3.23 \(\int \frac{(a+b \sec ^{-1}(c x))^2}{x^5} \, dx\)

Optimal. Leaf size=134 \[ \frac{3 b c^3 \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{16 x}+\frac{b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{8 x^3}+\frac{3}{16} a b c^4 \sec ^{-1}(c x)-\frac{\left (a+b \sec ^{-1}(c x)\right )^2}{4 x^4}+\frac{3 b^2 c^2}{32 x^2}+\frac{3}{32} b^2 c^4 \sec ^{-1}(c x)^2+\frac{b^2}{32 x^4} \]

[Out]

b^2/(32*x^4) + (3*b^2*c^2)/(32*x^2) + (3*a*b*c^4*ArcSec[c*x])/16 + (3*b^2*c^4*ArcSec[c*x]^2)/32 + (b*c*Sqrt[1
- 1/(c^2*x^2)]*(a + b*ArcSec[c*x]))/(8*x^3) + (3*b*c^3*Sqrt[1 - 1/(c^2*x^2)]*(a + b*ArcSec[c*x]))/(16*x) - (a
+ b*ArcSec[c*x])^2/(4*x^4)

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Rubi [A]  time = 0.111133, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {5222, 4405, 3310} \[ \frac{3 b c^3 \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{16 x}+\frac{b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{8 x^3}+\frac{3}{16} a b c^4 \sec ^{-1}(c x)-\frac{\left (a+b \sec ^{-1}(c x)\right )^2}{4 x^4}+\frac{3 b^2 c^2}{32 x^2}+\frac{3}{32} b^2 c^4 \sec ^{-1}(c x)^2+\frac{b^2}{32 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])^2/x^5,x]

[Out]

b^2/(32*x^4) + (3*b^2*c^2)/(32*x^2) + (3*a*b*c^4*ArcSec[c*x])/16 + (3*b^2*c^4*ArcSec[c*x]^2)/32 + (b*c*Sqrt[1
- 1/(c^2*x^2)]*(a + b*ArcSec[c*x]))/(8*x^3) + (3*b*c^3*Sqrt[1 - 1/(c^2*x^2)]*(a + b*ArcSec[c*x]))/(16*x) - (a
+ b*ArcSec[c*x])^2/(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 4405

Int[Cos[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)], x_Symbol] :> -Simp[((c +
d*x)^m*Cos[a + b*x]^(n + 1))/(b*(n + 1)), x] + Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Cos[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^5} \, dx &=c^4 \operatorname{Subst}\left (\int (a+b x)^2 \cos ^3(x) \sin (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \sec ^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{2} \left (b c^4\right ) \operatorname{Subst}\left (\int (a+b x) \cos ^4(x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{b^2}{32 x^4}+\frac{b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{8 x^3}-\frac{\left (a+b \sec ^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{8} \left (3 b c^4\right ) \operatorname{Subst}\left (\int (a+b x) \cos ^2(x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{b^2}{32 x^4}+\frac{3 b^2 c^2}{32 x^2}+\frac{b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{8 x^3}+\frac{3 b c^3 \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{16 x}-\frac{\left (a+b \sec ^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{16} \left (3 b c^4\right ) \operatorname{Subst}\left (\int (a+b x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{b^2}{32 x^4}+\frac{3 b^2 c^2}{32 x^2}+\frac{3}{16} a b c^4 \sec ^{-1}(c x)+\frac{3}{32} b^2 c^4 \sec ^{-1}(c x)^2+\frac{b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{8 x^3}+\frac{3 b c^3 \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{16 x}-\frac{\left (a+b \sec ^{-1}(c x)\right )^2}{4 x^4}\\ \end{align*}

Mathematica [A]  time = 0.173987, size = 148, normalized size = 1.1 \[ \frac{-8 a^2+6 a b c^3 x^3 \sqrt{1-\frac{1}{c^2 x^2}}+4 a b c x \sqrt{1-\frac{1}{c^2 x^2}}-6 a b c^4 x^4 \sin ^{-1}\left (\frac{1}{c x}\right )+2 b \sec ^{-1}(c x) \left (b c x \sqrt{1-\frac{1}{c^2 x^2}} \left (3 c^2 x^2+2\right )-8 a\right )+3 b^2 c^2 x^2+b^2 \left (3 c^4 x^4-8\right ) \sec ^{-1}(c x)^2+b^2}{32 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSec[c*x])^2/x^5,x]

[Out]

(-8*a^2 + b^2 + 4*a*b*c*Sqrt[1 - 1/(c^2*x^2)]*x + 3*b^2*c^2*x^2 + 6*a*b*c^3*Sqrt[1 - 1/(c^2*x^2)]*x^3 + 2*b*(-
8*a + b*c*Sqrt[1 - 1/(c^2*x^2)]*x*(2 + 3*c^2*x^2))*ArcSec[c*x] + b^2*(-8 + 3*c^4*x^4)*ArcSec[c*x]^2 - 6*a*b*c^
4*x^4*ArcSin[1/(c*x)])/(32*x^4)

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Maple [B]  time = 0.253, size = 265, normalized size = 2. \begin{align*} -{\frac{{a}^{2}}{4\,{x}^{4}}}-{\frac{{b}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}}{4\,{x}^{4}}}+{\frac{3\,{b}^{2}{c}^{4} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}}{32}}+{\frac{3\,{c}^{3}{b}^{2}{\rm arcsec} \left (cx\right )}{16\,x}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}+{\frac{c{b}^{2}{\rm arcsec} \left (cx\right )}{8\,{x}^{3}}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}+{\frac{{b}^{2}}{32\,{x}^{4}}}+{\frac{3\,{b}^{2}{c}^{2}}{32\,{x}^{2}}}-{\frac{ab{\rm arcsec} \left (cx\right )}{2\,{x}^{4}}}-{\frac{3\,a{c}^{3}b}{16\,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{3\,a{c}^{3}b}{16\,x}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{acb}{16\,{x}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{ab}{8\,c{x}^{5}}{\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^5,x)

[Out]

-1/4*a^2/x^4-1/4*b^2/x^4*arcsec(c*x)^2+3/32*b^2*c^4*arcsec(c*x)^2+3/16*c^3*b^2*arcsec(c*x)/x*((c^2*x^2-1)/c^2/
x^2)^(1/2)+1/8*c*b^2*arcsec(c*x)/x^3*((c^2*x^2-1)/c^2/x^2)^(1/2)+1/32*b^2/x^4+3/32*b^2*c^2/x^2-1/2*a*b/x^4*arc
sec(c*x)-3/16*c^3*a*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))+3/16*c^3*a*b
/((c^2*x^2-1)/c^2/x^2)^(1/2)/x-1/16*c*a*b/((c^2*x^2-1)/c^2/x^2)^(1/2)/x^3-1/8/c*a*b/((c^2*x^2-1)/c^2/x^2)^(1/2
)/x^5

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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^5,x, algorithm="maxima")

[Out]

1/16*a*b*((3*c^5*arctan(c*x*sqrt(-1/(c^2*x^2) + 1)) + (3*c^8*x^3*(-1/(c^2*x^2) + 1)^(3/2) + 5*c^6*x*sqrt(-1/(c
^2*x^2) + 1))/(c^4*x^4*(1/(c^2*x^2) - 1)^2 - 2*c^2*x^2*(1/(c^2*x^2) - 1) + 1))/c - 8*arcsec(c*x)/x^4) - 1/16*(
4*(2*(c^2*log(c*x + 1) + c^2*log(c*x - 1) - 2*c^2*log(x) + 1/x^2)*c^2*log(c)^2 - 16*c^2*integrate(1/4*x^2*log(
c^2*x^2)/(c^2*x^7 - x^5), x)*log(c) + 32*c^2*integrate(1/4*x^2*log(x)/(c^2*x^7 - x^5), x)*log(c) - 16*c^2*inte
grate(1/4*x^2*log(c^2*x^2)*log(x)/(c^2*x^7 - x^5), x) + 16*c^2*integrate(1/4*x^2*log(x)^2/(c^2*x^7 - x^5), x)
+ 4*c^2*integrate(1/4*x^2*log(c^2*x^2)/(c^2*x^7 - x^5), x) - (2*c^4*log(c*x + 1) + 2*c^4*log(c*x - 1) - 4*c^4*
log(x) + (2*c^2*x^2 + 1)/x^4)*log(c)^2 + 16*integrate(1/4*log(c^2*x^2)/(c^2*x^7 - x^5), x)*log(c) - 32*integra
te(1/4*log(x)/(c^2*x^7 - x^5), x)*log(c) - 8*integrate(1/4*sqrt(c*x + 1)*sqrt(c*x - 1)*arctan(sqrt(c*x + 1)*sq
rt(c*x - 1))/(c^2*x^7 - x^5), x) + 16*integrate(1/4*log(c^2*x^2)*log(x)/(c^2*x^7 - x^5), x) - 16*integrate(1/4
*log(x)^2/(c^2*x^7 - x^5), x) - 4*integrate(1/4*log(c^2*x^2)/(c^2*x^7 - x^5), x))*x^4 + 4*arctan(sqrt(c*x + 1)
*sqrt(c*x - 1))^2 - log(c^2*x^2)^2)*b^2/x^4 - 1/4*a^2/x^4

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Fricas [A]  time = 2.21055, size = 275, normalized size = 2.05 \begin{align*} \frac{3 \, b^{2} c^{2} x^{2} +{\left (3 \, b^{2} c^{4} x^{4} - 8 \, b^{2}\right )} \operatorname{arcsec}\left (c x\right )^{2} - 8 \, a^{2} + b^{2} + 2 \,{\left (3 \, a b c^{4} x^{4} - 8 \, a b\right )} \operatorname{arcsec}\left (c x\right ) + 2 \,{\left (3 \, a b c^{2} x^{2} + 2 \, a b +{\left (3 \, b^{2} c^{2} x^{2} + 2 \, b^{2}\right )} \operatorname{arcsec}\left (c x\right )\right )} \sqrt{c^{2} x^{2} - 1}}{32 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))^2/x^5,x, algorithm="fricas")

[Out]

1/32*(3*b^2*c^2*x^2 + (3*b^2*c^4*x^4 - 8*b^2)*arcsec(c*x)^2 - 8*a^2 + b^2 + 2*(3*a*b*c^4*x^4 - 8*a*b)*arcsec(c
*x) + 2*(3*a*b*c^2*x^2 + 2*a*b + (3*b^2*c^2*x^2 + 2*b^2)*arcsec(c*x))*sqrt(c^2*x^2 - 1))/x^4

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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^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asec(c*x))**2/x**5,x)

[Out]

Integral((a + b*asec(c*x))**2/x**5, 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^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))^2/x^5,x, algorithm="giac")

[Out]

integrate((b*arcsec(c*x) + a)^2/x^5, x)