3.22 \(\int \frac {(a+b \sec ^{-1}(c x))^2}{x^4} \, dx\)

Optimal. Leaf size=102 \[ \frac {2 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{9 x^2}+\frac {4}{9} b c^3 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )-\frac {\left (a+b \sec ^{-1}(c x)\right )^2}{3 x^3}+\frac {4 b^2 c^2}{9 x}+\frac {2 b^2}{27 x^3} \]

[Out]

2/27*b^2/x^3+4/9*b^2*c^2/x-1/3*(a+b*arcsec(c*x))^2/x^3+4/9*b*c^3*(a+b*arcsec(c*x))*(1-1/c^2/x^2)^(1/2)+2/9*b*c
*(a+b*arcsec(c*x))*(1-1/c^2/x^2)^(1/2)/x^2

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5222, 4405, 3310, 3296, 2638} \[ \frac {4}{9} b c^3 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )+\frac {2 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{9 x^2}-\frac {\left (a+b \sec ^{-1}(c x)\right )^2}{3 x^3}+\frac {4 b^2 c^2}{9 x}+\frac {2 b^2}{27 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])^2/x^4,x]

[Out]

(2*b^2)/(27*x^3) + (4*b^2*c^2)/(9*x) + (4*b*c^3*Sqrt[1 - 1/(c^2*x^2)]*(a + b*ArcSec[c*x]))/9 + (2*b*c*Sqrt[1 -
1/(c^2*x^2)]*(a + b*ArcSec[c*x]))/(9*x^2) - (a + b*ArcSec[c*x])^2/(3*x^3)

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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]

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 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])

Rubi steps

\begin {align*} \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^4} \, dx &=c^3 \operatorname {Subst}\left (\int (a+b x)^2 \cos ^2(x) \sin (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=-\frac {\left (a+b \sec ^{-1}(c x)\right )^2}{3 x^3}+\frac {1}{3} \left (2 b c^3\right ) \operatorname {Subst}\left (\int (a+b x) \cos ^3(x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {2 b^2}{27 x^3}+\frac {2 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{9 x^2}-\frac {\left (a+b \sec ^{-1}(c x)\right )^2}{3 x^3}+\frac {1}{9} \left (4 b c^3\right ) \operatorname {Subst}\left (\int (a+b x) \cos (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {2 b^2}{27 x^3}+\frac {4}{9} b c^3 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )+\frac {2 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{9 x^2}-\frac {\left (a+b \sec ^{-1}(c x)\right )^2}{3 x^3}-\frac {1}{9} \left (4 b^2 c^3\right ) \operatorname {Subst}\left (\int \sin (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {2 b^2}{27 x^3}+\frac {4 b^2 c^2}{9 x}+\frac {4}{9} b c^3 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )+\frac {2 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{9 x^2}-\frac {\left (a+b \sec ^{-1}(c x)\right )^2}{3 x^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.20, size = 108, normalized size = 1.06 \[ \frac {-9 a^2+6 a b c x \sqrt {1-\frac {1}{c^2 x^2}} \left (2 c^2 x^2+1\right )+6 b \sec ^{-1}(c x) \left (b c x \sqrt {1-\frac {1}{c^2 x^2}} \left (2 c^2 x^2+1\right )-3 a\right )+2 b^2 \left (6 c^2 x^2+1\right )-9 b^2 \sec ^{-1}(c x)^2}{27 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSec[c*x])^2/x^4,x]

[Out]

(-9*a^2 + 6*a*b*c*Sqrt[1 - 1/(c^2*x^2)]*x*(1 + 2*c^2*x^2) + 2*b^2*(1 + 6*c^2*x^2) + 6*b*(-3*a + b*c*Sqrt[1 - 1
/(c^2*x^2)]*x*(1 + 2*c^2*x^2))*ArcSec[c*x] - 9*b^2*ArcSec[c*x]^2)/(27*x^3)

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 93, normalized size = 0.91 \[ \frac {12 \, b^{2} c^{2} x^{2} - 9 \, b^{2} \operatorname {arcsec}\left (c x\right )^{2} - 18 \, a b \operatorname {arcsec}\left (c x\right ) - 9 \, a^{2} + 2 \, b^{2} + 6 \, {\left (2 \, a b c^{2} x^{2} + a b + {\left (2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \operatorname {arcsec}\left (c x\right )\right )} \sqrt {c^{2} x^{2} - 1}}{27 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))^2/x^4,x, algorithm="fricas")

[Out]

1/27*(12*b^2*c^2*x^2 - 9*b^2*arcsec(c*x)^2 - 18*a*b*arcsec(c*x) - 9*a^2 + 2*b^2 + 6*(2*a*b*c^2*x^2 + a*b + (2*
b^2*c^2*x^2 + b^2)*arcsec(c*x))*sqrt(c^2*x^2 - 1))/x^3

________________________________________________________________________________________

giac [A]  time = 0.16, size = 168, normalized size = 1.65 \[ \frac {1}{27} \, {\left (12 \, b^{2} c^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} \arccos \left (\frac {1}{c x}\right ) + 12 \, a b c^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} + \frac {12 \, b^{2} c}{x} + \frac {6 \, b^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} \arccos \left (\frac {1}{c x}\right )}{x^{2}} + \frac {6 \, a b \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{2}} - \frac {9 \, b^{2} \arccos \left (\frac {1}{c x}\right )^{2}}{c x^{3}} - \frac {18 \, a b \arccos \left (\frac {1}{c x}\right )}{c x^{3}} - \frac {9 \, a^{2}}{c x^{3}} + \frac {2 \, b^{2}}{c x^{3}}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))^2/x^4,x, algorithm="giac")

[Out]

1/27*(12*b^2*c^2*sqrt(-1/(c^2*x^2) + 1)*arccos(1/(c*x)) + 12*a*b*c^2*sqrt(-1/(c^2*x^2) + 1) + 12*b^2*c/x + 6*b
^2*sqrt(-1/(c^2*x^2) + 1)*arccos(1/(c*x))/x^2 + 6*a*b*sqrt(-1/(c^2*x^2) + 1)/x^2 - 9*b^2*arccos(1/(c*x))^2/(c*
x^3) - 18*a*b*arccos(1/(c*x))/(c*x^3) - 9*a^2/(c*x^3) + 2*b^2/(c*x^3))*c

________________________________________________________________________________________

maple [A]  time = 0.48, size = 154, normalized size = 1.51 \[ c^{3} \left (-\frac {a^{2}}{3 c^{3} x^{3}}+b^{2} \left (-\frac {\mathrm {arcsec}\left (c x \right )^{2}}{3 c^{3} x^{3}}+\frac {2 \,\mathrm {arcsec}\left (c x \right ) \left (2 c^{2} x^{2}+1\right ) \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{9 c^{2} x^{2}}+\frac {2}{27 c^{3} x^{3}}+\frac {4}{9 c x}\right )+2 a b \left (-\frac {\mathrm {arcsec}\left (c x \right )}{3 c^{3} x^{3}}+\frac {\left (c^{2} x^{2}-1\right ) \left (2 c^{2} x^{2}+1\right )}{9 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{4} x^{4}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsec(c*x))^2/x^4,x)

[Out]

c^3*(-1/3*a^2/c^3/x^3+b^2*(-1/3/c^3/x^3*arcsec(c*x)^2+2/9*arcsec(c*x)*(2*c^2*x^2+1)/c^2/x^2*((c^2*x^2-1)/c^2/x
^2)^(1/2)+2/27/c^3/x^3+4/9/c/x)+2*a*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 = 0.70, size = 164, normalized size = 1.61 \[ -\frac {2}{9} \, a b {\left (\frac {c^{4} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 3 \, c^{4} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c} + \frac {3 \, \operatorname {arcsec}\left (c x\right )}{x^{3}}\right )} - \frac {b^{2} \operatorname {arcsec}\left (c x\right )^{2}}{3 \, x^{3}} - \frac {a^{2}}{3 \, x^{3}} + \frac {2 \, {\left ({\left (6 \, c^{3} x^{2} + c\right )} \sqrt {c x + 1} \sqrt {c x - 1} + 3 \, {\left (2 \, c^{5} x^{4} - c^{3} x^{2} - c\right )} \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right )\right )} b^{2}}{27 \, \sqrt {c x + 1} \sqrt {c x - 1} c x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))^2/x^4,x, algorithm="maxima")

[Out]

-2/9*a*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*b^2*arcse
c(c*x)^2/x^3 - 1/3*a^2/x^3 + 2/27*((6*c^3*x^2 + c)*sqrt(c*x + 1)*sqrt(c*x - 1) + 3*(2*c^5*x^4 - c^3*x^2 - c)*a
rctan(sqrt(c*x + 1)*sqrt(c*x - 1)))*b^2/(sqrt(c*x + 1)*sqrt(c*x - 1)*c*x^3)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^2}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(1/(c*x)))^2/x^4,x)

[Out]

int((a + b*acos(1/(c*x)))^2/x^4, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asec}{\left (c x \right )}\right )^{2}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asec(c*x))**2/x**4,x)

[Out]

Integral((a + b*asec(c*x))**2/x**4, x)

________________________________________________________________________________________