\(\int \frac {\sec ^{-1}(a+b x)^3}{x^2} \, dx\) [37]

Optimal result

Integrand size = 12, antiderivative size = 362 \[ \int \frac {\sec ^{-1}(a+b x)^3}{x^2} \, dx=-\frac {b \sec ^{-1}(a+b x)^3}{a}-\frac {\sec ^{-1}(a+b x)^3}{x}-\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {3 i b \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {6 b \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {6 b \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {6 i b \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}+\frac {6 i b \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}} \] Output:

-b*arcsec(b*x+a)^3/a-arcsec(b*x+a)^3/x-3*I*b*arcsec(b*x+a)^2*ln(1-a*(1/(b* 
x+a)+I*(1-1/(b*x+a)^2)^(1/2))/(1-(-a^2+1)^(1/2)))/a/(-a^2+1)^(1/2)+3*I*b*a 
rcsec(b*x+a)^2*ln(1-a*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/(1+(-a^2+1)^(1/2 
)))/a/(-a^2+1)^(1/2)-6*b*arcsec(b*x+a)*polylog(2,a*(1/(b*x+a)+I*(1-1/(b*x+ 
a)^2)^(1/2))/(1-(-a^2+1)^(1/2)))/a/(-a^2+1)^(1/2)+6*b*arcsec(b*x+a)*polylo 
g(2,a*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/(1+(-a^2+1)^(1/2)))/a/(-a^2+1)^( 
1/2)-6*I*b*polylog(3,a*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/(1-(-a^2+1)^(1/ 
2)))/a/(-a^2+1)^(1/2)+6*I*b*polylog(3,a*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2) 
)/(1+(-a^2+1)^(1/2)))/a/(-a^2+1)^(1/2)
 

Mathematica [F(-1)]

Timed out. \[ \int \frac {\sec ^{-1}(a+b x)^3}{x^2} \, dx=\text {\$Aborted} \] Input:

Integrate[ArcSec[a + b*x]^3/x^2,x]
 

Output:

$Aborted
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5781, 4926, 3042, 4679, 2009}

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.

\(\displaystyle \int \frac {\sec ^{-1}(a+b x)^3}{x^2} \, dx\)

\(\Big \downarrow \) 5781

\(\displaystyle b \int \frac {(a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^3}{b^2 x^2}d\sec ^{-1}(a+b x)\)

\(\Big \downarrow \) 4926

\(\displaystyle b \left (-3 \int -\frac {\sec ^{-1}(a+b x)^2}{b x}d\sec ^{-1}(a+b x)-\frac {\sec ^{-1}(a+b x)^3}{b x}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b \left (-3 \int \frac {\sec ^{-1}(a+b x)^2}{a-\csc \left (\sec ^{-1}(a+b x)+\frac {\pi }{2}\right )}d\sec ^{-1}(a+b x)-\frac {\sec ^{-1}(a+b x)^3}{b x}\right )\)

\(\Big \downarrow \) 4679

\(\displaystyle b \left (-3 \int \left (\frac {\sec ^{-1}(a+b x)^2}{a}+\frac {\sec ^{-1}(a+b x)^2}{a \left (\frac {a}{a+b x}-1\right )}\right )d\sec ^{-1}(a+b x)-\frac {\sec ^{-1}(a+b x)^3}{b x}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle b \left (-\frac {\sec ^{-1}(a+b x)^3}{b x}-3 \left (\frac {2 \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {2 \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}+\frac {2 i \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {2 i \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}+\frac {i \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a \sqrt {1-a^2}}-\frac {i \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a \sqrt {1-a^2}}+\frac {\sec ^{-1}(a+b x)^3}{3 a}\right )\right )\)

Input:

Int[ArcSec[a + b*x]^3/x^2,x]
 

Output:

b*(-(ArcSec[a + b*x]^3/(b*x)) - 3*(ArcSec[a + b*x]^3/(3*a) + (I*ArcSec[a + 
 b*x]^2*Log[1 - (a*E^(I*ArcSec[a + b*x]))/(1 - Sqrt[1 - a^2])])/(a*Sqrt[1 
- a^2]) - (I*ArcSec[a + b*x]^2*Log[1 - (a*E^(I*ArcSec[a + b*x]))/(1 + Sqrt 
[1 - a^2])])/(a*Sqrt[1 - a^2]) + (2*ArcSec[a + b*x]*PolyLog[2, (a*E^(I*Arc 
Sec[a + b*x]))/(1 - Sqrt[1 - a^2])])/(a*Sqrt[1 - a^2]) - (2*ArcSec[a + b*x 
]*PolyLog[2, (a*E^(I*ArcSec[a + b*x]))/(1 + Sqrt[1 - a^2])])/(a*Sqrt[1 - a 
^2]) + ((2*I)*PolyLog[3, (a*E^(I*ArcSec[a + b*x]))/(1 - Sqrt[1 - a^2])])/( 
a*Sqrt[1 - a^2]) - ((2*I)*PolyLog[3, (a*E^(I*ArcSec[a + b*x]))/(1 + Sqrt[1 
 - a^2])])/(a*Sqrt[1 - a^2])))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4679

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si 
n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt 
Q[m, 0]
 

rule 4926

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]
 

rule 5781

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]
 

Maple [F]

\[\int \frac {\operatorname {arcsec}\left (b x +a \right )^{3}}{x^{2}}d x\]

Input:

int(arcsec(b*x+a)^3/x^2,x)
 

Output:

int(arcsec(b*x+a)^3/x^2,x)
 

Fricas [F]

\[ \int \frac {\sec ^{-1}(a+b x)^3}{x^2} \, dx=\int { \frac {\operatorname {arcsec}\left (b x + a\right )^{3}}{x^{2}} \,d x } \] Input:

integrate(arcsec(b*x+a)^3/x^2,x, algorithm="fricas")
 

Output:

integral(arcsec(b*x + a)^3/x^2, x)
 

Sympy [F]

\[ \int \frac {\sec ^{-1}(a+b x)^3}{x^2} \, dx=\int \frac {\operatorname {asec}^{3}{\left (a + b x \right )}}{x^{2}}\, dx \] Input:

integrate(asec(b*x+a)**3/x**2,x)
 

Output:

Integral(asec(a + b*x)**3/x**2, x)
 

Maxima [F]

\[ \int \frac {\sec ^{-1}(a+b x)^3}{x^2} \, dx=\int { \frac {\operatorname {arcsec}\left (b x + a\right )^{3}}{x^{2}} \,d x } \] Input:

integrate(arcsec(b*x+a)^3/x^2,x, algorithm="maxima")
 

Output:

-1/4*(4*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^3 - 3*arctan(sqrt(b*x 
+ a + 1)*sqrt(b*x + a - 1))*log(b^2*x^2 + 2*a*b*x + a^2)^2 - 4*x*integrate 
(3/4*((4*b*x*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1))^2 - b*x*log(b^2*x 
^2 + 2*a*b*x + a^2)^2)*sqrt(b*x + a + 1)*sqrt(b*x + a - 1) - 4*((b^3*x^3 + 
 3*a*b^2*x^2 + a^3 + (3*a^2 - 1)*b*x - a)*log(b*x + a)^2 + (b^3*x^3 + 2*a* 
b^2*x^2 + (a^2 - 1)*b*x - (b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2 - 1)*b*x - 
 a)*log(b*x + a))*log(b^2*x^2 + 2*a*b*x + a^2))*arctan(sqrt(b*x + a + 1)*s 
qrt(b*x + a - 1)))/(b^3*x^5 + 3*a*b^2*x^4 + (3*a^2 - 1)*b*x^3 + (a^3 - a)* 
x^2), x))/x
 

Giac [F]

\[ \int \frac {\sec ^{-1}(a+b x)^3}{x^2} \, dx=\int { \frac {\operatorname {arcsec}\left (b x + a\right )^{3}}{x^{2}} \,d x } \] Input:

integrate(arcsec(b*x+a)^3/x^2,x, algorithm="giac")
 

Output:

integrate(arcsec(b*x + a)^3/x^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^{-1}(a+b x)^3}{x^2} \, dx=\int \frac {{\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}^3}{x^2} \,d x \] Input:

int(acos(1/(a + b*x))^3/x^2,x)
 

Output:

int(acos(1/(a + b*x))^3/x^2, x)
 

Reduce [F]

\[ \int \frac {\sec ^{-1}(a+b x)^3}{x^2} \, dx=\int \frac {\mathit {asec} \left (b x +a \right )^{3}}{x^{2}}d x \] Input:

int(asec(b*x+a)^3/x^2,x)
 

Output:

int(asec(a + b*x)**3/x**2,x)