\(\int \frac {a+b \sec ^{-1}(c x)}{x (d+e x^2)} \, dx\) [94]

Optimal result

Integrand size = 21, antiderivative size = 459 \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )} \, dx=\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b d}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d}+\frac {i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d}+\frac {i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d} \] Output:

1/2*I*(a+b*arcsec(c*x))^2/b/d-1/2*(a+b*arcsec(c*x))*ln(1-c*(-d)^(1/2)*(1/c 
/x+I*(1-1/c^2/x^2)^(1/2))/(e^(1/2)-(c^2*d+e)^(1/2)))/d-1/2*(a+b*arcsec(c*x 
))*ln(1+c*(-d)^(1/2)*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/(e^(1/2)-(c^2*d+e)^(1/2 
)))/d-1/2*(a+b*arcsec(c*x))*ln(1-c*(-d)^(1/2)*(1/c/x+I*(1-1/c^2/x^2)^(1/2) 
)/(e^(1/2)+(c^2*d+e)^(1/2)))/d-1/2*(a+b*arcsec(c*x))*ln(1+c*(-d)^(1/2)*(1/ 
c/x+I*(1-1/c^2/x^2)^(1/2))/(e^(1/2)+(c^2*d+e)^(1/2)))/d+1/2*I*b*polylog(2, 
-c*(-d)^(1/2)*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/(e^(1/2)-(c^2*d+e)^(1/2)))/d+1 
/2*I*b*polylog(2,c*(-d)^(1/2)*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/(e^(1/2)-(c^2* 
d+e)^(1/2)))/d+1/2*I*b*polylog(2,-c*(-d)^(1/2)*(1/c/x+I*(1-1/c^2/x^2)^(1/2 
))/(e^(1/2)+(c^2*d+e)^(1/2)))/d+1/2*I*b*polylog(2,c*(-d)^(1/2)*(1/c/x+I*(1 
-1/c^2/x^2)^(1/2))/(e^(1/2)+(c^2*d+e)^(1/2)))/d
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 876, normalized size of antiderivative = 1.91 \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )} \, dx =\text {Too large to display} \] Input:

Integrate[(a + b*ArcSec[c*x])/(x*(d + e*x^2)),x]
 

Output:

((I/2)*(b*ArcSec[c*x]^2 - 4*b*ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqr 
t[2]]*ArcTan[(((-I)*c*Sqrt[d] + Sqrt[e])*Tan[ArcSec[c*x]/2])/Sqrt[c^2*d + 
e]] - 4*b*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*ArcTan[((I*c*S 
qrt[d] + Sqrt[e])*Tan[ArcSec[c*x]/2])/Sqrt[c^2*d + e]] + I*b*ArcSec[c*x]*L 
og[1 + (I*(Sqrt[e] - Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + (2 
*I)*b*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 + (I*(Sqrt[e 
] - Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + I*b*ArcSec[c*x]*Log 
[1 + (I*(-Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + (2* 
I)*b*ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 + (I*(-Sqrt[e 
] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + I*b*ArcSec[c*x]*Log 
[1 - (I*(Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - (2*I 
)*b*ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 - (I*(Sqrt[e] 
+ Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + I*b*ArcSec[c*x]*Log[1 
 + (I*(Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - (2*I)* 
b*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 + (I*(Sqrt[e] + 
Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - (2*I)*a*Log[x] + I*a*Lo 
g[d + e*x^2] + b*PolyLog[2, ((-I)*(-Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec 
[c*x]))/(c*Sqrt[d])] + b*PolyLog[2, (I*(-Sqrt[e] + Sqrt[c^2*d + e])*E^(I*A 
rcSec[c*x]))/(c*Sqrt[d])] + b*PolyLog[2, ((-I)*(Sqrt[e] + Sqrt[c^2*d + e]) 
*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + b*PolyLog[2, (I*(Sqrt[e] + Sqrt[c^2*...
 

Rubi [A] (verified)

Time = 1.38 (sec) , antiderivative size = 511, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5763, 5233, 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.

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

\(\Big \downarrow \) 5763

\(\displaystyle -\int \frac {a+b \arccos \left (\frac {1}{c x}\right )}{\left (\frac {d}{x^2}+e\right ) x}d\frac {1}{x}\)

\(\Big \downarrow \) 5233

\(\displaystyle -\int \left (\frac {\sqrt {-d} \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{2 d \left (\frac {\sqrt {-d}}{x}+\sqrt {e}\right )}-\frac {\sqrt {-d} \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{2 d \left (\sqrt {e}-\frac {\sqrt {-d}}{x}\right )}\right )d\frac {1}{x}\)

\(\Big \downarrow \) 2009

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

Input:

Int[(a + b*ArcSec[c*x])/(x*(d + e*x^2)),x]
 

Output:

((I/2)*(a + b*ArcCos[1/(c*x)])^2)/(b*d) - ((a + b*ArcCos[1/(c*x)])*Log[1 - 
 (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/(2*d) - 
((a + b*ArcCos[1/(c*x)])*Log[1 + (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[ 
e] - Sqrt[c^2*d + e])])/(2*d) - ((a + b*ArcCos[1/(c*x)])*Log[1 - (c*Sqrt[- 
d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e])])/(2*d) - ((a + b*Ar 
cCos[1/(c*x)])*Log[1 + (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[ 
c^2*d + e])])/(2*d) + ((I/2)*b*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcCos[1/(c*x 
)]))/(Sqrt[e] - Sqrt[c^2*d + e]))])/d + ((I/2)*b*PolyLog[2, (c*Sqrt[-d]*E^ 
(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/d + ((I/2)*b*PolyLog[2, 
 -((c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e]))])/d + ( 
(I/2)*b*PolyLog[2, (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2* 
d + e])])/d
 

Defintions of rubi rules used

rule 2009

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

rule 5233

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ 
.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcCos[c*x])^n, ( 
f*x)^m*(d + e*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[c^2*d + 
 e, 0] && IGtQ[n, 0] && IntegerQ[p] && IntegerQ[m]
 

rule 5763

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_.) + (e_.)*(x_) 
^2)^(p_.), x_Symbol] :> -Subst[Int[(e + d*x^2)^p*((a + b*ArcCos[x/c])^n/x^( 
m + 2*(p + 1))), x], x, 1/x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[n, 0] 
&& IntegerQ[m] && IntegerQ[p]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.97 (sec) , antiderivative size = 1970, normalized size of antiderivative = 4.29

method	result	size
parts	\(\text {Expression too large to display}\)	\(1970\)
derivativedivides	\(\text {Expression too large to display}\)	\(1997\)
default	\(\text {Expression too large to display}\)	\(1997\)

Input:

int((a+b*arcsec(c*x))/x/(e*x^2+d),x,method=_RETURNVERBOSE)
 

Output:

a/d*ln(x)-1/2*a/d*ln(e*x^2+d)+b*(1/2*I/d*sum((_R1^2*c^2*d+2*c^2*d+4*e)/(_R 
1^2*c^2*d+c^2*d+2*e)*(I*arcsec(c*x)*ln((_R1-1/c/x-I*(1-1/c^2/x^2)^(1/2))/_ 
R1)+dilog((_R1-1/c/x-I*(1-1/c^2/x^2)^(1/2))/_R1)),_R1=RootOf(c^2*d*_Z^4+(2 
*c^2*d+4*e)*_Z^2+c^2*d))-1/4*I*(e*(c^2*d+e))^(1/2)/e/(c^2*d+e)*arcsec(c*x) 
^2*c^2-I*(-(e*(c^2*d+e))^(1/2)*c^2*d+2*c^2*d*e-2*(e*(c^2*d+e))^(1/2)*e+2*e 
^2)*arcsec(c*x)^2/(c^2*d+e)/c^2/d^2-1/2*I*(-(e*(c^2*d+e))^(1/2)*c^2*d+2*c^ 
2*d*e-2*(e*(c^2*d+e))^(1/2)*e+2*e^2)*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2 
)^(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))/(c^2*d+e)/c^2/d^2-1/2*I*(e* 
(c^2*d+e))^(1/2)/d/(c^2*d+e)*arcsec(c*x)^2-1/4*I*(-(e*(c^2*d+e))^(1/2)*c^2 
*d+2*c^2*d*e-2*(e*(c^2*d+e))^(1/2)*e+2*e^2)*arcsec(c*x)^2/d/e/(c^2*d+e)+1/ 
2*(e*(c^2*d+e))^(1/2)/d/(c^2*d+e)*arcsec(c*x)*ln(1-d*c^2*(1/c/x+I*(1-1/c^2 
/x^2)^(1/2))^2/(-c^2*d+2*(e*(c^2*d+e))^(1/2)-2*e))+1/4*(e*(c^2*d+e))^(1/2) 
/e/(c^2*d+e)*c^2*arcsec(c*x)*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(- 
c^2*d+2*(e*(c^2*d+e))^(1/2)-2*e))-1/4*I*(e*(c^2*d+e))^(1/2)/d/(c^2*d+e)*po 
lylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2*d+2*(e*(c^2*d+e))^(1/2 
)-2*e))+1/2*I*(c^2*d-2*(e*(c^2*d+e))^(1/2)+2*e)*arcsec(c*x)^2/c^2/d^2-(c^2 
*d-2*(e*(c^2*d+e))^(1/2)+2*e)/c^4/d^3*e*ln(1-d*c^2*(1/c/x+I*(1-1/c^2/x^2)^ 
(1/2))^2/(-c^2*d-2*(e*(c^2*d+e))^(1/2)-2*e))*arcsec(c*x)-1/8*I*(e*(c^2*d+e 
))^(1/2)/e/(c^2*d+e)*polylog(2,d*c^2*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2/(-c^2 
*d+2*(e*(c^2*d+e))^(1/2)-2*e))*c^2+(-(e*(c^2*d+e))^(1/2)*c^2*d+2*c^2*d*...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-1/2*a*(log(e*x^2 + d)/d - 2*log(x)/d) + b*integrate(arctan(sqrt(c*x + 1)* 
sqrt(c*x - 1))/(e*x^3 + d*x), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*int(asec(c*x)/(d*x + e*x**3),x)*b*d - log(d + e*x**2)*a + 2*log(x)*a)/( 
2*d)