\(\int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx\) [24]

Optimal result

Integrand size = 10, antiderivative size = 70 \[ \int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx=-\frac {b \sec ^{-1}(a+b x)}{a}-\frac {\sec ^{-1}(a+b x)}{x}+\frac {2 b \arctan \left (\frac {\sqrt {1+a} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {1-a}}\right )}{a \sqrt {1-a^2}} \]

[Out]

-b*arcsec(b*x+a)/a-arcsec(b*x+a)/x+2*b*arctan((1+a)^(1/2)*tan(1/2*arcsec(b*x+a))/(1-a)^(1/2))/a/(-a^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5366, 4511, 3868, 2738, 211} \[ \int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx=\frac {2 b \arctan \left (\frac {\sqrt {a+1} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {1-a}}\right )}{a \sqrt {1-a^2}}-\frac {b \sec ^{-1}(a+b x)}{a}-\frac {\sec ^{-1}(a+b x)}{x} \]

[In]

Int[ArcSec[a + b*x]/x^2,x]

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3868

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a/b)*Sin[c
+ d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 4511

Int[((e_.) + (f_.)*(x_))^(m_.)*Sec[(c_.) + (d_.)*(x_)]*((a_) + (b_.)*Sec[(c_.) + (d_.)*(x_)])^(n_.)*Tan[(c_.)
+ (d_.)*(x_)], x_Symbol] :> Simp[(e + f*x)^m*((a + b*Sec[c + d*x])^(n + 1)/(b*d*(n + 1))), x] - Dist[f*(m/(b*d
*(n + 1))), Int[(e + f*x)^(m - 1)*(a + b*Sec[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && I
GtQ[m, 0] && NeQ[n, -1]

Rule 5366

Int[((a_.) + ArcSec[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d^(m + 1),
Subst[Int[(a + b*x)^p*Sec[x]*Tan[x]*(d*e - c*f + f*Sec[x])^m, x], x, ArcSec[c + d*x]], x] /; FreeQ[{a, b, c,
d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = b \text {Subst}\left (\int \frac {x \sec (x) \tan (x)}{(-a+\sec (x))^2} \, dx,x,\sec ^{-1}(a+b x)\right ) \\ & = -\frac {\sec ^{-1}(a+b x)}{x}+b \text {Subst}\left (\int \frac {1}{-a+\sec (x)} \, dx,x,\sec ^{-1}(a+b x)\right ) \\ & = -\frac {b \sec ^{-1}(a+b x)}{a}-\frac {\sec ^{-1}(a+b x)}{x}+\frac {b \text {Subst}\left (\int \frac {1}{1-a \cos (x)} \, dx,x,\sec ^{-1}(a+b x)\right )}{a} \\ & = -\frac {b \sec ^{-1}(a+b x)}{a}-\frac {\sec ^{-1}(a+b x)}{x}+\frac {(2 b) \text {Subst}\left (\int \frac {1}{1-a+(1+a) x^2} \, dx,x,\tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )\right )}{a} \\ & = -\frac {b \sec ^{-1}(a+b x)}{a}-\frac {\sec ^{-1}(a+b x)}{x}+\frac {2 b \arctan \left (\frac {\sqrt {1+a} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {1-a}}\right )}{a \sqrt {1-a^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.60 \[ \int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx=-\frac {\sec ^{-1}(a+b x)}{x}+\frac {b \left (\arcsin \left (\frac {1}{a+b x}\right )-\frac {i \log \left (\frac {2 \left (\frac {i a \left (-1+a^2+a b x\right )}{\sqrt {1-a^2}}+a (a+b x) \sqrt {\frac {-1+a^2+2 a b x+b^2 x^2}{(a+b x)^2}}\right )}{b x}\right )}{\sqrt {1-a^2}}\right )}{a} \]

[In]

Integrate[ArcSec[a + b*x]/x^2,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(125\) vs. \(2(62)=124\).

Time = 0.72 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.80

method	result	size
derivativedivides	\(b \left (-\frac {\operatorname {arcsec}\left (b x +a \right )}{b x}+\frac {\sqrt {\left (b x +a \right )^{2}-1}\, \left (\arctan \left (\frac {1}{\sqrt {\left (b x +a \right )^{2}-1}}\right ) \sqrt {a^{2}-1}-\ln \left (\frac {2 \sqrt {a^{2}-1}\, \sqrt {\left (b x +a \right )^{2}-1}+2 \left (b x +a \right ) a -2}{b x}\right )\right )}{\sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a \sqrt {a^{2}-1}}\right )\)	\(126\)
default	\(b \left (-\frac {\operatorname {arcsec}\left (b x +a \right )}{b x}+\frac {\sqrt {\left (b x +a \right )^{2}-1}\, \left (\arctan \left (\frac {1}{\sqrt {\left (b x +a \right )^{2}-1}}\right ) \sqrt {a^{2}-1}-\ln \left (\frac {2 \sqrt {a^{2}-1}\, \sqrt {\left (b x +a \right )^{2}-1}+2 \left (b x +a \right ) a -2}{b x}\right )\right )}{\sqrt {\frac {\left (b x +a \right )^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a \sqrt {a^{2}-1}}\right )\)	\(126\)
parts	\(-\frac {\operatorname {arcsec}\left (b x +a \right )}{x}+\frac {b \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \left (\arctan \left (\frac {1}{\sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}}\right ) \sqrt {a^{2}-1}-\ln \left (\frac {2 a^{2}-2+2 a b x +2 \sqrt {a^{2}-1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}}{x}\right )\right )}{\sqrt {\frac {b^{2} x^{2}+2 a b x +a^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a \sqrt {a^{2}-1}}\)	\(151\)

[In]

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

[Out]

b*(-1/b/x*arcsec(b*x+a)+((b*x+a)^2-1)^(1/2)*(arctan(1/((b*x+a)^2-1)^(1/2))*(a^2-1)^(1/2)-ln(2*((a^2-1)^(1/2)*(
(b*x+a)^2-1)^(1/2)+(b*x+a)*a-1)/b/x))/(((b*x+a)^2-1)/(b*x+a)^2)^(1/2)/(b*x+a)/a/(a^2-1)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (62) = 124\).

Time = 0.32 (sec) , antiderivative size = 281, normalized size of antiderivative = 4.01 \[ \int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx=\left [-\frac {2 \, {\left (a^{2} - 1\right )} b x \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt {a^{2} - 1} b x \log \left (\frac {a^{2} b x + a^{3} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} {\left (a^{2} - \sqrt {a^{2} - 1} a - 1\right )} - {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1} - a}{x}\right ) + {\left (a^{3} - a\right )} \operatorname {arcsec}\left (b x + a\right )}{{\left (a^{3} - a\right )} x}, -\frac {2 \, {\left (a^{2} - 1\right )} b x \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - 2 \, \sqrt {-a^{2} + 1} b x \arctan \left (-\frac {\sqrt {-a^{2} + 1} b x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} \sqrt {-a^{2} + 1}}{a^{2} - 1}\right ) + {\left (a^{3} - a\right )} \operatorname {arcsec}\left (b x + a\right )}{{\left (a^{3} - a\right )} x}\right ] \]

[In]

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

[Out]

[-(2*(a^2 - 1)*b*x*arctan(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)) - sqrt(a^2 - 1)*b*x*log((a^2*b*x + a^3
+ sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*(a^2 - sqrt(a^2 - 1)*a - 1) - (a*b*x + a^2 - 1)*sqrt(a^2 - 1) - a)/x) + (
a^3 - a)*arcsec(b*x + a))/((a^3 - a)*x), -(2*(a^2 - 1)*b*x*arctan(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)
) - 2*sqrt(-a^2 + 1)*b*x*arctan(-(sqrt(-a^2 + 1)*b*x - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*sqrt(-a^2 + 1))/(a^2
- 1)) + (a^3 - a)*arcsec(b*x + a))/((a^3 - a)*x)]

Sympy [F]

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

[In]

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

[Out]

Integral(asec(a + b*x)/x**2, x)

Maxima [F]

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

[In]

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

[Out]

(x*integrate((b^2*x + a*b)*e^(1/2*log(b*x + a + 1) + 1/2*log(b*x + a - 1))/(b^2*x^3 + 2*a*b*x^2 + (a^2 - 1)*x
+ (b^2*x^3 + 2*a*b*x^2 + (a^2 - 1)*x)*e^(log(b*x + a + 1) + log(b*x + a - 1))), x) - arctan(sqrt(b*x + a + 1)*
sqrt(b*x + a - 1)))/x

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.34 \[ \int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx=b {\left (\frac {2 \, \arctan \left (\frac {{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + a}{\sqrt {-a^{2} + 1}}\right )}{\sqrt {-a^{2} + 1} a} + \frac {\arccos \left (-\frac {1}{{\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a}\right )}{a {\left (\frac {a}{b x + a} - 1\right )}}\right )} \]

[In]

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

[Out]

b*(2*arctan(((b*x + a)*(sqrt(-1/(b*x + a)^2 + 1) - 1) + a)/sqrt(-a^2 + 1))/(sqrt(-a^2 + 1)*a) + arccos(-1/((b*
x + a)*(a/(b*x + a) - 1) - a))/(a*(a/(b*x + a) - 1)))

Mupad [F(-1)]

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

[In]

int(acos(1/(a + b*x))/x^2,x)

[Out]

int(acos(1/(a + b*x))/x^2, x)