\(\int \frac {a+b \sec ^{-1}(c x)}{x (d+e x^2)^2} \, dx\) [99]

Optimal result

Integrand size = 21, antiderivative size = 546 \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx=-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{2 d^2 \left (e+\frac {d}{x^2}\right )}+\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b d^2}-\frac {b \sqrt {e} \arctan \left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}} x}\right )}{2 d^2 \sqrt {c^2 d+e}}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^2}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^2}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}+\frac {i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^2}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^2}+\frac {i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^2} \] Output:

-1/2*e*(a+b*arcsec(c*x))/d^2/(e+d/x^2)+1/2*I*(a+b*arcsec(c*x))^2/b/d^2-1/2 
*b*e^(1/2)*arctan((c^2*d+e)^(1/2)/c/e^(1/2)/(1-1/c^2/x^2)^(1/2)/x)/d^2/(c^ 
2*d+e)^(1/2)-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^2-1/2*(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^2-1/2*(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^2-1/2*(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^2+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^2+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^2+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^2+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^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. \(1190\) vs. \(2(546)=1092\).

Time = 0.73 (sec) , antiderivative size = 1190, normalized size of antiderivative = 2.18 \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((2*a*d)/(d + e*x^2) + (b*Sqrt[d]*ArcSec[c*x])/(Sqrt[d] - I*Sqrt[e]*x) + ( 
b*Sqrt[d]*ArcSec[c*x])/(Sqrt[d] + I*Sqrt[e]*x) + (2*I)*b*ArcSec[c*x]^2 + 2 
*b*ArcSin[1/(c*x)] - (8*I)*b*ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt 
[2]]*ArcTan[(((-I)*c*Sqrt[d] + Sqrt[e])*Tan[ArcSec[c*x]/2])/Sqrt[c^2*d + e 
]] - (8*I)*b*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*ArcTan[((I* 
c*Sqrt[d] + Sqrt[e])*Tan[ArcSec[c*x]/2])/Sqrt[c^2*d + e]] - 2*b*ArcSec[c*x 
]*Log[1 + (I*(Sqrt[e] - Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - 
 4*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*b*ArcSec[c*x]*Log[ 
1 + (I*(-Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - 4*b* 
ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 + (I*(-Sqrt[e] + S 
qrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - 2*b*ArcSec[c*x]*Log[1 - 
(I*(Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + 4*b*ArcSi 
n[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*b*ArcSec[c*x]*Log[1 + (I*(Sq 
rt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + 4*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])] + 4*a*Log[x] - (b*Sqrt[e]*Log[(2*Sqrt[ 
d]*Sqrt[e]*(Sqrt[e] + c*(I*c*Sqrt[d] - Sqrt[-(c^2*d) - e]*Sqrt[1 - 1/(c^2* 
x^2)])*x))/(Sqrt[-(c^2*d) - e]*(Sqrt[d] - I*Sqrt[e]*x))])/Sqrt[-(c^2*d)...
 

Rubi [A] (verified)

Time = 1.58 (sec) , antiderivative size = 602, normalized size of antiderivative = 1.10, 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.

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

\(\Big \downarrow \) 5763

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

\(\Big \downarrow \) 5233

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/2*(e*(a + b*ArcCos[1/(c*x)]))/(d^2*(e + d/x^2)) + ((I/2)*(a + b*ArcCos[ 
1/(c*x)])^2)/(b*d^2) - (b*Sqrt[e]*ArcTan[Sqrt[c^2*d + e]/(c*Sqrt[e]*Sqrt[1 
 - 1/(c^2*x^2)]*x)])/(2*d^2*Sqrt[c^2*d + e]) - ((a + b*ArcCos[1/(c*x)])*Lo 
g[1 - (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/(2* 
d^2) - ((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^2) - ((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^2) - 
 ((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^2) + ((I/2)*b*PolyLog[2, -((c*Sqrt[-d]*E^(I* 
ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e]))])/d^2 + ((I/2)*b*PolyLog[2, 
 (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/d^2 + (( 
I/2)*b*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2 
*d + e]))])/d^2 + ((I/2)*b*PolyLog[2, (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/( 
Sqrt[e] + Sqrt[c^2*d + e])])/d^2
 

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 = 6.22 (sec) , antiderivative size = 2108, normalized size of antiderivative = 3.86

method	result	size
parts	\(\text {Expression too large to display}\)	\(2108\)
derivativedivides	\(\text {Expression too large to display}\)	\(2157\)
default	\(\text {Expression too large to display}\)	\(2157\)

Input:

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

Output:

a/d^2*ln(x)-1/2*a/d^2*ln(e*x^2+d)+1/2*a/d/(e*x^2+d)+b*(-1/8*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))/d^2/e/( 
c^2*d+e)+1/4*(-(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)/d^2/e/(c^2*d+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)+1/4*(e*(c^2*d+e))^(1/2)/d/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/e/(c^2*d+e)*arcsec(c 
*x)^2*c^2-1/8*I*(e*(c^2*d+e))^(1/2)/d/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-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)*e*arcsec(c*x) 
^2/d^4/(c^2*d+e)/c^4+(-(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)/d^4*e/(c^2*d+e)/c^4*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)-1/2*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)*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))/d^4/(c^2 
*d+e)/c^4+I*(c^2*d-2*(e*(c^2*d+e))^(1/2)+2*e)*arcsec(c*x)^2*e/d^4/c^4+(-(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)/d^3/(c^2* 
d+e)/c^2*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)-I*(-(e*(c^2*d+e))^(1/2)*c^2*d+2*c^2*d*e-2*(e*...
 

Fricas [F]

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

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

Output:

integral((b*arcsec(c*x) + a)/(e^2*x^5 + 2*d*e*x^3 + d^2*x), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

1/2*a*(1/(d*e*x^2 + d^2) - log(e*x^2 + d)/d^2 + 2*log(x)/d^2) + b*integrat 
e(arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(e^2*x^5 + 2*d*e*x^3 + d^2*x), x)
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

\[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^2} \, dx=\frac {2 \left (\int \frac {\mathit {asec} \left (c x \right )}{e^{2} x^{5}+2 d e \,x^{3}+d^{2} x}d x \right ) b \,d^{3}+2 \left (\int \frac {\mathit {asec} \left (c x \right )}{e^{2} x^{5}+2 d e \,x^{3}+d^{2} x}d x \right ) b \,d^{2} e \,x^{2}-\mathrm {log}\left (e \,x^{2}+d \right ) a d -\mathrm {log}\left (e \,x^{2}+d \right ) a e \,x^{2}+2 \,\mathrm {log}\left (x \right ) a d +2 \,\mathrm {log}\left (x \right ) a e \,x^{2}-a e \,x^{2}}{2 d^{2} \left (e \,x^{2}+d \right )} \] Input:

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

Output:

(2*int(asec(c*x)/(d**2*x + 2*d*e*x**3 + e**2*x**5),x)*b*d**3 + 2*int(asec( 
c*x)/(d**2*x + 2*d*e*x**3 + e**2*x**5),x)*b*d**2*e*x**2 - log(d + e*x**2)* 
a*d - log(d + e*x**2)*a*e*x**2 + 2*log(x)*a*d + 2*log(x)*a*e*x**2 - a*e*x* 
*2)/(2*d**2*(d + e*x**2))