\(\int \frac {x^4 (a+b \sec ^{-1}(c x))}{(d+e x^2)^2} \, dx\) [100]

Optimal result

Integrand size = 21, antiderivative size = 784 \[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx=-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{4 e^2 \left (\sqrt {-d} \sqrt {e}-\frac {d}{x}\right )}+\frac {d \left (a+b \sec ^{-1}(c x)\right )}{4 e^2 \left (\sqrt {-d} \sqrt {e}+\frac {d}{x}\right )}+\frac {x \left (a+b \sec ^{-1}(c x)\right )}{e^2}-\frac {b \text {arctanh}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{c e^2}-\frac {b \sqrt {d} \text {arctanh}\left (\frac {c^2 d-\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {c^2 d+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{4 e^2 \sqrt {c^2 d+e}}-\frac {b \sqrt {d} \text {arctanh}\left (\frac {c^2 d+\frac {\sqrt {-d} \sqrt {e}}{x}}{c \sqrt {d} \sqrt {c^2 d+e} \sqrt {1-\frac {1}{c^2 x^2}}}\right )}{4 e^2 \sqrt {c^2 d+e}}+\frac {3 \sqrt {-d} \left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}-\frac {3 \sqrt {-d} \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}+\frac {3 \sqrt {-d} \left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}-\frac {3 \sqrt {-d} \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}+\frac {3 i b \sqrt {-d} \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}-\frac {3 i b \sqrt {-d} \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}+\frac {3 i b \sqrt {-d} \operatorname {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{4 e^{5/2}}-\frac {3 i b \sqrt {-d} \operatorname {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{4 e^{5/2}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 1.43 (sec) , antiderivative size = 1331, normalized size of antiderivative = 1.70 \[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx =\text {Too large to display} \] Input:

Integrate[(x^4*(a + b*ArcSec[c*x]))/(d + e*x^2)^2,x]
 

Output:

(4*a*Sqrt[e]*x + (2*a*d*Sqrt[e]*x)/(d + e*x^2) - 6*a*Sqrt[d]*ArcTan[(Sqrt[ 
e]*x)/Sqrt[d]] + b*(4*Sqrt[e]*x*ArcSec[c*x] + (d*ArcSec[c*x])/((-I)*Sqrt[d 
] + Sqrt[e]*x) + (d*ArcSec[c*x])/(I*Sqrt[d] + Sqrt[e]*x) + 12*Sqrt[d]*ArcS 
in[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*ArcTan[(((-I)*c*Sqrt[d] + Sq 
rt[e])*Tan[ArcSec[c*x]/2])/Sqrt[c^2*d + e]] - 12*Sqrt[d]*ArcSin[Sqrt[1 + ( 
I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*ArcTan[((I*c*Sqrt[d] + Sqrt[e])*Tan[ArcSe 
c[c*x]/2])/Sqrt[c^2*d + e]] + (3*I)*Sqrt[d]*ArcSec[c*x]*Log[1 + (I*(Sqrt[e 
] - Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] + (6*I)*Sqrt[d]*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])] - (3*I)*Sqrt[d]*ArcSec[c*x]*Log[ 
1 + (I*(-Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - (6*I 
)*Sqrt[d]*ArcSin[Sqrt[1 - (I*Sqrt[e])/(c*Sqrt[d])]/Sqrt[2]]*Log[1 + (I*(-S 
qrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*Sqrt[d])] - (3*I)*Sqrt[d]* 
ArcSec[c*x]*Log[1 - (I*(Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcSec[c*x]))/(c*S 
qrt[d])] + (6*I)*Sqrt[d]*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[d]*ArcSec[c*x]*Log[1 + (I*(Sqrt[e] + Sqrt[c^2*d + e])*E^(I*ArcS 
ec[c*x]))/(c*Sqrt[d])] - (6*I)*Sqrt[d]*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*Sqrt[d]*Sqrt[e]*Log[(2*Sqrt[d]*Sqrt[e]*(Sqrt[e] + c*(I...
 

Rubi [A] (verified)

Time = 2.88 (sec) , antiderivative size = 844, 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.

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

\(\Big \downarrow \) 5763

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

\(\Big \downarrow \) 5233

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

\(\Big \downarrow \) 2009

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

Input:

Int[(x^4*(a + b*ArcSec[c*x]))/(d + e*x^2)^2,x]
 

Output:

-1/4*(d*(a + b*ArcCos[1/(c*x)]))/(e^2*(Sqrt[-d]*Sqrt[e] - d/x)) + (d*(a + 
b*ArcCos[1/(c*x)]))/(4*e^2*(Sqrt[-d]*Sqrt[e] + d/x)) + (x*(a + b*ArcCos[1/ 
(c*x)]))/e^2 - (b*ArcTanh[Sqrt[1 - 1/(c^2*x^2)]])/(c*e^2) - (b*Sqrt[d]*Arc 
Tanh[(c^2*d - (Sqrt[-d]*Sqrt[e])/x)/(c*Sqrt[d]*Sqrt[c^2*d + e]*Sqrt[1 - 1/ 
(c^2*x^2)])])/(4*e^2*Sqrt[c^2*d + e]) - (b*Sqrt[d]*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*e^2 
*Sqrt[c^2*d + e]) + (3*Sqrt[-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])])/(4*e^(5/2)) - (3*Sq 
rt[-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])])/(4*e^(5/2)) + (3*Sqrt[-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 
])])/(4*e^(5/2)) - (3*Sqrt[-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])])/(4*e^(5/2)) + (((3*I 
)/4)*b*Sqrt[-d]*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - 
 Sqrt[c^2*d + e]))])/e^(5/2) - (((3*I)/4)*b*Sqrt[-d]*PolyLog[2, (c*Sqrt[-d 
]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] - Sqrt[c^2*d + e])])/e^(5/2) + (((3*I)/4 
)*b*Sqrt[-d]*PolyLog[2, -((c*Sqrt[-d]*E^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sq 
rt[c^2*d + e]))])/e^(5/2) - (((3*I)/4)*b*Sqrt[-d]*PolyLog[2, (c*Sqrt[-d]*E 
^(I*ArcCos[1/(c*x)]))/(Sqrt[e] + Sqrt[c^2*d + e])])/e^(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 = 67.16 (sec) , antiderivative size = 941, normalized size of antiderivative = 1.20

method	result	size
parts	\(\text {Expression too large to display}\)	\(941\)
derivativedivides	\(\text {Expression too large to display}\)	\(966\)
default	\(\text {Expression too large to display}\)	\(966\)

Input:

int(x^4*(a+b*arcsec(c*x))/(e*x^2+d)^2,x,method=_RETURNVERBOSE)
 

Output:

a*(1/e^2*x-1/e^2*d*(-1/2*x/(e*x^2+d)+3/2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2 
))))+b/c^5*(1/2*x*c^5*arcsec(c*x)*(2*c^2*e*x^2+3*c^2*d)/(c^2*e*x^2+c^2*d)/ 
e^2-1/2*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)*c*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^2/(c^2*d+ 
e)/d^2+2*I/e^2*c^4*arctan(1/c/x+I*(1-1/c^2/x^2)^(1/2))-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)*c*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^2/(c^2*d+e)/d^2+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)*c*a 
rctanh(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^2/e^2+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)*c*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^2/e^2+3/16*I/e^3*c^6*d*sum( 
(_R1^2*c^2*d+c^2*d+4*e)/_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))-3/16*I/e^3*c^6*d*s 
um((_R1^2*c^2*d+4*_R1^2*e+c^2*d)/_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)))
 

Fricas [F]

\[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x^{4}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \] Input:

integrate(x^4*(a+b*arcsec(c*x))/(e*x^2+d)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(x**4*(a+b*asec(c*x))/(e*x**2+d)**2,x)
 

Output:

Integral(x**4*(a + b*asec(c*x))/(d + e*x**2)**2, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^4*(a+b*arcsec(c*x))/(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(-1)]

Timed out. \[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x^4*(a+b*arcsec(c*x))/(e*x^2+d)^2,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^2} \,d x \] Input:

int((x^4*(a + b*acos(1/(c*x))))/(d + e*x^2)^2,x)
 

Output:

int((x^4*(a + b*acos(1/(c*x))))/(d + e*x^2)^2, x)
 

Reduce [F]

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

int(x^4*(a+b*asec(c*x))/(e*x^2+d)^2,x)
                                                                                    
                                                                                    
 

Output:

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