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

Optimal result

Integrand size = 12, antiderivative size = 324 \[ \int \frac {\csc ^{-1}(a+b x)^2}{x} \, dx=\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\csc ^{-1}(a+b x)^2 \log \left (1-e^{2 i \csc ^{-1}(a+b x)}\right )-2 i \csc ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )-2 i \csc ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )+i \csc ^{-1}(a+b x) \operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}(a+b x)}\right )+2 \operatorname {PolyLog}\left (3,-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1-\sqrt {1-a^2}}\right )+2 \operatorname {PolyLog}\left (3,-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,e^{2 i \csc ^{-1}(a+b x)}\right ) \] Output:

arccsc(b*x+a)^2*ln(1+I*a*(I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2))/(1-(-a^2+1)^(1/ 
2)))+arccsc(b*x+a)^2*ln(1+I*a*(I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2))/(1+(-a^2+1 
)^(1/2)))-arccsc(b*x+a)^2*ln(1-(I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2))^2)-2*I*ar 
ccsc(b*x+a)*polylog(2,-I*a*(I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2))/(1-(-a^2+1)^( 
1/2)))-2*I*arccsc(b*x+a)*polylog(2,-I*a*(I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2))/ 
(1+(-a^2+1)^(1/2)))+I*arccsc(b*x+a)*polylog(2,(I/(b*x+a)+(1-1/(b*x+a)^2)^( 
1/2))^2)+2*polylog(3,-I*a*(I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2))/(1-(-a^2+1)^(1 
/2)))+2*polylog(3,-I*a*(I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2))/(1+(-a^2+1)^(1/2) 
))-1/2*polylog(3,(I/(b*x+a)+(1-1/(b*x+a)^2)^(1/2))^2)
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.26 \[ \int \frac {\csc ^{-1}(a+b x)^2}{x} \, dx=\frac {i \pi ^3}{6}-\frac {1}{3} i \csc ^{-1}(a+b x)^3-\csc ^{-1}(a+b x)^2 \log \left (1-e^{-i \csc ^{-1}(a+b x)}\right )-\csc ^{-1}(a+b x)^2 \log \left (1+e^{i \csc ^{-1}(a+b x)}\right )+\csc ^{-1}(a+b x)^2 \log \left (1-\frac {i a e^{i \csc ^{-1}(a+b x)}}{-1+\sqrt {1-a^2}}\right )+\csc ^{-1}(a+b x)^2 \log \left (1+\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-2 i \csc ^{-1}(a+b x) \operatorname {PolyLog}\left (2,e^{-i \csc ^{-1}(a+b x)}\right )+2 i \csc ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-e^{i \csc ^{-1}(a+b x)}\right )-2 i \csc ^{-1}(a+b x) \operatorname {PolyLog}\left (2,\frac {i a e^{i \csc ^{-1}(a+b x)}}{-1+\sqrt {1-a^2}}\right )-2 i \csc ^{-1}(a+b x) \operatorname {PolyLog}\left (2,-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right )-2 \operatorname {PolyLog}\left (3,e^{-i \csc ^{-1}(a+b x)}\right )-2 \operatorname {PolyLog}\left (3,-e^{i \csc ^{-1}(a+b x)}\right )+2 \operatorname {PolyLog}\left (3,\frac {i a e^{i \csc ^{-1}(a+b x)}}{-1+\sqrt {1-a^2}}\right )+2 \operatorname {PolyLog}\left (3,-\frac {i a e^{i \csc ^{-1}(a+b x)}}{1+\sqrt {1-a^2}}\right ) \] Input:

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

Output:

(I/6)*Pi^3 - (I/3)*ArcCsc[a + b*x]^3 - ArcCsc[a + b*x]^2*Log[1 - E^((-I)*A 
rcCsc[a + b*x])] - ArcCsc[a + b*x]^2*Log[1 + E^(I*ArcCsc[a + b*x])] + ArcC 
sc[a + b*x]^2*Log[1 - (I*a*E^(I*ArcCsc[a + b*x]))/(-1 + Sqrt[1 - a^2])] + 
ArcCsc[a + b*x]^2*Log[1 + (I*a*E^(I*ArcCsc[a + b*x]))/(1 + Sqrt[1 - a^2])] 
 - (2*I)*ArcCsc[a + b*x]*PolyLog[2, E^((-I)*ArcCsc[a + b*x])] + (2*I)*ArcC 
sc[a + b*x]*PolyLog[2, -E^(I*ArcCsc[a + b*x])] - (2*I)*ArcCsc[a + b*x]*Pol 
yLog[2, (I*a*E^(I*ArcCsc[a + b*x]))/(-1 + Sqrt[1 - a^2])] - (2*I)*ArcCsc[a 
 + b*x]*PolyLog[2, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 + Sqrt[1 - a^2])] - 2 
*PolyLog[3, E^((-I)*ArcCsc[a + b*x])] - 2*PolyLog[3, -E^(I*ArcCsc[a + b*x] 
)] + 2*PolyLog[3, (I*a*E^(I*ArcCsc[a + b*x]))/(-1 + Sqrt[1 - a^2])] + 2*Po 
lyLog[3, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 + Sqrt[1 - a^2])]
 

Rubi [A] (verified)

Time = 1.69 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.22, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.417, Rules used = {5782, 25, 5063, 5040, 25, 3042, 25, 4200, 25, 2620, 3011, 2720, 5030, 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.

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

\(\Big \downarrow \) 5782

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 5063

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

\(\Big \downarrow \) 5040

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4200

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 5030

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

(I/3)*ArcCsc[a + b*x]^3 - a*(((I/3)*ArcCsc[a + b*x]^3)/a - (ArcCsc[a + b*x 
]^2*Log[1 + (I*a*E^(I*ArcCsc[a + b*x]))/(1 - Sqrt[1 - a^2])])/a - (ArcCsc[ 
a + b*x]^2*Log[1 + (I*a*E^(I*ArcCsc[a + b*x]))/(1 + Sqrt[1 - a^2])])/a + ( 
2*(I*ArcCsc[a + b*x]*PolyLog[2, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 - Sqrt[1 
 - a^2])] - PolyLog[3, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 - Sqrt[1 - a^2])] 
))/a + (2*(I*ArcCsc[a + b*x]*PolyLog[2, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 
+ Sqrt[1 - a^2])] - PolyLog[3, ((-I)*a*E^(I*ArcCsc[a + b*x]))/(1 + Sqrt[1 
- a^2])]))/a) + (2*I)*((I/2)*ArcCsc[a + b*x]^2*Log[1 - E^((2*I)*ArcCsc[a + 
 b*x])] - I*((I/2)*ArcCsc[a + b*x]*PolyLog[2, E^((2*I)*ArcCsc[a + b*x])] - 
 PolyLog[3, E^((2*I)*ArcCsc[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 4200

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))), x] 
, x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]
 

rule 5030

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

rule 5040

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

rule 5063

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

rule 5782

Int[((a_.) + ArcCsc[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[-(d^(m + 1))^(-1)   Subst[Int[(a + b*x)^p*Csc[x]*Cot 
[x]*(d*e - c*f + f*Csc[x])^m, x], x, ArcCsc[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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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