3.1.42 \(\int \frac {1}{x^2 (a+b \sec ^{-1}(c x))^2} \, dx\) [42]

3.1.42.1 Optimal result

Integrand size = 14, antiderivative size = 75 \[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )^2} \, dx=-\frac {c \sqrt {1-\frac {1}{c^2 x^2}}}{b \left (a+b \sec ^{-1}(c x)\right )}+\frac {c \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{b^2}+\frac {c \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{b^2} \]

output

c*Ci(a/b+arcsec(c*x))*cos(a/b)/b^2+c*Si(a/b+arcsec(c*x))*sin(a/b)/b^2-c*(1 
-1/c^2/x^2)^(1/2)/b/(a+b*arcsec(c*x))
 

3.1.42.2 Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )^2} \, dx=\frac {c \left (-\frac {b \sqrt {1-\frac {1}{c^2 x^2}}}{a+b \sec ^{-1}(c x)}+\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )\right )}{b^2} \]

input

Integrate[1/(x^2*(a + b*ArcSec[c*x])^2),x]
 

output

(c*(-((b*Sqrt[1 - 1/(c^2*x^2)])/(a + b*ArcSec[c*x])) + Cos[a/b]*CosIntegra 
l[a/b + ArcSec[c*x]] + Sin[a/b]*SinIntegral[a/b + ArcSec[c*x]]))/b^2
 

3.1.42.3 Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5745, 3042, 3778, 3042, 3784, 3042, 3780, 3783}

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.

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

\(\Big \downarrow \) 5745

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3784

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

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

\(\Big \downarrow \) 3783

\(\displaystyle c \left (\frac {\frac {\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{b}+\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sec ^{-1}(c x)\right )}{b}}{b}-\frac {\sqrt {1-\frac {1}{c^2 x^2}}}{b \left (a+b \sec ^{-1}(c x)\right )}\right )\)

input

Int[1/(x^2*(a + b*ArcSec[c*x])^2),x]
 

output

c*(-(Sqrt[1 - 1/(c^2*x^2)]/(b*(a + b*ArcSec[c*x]))) + ((Cos[a/b]*CosIntegr 
al[a/b + ArcSec[c*x]])/b + (Sin[a/b]*SinIntegral[a/b + ArcSec[c*x]])/b)/b)
 

3.1.42.3.1 Defintions of rubi rules used

rule 3042

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

rule 3778

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

rule 3780

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte 
gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
 

rule 3783

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte 
gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - 
c*f, 0]
 

rule 3784

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

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])
 

3.1.42.4 Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(c \left (-\frac {\sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{\left (a +b \,\operatorname {arcsec}\left (c x \right )\right ) b}+\frac {\operatorname {Si}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) \sin \left (\frac {a}{b}\right )+\operatorname {Ci}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) \cos \left (\frac {a}{b}\right )}{b^{2}}\right )\)	\(78\)
default	\(c \left (-\frac {\sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{\left (a +b \,\operatorname {arcsec}\left (c x \right )\right ) b}+\frac {\operatorname {Si}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) \sin \left (\frac {a}{b}\right )+\operatorname {Ci}\left (\frac {a}{b}+\operatorname {arcsec}\left (c x \right )\right ) \cos \left (\frac {a}{b}\right )}{b^{2}}\right )\)	\(78\)

input

int(1/x^2/(a+b*arcsec(c*x))^2,x,method=_RETURNVERBOSE)
 

output

c*(-((c^2*x^2-1)/c^2/x^2)^(1/2)/(a+b*arcsec(c*x))/b+(Si(a/b+arcsec(c*x))*s 
in(a/b)+Ci(a/b+arcsec(c*x))*cos(a/b))/b^2)
 

3.1.42.5 Fricas [F]

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

input

integrate(1/x^2/(a+b*arcsec(c*x))^2,x, algorithm="fricas")
 

output

integral(1/(b^2*x^2*arcsec(c*x)^2 + 2*a*b*x^2*arcsec(c*x) + a^2*x^2), x)
 

3.1.42.6 Sympy [F]

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

input

integrate(1/x**2/(a+b*asec(c*x))**2,x)
 

output

Integral(1/(x**2*(a + b*asec(c*x))**2), x)
 

3.1.42.7 Maxima [F]

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

input

integrate(1/x^2/(a+b*arcsec(c*x))^2,x, algorithm="maxima")
 

output

-(4*sqrt(c*x + 1)*sqrt(c*x - 1)*(b*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) + a 
) - (4*b^3*x*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2 + b^3*x*log(c^2*x^2)^2 
+ 8*b^3*x*log(c)*log(x) + 4*b^3*x*log(x)^2 + 8*a*b^2*x*arctan(sqrt(c*x + 1 
)*sqrt(c*x - 1)) + 4*(b^3*log(c)^2 + a^2*b)*x - 4*(b^3*x*log(c) + b^3*x*lo 
g(x))*log(c^2*x^2))*integrate(4*sqrt(c*x + 1)*sqrt(c*x - 1)*(b*arctan(sqrt 
(c*x + 1)*sqrt(c*x - 1)) + a)/(4*(b^3*c^2*log(c)^2 + a^2*b*c^2)*x^4 - 4*(b 
^3*log(c)^2 + a^2*b)*x^2 + 4*(b^3*c^2*x^4 - b^3*x^2)*arctan(sqrt(c*x + 1)* 
sqrt(c*x - 1))^2 + (b^3*c^2*x^4 - b^3*x^2)*log(c^2*x^2)^2 + 4*(b^3*c^2*x^4 
 - b^3*x^2)*log(x)^2 + 8*(a*b^2*c^2*x^4 - a*b^2*x^2)*arctan(sqrt(c*x + 1)* 
sqrt(c*x - 1)) - 4*(b^3*c^2*x^4*log(c) - b^3*x^2*log(c) + (b^3*c^2*x^4 - b 
^3*x^2)*log(x))*log(c^2*x^2) + 8*(b^3*c^2*x^4*log(c) - b^3*x^2*log(c))*log 
(x)), x))/(4*b^3*x*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2 + b^3*x*log(c^2*x 
^2)^2 + 8*b^3*x*log(c)*log(x) + 4*b^3*x*log(x)^2 + 8*a*b^2*x*arctan(sqrt(c 
*x + 1)*sqrt(c*x - 1)) + 4*(b^3*log(c)^2 + a^2*b)*x - 4*(b^3*x*log(c) + b^ 
3*x*log(x))*log(c^2*x^2))
 

3.1.42.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (73) = 146\).

Time = 0.28 (sec) , antiderivative size = 226, normalized size of antiderivative = 3.01 \[ \int \frac {1}{x^2 \left (a+b \sec ^{-1}(c x)\right )^2} \, dx={\left (\frac {b \arccos \left (\frac {1}{c x}\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} + \frac {b \arccos \left (\frac {1}{c x}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} + \frac {a \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} + \frac {a \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} - \frac {b \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}}\right )} c \]

input

integrate(1/x^2/(a+b*arcsec(c*x))^2,x, algorithm="giac")
 

output

(b*arccos(1/(c*x))*cos(a/b)*cos_integral(a/b + arccos(1/(c*x)))/(b^3*arcco 
s(1/(c*x)) + a*b^2) + b*arccos(1/(c*x))*sin(a/b)*sin_integral(a/b + arccos 
(1/(c*x)))/(b^3*arccos(1/(c*x)) + a*b^2) + a*cos(a/b)*cos_integral(a/b + a 
rccos(1/(c*x)))/(b^3*arccos(1/(c*x)) + a*b^2) + a*sin(a/b)*sin_integral(a/ 
b + arccos(1/(c*x)))/(b^3*arccos(1/(c*x)) + a*b^2) - b*sqrt(-1/(c^2*x^2) + 
 1)/(b^3*arccos(1/(c*x)) + a*b^2))*c
 

3.1.42.9 Mupad [F(-1)]

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

input

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

output

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