3.1.36 \(\int \frac {\sec ^{-1}(a+b x)^3}{x} \, dx\) [36]

3.1.36.1 Optimal result

Integrand size = 12, antiderivative size = 430 \[ \int \frac {\sec ^{-1}(a+b x)^3}{x} \, dx=\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-3 i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-3 i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+\frac {3}{2} i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )+6 \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+6 \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\frac {3}{2} \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,-e^{2 i \sec ^{-1}(a+b x)}\right )+6 i \operatorname {PolyLog}\left (4,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+6 i \operatorname {PolyLog}\left (4,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\frac {3}{4} i \operatorname {PolyLog}\left (4,-e^{2 i \sec ^{-1}(a+b x)}\right ) \]

output

-arcsec(b*x+a)^3*ln(1+(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2)+arcsec(b*x+a) 
^3*ln(1-a*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/(1-(-a^2+1)^(1/2)))+arcsec(b 
*x+a)^3*ln(1-a*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/(1+(-a^2+1)^(1/2)))+3/2 
*I*arcsec(b*x+a)^2*polylog(2,-(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2)-3*I*a 
rcsec(b*x+a)^2*polylog(2,a*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/(1-(-a^2+1) 
^(1/2)))-3*I*arcsec(b*x+a)^2*polylog(2,a*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2 
))/(1+(-a^2+1)^(1/2)))-3/2*arcsec(b*x+a)*polylog(3,-(1/(b*x+a)+I*(1-1/(b*x 
+a)^2)^(1/2))^2)+6*arcsec(b*x+a)*polylog(3,a*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^ 
(1/2))/(1-(-a^2+1)^(1/2)))+6*arcsec(b*x+a)*polylog(3,a*(1/(b*x+a)+I*(1-1/( 
b*x+a)^2)^(1/2))/(1+(-a^2+1)^(1/2)))-3/4*I*polylog(4,-(1/(b*x+a)+I*(1-1/(b 
*x+a)^2)^(1/2))^2)+6*I*polylog(4,a*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/(1- 
(-a^2+1)^(1/2)))+6*I*polylog(4,a*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/(1+(- 
a^2+1)^(1/2)))
 

3.1.36.2 Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1058\) vs. \(2(430)=860\).

Time = 2.92 (sec) , antiderivative size = 1058, normalized size of antiderivative = 2.46 \[ \int \frac {\sec ^{-1}(a+b x)^3}{x} \, dx=2 \sec ^{-1}(a+b x)^3 \log \left (1+\frac {a e^{i \sec ^{-1}(a+b x)}}{-1+\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^3 \log \left (1+\frac {\left (-1+\sqrt {1-a^2}\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )-6 \sec ^{-1}(a+b x)^2 \arcsin \left (\frac {\sqrt {\frac {-1+a}{a}}}{\sqrt {2}}\right ) \log \left (1+\frac {\left (-1+\sqrt {1-a^2}\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )+2 \sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^3 \log \left (1-\frac {\left (1+\sqrt {1-a^2}\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )+6 \sec ^{-1}(a+b x)^2 \arcsin \left (\frac {\sqrt {\frac {-1+a}{a}}}{\sqrt {2}}\right ) \log \left (1-\frac {\left (1+\sqrt {1-a^2}\right ) e^{i \sec ^{-1}(a+b x)}}{a}\right )-3 \sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )+2 \sec ^{-1}(a+b x)^3 \log \left (\frac {2 \left (\frac {1}{a+b x}+i \sqrt {1-\frac {1}{(a+b x)^2}}\right )}{a+b x}\right )-\sec ^{-1}(a+b x)^3 \log \left (1+\frac {a \left (\frac {1}{a+b x}+i \sqrt {1-\frac {1}{(a+b x)^2}}\right )}{-1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x)^3 \log \left (1+\frac {\left (-1+\sqrt {1-a^2}\right ) \left (\frac {1}{a+b x}+i \sqrt {1-\frac {1}{(a+b x)^2}}\right )}{a}\right )+6 \sec ^{-1}(a+b x)^2 \arcsin \left (\frac {\sqrt {\frac {-1+a}{a}}}{\sqrt {2}}\right ) \log \left (1+\frac {\left (-1+\sqrt {1-a^2}\right ) \left (\frac {1}{a+b x}+i \sqrt {1-\frac {1}{(a+b x)^2}}\right )}{a}\right )-\sec ^{-1}(a+b x)^3 \log \left (1-\frac {a \left (\frac {1}{a+b x}+i \sqrt {1-\frac {1}{(a+b x)^2}}\right )}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x)^3 \log \left (1-\frac {\left (1+\sqrt {1-a^2}\right ) \left (\frac {1}{a+b x}+i \sqrt {1-\frac {1}{(a+b x)^2}}\right )}{a}\right )-6 \sec ^{-1}(a+b x)^2 \arcsin \left (\frac {\sqrt {\frac {-1+a}{a}}}{\sqrt {2}}\right ) \log \left (1-\frac {\left (1+\sqrt {1-a^2}\right ) \left (\frac {1}{a+b x}+i \sqrt {1-\frac {1}{(a+b x)^2}}\right )}{a}\right )-3 i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i \sec ^{-1}(a+b x)}}{-1+\sqrt {1-a^2}}\right )-3 i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+\frac {3}{2} i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )+6 \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,-\frac {a e^{i \sec ^{-1}(a+b x)}}{-1+\sqrt {1-a^2}}\right )+6 \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\frac {3}{2} \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,-e^{2 i \sec ^{-1}(a+b x)}\right )+6 i \operatorname {PolyLog}\left (4,-\frac {a e^{i \sec ^{-1}(a+b x)}}{-1+\sqrt {1-a^2}}\right )+6 i \operatorname {PolyLog}\left (4,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\frac {3}{4} i \operatorname {PolyLog}\left (4,-e^{2 i \sec ^{-1}(a+b x)}\right ) \]

input

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

output

2*ArcSec[a + b*x]^3*Log[1 + (a*E^(I*ArcSec[a + b*x]))/(-1 + Sqrt[1 - a^2]) 
] + ArcSec[a + b*x]^3*Log[1 + ((-1 + Sqrt[1 - a^2])*E^(I*ArcSec[a + b*x])) 
/a] - 6*ArcSec[a + b*x]^2*ArcSin[Sqrt[(-1 + a)/a]/Sqrt[2]]*Log[1 + ((-1 + 
Sqrt[1 - a^2])*E^(I*ArcSec[a + b*x]))/a] + 2*ArcSec[a + b*x]^3*Log[1 - (a* 
E^(I*ArcSec[a + b*x]))/(1 + Sqrt[1 - a^2])] + ArcSec[a + b*x]^3*Log[1 - (( 
1 + Sqrt[1 - a^2])*E^(I*ArcSec[a + b*x]))/a] + 6*ArcSec[a + b*x]^2*ArcSin[ 
Sqrt[(-1 + a)/a]/Sqrt[2]]*Log[1 - ((1 + Sqrt[1 - a^2])*E^(I*ArcSec[a + b*x 
]))/a] - 3*ArcSec[a + b*x]^3*Log[1 + E^((2*I)*ArcSec[a + b*x])] + 2*ArcSec 
[a + b*x]^3*Log[(2*((a + b*x)^(-1) + I*Sqrt[1 - (a + b*x)^(-2)]))/(a + b*x 
)] - ArcSec[a + b*x]^3*Log[1 + (a*((a + b*x)^(-1) + I*Sqrt[1 - (a + b*x)^( 
-2)]))/(-1 + Sqrt[1 - a^2])] - ArcSec[a + b*x]^3*Log[1 + ((-1 + Sqrt[1 - a 
^2])*((a + b*x)^(-1) + I*Sqrt[1 - (a + b*x)^(-2)]))/a] + 6*ArcSec[a + b*x] 
^2*ArcSin[Sqrt[(-1 + a)/a]/Sqrt[2]]*Log[1 + ((-1 + Sqrt[1 - a^2])*((a + b* 
x)^(-1) + I*Sqrt[1 - (a + b*x)^(-2)]))/a] - ArcSec[a + b*x]^3*Log[1 - (a*( 
(a + b*x)^(-1) + I*Sqrt[1 - (a + b*x)^(-2)]))/(1 + Sqrt[1 - a^2])] - ArcSe 
c[a + b*x]^3*Log[1 - ((1 + Sqrt[1 - a^2])*((a + b*x)^(-1) + I*Sqrt[1 - (a 
+ b*x)^(-2)]))/a] - 6*ArcSec[a + b*x]^2*ArcSin[Sqrt[(-1 + a)/a]/Sqrt[2]]*L 
og[1 - ((1 + Sqrt[1 - a^2])*((a + b*x)^(-1) + I*Sqrt[1 - (a + b*x)^(-2)])) 
/a] - (3*I)*ArcSec[a + b*x]^2*PolyLog[2, -((a*E^(I*ArcSec[a + b*x]))/(-1 + 
 Sqrt[1 - a^2]))] - (3*I)*ArcSec[a + b*x]^2*PolyLog[2, (a*E^(I*ArcSec[a...
 

3.1.36.3 Rubi [A] (verified)

Time = 1.85 (sec) , antiderivative size = 529, normalized size of antiderivative = 1.23, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {5781, 25, 5062, 5041, 25, 3042, 4202, 2620, 3011, 5031, 2620, 3011, 7163, 2720, 7143}

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.

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

\(\Big \downarrow \) 5781

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 5062

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

\(\Big \downarrow \) 5041

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4202

\(\displaystyle -2 i \int \frac {e^{2 i \sec ^{-1}(a+b x)} \sec ^{-1}(a+b x)^3}{1+e^{2 i \sec ^{-1}(a+b x)}}d\sec ^{-1}(a+b x)+a \int \frac {\sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^3}{1-\frac {a}{a+b x}}d\sec ^{-1}(a+b x)+\frac {1}{4} i \sec ^{-1}(a+b x)^4\)

\(\Big \downarrow \) 2620

\(\displaystyle a \int \frac {\sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^3}{1-\frac {a}{a+b x}}d\sec ^{-1}(a+b x)-2 i \left (\frac {3}{2} i \int \sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )d\sec ^{-1}(a+b x)-\frac {1}{2} i \sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )\right )+\frac {1}{4} i \sec ^{-1}(a+b x)^4\)

\(\Big \downarrow \) 3011

\(\displaystyle -2 i \left (\frac {3}{2} i \left (\frac {1}{2} i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )-i \int \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )d\sec ^{-1}(a+b x)\right )-\frac {1}{2} i \sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )\right )+a \int \frac {\sqrt {1-\frac {1}{(a+b x)^2}} \sec ^{-1}(a+b x)^3}{1-\frac {a}{a+b x}}d\sec ^{-1}(a+b x)+\frac {1}{4} i \sec ^{-1}(a+b x)^4\)

\(\Big \downarrow \) 5031

\(\displaystyle a \left (-i \int \frac {e^{i \sec ^{-1}(a+b x)} \sec ^{-1}(a+b x)^3}{-e^{i \sec ^{-1}(a+b x)} a-\sqrt {1-a^2}+1}d\sec ^{-1}(a+b x)-i \int \frac {e^{i \sec ^{-1}(a+b x)} \sec ^{-1}(a+b x)^3}{-e^{i \sec ^{-1}(a+b x)} a+\sqrt {1-a^2}+1}d\sec ^{-1}(a+b x)-\frac {i \sec ^{-1}(a+b x)^4}{4 a}\right )-2 i \left (\frac {3}{2} i \left (\frac {1}{2} i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )-i \int \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )d\sec ^{-1}(a+b x)\right )-\frac {1}{2} i \sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )\right )+\frac {1}{4} i \sec ^{-1}(a+b x)^4\)

\(\Big \downarrow \) 2620

\(\displaystyle a \left (-i \left (\frac {i \sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a}-\frac {3 i \int \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )d\sec ^{-1}(a+b x)}{a}\right )-i \left (\frac {i \sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a}-\frac {3 i \int \sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )d\sec ^{-1}(a+b x)}{a}\right )-\frac {i \sec ^{-1}(a+b x)^4}{4 a}\right )-2 i \left (\frac {3}{2} i \left (\frac {1}{2} i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )-i \int \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )d\sec ^{-1}(a+b x)\right )-\frac {1}{2} i \sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )\right )+\frac {1}{4} i \sec ^{-1}(a+b x)^4\)

\(\Big \downarrow \) 3011

\(\displaystyle a \left (-i \left (\frac {i \sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a}-\frac {3 i \left (i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \int \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )d\sec ^{-1}(a+b x)\right )}{a}\right )-i \left (\frac {i \sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a}-\frac {3 i \left (i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )-2 i \int \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )d\sec ^{-1}(a+b x)\right )}{a}\right )-\frac {i \sec ^{-1}(a+b x)^4}{4 a}\right )-2 i \left (\frac {3}{2} i \left (\frac {1}{2} i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )-i \int \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )d\sec ^{-1}(a+b x)\right )-\frac {1}{2} i \sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )\right )+\frac {1}{4} i \sec ^{-1}(a+b x)^4\)

\(\Big \downarrow \) 7163

\(\displaystyle a \left (-i \left (\frac {i \sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a}-\frac {3 i \left (i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \left (i \int \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )d\sec ^{-1}(a+b x)-i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )\right )\right )}{a}\right )-i \left (\frac {i \sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a}-\frac {3 i \left (i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )-2 i \left (i \int \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )d\sec ^{-1}(a+b x)-i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )\right )\right )}{a}\right )-\frac {i \sec ^{-1}(a+b x)^4}{4 a}\right )-2 i \left (\frac {3}{2} i \left (\frac {1}{2} i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )-i \left (\frac {1}{2} i \int \operatorname {PolyLog}\left (3,-e^{2 i \sec ^{-1}(a+b x)}\right )d\sec ^{-1}(a+b x)-\frac {1}{2} i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,-e^{2 i \sec ^{-1}(a+b x)}\right )\right )\right )-\frac {1}{2} i \sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )\right )+\frac {1}{4} i \sec ^{-1}(a+b x)^4\)

\(\Big \downarrow \) 2720

\(\displaystyle a \left (-i \left (\frac {i \sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a}-\frac {3 i \left (i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \left (\int e^{-i \sec ^{-1}(a+b x)} \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )de^{i \sec ^{-1}(a+b x)}-i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )\right )\right )}{a}\right )-i \left (\frac {i \sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a}-\frac {3 i \left (i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )-2 i \left (\int e^{-i \sec ^{-1}(a+b x)} \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )de^{i \sec ^{-1}(a+b x)}-i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )\right )\right )}{a}\right )-\frac {i \sec ^{-1}(a+b x)^4}{4 a}\right )-2 i \left (\frac {3}{2} i \left (\frac {1}{2} i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )-i \left (\frac {1}{4} \int e^{-2 i \sec ^{-1}(a+b x)} \operatorname {PolyLog}\left (3,-e^{2 i \sec ^{-1}(a+b x)}\right )de^{2 i \sec ^{-1}(a+b x)}-\frac {1}{2} i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,-e^{2 i \sec ^{-1}(a+b x)}\right )\right )\right )-\frac {1}{2} i \sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )\right )+\frac {1}{4} i \sec ^{-1}(a+b x)^4\)

\(\Big \downarrow \) 7143

\(\displaystyle a \left (-i \left (\frac {i \sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )}{a}-\frac {3 i \left (i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \left (\operatorname {PolyLog}\left (4,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )\right )\right )}{a}\right )-i \left (\frac {i \sec ^{-1}(a+b x)^3 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )}{a}-\frac {3 i \left (i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )-2 i \left (\operatorname {PolyLog}\left (4,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )-i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{\sqrt {1-a^2}+1}\right )\right )\right )}{a}\right )-\frac {i \sec ^{-1}(a+b x)^4}{4 a}\right )-2 i \left (\frac {3}{2} i \left (\frac {1}{2} i \sec ^{-1}(a+b x)^2 \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )-i \left (\frac {1}{4} \operatorname {PolyLog}\left (4,-e^{2 i \sec ^{-1}(a+b x)}\right )-\frac {1}{2} i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (3,-e^{2 i \sec ^{-1}(a+b x)}\right )\right )\right )-\frac {1}{2} i \sec ^{-1}(a+b x)^3 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )\right )+\frac {1}{4} i \sec ^{-1}(a+b x)^4\)

input

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

output

(I/4)*ArcSec[a + b*x]^4 + a*(((-1/4*I)*ArcSec[a + b*x]^4)/a - I*((I*ArcSec 
[a + b*x]^3*Log[1 - (a*E^(I*ArcSec[a + b*x]))/(1 - Sqrt[1 - a^2])])/a - (( 
3*I)*(I*ArcSec[a + b*x]^2*PolyLog[2, (a*E^(I*ArcSec[a + b*x]))/(1 - Sqrt[1 
 - a^2])] - (2*I)*((-I)*ArcSec[a + b*x]*PolyLog[3, (a*E^(I*ArcSec[a + b*x] 
))/(1 - Sqrt[1 - a^2])] + PolyLog[4, (a*E^(I*ArcSec[a + b*x]))/(1 - Sqrt[1 
 - a^2])])))/a) - I*((I*ArcSec[a + b*x]^3*Log[1 - (a*E^(I*ArcSec[a + b*x]) 
)/(1 + Sqrt[1 - a^2])])/a - ((3*I)*(I*ArcSec[a + b*x]^2*PolyLog[2, (a*E^(I 
*ArcSec[a + b*x]))/(1 + Sqrt[1 - a^2])] - (2*I)*((-I)*ArcSec[a + b*x]*Poly 
Log[3, (a*E^(I*ArcSec[a + b*x]))/(1 + Sqrt[1 - a^2])] + PolyLog[4, (a*E^(I 
*ArcSec[a + b*x]))/(1 + Sqrt[1 - a^2])])))/a)) - (2*I)*((-1/2*I)*ArcSec[a 
+ b*x]^3*Log[1 + E^((2*I)*ArcSec[a + b*x])] + ((3*I)/2)*((I/2)*ArcSec[a + 
b*x]^2*PolyLog[2, -E^((2*I)*ArcSec[a + b*x])] - I*((-1/2*I)*ArcSec[a + b*x 
]*PolyLog[3, -E^((2*I)*ArcSec[a + b*x])] + PolyLog[4, -E^((2*I)*ArcSec[a + 
 b*x])]/4)))
 

3.1.36.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

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

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^m*(E^(2*I*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 5031

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)])/(Cos[(c_.) + (d_.) 
*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[I*((e + f*x)^(m + 1)/(b*f*(m + 1))) 
, x] + (-Simp[I   Int[(e + f*x)^m*(E^(I*(c + d*x))/(a - Rt[a^2 - b^2, 2] + 
b*E^(I*(c + d*x)))), x], x] - Simp[I   Int[(e + f*x)^m*(E^(I*(c + d*x))/(a 
+ Rt[a^2 - b^2, 2] + b*E^(I*(c + d*x)))), x], x]) /; FreeQ[{a, b, c, d, e, 
f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]
 

rule 5041

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

rule 5062

Int[(((e_.) + (f_.)*(x_))^(m_.)*(F_)[(c_.) + (d_.)*(x_)]^(n_.)*(G_)[(c_.) + 
 (d_.)*(x_)]^(p_.))/((a_) + (b_.)*Sec[(c_.) + (d_.)*(x_)]), x_Symbol] :> In 
t[(e + f*x)^m*Cos[c + d*x]*F[c + d*x]^n*(G[c + d*x]^p/(b + a*Cos[c + d*x])) 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && TrigQ[F] && TrigQ[G] && IntegersQ[m 
, n, p]
 

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]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

3.1.36.4 Maple [F]

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

input

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

output

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

3.1.36.5 Fricas [F]

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

input

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

output

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

3.1.36.6 Sympy [F]

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

input

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

output

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

3.1.36.7 Maxima [F]

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

input

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

output

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

3.1.36.8 Giac [F]

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

input

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

output

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

3.1.36.9 Mupad [F(-1)]

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

input

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

output

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