\(\int \frac {x^4 (a+b \arccos (c x))}{d+e x^2} \, dx\) [626]

Optimal result

Integrand size = 21, antiderivative size = 653 \[ \int \frac {x^4 (a+b \arccos (c x))}{d+e x^2} \, dx=-\frac {a d x}{e^2}-\frac {b d \sqrt {1-c^2 x^2}}{c e^2}+\frac {b \sqrt {1-c^2 x^2}}{3 c^3 e}-\frac {b \left (1-c^2 x^2\right )^{3/2}}{9 c^3 e}-\frac {b d x \arccos (c x)}{e^2}+\frac {x^3 (a+b \arccos (c x))}{3 e}+\frac {(-d)^{3/2} (a+b \arccos (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arccos (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^{5/2}}-\frac {(-d)^{3/2} (a+b \arccos (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arccos (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^{5/2}}+\frac {(-d)^{3/2} (a+b \arccos (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arccos (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^{5/2}}-\frac {(-d)^{3/2} (a+b \arccos (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arccos (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^{5/2}}+\frac {i b (-d)^{3/2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arccos (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^{5/2}}-\frac {i b (-d)^{3/2} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arccos (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e^{5/2}}+\frac {i b (-d)^{3/2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arccos (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^{5/2}}-\frac {i b (-d)^{3/2} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arccos (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e^{5/2}} \] Output:

-a*d*x/e^2-b*d*(-c^2*x^2+1)^(1/2)/c/e^2+1/3*b*(-c^2*x^2+1)^(1/2)/c^3/e-1/9 
*b*(-c^2*x^2+1)^(3/2)/c^3/e-b*d*x*arccos(c*x)/e^2+1/3*x^3*(a+b*arccos(c*x) 
)/e+1/2*(-d)^(3/2)*(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)))/e^(5/2)-1/2*(-d)^(3/2)*(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 
)))/e^(5/2)+1/2*(-d)^(3/2)*(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)))/e^(5/2)-1/2*(-d)^(3/2)*(a+b*a 
rccos(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)))/e^(5/2)+1/2*I*b*(-d)^(3/2)*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)))/e^(5/2)-1/2*I*b*(-d)^(3/2)*po 
lylog(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) 
))/e^(5/2)+1/2*I*b*(-d)^(3/2)*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)))/e^(5/2)-1/2*I*b*(-d)^(3/2)*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)))/e^(5/2 
)
 

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 968, normalized size of antiderivative = 1.48 \[ \int \frac {x^4 (a+b \arccos (c x))}{d+e x^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-((a*d*x)/e^2) + (a*x^3)/(3*e) + (a*d^(3/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/e 
^(5/2) + (b*((-4*e^(3/2)*Sqrt[1 - c^2*x^2]*(2 + c^2*x^2))/(9*c^3) + (4*e^( 
3/2)*x^3*ArcCos[c*x])/3 + (4*d*Sqrt[e]*(Sqrt[1 - c^2*x^2] - c*x*ArcCos[c*x 
]))/c + d^(3/2)*(ArcCos[c*x]^2 - 8*ArcSin[Sqrt[1 + (I*c*Sqrt[d])/Sqrt[e]]/ 
Sqrt[2]]*ArcTan[((c*Sqrt[d] + I*Sqrt[e])*Tan[ArcCos[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*A 
rcCos[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] + 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]]*L 
og[1 + (I*(c*Sqrt[d] + Sqrt[c^2*d + e])*E^(I*ArcCos[c*x]))/Sqrt[e]] + 2*Po 
lyLog[2, (I*(-(c*Sqrt[d]) + Sqrt[c^2*d + e])*E^(I*ArcCos[c*x]))/Sqrt[e]] + 
 2*PolyLog[2, ((-I)*(c*Sqrt[d] + Sqrt[c^2*d + e])*E^(I*ArcCos[c*x]))/Sqrt[ 
e]]) - d^(3/2)*(ArcCos[c*x]^2 - 8*ArcSin[Sqrt[1 - (I*c*Sqrt[d])/Sqrt[e]]/S 
qrt[2]]*ArcTan[((c*Sqrt[d] - I*Sqrt[e])*Tan[ArcCos[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*Ar 
cCos[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] + 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]]...
 

Rubi [A] (verified)

Time = 1.54 (sec) , antiderivative size = 652, 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.

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

\(\Big \downarrow \) 5233

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(-d)^{3/2} (a+b \arccos (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arccos (c x)}}{c \sqrt {-d}-i \sqrt {c^2 d+e}}\right )}{2 e^{5/2}}-\frac {(-d)^{3/2} (a+b \arccos (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arccos (c x)}}{c \sqrt {-d}-i \sqrt {c^2 d+e}}\right )}{2 e^{5/2}}+\frac {(-d)^{3/2} (a+b \arccos (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arccos (c x)}}{c \sqrt {-d}+i \sqrt {c^2 d+e}}\right )}{2 e^{5/2}}-\frac {(-d)^{3/2} (a+b \arccos (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arccos (c x)}}{c \sqrt {-d}+i \sqrt {c^2 d+e}}\right )}{2 e^{5/2}}+\frac {x^3 (a+b \arccos (c x))}{3 e}-\frac {a d x}{e^2}+\frac {i b (-d)^{3/2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arccos (c x)}}{c \sqrt {-d}-i \sqrt {d c^2+e}}\right )}{2 e^{5/2}}-\frac {i b (-d)^{3/2} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arccos (c x)}}{c \sqrt {-d}-i \sqrt {d c^2+e}}\right )}{2 e^{5/2}}+\frac {i b (-d)^{3/2} \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arccos (c x)}}{\sqrt {-d} c+i \sqrt {d c^2+e}}\right )}{2 e^{5/2}}-\frac {i b (-d)^{3/2} \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arccos (c x)}}{\sqrt {-d} c+i \sqrt {d c^2+e}}\right )}{2 e^{5/2}}-\frac {b d x \arccos (c x)}{e^2}+\frac {b d \sqrt {1-c^2 x^2}}{c e^2}+\frac {b \left (1-c^2 x^2\right )^{3/2}}{9 c^3 e}-\frac {b \sqrt {1-c^2 x^2}}{3 c^3 e}\)

Input:

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

Output:

-((a*d*x)/e^2) + (b*d*Sqrt[1 - c^2*x^2])/(c*e^2) - (b*Sqrt[1 - c^2*x^2])/( 
3*c^3*e) + (b*(1 - c^2*x^2)^(3/2))/(9*c^3*e) - (b*d*x*ArcCos[c*x])/e^2 + ( 
x^3*(a + b*ArcCos[c*x]))/(3*e) + ((-d)^(3/2)*(a + b*ArcCos[c*x])*Log[1 - ( 
Sqrt[e]*E^(I*ArcCos[c*x]))/(c*Sqrt[-d] - I*Sqrt[c^2*d + e])])/(2*e^(5/2)) 
- ((-d)^(3/2)*(a + b*ArcCos[c*x])*Log[1 + (Sqrt[e]*E^(I*ArcCos[c*x]))/(c*S 
qrt[-d] - I*Sqrt[c^2*d + e])])/(2*e^(5/2)) + ((-d)^(3/2)*(a + b*ArcCos[c*x 
])*Log[1 - (Sqrt[e]*E^(I*ArcCos[c*x]))/(c*Sqrt[-d] + I*Sqrt[c^2*d + e])])/ 
(2*e^(5/2)) - ((-d)^(3/2)*(a + b*ArcCos[c*x])*Log[1 + (Sqrt[e]*E^(I*ArcCos 
[c*x]))/(c*Sqrt[-d] + I*Sqrt[c^2*d + e])])/(2*e^(5/2)) + ((I/2)*b*(-d)^(3/ 
2)*PolyLog[2, -((Sqrt[e]*E^(I*ArcCos[c*x]))/(c*Sqrt[-d] - I*Sqrt[c^2*d + e 
]))])/e^(5/2) - ((I/2)*b*(-d)^(3/2)*PolyLog[2, (Sqrt[e]*E^(I*ArcCos[c*x])) 
/(c*Sqrt[-d] - I*Sqrt[c^2*d + e])])/e^(5/2) + ((I/2)*b*(-d)^(3/2)*PolyLog[ 
2, -((Sqrt[e]*E^(I*ArcCos[c*x]))/(c*Sqrt[-d] + I*Sqrt[c^2*d + e]))])/e^(5/ 
2) - ((I/2)*b*(-d)^(3/2)*PolyLog[2, (Sqrt[e]*E^(I*ArcCos[c*x]))/(c*Sqrt[-d 
] + I*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]
 

Maple [C] (verified)

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

Time = 137.58 (sec) , antiderivative size = 373, normalized size of antiderivative = 0.57

method	result	size
parts	\(\frac {a \,x^{3}}{3 e}-\frac {a d x}{e^{2}}+\frac {a \,d^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{e^{2} \sqrt {d e}}+\frac {b d \sqrt {-c^{2} x^{2}+1}}{c \,e^{2}}-\frac {b d x \arccos \left (c x \right )}{e^{2}}-\frac {b \sqrt {-c^{2} x^{2}+1}}{4 c^{3} e}+\frac {b \arccos \left (c x \right ) x}{4 c^{2} e}-\frac {i b c \,d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (e \,\textit {\_Z}^{4}+\left (4 c^{2} d +2 e \right ) \textit {\_Z}^{2}+e \right )}{\sum }\frac {\textit {\_R1} \left (i \arccos \left (c x \right ) \ln \left (\frac {\textit {\_R1} -c x -i \sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -c x -i \sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )\right )}{\textit {\_R1}^{2} e +2 c^{2} d +e}\right )}{2 e^{2}}+\frac {i b c \,d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (e \,\textit {\_Z}^{4}+\left (4 c^{2} d +2 e \right ) \textit {\_Z}^{2}+e \right )}{\sum }\frac {i \arccos \left (c x \right ) \ln \left (\frac {\textit {\_R1} -c x -i \sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -c x -i \sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )}{\textit {\_R1} \left (\textit {\_R1}^{2} e +2 c^{2} d +e \right )}\right )}{2 e^{2}}+\frac {b \arccos \left (c x \right ) \cos \left (3 \arccos \left (c x \right )\right )}{12 c^{3} e}-\frac {b \sin \left (3 \arccos \left (c x \right )\right )}{36 c^{3} e}\)	\(373\)
derivativedivides	\(\frac {-\frac {a \,c^{5} d x}{e^{2}}+\frac {a \,c^{5} x^{3}}{3 e}+\frac {a \,c^{5} d^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{e^{2} \sqrt {d e}}+\frac {b \,c^{4} \sqrt {-c^{2} x^{2}+1}\, d}{e^{2}}-\frac {b \,c^{5} \arccos \left (c x \right ) d x}{e^{2}}-\frac {b \,c^{2} \sqrt {-c^{2} x^{2}+1}}{4 e}+\frac {b \,c^{3} \arccos \left (c x \right ) x}{4 e}-\frac {i b \,c^{6} d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (e \,\textit {\_Z}^{4}+\left (4 c^{2} d +2 e \right ) \textit {\_Z}^{2}+e \right )}{\sum }\frac {\textit {\_R1} \left (i \arccos \left (c x \right ) \ln \left (\frac {\textit {\_R1} -c x -i \sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -c x -i \sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )\right )}{\textit {\_R1}^{2} e +2 c^{2} d +e}\right )}{2 e^{2}}+\frac {i b \,c^{6} d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (e \,\textit {\_Z}^{4}+\left (4 c^{2} d +2 e \right ) \textit {\_Z}^{2}+e \right )}{\sum }\frac {i \arccos \left (c x \right ) \ln \left (\frac {\textit {\_R1} -c x -i \sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -c x -i \sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )}{\textit {\_R1} \left (\textit {\_R1}^{2} e +2 c^{2} d +e \right )}\right )}{2 e^{2}}+\frac {b \,c^{2} \arccos \left (c x \right ) \cos \left (3 \arccos \left (c x \right )\right )}{12 e}-\frac {b \,c^{2} \sin \left (3 \arccos \left (c x \right )\right )}{36 e}}{c^{5}}\)	\(393\)
default	\(\frac {-\frac {a \,c^{5} d x}{e^{2}}+\frac {a \,c^{5} x^{3}}{3 e}+\frac {a \,c^{5} d^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{e^{2} \sqrt {d e}}+\frac {b \,c^{4} \sqrt {-c^{2} x^{2}+1}\, d}{e^{2}}-\frac {b \,c^{5} \arccos \left (c x \right ) d x}{e^{2}}-\frac {b \,c^{2} \sqrt {-c^{2} x^{2}+1}}{4 e}+\frac {b \,c^{3} \arccos \left (c x \right ) x}{4 e}-\frac {i b \,c^{6} d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (e \,\textit {\_Z}^{4}+\left (4 c^{2} d +2 e \right ) \textit {\_Z}^{2}+e \right )}{\sum }\frac {\textit {\_R1} \left (i \arccos \left (c x \right ) \ln \left (\frac {\textit {\_R1} -c x -i \sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -c x -i \sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )\right )}{\textit {\_R1}^{2} e +2 c^{2} d +e}\right )}{2 e^{2}}+\frac {i b \,c^{6} d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (e \,\textit {\_Z}^{4}+\left (4 c^{2} d +2 e \right ) \textit {\_Z}^{2}+e \right )}{\sum }\frac {i \arccos \left (c x \right ) \ln \left (\frac {\textit {\_R1} -c x -i \sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -c x -i \sqrt {-c^{2} x^{2}+1}}{\textit {\_R1}}\right )}{\textit {\_R1} \left (\textit {\_R1}^{2} e +2 c^{2} d +e \right )}\right )}{2 e^{2}}+\frac {b \,c^{2} \arccos \left (c x \right ) \cos \left (3 \arccos \left (c x \right )\right )}{12 e}-\frac {b \,c^{2} \sin \left (3 \arccos \left (c x \right )\right )}{36 e}}{c^{5}}\)	\(393\)

Input:

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

Output:

1/3*a/e*x^3-a*d*x/e^2+a*d^2/e^2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))+b*d*(- 
c^2*x^2+1)^(1/2)/c/e^2-b*d*x*arccos(c*x)/e^2-1/4*b*(-c^2*x^2+1)^(1/2)/c^3/ 
e+1/4*b/c^2/e*arccos(c*x)*x-1/2*I*b*c*d^2/e^2*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))+1/2*I*b*c* 
d^2/e^2*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))+1/12*b/c^3*arccos(c*x)/e*cos(3*arccos(c*x))-1/ 
36*b/c^3/e*sin(3*arccos(c*x))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x^4*(a+b*arccos(c*x))/(e*x^2+d),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. \[ \int \frac {x^4 (a+b \arccos (c x))}{d+e x^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

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

Output:

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

Reduce [F]

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

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

Output:

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