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(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)

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Rubi [A]  time = 0.10, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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]

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 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 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 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]

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*}

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Mathematica [C]  time = 0.31, size = 112, normalized size = 1.60 \[ -\frac {\sec ^{-1}(a+b x)}{x}+\frac {b \left (\sin ^{-1}\left (\frac {1}{a+b x}\right )-\frac {i \log \left (\frac {2 \left (a \sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} (a+b x)+\frac {i a \left (a^2+a b x-1\right )}{\sqrt {1-a^2}}\right )}{b x}\right )}{\sqrt {1-a^2}}\right )}{a} \]

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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fricas [B]  time = 1.01, size = 281, normalized size = 4.01 \[ \left [-\frac {2 \, {\left (a^{2} - 1\right )} b x \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt {a^{2} - 1} b x \log \left (\frac {a^{2} b x + a^{3} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} {\left (a^{2} - \sqrt {a^{2} - 1} a - 1\right )} - {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1} - a}{x}\right ) + {\left (a^{3} - a\right )} \operatorname {arcsec}\left (b x + a\right )}{{\left (a^{3} - a\right )} x}, -\frac {2 \, {\left (a^{2} - 1\right )} b x \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - 2 \, \sqrt {-a^{2} + 1} b x \arctan \left (-\frac {\sqrt {-a^{2} + 1} b x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} \sqrt {-a^{2} + 1}}{a^{2} - 1}\right ) + {\left (a^{3} - a\right )} \operatorname {arcsec}\left (b x + a\right )}{{\left (a^{3} - a\right )} x}\right ] \]

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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giac [A]  time = 0.19, size = 94, normalized size = 1.34 \[ b {\left (\frac {2 \, \arctan \left (\frac {{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + a}{\sqrt {-a^{2} + 1}}\right )}{\sqrt {-a^{2} + 1} a} + \frac {\arccos \left (-\frac {1}{{\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a}\right )}{a {\left (\frac {a}{b x + a} - 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maple [B]  time = 0.06, size = 154, normalized size = 2.20 \[ -\frac {\mathrm {arcsec}\left (b x +a \right )}{x}+\frac {b \sqrt {-1+\left (b x +a \right )^{2}}\, \arctan \left (\frac {1}{\sqrt {-1+\left (b x +a \right )^{2}}}\right )}{\sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a}-\frac {b \sqrt {-1+\left (b x +a \right )^{2}}\, \ln \left (\frac {2 \sqrt {a^{2}-1}\, \sqrt {-1+\left (b x +a \right )^{2}}+2 a \left (b x +a \right )-2}{b x}\right )}{\sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a \sqrt {a^{2}-1}} \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {b x \int \frac {1}{\sqrt {b x + a + 1} \sqrt {b x + a - 1} b x^{2} + \sqrt {b x + a + 1} \sqrt {b x + a - 1} a x}\,{d x} - \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(1/(a + b*x))/x^2,x)

[Out]

int(acos(1/(a + b*x))/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asec}{\left (a + b x \right )}}{x^{2}}\, dx \]

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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