\(\int \frac {x^4 (a+b \arccos (c x))}{(d+e x^2)^3} \, dx\) [648]

Optimal result

Integrand size = 21, antiderivative size = 1082 \[ \int \frac {x^4 (a+b \arccos (c x))}{\left (d+e x^2\right )^3} \, dx =\text {Too large to display} \] Output:

1/16*b*c*(-d)^(1/2)*(-c^2*x^2+1)^(1/2)/e^2/(c^2*d+e)/((-d)^(1/2)-e^(1/2)*x 
)+1/16*b*c*(-d)^(1/2)*(-c^2*x^2+1)^(1/2)/e^2/(c^2*d+e)/((-d)^(1/2)+e^(1/2) 
*x)-1/16*(-d)^(1/2)*(a+b*arccos(c*x))/e^(5/2)/((-d)^(1/2)-e^(1/2)*x)^2+5/1 
6*(a+b*arccos(c*x))/e^(5/2)/((-d)^(1/2)-e^(1/2)*x)+1/16*(-d)^(1/2)*(a+b*ar 
ccos(c*x))/e^(5/2)/((-d)^(1/2)+e^(1/2)*x)^2-5/16*(a+b*arccos(c*x))/e^(5/2) 
/((-d)^(1/2)+e^(1/2)*x)+1/16*b*c^3*d*arctanh((e^(1/2)-c^2*(-d)^(1/2)*x)/(c 
^2*d+e)^(1/2)/(-c^2*x^2+1)^(1/2))/e^(5/2)/(c^2*d+e)^(3/2)-5/16*b*c*arctanh 
((e^(1/2)-c^2*(-d)^(1/2)*x)/(c^2*d+e)^(1/2)/(-c^2*x^2+1)^(1/2))/e^(5/2)/(c 
^2*d+e)^(1/2)+1/16*b*c^3*d*arctanh((e^(1/2)+c^2*(-d)^(1/2)*x)/(c^2*d+e)^(1 
/2)/(-c^2*x^2+1)^(1/2))/e^(5/2)/(c^2*d+e)^(3/2)-5/16*b*c*arctanh((e^(1/2)+ 
c^2*(-d)^(1/2)*x)/(c^2*d+e)^(1/2)/(-c^2*x^2+1)^(1/2))/e^(5/2)/(c^2*d+e)^(1 
/2)+3/16*(a+b*arccos(c*x))*ln(1-e^(1/2)*(c*x+I*(-c^2*x^2+1)^(1/2))/(I*c*(- 
d)^(1/2)-(c^2*d+e)^(1/2)))/(-d)^(1/2)/e^(5/2)-3/16*(a+b*arccos(c*x))*ln(1+ 
e^(1/2)*(c*x+I*(-c^2*x^2+1)^(1/2))/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2)))/(-d)^ 
(1/2)/e^(5/2)+3/16*(a+b*arccos(c*x))*ln(1-e^(1/2)*(c*x+I*(-c^2*x^2+1)^(1/2 
))/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/2)))/(-d)^(1/2)/e^(5/2)-3/16*(a+b*arccos(c 
*x))*ln(1+e^(1/2)*(c*x+I*(-c^2*x^2+1)^(1/2))/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/ 
2)))/(-d)^(1/2)/e^(5/2)+3/16*I*b*polylog(2,-e^(1/2)*(c*x+I*(-c^2*x^2+1)^(1 
/2))/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/2)))/(-d)^(1/2)/e^(5/2)-3/16*I*b*polylog 
(2,e^(1/2)*(c*x+I*(-c^2*x^2+1)^(1/2))/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2)))...
 

Mathematica [A] (warning: unable to verify)

Time = 6.02 (sec) , antiderivative size = 1527, normalized size of antiderivative = 1.41 \[ \int \frac {x^4 (a+b \arccos (c x))}{\left (d+e x^2\right )^3} \, dx =\text {Too large to display} \] Input:

Integrate[(x^4*(a + b*ArcCos[c*x]))/(d + e*x^2)^3,x]
 

Output:

((8*a*d*Sqrt[e]*x)/(d + e*x^2)^2 - (20*a*Sqrt[e]*x)/(d + e*x^2) + (12*a*Ar 
cTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[d] - 10*b*(ArcCos[c*x]/((-I)*Sqrt[d] + Sqr 
t[e]*x) - (c*Log[(2*e*(Sqrt[e] - I*c^2*Sqrt[d]*x + Sqrt[c^2*d + e]*Sqrt[1 
- c^2*x^2]))/(c*Sqrt[c^2*d + e]*((-I)*Sqrt[d] + Sqrt[e]*x))])/Sqrt[c^2*d + 
 e]) + 2*b*Sqrt[d]*((I*c*Sqrt[e]*Sqrt[1 - c^2*x^2])/((c^2*d + e)*((-I)*Sqr 
t[d] + Sqrt[e]*x)) + (I*ArcCos[c*x])/(Sqrt[d] + I*Sqrt[e]*x)^2 - (c^3*Sqrt 
[d]*Log[(-4*e*Sqrt[c^2*d + e]*(Sqrt[e] - I*c^2*Sqrt[d]*x + Sqrt[c^2*d + e] 
*Sqrt[1 - c^2*x^2]))/(c^3*(d + I*Sqrt[d]*Sqrt[e]*x))])/(c^2*d + e)^(3/2)) 
+ 10*b*(-(ArcCos[c*x]/(I*Sqrt[d] + Sqrt[e]*x)) + (c*Log[(-2*e*(Sqrt[e] + I 
*c^2*Sqrt[d]*x + Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2]))/(c*Sqrt[c^2*d + e]*(I 
*Sqrt[d] + Sqrt[e]*x))])/Sqrt[c^2*d + e]) + 2*b*Sqrt[d]*(((-I)*c*Sqrt[e]*S 
qrt[1 - c^2*x^2])/((c^2*d + e)*(I*Sqrt[d] + Sqrt[e]*x)) + (I*ArcCos[c*x])/ 
(I*Sqrt[d] + Sqrt[e]*x)^2 - (c^3*Sqrt[d]*Log[(-4*e*Sqrt[c^2*d + e]*(Sqrt[e 
] + I*c^2*Sqrt[d]*x + Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2]))/(c^3*(d - I*Sqrt 
[d]*Sqrt[e]*x))])/(c^2*d + e)^(3/2)) + (3*b*(ArcCos[c*x]^2 - 8*ArcSin[Sqrt 
[1 + (I*c*Sqrt[d])/Sqrt[e]]/Sqrt[2]]*ArcTan[((c*Sqrt[d] + I*Sqrt[e])*Tan[A 
rcCos[c*x]/2])/Sqrt[c^2*d + e]] + (2*I)*ArcCos[c*x]*Log[1 - (I*(-(c*Sqrt[d 
]) + Sqrt[c^2*d + e])*E^(I*ArcCos[c*x]))/Sqrt[e]] + (4*I)*ArcSin[Sqrt[1 + 
(I*c*Sqrt[d])/Sqrt[e]]/Sqrt[2]]*Log[1 - (I*(-(c*Sqrt[d]) + Sqrt[c^2*d + e] 
)*E^(I*ArcCos[c*x]))/Sqrt[e]] + (2*I)*ArcCos[c*x]*Log[1 + (I*(c*Sqrt[d]...
 

Rubi [A] (verified)

Time = 4.11 (sec) , antiderivative size = 1082, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {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.

\(\displaystyle \int \frac {x^4 (a+b \arccos (c x))}{\left (d+e x^2\right )^3} \, dx\)

\(\Big \downarrow \) 5233

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {b d \text {arctanh}\left (\frac {\sqrt {e}-c^2 \sqrt {-d} x}{\sqrt {d c^2+e} \sqrt {1-c^2 x^2}}\right ) c^3}{16 e^{5/2} \left (d c^2+e\right )^{3/2}}-\frac {b d \text {arctanh}\left (\frac {\sqrt {-d} x c^2+\sqrt {e}}{\sqrt {d c^2+e} \sqrt {1-c^2 x^2}}\right ) c^3}{16 e^{5/2} \left (d c^2+e\right )^{3/2}}+\frac {5 b \text {arctanh}\left (\frac {\sqrt {e}-c^2 \sqrt {-d} x}{\sqrt {d c^2+e} \sqrt {1-c^2 x^2}}\right ) c}{16 e^{5/2} \sqrt {d c^2+e}}+\frac {5 b \text {arctanh}\left (\frac {\sqrt {-d} x c^2+\sqrt {e}}{\sqrt {d c^2+e} \sqrt {1-c^2 x^2}}\right ) c}{16 e^{5/2} \sqrt {d c^2+e}}-\frac {b \sqrt {-d} \sqrt {1-c^2 x^2} c}{16 e^2 \left (d c^2+e\right ) \left (\sqrt {-d}-\sqrt {e} x\right )}-\frac {b \sqrt {-d} \sqrt {1-c^2 x^2} c}{16 e^2 \left (d c^2+e\right ) \left (\sqrt {e} x+\sqrt {-d}\right )}+\frac {5 (a+b \arccos (c x))}{16 e^{5/2} \left (\sqrt {-d}-\sqrt {e} x\right )}-\frac {5 (a+b \arccos (c x))}{16 e^{5/2} \left (\sqrt {e} x+\sqrt {-d}\right )}-\frac {\sqrt {-d} (a+b \arccos (c x))}{16 e^{5/2} \left (\sqrt {-d}-\sqrt {e} x\right )^2}+\frac {\sqrt {-d} (a+b \arccos (c x))}{16 e^{5/2} \left (\sqrt {e} x+\sqrt {-d}\right )^2}+\frac {3 (a+b \arccos (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arccos (c x)}}{c \sqrt {-d}-i \sqrt {d c^2+e}}\right )}{16 \sqrt {-d} e^{5/2}}-\frac {3 (a+b \arccos (c x)) \log \left (\frac {e^{i \arccos (c x)} \sqrt {e}}{c \sqrt {-d}-i \sqrt {d c^2+e}}+1\right )}{16 \sqrt {-d} e^{5/2}}+\frac {3 (a+b \arccos (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arccos (c x)}}{\sqrt {-d} c+i \sqrt {d c^2+e}}\right )}{16 \sqrt {-d} e^{5/2}}-\frac {3 (a+b \arccos (c x)) \log \left (\frac {e^{i \arccos (c x)} \sqrt {e}}{\sqrt {-d} c+i \sqrt {d c^2+e}}+1\right )}{16 \sqrt {-d} e^{5/2}}+\frac {3 i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arccos (c x)}}{c \sqrt {-d}-i \sqrt {d c^2+e}}\right )}{16 \sqrt {-d} e^{5/2}}-\frac {3 i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arccos (c x)}}{c \sqrt {-d}-i \sqrt {d c^2+e}}\right )}{16 \sqrt {-d} e^{5/2}}+\frac {3 i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arccos (c x)}}{\sqrt {-d} c+i \sqrt {d c^2+e}}\right )}{16 \sqrt {-d} e^{5/2}}-\frac {3 i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arccos (c x)}}{\sqrt {-d} c+i \sqrt {d c^2+e}}\right )}{16 \sqrt {-d} e^{5/2}}\)

Input:

Int[(x^4*(a + b*ArcCos[c*x]))/(d + e*x^2)^3,x]
 

Output:

-1/16*(b*c*Sqrt[-d]*Sqrt[1 - c^2*x^2])/(e^2*(c^2*d + e)*(Sqrt[-d] - Sqrt[e 
]*x)) - (b*c*Sqrt[-d]*Sqrt[1 - c^2*x^2])/(16*e^2*(c^2*d + e)*(Sqrt[-d] + S 
qrt[e]*x)) - (Sqrt[-d]*(a + b*ArcCos[c*x]))/(16*e^(5/2)*(Sqrt[-d] - Sqrt[e 
]*x)^2) + (5*(a + b*ArcCos[c*x]))/(16*e^(5/2)*(Sqrt[-d] - Sqrt[e]*x)) + (S 
qrt[-d]*(a + b*ArcCos[c*x]))/(16*e^(5/2)*(Sqrt[-d] + Sqrt[e]*x)^2) - (5*(a 
 + b*ArcCos[c*x]))/(16*e^(5/2)*(Sqrt[-d] + Sqrt[e]*x)) - (b*c^3*d*ArcTanh[ 
(Sqrt[e] - c^2*Sqrt[-d]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2])])/(16*e^(5/ 
2)*(c^2*d + e)^(3/2)) + (5*b*c*ArcTanh[(Sqrt[e] - c^2*Sqrt[-d]*x)/(Sqrt[c^ 
2*d + e]*Sqrt[1 - c^2*x^2])])/(16*e^(5/2)*Sqrt[c^2*d + e]) - (b*c^3*d*ArcT 
anh[(Sqrt[e] + c^2*Sqrt[-d]*x)/(Sqrt[c^2*d + e]*Sqrt[1 - c^2*x^2])])/(16*e 
^(5/2)*(c^2*d + e)^(3/2)) + (5*b*c*ArcTanh[(Sqrt[e] + c^2*Sqrt[-d]*x)/(Sqr 
t[c^2*d + e]*Sqrt[1 - c^2*x^2])])/(16*e^(5/2)*Sqrt[c^2*d + e]) + (3*(a + b 
*ArcCos[c*x])*Log[1 - (Sqrt[e]*E^(I*ArcCos[c*x]))/(c*Sqrt[-d] - I*Sqrt[c^2 
*d + e])])/(16*Sqrt[-d]*e^(5/2)) - (3*(a + b*ArcCos[c*x])*Log[1 + (Sqrt[e] 
*E^(I*ArcCos[c*x]))/(c*Sqrt[-d] - I*Sqrt[c^2*d + e])])/(16*Sqrt[-d]*e^(5/2 
)) + (3*(a + b*ArcCos[c*x])*Log[1 - (Sqrt[e]*E^(I*ArcCos[c*x]))/(c*Sqrt[-d 
] + I*Sqrt[c^2*d + e])])/(16*Sqrt[-d]*e^(5/2)) - (3*(a + b*ArcCos[c*x])*Lo 
g[1 + (Sqrt[e]*E^(I*ArcCos[c*x]))/(c*Sqrt[-d] + I*Sqrt[c^2*d + e])])/(16*S 
qrt[-d]*e^(5/2)) + (((3*I)/16)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcCos[c*x]))/ 
(c*Sqrt[-d] - I*Sqrt[c^2*d + e]))])/(Sqrt[-d]*e^(5/2)) - (((3*I)/16)*b*...
 

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]
 

Maple [C] (warning: unable to verify)

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

Time = 11.38 (sec) , antiderivative size = 1760, normalized size of antiderivative = 1.63

method	result	size
parts	\(\text {Expression too large to display}\)	\(1760\)
derivativedivides	\(\text {Expression too large to display}\)	\(1762\)
default	\(\text {Expression too large to display}\)	\(1762\)

Input:

int(x^4*(a+b*arccos(c*x))/(e*x^2+d)^3,x,method=_RETURNVERBOSE)
 

Output:

a*((-5/8/e*x^3-3/8*d/e^2*x)/(e*x^2+d)^2+3/8/e^2/(d*e)^(1/2)*arctan(e*x/(d* 
e)^(1/2)))+b/c^5*(-1/8*c^6*(3*arccos(c*x)*d^2*c^5*x+5*arccos(c*x)*c^5*d*e* 
x^3+d^2*c^4*(-c^2*x^2+1)^(1/2)+c^4*d*e*x^2*(-c^2*x^2+1)^(1/2)+3*arccos(c*x 
)*c^3*d*e*x+5*arccos(c*x)*e^2*c^3*x^3)/e^2/(c^2*d+e)/(c^2*e*x^2+c^2*d)^2-5 
/8*I*(-(2*c^2*d-2*(c^2*d*(c^2*d+e))^(1/2)+e)*e)^(1/2)*(2*c^2*d+2*(c^2*d*(c 
^2*d+e))^(1/2)+e)*arctanh(e*(c*x+I*(-c^2*x^2+1)^(1/2))/((-2*c^2*d+2*(c^2*d 
*(c^2*d+e))^(1/2)-e)*e)^(1/2))*c^6/e^4/(c^2*d+e)-1/2*I*((2*c^2*d+2*(c^2*d* 
(c^2*d+e))^(1/2)+e)*e)^(1/2)*(2*c^2*d-2*(c^2*d*(c^2*d+e))^(1/2)+e)*arctan( 
e*(c*x+I*(-c^2*x^2+1)^(1/2))/((2*c^2*d+2*(c^2*d*(c^2*d+e))^(1/2)+e)*e)^(1/ 
2))*c^8*d/e^5/(c^2*d+e)-3/16*I/(c^2*d+e)*c^6/e*sum(_R1/(_R1^2*e+2*c^2*d+e) 
*(I*arccos(c*x)*ln((_R1-c*x-I*(-c^2*x^2+1)^(1/2))/_R1)+dilog((_R1-c*x-I*(- 
c^2*x^2+1)^(1/2))/_R1)),_R1=RootOf(e*_Z^4+(4*c^2*d+2*e)*_Z^2+e))-3/16*I/(c 
^2*d+e)*c^8/e^2*d*sum(_R1/(_R1^2*e+2*c^2*d+e)*(I*arccos(c*x)*ln((_R1-c*x-I 
*(-c^2*x^2+1)^(1/2))/_R1)+dilog((_R1-c*x-I*(-c^2*x^2+1)^(1/2))/_R1)),_R1=R 
ootOf(e*_Z^4+(4*c^2*d+2*e)*_Z^2+e))+3/16*I/(c^2*d+e)*c^6/e*sum(1/_R1/(_R1^ 
2*e+2*c^2*d+e)*(I*arccos(c*x)*ln((_R1-c*x-I*(-c^2*x^2+1)^(1/2))/_R1)+dilog 
((_R1-c*x-I*(-c^2*x^2+1)^(1/2))/_R1)),_R1=RootOf(e*_Z^4+(4*c^2*d+2*e)*_Z^2 
+e))+1/2*I*(-(2*c^2*d-2*(c^2*d*(c^2*d+e))^(1/2)+e)*e)^(1/2)*(2*c^4*d^2+2*d 
*c^2*(c^2*d*(c^2*d+e))^(1/2)+2*c^2*d*e+(c^2*d*(c^2*d+e))^(1/2)*e)*c^8*d*ar 
ctanh(e*(c*x+I*(-c^2*x^2+1)^(1/2))/((-2*c^2*d+2*(c^2*d*(c^2*d+e))^(1/2)...
 

Fricas [F]

\[ \int \frac {x^4 (a+b \arccos (c x))}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \arccos \left (c x\right ) + a\right )} x^{4}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \] Input:

integrate(x^4*(a+b*arccos(c*x))/(e*x^2+d)^3,x, algorithm="fricas")
 

Output:

integral((b*x^4*arccos(c*x) + a*x^4)/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 
+ d^3), x)
 

Sympy [F]

\[ \int \frac {x^4 (a+b \arccos (c x))}{\left (d+e x^2\right )^3} \, dx=\int \frac {x^{4} \left (a + b \operatorname {acos}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{3}}\, dx \] Input:

integrate(x**4*(a+b*acos(c*x))/(e*x**2+d)**3,x)
 

Output:

Integral(x**4*(a + b*acos(c*x))/(d + e*x**2)**3, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^4 (a+b \arccos (c x))}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^4*(a+b*arccos(c*x))/(e*x^2+d)^3,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]

\[ \int \frac {x^4 (a+b \arccos (c x))}{\left (d+e x^2\right )^3} \, dx=\int { \frac {{\left (b \arccos \left (c x\right ) + a\right )} x^{4}}{{\left (e x^{2} + d\right )}^{3}} \,d x } \] Input:

integrate(x^4*(a+b*arccos(c*x))/(e*x^2+d)^3,x, algorithm="giac")
 

Output:

integrate((b*arccos(c*x) + a)*x^4/(e*x^2 + d)^3, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (a+b \arccos (c x))}{\left (d+e x^2\right )^3} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}{{\left (e\,x^2+d\right )}^3} \,d x \] Input:

int((x^4*(a + b*acos(c*x)))/(d + e*x^2)^3,x)
 

Output:

int((x^4*(a + b*acos(c*x)))/(d + e*x^2)^3, x)
 

Reduce [F]

\[ \int \frac {x^4 (a+b \arccos (c x))}{\left (d+e x^2\right )^3} \, dx=\frac {3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a \,d^{2}+6 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a d e \,x^{2}+3 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) a \,e^{2} x^{4}+8 \left (\int \frac {\mathit {acos} \left (c x \right ) x^{4}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) b \,d^{3} e^{3}+16 \left (\int \frac {\mathit {acos} \left (c x \right ) x^{4}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) b \,d^{2} e^{4} x^{2}+8 \left (\int \frac {\mathit {acos} \left (c x \right ) x^{4}}{e^{3} x^{6}+3 d \,e^{2} x^{4}+3 d^{2} e \,x^{2}+d^{3}}d x \right ) b d \,e^{5} x^{4}-3 a \,d^{2} e x -5 a d \,e^{2} x^{3}}{8 d \,e^{3} \left (e^{2} x^{4}+2 d e \,x^{2}+d^{2}\right )} \] Input:

int(x^4*(a+b*acos(c*x))/(e*x^2+d)^3,x)
 

Output:

(3*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d**2 + 6*sqrt(e)*sqrt(d 
)*atan((e*x)/(sqrt(e)*sqrt(d)))*a*d*e*x**2 + 3*sqrt(e)*sqrt(d)*atan((e*x)/ 
(sqrt(e)*sqrt(d)))*a*e**2*x**4 + 8*int((acos(c*x)*x**4)/(d**3 + 3*d**2*e*x 
**2 + 3*d*e**2*x**4 + e**3*x**6),x)*b*d**3*e**3 + 16*int((acos(c*x)*x**4)/ 
(d**3 + 3*d**2*e*x**2 + 3*d*e**2*x**4 + e**3*x**6),x)*b*d**2*e**4*x**2 + 8 
*int((acos(c*x)*x**4)/(d**3 + 3*d**2*e*x**2 + 3*d*e**2*x**4 + e**3*x**6),x 
)*b*d*e**5*x**4 - 3*a*d**2*e*x - 5*a*d*e**2*x**3)/(8*d*e**3*(d**2 + 2*d*e* 
x**2 + e**2*x**4))