Integrand size = 10, antiderivative size = 70 \[ \int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx=-\frac {b \sec ^{-1}(a+b x)}{a}-\frac {\sec ^{-1}(a+b x)}{x}+\frac {2 b \arctan \left (\frac {\sqrt {1+a} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {1-a}}\right )}{a \sqrt {1-a^2}} \]
-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)
Result contains complex when optimal does not.
Time = 0.21 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.60 \[ \int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx=-\frac {\sec ^{-1}(a+b x)}{x}+\frac {b \left (\arcsin \left (\frac {1}{a+b x}\right )-\frac {i \log \left (\frac {2 \left (\frac {i a \left (-1+a^2+a b x\right )}{\sqrt {1-a^2}}+a (a+b x) \sqrt {\frac {-1+a^2+2 a b x+b^2 x^2}{(a+b x)^2}}\right )}{b x}\right )}{\sqrt {1-a^2}}\right )}{a} \]
-(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
Time = 0.41 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5781, 4926, 3042, 4270, 3042, 3138, 218}
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.
\(\displaystyle \int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx\) |
\(\Big \downarrow \) 5781 |
\(\displaystyle b \int \frac {(a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)}{b^2 x^2}d\sec ^{-1}(a+b x)\) |
\(\Big \downarrow \) 4926 |
\(\displaystyle b \left (-\int -\frac {1}{b x}d\sec ^{-1}(a+b x)-\frac {\sec ^{-1}(a+b x)}{b x}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (-\int \frac {1}{a-\csc \left (\sec ^{-1}(a+b x)+\frac {\pi }{2}\right )}d\sec ^{-1}(a+b x)-\frac {\sec ^{-1}(a+b x)}{b x}\right )\) |
\(\Big \downarrow \) 4270 |
\(\displaystyle b \left (\frac {\int \frac {1}{1-\frac {a}{a+b x}}d\sec ^{-1}(a+b x)}{a}-\frac {\sec ^{-1}(a+b x)}{a}-\frac {\sec ^{-1}(a+b x)}{b x}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (\frac {\int \frac {1}{1-a \sin \left (\sec ^{-1}(a+b x)+\frac {\pi }{2}\right )}d\sec ^{-1}(a+b x)}{a}-\frac {\sec ^{-1}(a+b x)}{a}-\frac {\sec ^{-1}(a+b x)}{b x}\right )\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle b \left (\frac {2 \int \frac {1}{(a+1) \tan ^2\left (\frac {1}{2} \sec ^{-1}(a+b x)\right )-a+1}d\tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{a}-\frac {\sec ^{-1}(a+b x)}{a}-\frac {\sec ^{-1}(a+b x)}{b x}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle b \left (\frac {2 \arctan \left (\frac {\sqrt {a+1} \tan \left (\frac {1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt {1-a}}\right )}{a \sqrt {1-a^2}}-\frac {\sec ^{-1}(a+b x)}{a}-\frac {\sec ^{-1}(a+b x)}{b x}\right )\) |
b*(-(ArcSec[a + b*x]/a) - ArcSec[a + b*x]/(b*x) + (2*ArcTan[(Sqrt[1 + a]*T an[ArcSec[a + b*x]/2])/Sqrt[1 - a]])/(a*Sqrt[1 - a^2]))
3.1.24.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[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]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Simp[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]
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] - Simp[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] && IGtQ[m, 0] && NeQ[n, -1]
Int[((a_.) + ArcSec[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[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]
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 | \(-\frac {\operatorname {arcsec}\left (b x +a \right )}{x}+\frac {b \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}\, \left (\arctan \left (\frac {1}{\sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}}\right ) \sqrt {a^{2}-1}-\ln \left (\frac {2 a^{2}-2+2 a b x +2 \sqrt {a^{2}-1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}-1}}{x}\right )\right )}{\sqrt {\frac {b^{2} x^{2}+2 a b x +a^{2}-1}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a \sqrt {a^{2}-1}}\) | \(151\) |
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))
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 \[ \int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx=\left [-\frac {2 \, {\left (a^{2} - 1\right )} b x \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt {a^{2} - 1} b x \log \left (\frac {a^{2} b x + a^{3} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} {\left (a^{2} - \sqrt {a^{2} - 1} a - 1\right )} - {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1} - a}{x}\right ) + {\left (a^{3} - a\right )} \operatorname {arcsec}\left (b x + a\right )}{{\left (a^{3} - a\right )} x}, -\frac {2 \, {\left (a^{2} - 1\right )} b x \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - 2 \, \sqrt {-a^{2} + 1} b x \arctan \left (-\frac {\sqrt {-a^{2} + 1} b x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} \sqrt {-a^{2} + 1}}{a^{2} - 1}\right ) + {\left (a^{3} - a\right )} \operatorname {arcsec}\left (b x + a\right )}{{\left (a^{3} - a\right )} x}\right ] \]
[-(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)]
\[ \int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx=\int \frac {\operatorname {asec}{\left (a + b x \right )}}{x^{2}}\, dx \]
\[ \int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx=\int { \frac {\operatorname {arcsec}\left (b x + a\right )}{x^{2}} \,d x } \]
(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)*sq rt(b*x + a - 1)))/x
Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.34 \[ \int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx=b {\left (\frac {2 \, \arctan \left (\frac {{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + a}{\sqrt {-a^{2} + 1}}\right )}{\sqrt {-a^{2} + 1} a} + \frac {\arccos \left (-\frac {1}{{\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a}\right )}{a {\left (\frac {a}{b x + a} - 1\right )}}\right )} \]
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)))
Timed out. \[ \int \frac {\sec ^{-1}(a+b x)}{x^2} \, dx=\int \frac {\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}{x^2} \,d x \]