\(\int \frac {\sec ^{-1}(a+b x)^2}{x} \, dx\) [31]

Optimal result

Integrand size = 12, antiderivative size = 310 \[ \int \frac {\sec ^{-1}(a+b x)^2}{x} \, dx=\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\sec ^{-1}(a+b x)^2 \log \left (1-\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\sec ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sec ^{-1}(a+b x)}\right )-2 i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+i \sec ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(a+b x)}\right )+2 \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+2 \operatorname {PolyLog}\left (3,\frac {a e^{i \sec ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,-e^{2 i \sec ^{-1}(a+b x)}\right ) \] Output:

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)))+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)))-arcsec(b*x+a)^2*ln(1+(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^2)-2*I* 
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)))-2*I*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)))+I*arcsec(b*x+a)*polylog(2,-(1/(b*x+a)+I*(1-1/(b*x+a)^2 
)^(1/2))^2)+2*polylog(3,a*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/(1-(-a^2+1)^ 
(1/2)))+2*polylog(3,a*(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))/(1+(-a^2+1)^(1/2 
)))-1/2*polylog(3,-(1/(b*x+a)+I*(1-1/(b*x+a)^2)^(1/2))^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. \(813\) vs. \(2(310)=620\).

Time = 1.52 (sec) , antiderivative size = 813, normalized size of antiderivative = 2.62 \[ \int \frac {\sec ^{-1}(a+b x)^2}{x} \, dx =\text {Too large to display} \] Input:

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

Output:

ArcSec[a + b*x]^2*Log[1 + (a*E^(I*ArcSec[a + b*x]))/(-1 + Sqrt[1 - a^2])] 
+ ArcSec[a + b*x]^2*Log[1 + ((-1 + Sqrt[1 - a^2])*E^(I*ArcSec[a + b*x]))/a 
] - 4*ArcSec[a + b*x]*ArcSin[Sqrt[(-1 + a)/a]/Sqrt[2]]*Log[1 + ((-1 + Sqrt 
[1 - a^2])*E^(I*ArcSec[a + b*x]))/a] + ArcSec[a + b*x]^2*Log[1 - (a*E^(I*A 
rcSec[a + b*x]))/(1 + Sqrt[1 - a^2])] + ArcSec[a + b*x]^2*Log[1 - ((1 + Sq 
rt[1 - a^2])*E^(I*ArcSec[a + b*x]))/a] + 4*ArcSec[a + b*x]*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]^2*Log[1 + E^((2*I)*ArcSec[a + b*x])] + ArcSec[a + b*x]^ 
2*Log[(2*((a + b*x)^(-1) + I*Sqrt[1 - (a + b*x)^(-2)]))/(a + b*x)] - ArcSe 
c[a + b*x]^2*Log[1 + ((-1 + Sqrt[1 - a^2])*((a + b*x)^(-1) + I*Sqrt[1 - (a 
 + b*x)^(-2)]))/a] + 4*ArcSec[a + b*x]*ArcSin[Sqrt[(-1 + a)/a]/Sqrt[2]]*Lo 
g[1 + ((-1 + Sqrt[1 - a^2])*((a + b*x)^(-1) + I*Sqrt[1 - (a + b*x)^(-2)])) 
/a] - ArcSec[a + b*x]^2*Log[1 - ((1 + Sqrt[1 - a^2])*((a + b*x)^(-1) + I*S 
qrt[1 - (a + b*x)^(-2)]))/a] - 4*ArcSec[a + b*x]*ArcSin[Sqrt[(-1 + a)/a]/S 
qrt[2]]*Log[1 - ((1 + Sqrt[1 - a^2])*((a + b*x)^(-1) + I*Sqrt[1 - (a + b*x 
)^(-2)]))/a] - (2*I)*ArcSec[a + b*x]*PolyLog[2, -((a*E^(I*ArcSec[a + b*x]) 
)/(-1 + Sqrt[1 - a^2]))] - (2*I)*ArcSec[a + b*x]*PolyLog[2, (a*E^(I*ArcSec 
[a + b*x]))/(1 + Sqrt[1 - a^2])] + I*ArcSec[a + b*x]*PolyLog[2, -E^((2*I)* 
ArcSec[a + b*x])] + 2*PolyLog[3, -((a*E^(I*ArcSec[a + b*x]))/(-1 + Sqrt[1 
- a^2]))] + 2*PolyLog[3, (a*E^(I*ArcSec[a + b*x]))/(1 + Sqrt[1 - a^2])]...
 

Rubi [A] (verified)

Time = 1.58 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.28, 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, 2720, 5031, 2620, 3011, 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.

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

\(\Big \downarrow \) 5781

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 5062

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

\(\Big \downarrow \) 5041

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 5031

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

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

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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