\(\int \frac {a+b \sec ^{-1}(c x)}{x^2 (d+e x^2)^2} \, dx\) [103]

Optimal result

Integrand size = 21, antiderivative size = 785 \[ \int \frac {a+b \sec ^{-1}(c x)}{x^2 \left (d+e x^2\right )^2} \, dx=\frac {b c \sqrt {1-\frac {1}{c^2 x^2}}}{d^2}-\frac {a}{d^2 x}-\frac {b \sec ^{-1}(c x)}{d^2 x}+\frac {e \left (a+b \sec ^{-1}(c x)\right )}{4 d^2 \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{4 d^2 \left (\sqrt {-d} \sqrt {e}+\frac {d}{x}\right )}+\frac {b e \text {arctanh}\left (\frac {c^2 d-\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {c^2 d+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{4 d^{5/2} \sqrt {c^2 d+e}}+\frac {b e \text {arctanh}\left (\frac {c^2 d+\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {c^2 d+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{4 d^{5/2} \sqrt {c^2 d+e}}-\frac {3 \sqrt {e} \left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{4 (-d)^{5/2}}+\frac {3 \sqrt {e} \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{4 (-d)^{5/2}}-\frac {3 \sqrt {e} \left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{4 (-d)^{5/2}}+\frac {3 \sqrt {e} \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{4 (-d)^{5/2}}-\frac {3 i b \sqrt {e} \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{4 (-d)^{5/2}}+\frac {3 i b \sqrt {e} \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{4 (-d)^{5/2}}-\frac {3 i b \sqrt {e} \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{4 (-d)^{5/2}}+\frac {3 i b \sqrt {e} \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{4 (-d)^{5/2}} \] Output:

b*c*(1-1/c^2/x^2)^(1/2)/d^2-a/d^2/x-b*arcsec(c*x)/d^2/x+1/4*e*(a+b*arcsec( 
c*x))/d^2/((-d)^(1/2)*e^(1/2)-d/x)-1/4*e*(a+b*arcsec(c*x))/d^2/((-d)^(1/2) 
*e^(1/2)+d/x)+1/4*b*e*arctanh((c^2*d-(-d)^(1/2)*e^(1/2)/x)/c/d^(1/2)/(c^2* 
d+e)^(1/2)/(1-1/c^2/x^2)^(1/2))/d^(5/2)/(c^2*d+e)^(1/2)+1/4*b*e*arctanh((c 
^2*d+(-d)^(1/2)*e^(1/2)/x)/c/d^(1/2)/(c^2*d+e)^(1/2)/(1-1/c^2/x^2)^(1/2))/ 
d^(5/2)/(c^2*d+e)^(1/2)-3/4*e^(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)^(5/2)+3/4*e^(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)^(5/2)-3/4*e^(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)^(5/2 
)+3/4*e^(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)^(5/2)-3/4*I*b*e^(1/2)*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)^(5 
/2)+3/4*I*b*e^(1/2)*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)^(5/2)-3/4*I*b*e^(1/2)*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)^(5/2)+3/4 
*I*b*e^(1/2)*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)^(5/2)
 

Mathematica [A] (warning: unable to verify)

Time = 1.34 (sec) , antiderivative size = 1291, normalized size of antiderivative = 1.64 \[ \int \frac {a+b \sec ^{-1}(c x)}{x^2 \left (d+e x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((-4*a*Sqrt[d])/x - (2*a*Sqrt[d]*e*x)/(d + e*x^2) - 6*a*Sqrt[e]*ArcTan[(Sq 
rt[e]*x)/Sqrt[d]] + b*(4*c*Sqrt[d]*Sqrt[1 - 1/(c^2*x^2)] - (4*Sqrt[d]*ArcS 
ec[c*x])/x - (Sqrt[d]*e*ArcSec[c*x])/((-I)*Sqrt[d]*Sqrt[e] + e*x) - (Sqrt[ 
d]*e*ArcSec[c*x])/(I*Sqrt[d]*Sqrt[e] + e*x) + 12*Sqrt[e]*ArcSin[Sqrt[1 - ( 
I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*ArcTan[(((-I)*c*Sqrt[d] + Sqrt[e])*Tan[Ar 
cSec[c*x]/2])/Sqrt[c^2*d + e]] - 12*Sqrt[e]*ArcSin[Sqrt[1 + (I*Sqrt[e])/(c 
*Sqrt[d])]/Sqrt[2]]*ArcTan[((I*c*Sqrt[d] + Sqrt[e])*Tan[ArcSec[c*x]/2])/Sq 
rt[c^2*d + e]] + (3*I)*Sqrt[e]*ArcSec[c*x]*Log[1 + (I*(Sqrt[e] - Sqrt[c^2* 
d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + (6*I)*Sqrt[e]*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])] - (3*I)*Sqrt[e]*ArcSec[c*x]*Log[1 + (I*(-Sqrt 
[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - (6*I)*Sqrt[e]*Arc 
Sin[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])] - (3*I)*Sqrt[e]*ArcSec[c*x]*L 
og[1 - (I*(Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + (6 
*I)*Sqrt[e]*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])] + (3*I)*Sqrt[e] 
*ArcSec[c*x]*Log[1 + (I*(Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c* 
Sqrt[d])] - (6*I)*Sqrt[e]*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])...
 

Rubi [A] (verified)

Time = 2.79 (sec) , antiderivative size = 845, normalized size of antiderivative = 1.08, 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.

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

\(\Big \downarrow \) 5763

\(\displaystyle -\int \frac {a+b \arccos \left (\frac {1}{c x}\right )}{\left (\frac {d}{x^2}+e\right )^2 x^4}d\frac {1}{x}\)

\(\Big \downarrow \) 5233

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a}{d^2 x}-\frac {b \arccos \left (\frac {1}{c x}\right )}{d^2 x}+\frac {e \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{4 d^2 \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )}-\frac {e \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{4 d^2 \left (\frac {d}{x}+\sqrt {-d} \sqrt {e}\right )}+\frac {b e \text {arctanh}\left (\frac {c^2 d-\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {d c^2+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{4 d^{5/2} \sqrt {d c^2+e}}+\frac {b e \text {arctanh}\left (\frac {d c^2+\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {d c^2+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{4 d^{5/2} \sqrt {d c^2+e}}-\frac {3 \sqrt {e} \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 {d c^2+e}}\right )}{4 (-d)^{5/2}}+\frac {3 \sqrt {e} \left (a+b \arccos \left (\frac {1}{c x}\right )\right ) \log \left (\frac {\sqrt {-d} e^{i \arccos \left (\frac {1}{c x}\right )} c}{\sqrt {e}-\sqrt {d c^2+e}}+1\right )}{4 (-d)^{5/2}}-\frac {3 \sqrt {e} \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 {d c^2+e}}\right )}{4 (-d)^{5/2}}+\frac {3 \sqrt {e} \left (a+b \arccos \left (\frac {1}{c x}\right )\right ) \log \left (\frac {\sqrt {-d} e^{i \arccos \left (\frac {1}{c x}\right )} c}{\sqrt {e}+\sqrt {d c^2+e}}+1\right )}{4 (-d)^{5/2}}-\frac {3 i b \sqrt {e} \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 )}{4 (-d)^{5/2}}+\frac {3 i b \sqrt {e} \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 )}{4 (-d)^{5/2}}-\frac {3 i b \sqrt {e} \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 )}{4 (-d)^{5/2}}+\frac {3 i b \sqrt {e} \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 )}{4 (-d)^{5/2}}+\frac {b c \sqrt {1-\frac {1}{c^2 x^2}}}{d^2}\)

Input:

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

Output:

(b*c*Sqrt[1 - 1/(c^2*x^2)])/d^2 - a/(d^2*x) - (b*ArcCos[1/(c*x)])/(d^2*x) 
+ (e*(a + b*ArcCos[1/(c*x)]))/(4*d^2*(Sqrt[-d]*Sqrt[e] - d/x)) - (e*(a + b 
*ArcCos[1/(c*x)]))/(4*d^2*(Sqrt[-d]*Sqrt[e] + d/x)) + (b*e*ArcTanh[(c^2*d 
- (Sqrt[-d]*Sqrt[e])/x)/(c*Sqrt[d]*Sqrt[c^2*d + e]*Sqrt[1 - 1/(c^2*x^2)])] 
)/(4*d^(5/2)*Sqrt[c^2*d + e]) + (b*e*ArcTanh[(c^2*d + (Sqrt[-d]*Sqrt[e])/x 
)/(c*Sqrt[d]*Sqrt[c^2*d + e]*Sqrt[1 - 1/(c^2*x^2)])])/(4*d^(5/2)*Sqrt[c^2* 
d + e]) - (3*Sqrt[e]*(a + b*ArcCos[1/(c*x)])*Log[1 - (c*Sqrt[-d]*E^(I*ArcC 
os[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/(4*(-d)^(5/2)) + (3*Sqrt[e]*(a 
 + b*ArcCos[1/(c*x)])*Log[1 + (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] 
- Sqrt[c^2*d + e])])/(4*(-d)^(5/2)) - (3*Sqrt[e]*(a + b*ArcCos[1/(c*x)])*L 
og[1 - (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e])])/(4 
*(-d)^(5/2)) + (3*Sqrt[e]*(a + b*ArcCos[1/(c*x)])*Log[1 + (c*Sqrt[-d]*E^(I 
*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e])])/(4*(-d)^(5/2)) - (((3*I)/ 
4)*b*Sqrt[e]*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sq 
rt[c^2*d + e]))])/(-d)^(5/2) + (((3*I)/4)*b*Sqrt[e]*PolyLog[2, (c*Sqrt[-d] 
*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/(-d)^(5/2) - (((3*I) 
/4)*b*Sqrt[e]*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + S 
qrt[c^2*d + e]))])/(-d)^(5/2) + (((3*I)/4)*b*Sqrt[e]*PolyLog[2, (c*Sqrt[-d 
]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e])])/(-d)^(5/2)
 

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 = 67.93 (sec) , antiderivative size = 933, normalized size of antiderivative = 1.19

method	result	size
parts	\(\text {Expression too large to display}\)	\(933\)
derivativedivides	\(\text {Expression too large to display}\)	\(960\)
default	\(\text {Expression too large to display}\)	\(960\)

Input:

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

Output:

a*(-1/d^2/x-e/d^2*(1/2*x/(e*x^2+d)+3/2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2)) 
))+b*c*(-1/2*(arcsec(c*x)+I)/d^2*(I*((c^2*x^2-1)/c^2/x^2)^(1/2)*c*x+1)/x/c 
+1/2*(I*((c^2*x^2-1)/c^2/x^2)^(1/2)*c*x-1)*(arcsec(c*x)-I)/d^2/x/c-1/2*arc 
sec(c*x)/d^2*e*x*c/(c^2*e*x^2+c^2*d)-1/2*I*(-(c^2*d-2*(e*(c^2*d+e))^(1/2)+ 
2*e)*d)^(1/2)*(c^2*d+2*(e*(c^2*d+e))^(1/2)+2*e)*arctanh(c*d*(1/c/x+I*(1-1/ 
c^2/x^2)^(1/2))/((-c^2*d+2*(e*(c^2*d+e))^(1/2)-2*e)*d)^(1/2))*e/d^5/c^5+1/ 
2*I*(-(c^2*d-2*(e*(c^2*d+e))^(1/2)+2*e)*d)^(1/2)*((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*arctanh(c*d*(1/c/x+I*(1-1/c^2 
/x^2)^(1/2))/((-c^2*d+2*(e*(c^2*d+e))^(1/2)-2*e)*d)^(1/2))/d^5/c^5/(c^2*d+ 
e)-1/2*I*((c^2*d+2*(e*(c^2*d+e))^(1/2)+2*e)*d)^(1/2)*(c^2*d-2*(e*(c^2*d+e) 
)^(1/2)+2*e)*arctan(c*d*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/((c^2*d+2*(e*(c^2*d+ 
e))^(1/2)+2*e)*d)^(1/2))*e/d^5/c^5+1/2*I*((c^2*d+2*(e*(c^2*d+e))^(1/2)+2*e 
)*d)^(1/2)*(-(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*arctan(c*d*(1/c/x+I*(1-1/c^2/x^2)^(1/2))/((c^2*d+2*(e*(c^2*d+e))^( 
1/2)+2*e)*d)^(1/2))/d^5/c^5/(c^2*d+e)+3/4*I*e/d^2*sum(1/_R1/(_R1^2*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))-3/4*I*e/d^2*sum(_R1/(_R1^2*c^2*d+c^2*d+2*e)*(I*arcsec(c*x)*l 
n((_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)))
 

Fricas [F]

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

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

Output:

integral((b*arcsec(c*x) + a)/(e^2*x^6 + 2*d*e*x^4 + d^2*x^2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

\[ \int \frac {a+b \sec ^{-1}(c x)}{x^2 \left (d+e x^2\right )^2} \, dx=\frac {-3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a d x -3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a e \,x^{3}+2 \left (\int \frac {\mathit {asec} \left (c x \right )}{e^{2} x^{6}+2 d e \,x^{4}+d^{2} x^{2}}d x \right ) b \,d^{4} x +2 \left (\int \frac {\mathit {asec} \left (c x \right )}{e^{2} x^{6}+2 d e \,x^{4}+d^{2} x^{2}}d x \right ) b \,d^{3} e \,x^{3}-2 a \,d^{2}-3 a d e \,x^{2}}{2 d^{3} x \left (e \,x^{2}+d \right )} \] Input:

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

Output:

( - 3*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d*x - 3*sqrt(e)*sqrt 
(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*e*x**3 + 2*int(asec(c*x)/(d**2*x**2 + 
2*d*e*x**4 + e**2*x**6),x)*b*d**4*x + 2*int(asec(c*x)/(d**2*x**2 + 2*d*e*x 
**4 + e**2*x**6),x)*b*d**3*e*x**3 - 2*a*d**2 - 3*a*d*e*x**2)/(2*d**3*x*(d 
+ e*x**2))