\(\int \frac {a+b \sec ^{-1}(c x)}{x (d+e x^2)^3} \, dx\) [107]

Optimal result
Mathematica [B] (warning: unable to verify)
Rubi [A] (verified)
Maple [C] (warning: unable to verify)
Fricas [F]
Sympy [F(-1)]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 21, antiderivative size = 685 \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^3} \, dx=\frac {b c e \sqrt {1-\frac {1}{c^2 x^2}}}{8 d^2 \left (c^2 d+e\right ) \left (e+\frac {d}{x^2}\right ) x}+\frac {e^2 \left (a+b \sec ^{-1}(c x)\right )}{4 d^3 \left (e+\frac {d}{x^2}\right )^2}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{d^3 \left (e+\frac {d}{x^2}\right )}+\frac {i \left (a+b \sec ^{-1}(c x)\right )^2}{2 b d^3}-\frac {b \sqrt {e} \arctan \left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}} x}\right )}{d^3 \sqrt {c^2 d+e}}+\frac {b \sqrt {e} \left (c^2 d+2 e\right ) \arctan \left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} \sqrt {1-\frac {1}{c^2 x^2}} x}\right )}{8 d^3 \left (c^2 d+e\right )^{3/2}}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^3}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^3}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^3}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^3}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^3}+\frac {i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 d^3}+\frac {i b \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^3}+\frac {i b \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 d^3} \] Output:

1/8*b*c*e*(1-1/c^2/x^2)^(1/2)/d^2/(c^2*d+e)/(e+d/x^2)/x+1/4*e^2*(a+b*arcse 
c(c*x))/d^3/(e+d/x^2)^2-e*(a+b*arcsec(c*x))/d^3/(e+d/x^2)+1/2*I*(a+b*arcse 
c(c*x))^2/b/d^3-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^3/(c^2*d+e)^(1/2)+1/8*b*e^(1/2)*(c^2*d+2*e)*arctan((c^2*d+e)^(1/ 
2)/c/e^(1/2)/(1-1/c^2/x^2)^(1/2)/x)/d^3/(c^2*d+e)^(3/2)-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^3-1/2*(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^3-1/2*(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^3-1/2*(a+b*ar 
csec(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^3+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^3+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^3+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^3+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^3
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1871\) vs. \(2(685)=1370\).

Time = 6.05 (sec) , antiderivative size = 1871, normalized size of antiderivative = 2.73 \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^3} \, dx =\text {Too large to display} \] Input:

Integrate[(a + b*ArcSec[c*x])/(x*(d + e*x^2)^3),x]
 

Output:

a/(4*d*(d + e*x^2)^2) + a/(2*d^2*(d + e*x^2)) + (a*Log[x])/d^3 - (a*Log[d 
+ e*x^2])/(2*d^3) + b*((((-5*I)/16)*Sqrt[e]*(-(ArcSec[c*x]/(I*Sqrt[d]*Sqrt 
[e] + e*x)) + (I*(ArcSin[1/(c*x)]/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) - e]))/Sqrt[d]))/d^(5 
/2) + (((5*I)/16)*Sqrt[e]*(-(ArcSec[c*x]/((-I)*Sqrt[d]*Sqrt[e] + e*x)) - ( 
I*(ArcSin[1/(c*x)]/Sqrt[e] - Log[(2*Sqrt[d]*Sqrt[e]*(-Sqrt[e] + c*(I*c*Sqr 
t[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) - e]))/Sqrt[d]))/d^(5/2) + (Sqrt[e] 
*(-(ArcSec[c*x]/(Sqrt[e]*((-I)*Sqrt[d] + Sqrt[e]*x)^2)) + (ArcSin[1/(c*x)] 
/Sqrt[e] - I*((c*Sqrt[d]*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x)/((c^2*d + e)*((- 
I)*Sqrt[d] + Sqrt[e]*x)) + ((2*c^2*d + e)*Log[(-4*d*Sqrt[e]*Sqrt[c^2*d + e 
]*(I*Sqrt[e] + c*(c*Sqrt[d] - Sqrt[c^2*d + e]*Sqrt[1 - 1/(c^2*x^2)])*x))/( 
(2*c^2*d + e)*((-I)*Sqrt[d] + Sqrt[e]*x))])/(c^2*d + e)^(3/2)))/d))/(16*d^ 
2) + (Sqrt[e]*((I*c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x)/(Sqrt[d]*(c^2*d + e)* 
(I*Sqrt[d] + Sqrt[e]*x)) - ArcSec[c*x]/(Sqrt[e]*(I*Sqrt[d] + Sqrt[e]*x)^2) 
 + ArcSin[1/(c*x)]/(d*Sqrt[e]) - (I*(2*c^2*d + e)*Log[(4*d*Sqrt[e]*Sqrt[c^ 
2*d + e]*((-I)*Sqrt[e] + c*(c*Sqrt[d] + Sqrt[c^2*d + e]*Sqrt[1 - 1/(c^2*x^ 
2)])*x))/((2*c^2*d + e)*(I*Sqrt[d] + Sqrt[e]*x))])/(d*(c^2*d + e)^(3/2)))) 
/(16*d^2) + ((I/2)*ArcSec[c*x]^2 - ArcSec[c*x]*Log[1 + E^((2*I)*ArcSec[...
 

Rubi [A] (verified)

Time = 1.82 (sec) , antiderivative size = 745, normalized size of antiderivative = 1.09, 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.

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

\(\Big \downarrow \) 5763

\(\displaystyle -\int \frac {a+b \arccos \left (\frac {1}{c x}\right )}{\left (\frac {d}{x^2}+e\right )^3 x^5}d\frac {1}{x}\)

\(\Big \downarrow \) 5233

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

\(\Big \downarrow \) 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^3}-\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^3}-\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^3}-\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^3}+\frac {e^2 \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{4 d^3 \left (\frac {d}{x^2}+e\right )^2}-\frac {e \left (a+b \arccos \left (\frac {1}{c x}\right )\right )}{d^3 \left (\frac {d}{x^2}+e\right )}+\frac {i \left (a+b \arccos \left (\frac {1}{c x}\right )\right )^2}{2 b d^3}+\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^3}+\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^3}+\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^3}+\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^3}+\frac {b \sqrt {e} \left (c^2 d+2 e\right ) \arctan \left (\frac {\sqrt {c^2 d+e}}{c \sqrt {e} x \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{8 d^3 \left (c^2 d+e\right )^{3/2}}-\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 )}{d^3 \sqrt {c^2 d+e}}+\frac {b c e \sqrt {1-\frac {1}{c^2 x^2}}}{8 d^2 x \left (c^2 d+e\right ) \left (\frac {d}{x^2}+e\right )}\)

Input:

Int[(a + b*ArcSec[c*x])/(x*(d + e*x^2)^3),x]
 

Output:

(b*c*e*Sqrt[1 - 1/(c^2*x^2)])/(8*d^2*(c^2*d + e)*(e + d/x^2)*x) + (e^2*(a 
+ b*ArcCos[1/(c*x)]))/(4*d^3*(e + d/x^2)^2) - (e*(a + b*ArcCos[1/(c*x)]))/ 
(d^3*(e + d/x^2)) + ((I/2)*(a + b*ArcCos[1/(c*x)])^2)/(b*d^3) - (b*Sqrt[e] 
*ArcTan[Sqrt[c^2*d + e]/(c*Sqrt[e]*Sqrt[1 - 1/(c^2*x^2)]*x)])/(d^3*Sqrt[c^ 
2*d + e]) + (b*Sqrt[e]*(c^2*d + 2*e)*ArcTan[Sqrt[c^2*d + e]/(c*Sqrt[e]*Sqr 
t[1 - 1/(c^2*x^2)]*x)])/(8*d^3*(c^2*d + e)^(3/2)) - ((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^3) - ((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^3) - ((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 
^3) - ((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^3) + ((I/2)*b*PolyLog[2, -((c*Sqrt[-d]* 
E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e]))])/d^3 + ((I/2)*b*PolyL 
og[2, (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/d^3 
 + ((I/2)*b*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqr 
t[c^2*d + e]))])/d^3 + ((I/2)*b*PolyLog[2, (c*Sqrt[-d]*E^(I*ArcCos[1/(c*x) 
]))/(Sqrt[e] + Sqrt[c^2*d + e])])/d^3
 

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 = 2.71 (sec) , antiderivative size = 3533, normalized size of antiderivative = 5.16

method result size
parts \(\text {Expression too large to display}\) \(3533\)
derivativedivides \(\text {Expression too large to display}\) \(3607\)
default \(\text {Expression too large to display}\) \(3607\)

Input:

int((a+b*arcsec(c*x))/x/(e*x^2+d)^3,x,method=_RETURNVERBOSE)
 

Output:

a/d^3*ln(x)-1/2*a/d^3*ln(e*x^2+d)+1/2*a/d^2/(e*x^2+d)+1/4*a/d/(e*x^2+d)^2+ 
b*(3/4*I*(e*(c^2*d+e))^(1/2)/(c^2*d+e)^2/d^3*e*arctanh(1/4*(2*c^2*d*(1/c/x 
+I*(1-1/c^2/x^2)^(1/2))^2+2*c^2*d+4*e)/(c^2*d*e+e^2)^(1/2))+1/2*(e*(c^2*d+ 
e))^(1/2)/(c^2*d+e)^2/d^3*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))-1/2*I*(e*(c^2*d+e))^(1/2)/(c^ 
2*d+e)^2/d^3*e*arcsec(c*x)^2+3/4*(e*(c^2*d+e))^(1/2)/(c^2*d+e)^2/d^2*c^2*a 
rcsec(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)/(c^2*d+e)^2/d^3*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))-3/4* 
I*(e*(c^2*d+e))^(1/2)/(c^2*d+e)^2/d^2*arcsec(c*x)^2*c^2-3/8*I*(e*(c^2*d+e) 
)^(1/2)/(c^2*d+e)^2/d^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+7/8*I*(e*(c^2*d+e))^(1/2)/(c^2*d+e)^ 
2/d^2*arctanh(1/4*(2*c^2*d*(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2+2*c^2*d+4*e)/(c 
^2*d*e+e^2)^(1/2))*c^2-1/8*I*(e*(c^2*d+e))^(1/2)/(c^2*d+e)^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^4+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)/e/(c^4*d^2+2*c^2*d*e+e^2)*c^2/d^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)-1/2*I*arcsec(c*x)^2 
/d^3+1/2*I*(c^2*d-2*(e*(c^2*d+e))^(1/2)+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))*e^2/c^4/d^5/(c^2...
 

Fricas [F]

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

integrate((a+b*arcsec(c*x))/x/(e*x^2+d)^3,x, algorithm="fricas")
 

Output:

integral((b*arcsec(c*x) + a)/(e^3*x^7 + 3*d*e^2*x^5 + 3*d^2*e*x^3 + d^3*x) 
, x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate((a+b*asec(c*x))/x/(e*x**2+d)**3,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+b*arcsec(c*x))/x/(e*x^2+d)^3,x, algorithm="maxima")
 

Output:

1/4*a*((2*e*x^2 + 3*d)/(d^2*e^2*x^4 + 2*d^3*e*x^2 + d^4) - 2*log(e*x^2 + d 
)/d^3 + 4*log(x)/d^3) + b*integrate(arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(e 
^3*x^7 + 3*d*e^2*x^5 + 3*d^2*e*x^3 + d^3*x), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((a+b*arcsec(c*x))/x/(e*x^2+d)^3,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. \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^3} \, dx=\int \frac {a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )}{x\,{\left (e\,x^2+d\right )}^3} \,d x \] Input:

int((a + b*acos(1/(c*x)))/(x*(d + e*x^2)^3),x)
 

Output:

int((a + b*acos(1/(c*x)))/(x*(d + e*x^2)^3), x)
 

Reduce [F]

\[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^3} \, dx=\frac {4 \left (\int \frac {\mathit {asec} \left (c x \right )}{e^{3} x^{7}+3 d \,e^{2} x^{5}+3 d^{2} e \,x^{3}+d^{3} x}d x \right ) b \,d^{5}+8 \left (\int \frac {\mathit {asec} \left (c x \right )}{e^{3} x^{7}+3 d \,e^{2} x^{5}+3 d^{2} e \,x^{3}+d^{3} x}d x \right ) b \,d^{4} e \,x^{2}+4 \left (\int \frac {\mathit {asec} \left (c x \right )}{e^{3} x^{7}+3 d \,e^{2} x^{5}+3 d^{2} e \,x^{3}+d^{3} x}d x \right ) b \,d^{3} e^{2} x^{4}-2 \,\mathrm {log}\left (e \,x^{2}+d \right ) a \,d^{2}-4 \,\mathrm {log}\left (e \,x^{2}+d \right ) a d e \,x^{2}-2 \,\mathrm {log}\left (e \,x^{2}+d \right ) a \,e^{2} x^{4}+4 \,\mathrm {log}\left (x \right ) a \,d^{2}+8 \,\mathrm {log}\left (x \right ) a d e \,x^{2}+4 \,\mathrm {log}\left (x \right ) a \,e^{2} x^{4}+2 a \,d^{2}-a \,e^{2} x^{4}}{4 d^{3} \left (e^{2} x^{4}+2 d e \,x^{2}+d^{2}\right )} \] Input:

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

Output:

(4*int(asec(c*x)/(d**3*x + 3*d**2*e*x**3 + 3*d*e**2*x**5 + e**3*x**7),x)*b 
*d**5 + 8*int(asec(c*x)/(d**3*x + 3*d**2*e*x**3 + 3*d*e**2*x**5 + e**3*x** 
7),x)*b*d**4*e*x**2 + 4*int(asec(c*x)/(d**3*x + 3*d**2*e*x**3 + 3*d*e**2*x 
**5 + e**3*x**7),x)*b*d**3*e**2*x**4 - 2*log(d + e*x**2)*a*d**2 - 4*log(d 
+ e*x**2)*a*d*e*x**2 - 2*log(d + e*x**2)*a*e**2*x**4 + 4*log(x)*a*d**2 + 8 
*log(x)*a*d*e*x**2 + 4*log(x)*a*e**2*x**4 + 2*a*d**2 - a*e**2*x**4)/(4*d** 
3*(d**2 + 2*d*e*x**2 + e**2*x**4))