\(\int \frac {(a+b \sec ^{-1}(c x))^2}{x^5} \, dx\) [23]

Optimal result

Integrand size = 14, antiderivative size = 134 \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^5} \, dx=\frac {b^2}{32 x^4}+\frac {3 b^2 c^2}{32 x^2}+\frac {3}{16} a b c^4 \sec ^{-1}(c x)+\frac {3}{32} b^2 c^4 \sec ^{-1}(c x)^2+\frac {b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{8 x^3}+\frac {3 b c^3 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{16 x}-\frac {\left (a+b \sec ^{-1}(c x)\right )^2}{4 x^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5330, 4490, 3391} \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^5} \, dx=\frac {3}{16} a b c^4 \sec ^{-1}(c x)+\frac {b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{8 x^3}+\frac {3 b c^3 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{16 x}-\frac {\left (a+b \sec ^{-1}(c x)\right )^2}{4 x^4}+\frac {3}{32} b^2 c^4 \sec ^{-1}(c x)^2+\frac {3 b^2 c^2}{32 x^2}+\frac {b^2}{32 x^4} \]

[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 3391

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 4490

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 5330

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*} \text {integral}& = c^4 \text {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 ) \text {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 ) \text {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 ) \text {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] (verified)

Time = 0.12 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^5} \, dx=\frac {-8 a^2+b^2+4 a b c \sqrt {1-\frac {1}{c^2 x^2}} x+3 b^2 c^2 x^2+6 a b c^3 \sqrt {1-\frac {1}{c^2 x^2}} x^3+2 b \left (-8 a+b c \sqrt {1-\frac {1}{c^2 x^2}} x \left (2+3 c^2 x^2\right )\right ) \sec ^{-1}(c x)+b^2 \left (-8+3 c^4 x^4\right ) \sec ^{-1}(c x)^2-6 a b c^4 x^4 \arcsin \left (\frac {1}{c x}\right )}{32 x^4} \]

[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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(264\) vs. \(2(116)=232\).

Time = 0.82 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.98

method	result	size
parts	\(-\frac {a^{2}}{4 x^{4}}+b^{2} c^{4} \left (-\frac {\operatorname {arcsec}\left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {\operatorname {arcsec}\left (c x \right ) \left (3 c^{3} x^{3} \operatorname {arcsec}\left (c x \right )+3 c^{2} x^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}+2 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\right )}{16 c^{3} x^{3}}-\frac {3 \operatorname {arcsec}\left (c x \right )^{2}}{32}+\frac {\left (3 c^{2} x^{2}+2\right )^{2}}{128 c^{4} x^{4}}\right )-\frac {a b \,\operatorname {arcsec}\left (c x \right )}{2 x^{4}}-\frac {3 a b \,c^{3} \sqrt {c^{2} x^{2}-1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{16 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x}+\frac {3 a b c \left (c^{2} x^{2}-1\right )}{16 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x^{3}}+\frac {a b \left (c^{2} x^{2}-1\right )}{8 c \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x^{5}}\)	\(265\)
derivativedivides	\(c^{4} \left (-\frac {a^{2}}{4 c^{4} x^{4}}+b^{2} \left (-\frac {\operatorname {arcsec}\left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {\operatorname {arcsec}\left (c x \right ) \left (3 c^{3} x^{3} \operatorname {arcsec}\left (c x \right )+3 c^{2} x^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}+2 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\right )}{16 c^{3} x^{3}}-\frac {3 \operatorname {arcsec}\left (c x \right )^{2}}{32}+\frac {\left (3 c^{2} x^{2}+2\right )^{2}}{128 c^{4} x^{4}}\right )-\frac {a b \,\operatorname {arcsec}\left (c x \right )}{2 c^{4} x^{4}}-\frac {3 a b \sqrt {c^{2} x^{2}-1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{16 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}+\frac {3 a b \left (c^{2} x^{2}-1\right )}{16 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{3} x^{3}}+\frac {a b \left (c^{2} x^{2}-1\right )}{8 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{5} x^{5}}\right )\)	\(274\)
default	\(c^{4} \left (-\frac {a^{2}}{4 c^{4} x^{4}}+b^{2} \left (-\frac {\operatorname {arcsec}\left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {\operatorname {arcsec}\left (c x \right ) \left (3 c^{3} x^{3} \operatorname {arcsec}\left (c x \right )+3 c^{2} x^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}+2 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\right )}{16 c^{3} x^{3}}-\frac {3 \operatorname {arcsec}\left (c x \right )^{2}}{32}+\frac {\left (3 c^{2} x^{2}+2\right )^{2}}{128 c^{4} x^{4}}\right )-\frac {a b \,\operatorname {arcsec}\left (c x \right )}{2 c^{4} x^{4}}-\frac {3 a b \sqrt {c^{2} x^{2}-1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{16 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}+\frac {3 a b \left (c^{2} x^{2}-1\right )}{16 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{3} x^{3}}+\frac {a b \left (c^{2} x^{2}-1\right )}{8 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{5} x^{5}}\right )\)	\(274\)

[In]

int((a+b*arcsec(c*x))^2/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/4*a^2/x^4+b^2*c^4*(-1/4/c^4/x^4*arcsec(c*x)^2+1/16*arcsec(c*x)*(3*c^3*x^3*arcsec(c*x)+3*c^2*x^2*((c^2*x^2-1
)/c^2/x^2)^(1/2)+2*((c^2*x^2-1)/c^2/x^2)^(1/2))/c^3/x^3-3/32*arcsec(c*x)^2+1/128*(3*c^2*x^2+2)^2/c^4/x^4)-1/2*
a*b/x^4*arcsec(c*x)-3/16*a*b*c^3*(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*a*b*c*(c^2*x^2-1)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x^3+1/8*a*b/c*(c^2*x^2-1)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x^5

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^5} \, dx=\frac {3 \, b^{2} c^{2} x^{2} + {\left (3 \, b^{2} c^{4} x^{4} - 8 \, b^{2}\right )} \operatorname {arcsec}\left (c x\right )^{2} - 8 \, a^{2} + b^{2} + 2 \, {\left (3 \, a b c^{4} x^{4} - 8 \, a b\right )} \operatorname {arcsec}\left (c x\right ) + 2 \, {\left (3 \, a b c^{2} x^{2} + 2 \, a b + {\left (3 \, b^{2} c^{2} x^{2} + 2 \, b^{2}\right )} \operatorname {arcsec}\left (c x\right )\right )} \sqrt {c^{2} x^{2} - 1}}{32 \, x^{4}} \]

[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

Sympy [F]

\[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^5} \, dx=\int \frac {\left (a + b \operatorname {asec}{\left (c x \right )}\right )^{2}}{x^{5}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^5} \, dx=\int { \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{2}}{x^{5}} \,d x } \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.60 \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^5} \, dx=\frac {1}{256} \, {\left (24 \, b^{2} c^{3} \arccos \left (\frac {1}{c x}\right )^{2} + 48 \, a b c^{3} \arccos \left (\frac {1}{c x}\right ) - 15 \, b^{2} c^{3} + \frac {48 \, b^{2} c^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} \arccos \left (\frac {1}{c x}\right )}{x} + \frac {48 \, a b c^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x} + \frac {24 \, b^{2} c}{x^{2}} + \frac {32 \, b^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} \arccos \left (\frac {1}{c x}\right )}{x^{3}} + \frac {32 \, a b \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{3}} - \frac {64 \, b^{2} \arccos \left (\frac {1}{c x}\right )^{2}}{c x^{4}} - \frac {128 \, a b \arccos \left (\frac {1}{c x}\right )}{c x^{4}} - \frac {64 \, a^{2}}{c x^{4}} + \frac {8 \, b^{2}}{c x^{4}}\right )} c \]

[In]

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

[Out]

1/256*(24*b^2*c^3*arccos(1/(c*x))^2 + 48*a*b*c^3*arccos(1/(c*x)) - 15*b^2*c^3 + 48*b^2*c^2*sqrt(-1/(c^2*x^2) +
1)*arccos(1/(c*x))/x + 48*a*b*c^2*sqrt(-1/(c^2*x^2) + 1)/x + 24*b^2*c/x^2 + 32*b^2*sqrt(-1/(c^2*x^2) + 1)*arc
cos(1/(c*x))/x^3 + 32*a*b*sqrt(-1/(c^2*x^2) + 1)/x^3 - 64*b^2*arccos(1/(c*x))^2/(c*x^4) - 128*a*b*arccos(1/(c*
x))/(c*x^4) - 64*a^2/(c*x^4) + 8*b^2/(c*x^4))*c

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^5} \, dx=\int \frac {{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^2}{x^5} \,d x \]

[In]

int((a + b*acos(1/(c*x)))^2/x^5,x)

[Out]

int((a + b*acos(1/(c*x)))^2/x^5, x)