \cos ^4(x) \, dx,x,\sec ^{-1}(x)\right )+\frac{3}{4} \operatorname{Subst}\left (\int x^3 \cos ^2(x) \, dx,x,\sec ^{-1}(x)\right )\\ &=-\frac{3}{128 x^4}-\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)}{32 x^3}+\frac{3 \sec ^{-1}(x)^2}{16 x^4}+\frac{9 \sec ^{-1}(x)^2}{16 x^2}+\frac{\sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^3}{8 x}-\frac{\sec ^{-1}(x)^4}{4 x^4}-\frac{9}{32} \operatorname{Subst}\left (\int x \cos ^2(x) \, dx,x,\sec ^{-1}(x)\right )+\frac{3}{8} \operatorname{Subst}\left (\int x^3 \, dx,x,\sec ^{-1}(x)\right )-\frac{9}{8} \operatorname{Subst}\left (\int x \cos ^2(x) \, dx,x,\sec ^{-1}(x)\right )\\ &=-\frac{3}{128 x^4}-\frac{45}{128 x^2}-\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)}{32 x^3}-\frac{45 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)}{64 x}+\frac{3 \sec ^{-1}(x)^2}{16 x^4}+\frac{9 \sec ^{-1}(x)^2}{16 x^2}+\frac{\sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac{3}{32} \sec ^{-1}(x)^4-\frac{\sec ^{-1}(x)^4}{4 x^4}-\frac{9}{64} \operatorname{Subst}\left (\int x \, dx,x,\sec ^{-1}(x)\right )-\frac{9}{16} \operatorname{Subst}\left (\int x \, dx,x,\sec ^{-1}(x)\right )\\ &=-\frac{3}{128 x^4}-\frac{45}{128 x^2}-\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)}{32 x^3}-\frac{45 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)}{64 x}-\frac{45}{128} \sec ^{-1}(x)^2+\frac{3 \sec ^{-1}(x)^2}{16 x^4}+\frac{9 \sec ^{-1}(x)^2}{16 x^2}+\frac{\sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^3}{4 x^3}+\frac{3 \sqrt{1-\frac{1}{x^2}} \sec ^{-1}(x)^3}{8 x}+\frac{3}{32} \sec ^{-1}(x)^4-\frac{\sec ^{-1}(x)^4}{4 x^4}\\ \end{align*}

Mathematica [A]  time = 0.0738931, size = 92, normalized size = 0.62 \[ \frac{-45 x^2+4 \left (3 x^4-8\right ) \sec ^{-1}(x)^4+16 \sqrt{1-\frac{1}{x^2}} x \left (3 x^2+2\right ) \sec ^{-1}(x)^3+\left (-45 x^4+72 x^2+24\right ) \sec ^{-1}(x)^2-6 \sqrt{1-\frac{1}{x^2}} x \left (15 x^2+2\right ) \sec ^{-1}(x)-3}{128 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[x]^4/x^5,x]

[Out]

(-3 - 45*x^2 - 6*Sqrt[1 - x^(-2)]*x*(2 + 15*x^2)*ArcSec[x] + (24 + 72*x^2 - 45*x^4)*ArcSec[x]^2 + 16*Sqrt[1 -
x^(-2)]*x*(2 + 3*x^2)*ArcSec[x]^3 + 4*(-8 + 3*x^4)*ArcSec[x]^4)/(128*x^4)

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Maple [A]  time = 0.062, size = 165, normalized size = 1.1 \begin{align*} -{\frac{ \left ({\rm arcsec} \left (x\right ) \right ) ^{4}}{4\,{x}^{4}}}+{\frac{ \left ({\rm arcsec} \left (x\right ) \right ) ^{3}}{8\,{x}^{3}} \left ( 3\,{\rm arcsec} \left (x\right ){x}^{3}+3\,{x}^{2}\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}+2\,\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}} \right ) }+{\frac{3\, \left ({\rm arcsec} \left (x\right ) \right ) ^{2}}{16\,{x}^{4}}}-{\frac{3\,{\rm arcsec} \left (x\right )}{64\,{x}^{3}} \left ( 3\,{\rm arcsec} \left (x\right ){x}^{3}+3\,{x}^{2}\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}}+2\,\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}} \right ) }+{\frac{45\, \left ({\rm arcsec} \left (x\right ) \right ) ^{2}}{128}}-{\frac{3}{128\,{x}^{4}}}-{\frac{45}{128\,{x}^{2}}}+{\frac{9\, \left ({\rm arcsec} \left (x\right ) \right ) ^{2}}{16\,{x}^{2}}}-{\frac{9\,{\rm arcsec} \left (x\right )}{16\,x} \left ({\rm arcsec} \left (x\right )x+\sqrt{{\frac{{x}^{2}-1}{{x}^{2}}}} \right ) }+{\frac{9}{32}}-{\frac{9\, \left ({\rm arcsec} \left (x\right ) \right ) ^{4}}{32}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(x)^4/x^5,x)

[Out]

-1/4*arcsec(x)^4/x^4+1/8*arcsec(x)^3*(3*arcsec(x)*x^3+3*x^2*((x^2-1)/x^2)^(1/2)+2*((x^2-1)/x^2)^(1/2))/x^3+3/1
6*arcsec(x)^2/x^4-3/64*arcsec(x)*(3*arcsec(x)*x^3+3*x^2*((x^2-1)/x^2)^(1/2)+2*((x^2-1)/x^2)^(1/2))/x^3+45/128*
arcsec(x)^2-3/128/x^4-45/128/x^2+9/16*arcsec(x)^2/x^2-9/16*arcsec(x)*(arcsec(x)*x+((x^2-1)/x^2)^(1/2))/x+9/32-
9/32*arcsec(x)^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{8 \, x^{4} \int \frac{12 \,{\left (x^{2} - 1\right )} \log \left (x^{2}\right )^{2} \log \left (x\right )^{2} - 16 \,{\left (x^{2} - 1\right )} \log \left (x^{2}\right ) \log \left (x\right )^{3} + 8 \,{\left (x^{2} - 1\right )} \log \left (x\right )^{4} +{\left (x^{2} - 4 \,{\left (x^{2} - 1\right )} \log \left (x\right ) - 1\right )} \log \left (x^{2}\right )^{3} - 12 \,{\left (4 \,{\left (x^{2} - 1\right )} \log \left (x\right )^{2} +{\left (x^{2} - 4 \,{\left (x^{2} - 1\right )} \log \left (x\right ) - 1\right )} \log \left (x^{2}\right )\right )} \arctan \left (\sqrt{x + 1} \sqrt{x - 1}\right )^{2} + 2 \,{\left (4 \, \arctan \left (\sqrt{x + 1} \sqrt{x - 1}\right )^{3} - 3 \, \arctan \left (\sqrt{x + 1} \sqrt{x - 1}\right ) \log \left (x^{2}\right )^{2}\right )} \sqrt{x + 1} \sqrt{x - 1}}{x^{7} - x^{5}}\,{d x} - 16 \, \arctan \left (\sqrt{x + 1} \sqrt{x - 1}\right )^{4} + 24 \, \arctan \left (\sqrt{x + 1} \sqrt{x - 1}\right )^{2} \log \left (x^{2}\right )^{2} - \log \left (x^{2}\right )^{4}}{64 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)^4/x^5,x, algorithm="maxima")

[Out]

1/64*(64*x^4*integrate(1/8*(12*(x^2 - 1)*log(x^2)^2*log(x)^2 - 16*(x^2 - 1)*log(x^2)*log(x)^3 + 8*(x^2 - 1)*lo
g(x)^4 + (x^2 - 4*(x^2 - 1)*log(x) - 1)*log(x^2)^3 - 12*(4*(x^2 - 1)*log(x)^2 + (x^2 - 4*(x^2 - 1)*log(x) - 1)
*log(x^2))*arctan(sqrt(x + 1)*sqrt(x - 1))^2 + 2*(4*arctan(sqrt(x + 1)*sqrt(x - 1))^3 - 3*arctan(sqrt(x + 1)*s
qrt(x - 1))*log(x^2)^2)*sqrt(x + 1)*sqrt(x - 1))/(x^7 - x^5), x) - 16*arctan(sqrt(x + 1)*sqrt(x - 1))^4 + 24*a
rctan(sqrt(x + 1)*sqrt(x - 1))^2*log(x^2)^2 - log(x^2)^4)/x^4

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Fricas [A]  time = 2.7025, size = 220, normalized size = 1.49 \begin{align*} \frac{4 \,{\left (3 \, x^{4} - 8\right )} \operatorname{arcsec}\left (x\right )^{4} - 3 \,{\left (15 \, x^{4} - 24 \, x^{2} - 8\right )} \operatorname{arcsec}\left (x\right )^{2} - 45 \, x^{2} + 2 \,{\left (8 \,{\left (3 \, x^{2} + 2\right )} \operatorname{arcsec}\left (x\right )^{3} - 3 \,{\left (15 \, x^{2} + 2\right )} \operatorname{arcsec}\left (x\right )\right )} \sqrt{x^{2} - 1} - 3}{128 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)^4/x^5,x, algorithm="fricas")

[Out]

1/128*(4*(3*x^4 - 8)*arcsec(x)^4 - 3*(15*x^4 - 24*x^2 - 8)*arcsec(x)^2 - 45*x^2 + 2*(8*(3*x^2 + 2)*arcsec(x)^3
- 3*(15*x^2 + 2)*arcsec(x))*sqrt(x^2 - 1) - 3)/x^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asec}^{4}{\left (x \right )}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asec(x)**4/x**5,x)

[Out]

Integral(asec(x)**4/x**5, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcsec}\left (x\right )^{4}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(x)^4/x^5,x, algorithm="giac")

[Out]

integrate(arcsec(x)^4/x^5, x)