\(\int \frac {(a+b \sec ^{-1}(c x))^2}{x^3} \, dx\) [21]

Optimal result

Integrand size = 14, antiderivative size = 82 \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^3} \, dx=\frac {b^2}{4 x^2}+\frac {b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{2 x}-\frac {1}{4} c^2 \left (a+b \sec ^{-1}(c x)\right )^2+\frac {1}{2} \left (c^2-\frac {1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )^2 \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^3} \, dx=\frac {-2 a^2+b^2+2 a b c \sqrt {1-\frac {1}{c^2 x^2}} x+2 b \left (-2 a+b c \sqrt {1-\frac {1}{c^2 x^2}} x\right ) \sec ^{-1}(c x)+b^2 \left (-2+c^2 x^2\right ) \sec ^{-1}(c x)^2-2 a b c^2 x^2 \arcsin \left (\frac {1}{c x}\right )}{4 x^2} \] Input:

Integrate[(a + b*ArcSec[c*x])^2/x^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5745, 4904, 3042, 3791, 17}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^3} \, dx\)

\(\Big \downarrow \) 5745

\(\displaystyle c^2 \int \frac {\sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{c x}d\sec ^{-1}(c x)\)

\(\Big \downarrow \) 4904

\(\displaystyle c^2 \left (\frac {1}{2} \left (1-\frac {1}{c^2 x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )^2-b \int \left (1-\frac {1}{c^2 x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )d\sec ^{-1}(c x)\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle c^2 \left (\frac {1}{2} \left (1-\frac {1}{c^2 x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )^2-b \int \left (a+b \sec ^{-1}(c x)\right ) \sin \left (\sec ^{-1}(c x)\right )^2d\sec ^{-1}(c x)\right )\)

\(\Big \downarrow \) 3791

\(\displaystyle c^2 \left (\frac {1}{2} \left (1-\frac {1}{c^2 x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )^2-b \left (\frac {1}{2} \int \left (a+b \sec ^{-1}(c x)\right )d\sec ^{-1}(c x)-\frac {\sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{2 c x}+\frac {1}{4} b \left (1-\frac {1}{c^2 x^2}\right )\right )\right )\)

\(\Big \downarrow \) 17

\(\displaystyle c^2 \left (\frac {1}{2} \left (1-\frac {1}{c^2 x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )^2-b \left (-\frac {\sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )}{2 c x}+\frac {\left (a+b \sec ^{-1}(c x)\right )^2}{4 b}+\frac {1}{4} b \left (1-\frac {1}{c^2 x^2}\right )\right )\right )\)

Input:

Int[(a + b*ArcSec[c*x])^2/x^3,x]
 

Output:

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

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3791

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] + (-Simp[b*(c + d*x)*Cos[e + f*x 
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^2*((n - 1)/n)   Int[(c + d* 
x)*(b*Sin[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 
 1]
 

rule 4904

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x 
_)]^(n_.), x_Symbol] :> Simp[(c + d*x)^m*(Sin[a + b*x]^(n + 1)/(b*(n + 1))) 
, x] - Simp[d*(m/(b*(n + 1)))   Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n + 1), 
 x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]
 

rule 5745

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[1 
/c^(m + 1)   Subst[Int[(a + b*x)^n*Sec[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])
 

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.77

method	result	size
parts	\(-\frac {a^{2}}{2 x^{2}}+b^{2} c^{2} \left (-\frac {\cos \left (2 \,\operatorname {arcsec}\left (c x \right )\right ) \operatorname {arcsec}\left (c x \right )^{2}}{4}+\frac {\cos \left (2 \,\operatorname {arcsec}\left (c x \right )\right )}{8}+\frac {\sin \left (2 \,\operatorname {arcsec}\left (c x \right )\right ) \operatorname {arcsec}\left (c x \right )}{4}\right )+2 a b \,c^{2} \left (-\frac {\operatorname {arcsec}\left (c x \right )}{2 c^{2} x^{2}}-\frac {\sqrt {c^{2} x^{2}-1}\, \left (\arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right ) c^{2} x^{2}-\sqrt {c^{2} x^{2}-1}\right )}{4 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{3} x^{3}}\right )\)	\(145\)
derivativedivides	\(c^{2} \left (-\frac {a^{2}}{2 c^{2} x^{2}}+b^{2} \left (-\frac {\cos \left (2 \,\operatorname {arcsec}\left (c x \right )\right ) \operatorname {arcsec}\left (c x \right )^{2}}{4}+\frac {\cos \left (2 \,\operatorname {arcsec}\left (c x \right )\right )}{8}+\frac {\sin \left (2 \,\operatorname {arcsec}\left (c x \right )\right ) \operatorname {arcsec}\left (c x \right )}{4}\right )+2 a b \left (-\frac {\operatorname {arcsec}\left (c x \right )}{2 c^{2} x^{2}}-\frac {\sqrt {c^{2} x^{2}-1}\, \left (\arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right ) c^{2} x^{2}-\sqrt {c^{2} x^{2}-1}\right )}{4 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{3} x^{3}}\right )\right )\)	\(146\)
default	\(c^{2} \left (-\frac {a^{2}}{2 c^{2} x^{2}}+b^{2} \left (-\frac {\cos \left (2 \,\operatorname {arcsec}\left (c x \right )\right ) \operatorname {arcsec}\left (c x \right )^{2}}{4}+\frac {\cos \left (2 \,\operatorname {arcsec}\left (c x \right )\right )}{8}+\frac {\sin \left (2 \,\operatorname {arcsec}\left (c x \right )\right ) \operatorname {arcsec}\left (c x \right )}{4}\right )+2 a b \left (-\frac {\operatorname {arcsec}\left (c x \right )}{2 c^{2} x^{2}}-\frac {\sqrt {c^{2} x^{2}-1}\, \left (\arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right ) c^{2} x^{2}-\sqrt {c^{2} x^{2}-1}\right )}{4 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{3} x^{3}}\right )\right )\)	\(146\)

Input:

int((a+b*arcsec(c*x))^2/x^3,x,method=_RETURNVERBOSE)
 

Output:

-1/2*a^2/x^2+b^2*c^2*(-1/4*cos(2*arcsec(c*x))*arcsec(c*x)^2+1/8*cos(2*arcs 
ec(c*x))+1/4*sin(2*arcsec(c*x))*arcsec(c*x))+2*a*b*c^2*(-1/2/c^2/x^2*arcse 
c(c*x)-1/4*(c^2*x^2-1)^(1/2)*(arctan(1/(c^2*x^2-1)^(1/2))*c^2*x^2-(c^2*x^2 
-1)^(1/2))/((c^2*x^2-1)/c^2/x^2)^(1/2)/c^3/x^3)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^3} \, dx=\frac {{\left (b^{2} c^{2} x^{2} - 2 \, b^{2}\right )} \operatorname {arcsec}\left (c x\right )^{2} - 2 \, a^{2} + b^{2} + 2 \, {\left (a b c^{2} x^{2} - 2 \, a b\right )} \operatorname {arcsec}\left (c x\right ) + 2 \, \sqrt {c^{2} x^{2} - 1} {\left (b^{2} \operatorname {arcsec}\left (c x\right ) + a b\right )}}{4 \, x^{2}} \] Input:

integrate((a+b*arcsec(c*x))^2/x^3,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^3} \, dx=\int \frac {\left (a + b \operatorname {asec}{\left (c x \right )}\right )^{2}}{x^{3}}\, dx \] Input:

integrate((a+b*asec(c*x))**2/x**3,x)
 

Output:

Integral((a + b*asec(c*x))**2/x**3, x)
 

Maxima [F]

\[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^3} \, dx=\int { \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}^{2}}{x^{3}} \,d x } \] Input:

integrate((a+b*arcsec(c*x))^2/x^3,x, algorithm="maxima")
 

Output:

-1/2*a*b*((c^4*x*sqrt(-1/(c^2*x^2) + 1)/(c^2*x^2*(1/(c^2*x^2) - 1) - 1) - 
c^3*arctan(c*x*sqrt(-1/(c^2*x^2) + 1)))/c + 2*arcsec(c*x)/x^2) - 1/8*(4*(c 
^2*(log(c*x + 1) + log(c*x - 1) - 2*log(x))*log(c)^2 - 4*c^2*integrate(1/2 
*x^2*log(c^2*x^2)/(c^2*x^5 - x^3), x)*log(c) + 8*c^2*integrate(1/2*x^2*log 
(x)/(c^2*x^5 - x^3), x)*log(c) - 4*c^2*integrate(1/2*x^2*log(c^2*x^2)*log( 
x)/(c^2*x^5 - x^3), x) + 4*c^2*integrate(1/2*x^2*log(x)^2/(c^2*x^5 - x^3), 
 x) + 2*c^2*integrate(1/2*x^2*log(c^2*x^2)/(c^2*x^5 - x^3), x) - (c^2*log( 
c*x + 1) + c^2*log(c*x - 1) - 2*c^2*log(x) + 1/x^2)*log(c)^2 + 4*integrate 
(1/2*log(c^2*x^2)/(c^2*x^5 - x^3), x)*log(c) - 8*integrate(1/2*log(x)/(c^2 
*x^5 - x^3), x)*log(c) - 4*integrate(1/2*sqrt(c*x + 1)*sqrt(c*x - 1)*arcta 
n(sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x^5 - x^3), x) + 4*integrate(1/2*log(c 
^2*x^2)*log(x)/(c^2*x^5 - x^3), x) - 4*integrate(1/2*log(x)^2/(c^2*x^5 - x 
^3), x) - 2*integrate(1/2*log(c^2*x^2)/(c^2*x^5 - x^3), x))*x^2 + 4*arctan 
(sqrt(c*x + 1)*sqrt(c*x - 1))^2 - log(c^2*x^2)^2)*b^2/x^2 - 1/2*a^2/x^2
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (72) = 144\).

Time = 0.14 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.79 \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^3} \, dx=\frac {1}{8} \, {\left (2 \, b^{2} c \arccos \left (\frac {1}{c x}\right )^{2} + 4 \, a b c \arccos \left (\frac {1}{c x}\right ) - b^{2} c + \frac {4 \, b^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} \arccos \left (\frac {1}{c x}\right )}{x} + \frac {4 \, a b \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x} - \frac {4 \, b^{2} \arccos \left (\frac {1}{c x}\right )^{2}}{c x^{2}} - \frac {8 \, a b \arccos \left (\frac {1}{c x}\right )}{c x^{2}} - \frac {4 \, a^{2}}{c x^{2}} + \frac {2 \, b^{2}}{c x^{2}}\right )} c \] Input:

integrate((a+b*arcsec(c*x))^2/x^3,x, algorithm="giac")
 

Output:

1/8*(2*b^2*c*arccos(1/(c*x))^2 + 4*a*b*c*arccos(1/(c*x)) - b^2*c + 4*b^2*s 
qrt(-1/(c^2*x^2) + 1)*arccos(1/(c*x))/x + 4*a*b*sqrt(-1/(c^2*x^2) + 1)/x - 
 4*b^2*arccos(1/(c*x))^2/(c*x^2) - 8*a*b*arccos(1/(c*x))/(c*x^2) - 4*a^2/( 
c*x^2) + 2*b^2/(c*x^2))*c
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^3} \, dx=\int \frac {{\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}^2}{x^3} \,d x \] Input:

int((a + b*acos(1/(c*x)))^2/x^3,x)
 

Output:

int((a + b*acos(1/(c*x)))^2/x^3, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \sec ^{-1}(c x)\right )^2}{x^3} \, dx=\frac {4 \left (\int \frac {\mathit {asec} \left (c x \right )}{x^{3}}d x \right ) a b \,x^{2}+2 \left (\int \frac {\mathit {asec} \left (c x \right )^{2}}{x^{3}}d x \right ) b^{2} x^{2}-a^{2}}{2 x^{2}} \] Input:

int((a+b*asec(c*x))^2/x^3,x)
 

Output:

(4*int(asec(c*x)/x**3,x)*a*b*x**2 + 2*int(asec(c*x)**2/x**3,x)*b**2*x**2 - 
 a**2)/(2*x**2)