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

Optimal. Leaf size=70 \[ \frac{2 b \tan ^{-1}\left (\frac{\sqrt{a+1} \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{1-a}}\right )}{a \sqrt{1-a^2}}-\frac{b \sec ^{-1}(a+b x)}{a}-\frac{\sec ^{-1}(a+b x)}{x} \]

[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])

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Rubi [A]  time = 0.0974052, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5258, 4426, 3783, 2659, 205} \[ \frac{2 b \tan ^{-1}\left (\frac{\sqrt{a+1} \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{1-a}}\right )}{a \sqrt{1-a^2}}-\frac{b \sec ^{-1}(a+b x)}{a}-\frac{\sec ^{-1}(a+b x)}{x} \]

Antiderivative was successfully verified.

[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 5258

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]

Rule 4426

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 3783

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

Rule 2659

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 205

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

Rubi steps

\begin{align*} \int \frac{\sec ^{-1}(a+b x)}{x^2} \, dx &=b \operatorname{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 \operatorname{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 \operatorname{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) \operatorname{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 \tan ^{-1}\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]  time = 0.301405, size = 112, normalized size = 1.6 \[ -\frac{\sec ^{-1}(a+b x)}{x}+\frac{b \left (\sin ^{-1}\left (\frac{1}{a+b x}\right )-\frac{i \log \left (\frac{2 \left (a \sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} (a+b x)+\frac{i a \left (a^2+a b x-1\right )}{\sqrt{1-a^2}}\right )}{b x}\right )}{\sqrt{1-a^2}}\right )}{a} \]

Antiderivative was successfully verified.

[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

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Maple [B]  time = 0.219, size = 154, normalized size = 2.2 \begin{align*} -{\frac{{\rm arcsec} \left (bx+a\right )}{x}}+{\frac{b}{a \left ( bx+a \right ) }\sqrt{-1+ \left ( bx+a \right ) ^{2}}\arctan \left ({\frac{1}{\sqrt{-1+ \left ( bx+a \right ) ^{2}}}} \right ){\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}}-{\frac{b}{a \left ( bx+a \right ) }\sqrt{-1+ \left ( bx+a \right ) ^{2}}\ln \left ( 2\,{\frac{\sqrt{{a}^{2}-1}\sqrt{-1+ \left ( bx+a \right ) ^{2}}+a \left ( bx+a \right ) -1}{bx}} \right ){\frac{1}{\sqrt{{\frac{-1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}{\frac{1}{\sqrt{{a}^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [B]  time = 2.58253, size = 666, normalized size = 9.51 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asec}{\left (a + b x \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asec(a + b*x)/x**2, x)

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Giac [B]  time = 1.63546, size = 174, normalized size = 2.49 \begin{align*} -2 \, b{\left (\frac{\arctan \left (-\frac{{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} b + a{\left | b \right |}}{b}\right )}{a \mathrm{sgn}\left (b x + a\right )} - \frac{\arctan \left (-\frac{x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{\sqrt{-a^{2} + 1}}\right )}{\sqrt{-a^{2} + 1} a \mathrm{sgn}\left (b x + a\right )}\right )} - \frac{\arccos \left (\frac{1}{b x + a}\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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