\(\int \frac {x^4}{\arccos (a x)^3} \, dx\) [60]

Optimal result

Integrand size = 10, antiderivative size = 98 \[ \int \frac {x^4}{\arccos (a x)^3} \, dx=\frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}-\frac {2 x^3}{a^2 \arccos (a x)}+\frac {5 x^5}{2 \arccos (a x)}+\frac {\text {Si}(\arccos (a x))}{16 a^5}+\frac {27 \text {Si}(3 \arccos (a x))}{32 a^5}+\frac {25 \text {Si}(5 \arccos (a x))}{32 a^5} \] Output:

1/2*x^4*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)^2-2*x^3/a^2/arccos(a*x)+5/2*x^5/a 
rccos(a*x)+1/16*Si(arccos(a*x))/a^5+27/32*Si(3*arccos(a*x))/a^5+25/32*Si(5 
*arccos(a*x))/a^5
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05 \[ \int \frac {x^4}{\arccos (a x)^3} \, dx=\frac {16 a^4 x^4 \sqrt {1-a^2 x^2}-64 a^3 x^3 \arccos (a x)+80 a^5 x^5 \arccos (a x)+2 \arccos (a x)^2 \text {Si}(\arccos (a x))+27 \arccos (a x)^2 \text {Si}(3 \arccos (a x))+25 \arccos (a x)^2 \text {Si}(5 \arccos (a x))}{32 a^5 \arccos (a x)^2} \] Input:

Integrate[x^4/ArcCos[a*x]^3,x]
 

Output:

(16*a^4*x^4*Sqrt[1 - a^2*x^2] - 64*a^3*x^3*ArcCos[a*x] + 80*a^5*x^5*ArcCos 
[a*x] + 2*ArcCos[a*x]^2*SinIntegral[ArcCos[a*x]] + 27*ArcCos[a*x]^2*SinInt 
egral[3*ArcCos[a*x]] + 25*ArcCos[a*x]^2*SinIntegral[5*ArcCos[a*x]])/(32*a^ 
5*ArcCos[a*x]^2)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.35, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5145, 5223, 5147, 4906, 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.

\(\displaystyle \int \frac {x^4}{\arccos (a x)^3} \, dx\)

\(\Big \downarrow \) 5145

\(\displaystyle \frac {5}{2} a \int \frac {x^5}{\sqrt {1-a^2 x^2} \arccos (a x)^2}dx-\frac {2 \int \frac {x^3}{\sqrt {1-a^2 x^2} \arccos (a x)^2}dx}{a}+\frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}\)

\(\Big \downarrow \) 5223

\(\displaystyle \frac {5}{2} a \left (\frac {x^5}{a \arccos (a x)}-\frac {5 \int \frac {x^4}{\arccos (a x)}dx}{a}\right )-\frac {2 \left (\frac {x^3}{a \arccos (a x)}-\frac {3 \int \frac {x^2}{\arccos (a x)}dx}{a}\right )}{a}+\frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}\)

\(\Big \downarrow \) 5147

\(\displaystyle -\frac {2 \left (\frac {3 \int \frac {a^2 x^2 \sqrt {1-a^2 x^2}}{\arccos (a x)}d\arccos (a x)}{a^4}+\frac {x^3}{a \arccos (a x)}\right )}{a}+\frac {5}{2} a \left (\frac {5 \int \frac {a^4 x^4 \sqrt {1-a^2 x^2}}{\arccos (a x)}d\arccos (a x)}{a^6}+\frac {x^5}{a \arccos (a x)}\right )+\frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}\)

\(\Big \downarrow \) 4906

\(\displaystyle \frac {5}{2} a \left (\frac {5 \int \left (\frac {3 \sin (3 \arccos (a x))}{16 \arccos (a x)}+\frac {\sin (5 \arccos (a x))}{16 \arccos (a x)}+\frac {\sqrt {1-a^2 x^2}}{8 \arccos (a x)}\right )d\arccos (a x)}{a^6}+\frac {x^5}{a \arccos (a x)}\right )-\frac {2 \left (\frac {3 \int \left (\frac {\sin (3 \arccos (a x))}{4 \arccos (a x)}+\frac {\sqrt {1-a^2 x^2}}{4 \arccos (a x)}\right )d\arccos (a x)}{a^4}+\frac {x^3}{a \arccos (a x)}\right )}{a}+\frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {5}{2} a \left (\frac {5 \left (\frac {1}{8} \text {Si}(\arccos (a x))+\frac {3}{16} \text {Si}(3 \arccos (a x))+\frac {1}{16} \text {Si}(5 \arccos (a x))\right )}{a^6}+\frac {x^5}{a \arccos (a x)}\right )-\frac {2 \left (\frac {3 \left (\frac {1}{4} \text {Si}(\arccos (a x))+\frac {1}{4} \text {Si}(3 \arccos (a x))\right )}{a^4}+\frac {x^3}{a \arccos (a x)}\right )}{a}+\frac {x^4 \sqrt {1-a^2 x^2}}{2 a \arccos (a x)^2}\)

Input:

Int[x^4/ArcCos[a*x]^3,x]
 

Output:

(x^4*Sqrt[1 - a^2*x^2])/(2*a*ArcCos[a*x]^2) - (2*(x^3/(a*ArcCos[a*x]) + (3 
*(SinIntegral[ArcCos[a*x]]/4 + SinIntegral[3*ArcCos[a*x]]/4))/a^4))/a + (5 
*a*(x^5/(a*ArcCos[a*x]) + (5*(SinIntegral[ArcCos[a*x]]/8 + (3*SinIntegral[ 
3*ArcCos[a*x]])/16 + SinIntegral[5*ArcCos[a*x]]/16))/a^6))/2
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x 
]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG 
tQ[p, 0]
 

rule 5145

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[( 
-x^m)*Sqrt[1 - c^2*x^2]*((a + b*ArcCos[c*x])^(n + 1)/(b*c*(n + 1))), x] + ( 
-Simp[c*((m + 1)/(b*(n + 1)))   Int[x^(m + 1)*((a + b*ArcCos[c*x])^(n + 1)/ 
Sqrt[1 - c^2*x^2]), x], x] + Simp[m/(b*c*(n + 1))   Int[x^(m - 1)*((a + b*A 
rcCos[c*x])^(n + 1)/Sqrt[1 - c^2*x^2]), x], x]) /; FreeQ[{a, b, c}, x] && I 
GtQ[m, 0] && LtQ[n, -2]
 

rule 5147

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[- 
(b*c^(m + 1))^(-1)   Subst[Int[x^n*Cos[-a/b + x/b]^m*Sin[-a/b + x/b], x], x 
, a + b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]
 

rule 5223

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) 
+ (e_.)*(x_)^2], x_Symbol] :> Simp[(-(f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 - c 
^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcCos[c*x])^(n + 1), x] + Simp[f*(m/(b*c*( 
n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]   Int[(f*x)^(m - 1)*(a + b 
*ArcCos[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2 
*d + e, 0] && LtQ[n, -1]
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\frac {\frac {\sqrt {-a^{2} x^{2}+1}}{16 \arccos \left (a x \right )^{2}}+\frac {a x}{16 \arccos \left (a x \right )}+\frac {\operatorname {Si}\left (\arccos \left (a x \right )\right )}{16}+\frac {3 \sin \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )^{2}}+\frac {9 \cos \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}+\frac {27 \,\operatorname {Si}\left (3 \arccos \left (a x \right )\right )}{32}+\frac {\sin \left (5 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )^{2}}+\frac {5 \cos \left (5 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}+\frac {25 \,\operatorname {Si}\left (5 \arccos \left (a x \right )\right )}{32}}{a^{5}}\)	\(121\)
default	\(\frac {\frac {\sqrt {-a^{2} x^{2}+1}}{16 \arccos \left (a x \right )^{2}}+\frac {a x}{16 \arccos \left (a x \right )}+\frac {\operatorname {Si}\left (\arccos \left (a x \right )\right )}{16}+\frac {3 \sin \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )^{2}}+\frac {9 \cos \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}+\frac {27 \,\operatorname {Si}\left (3 \arccos \left (a x \right )\right )}{32}+\frac {\sin \left (5 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )^{2}}+\frac {5 \cos \left (5 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}+\frac {25 \,\operatorname {Si}\left (5 \arccos \left (a x \right )\right )}{32}}{a^{5}}\)	\(121\)

Input:

int(x^4/arccos(a*x)^3,x,method=_RETURNVERBOSE)
 

Output:

1/a^5*(1/16/arccos(a*x)^2*(-a^2*x^2+1)^(1/2)+1/16*a*x/arccos(a*x)+1/16*Si( 
arccos(a*x))+3/32/arccos(a*x)^2*sin(3*arccos(a*x))+9/32*cos(3*arccos(a*x)) 
/arccos(a*x)+27/32*Si(3*arccos(a*x))+1/32/arccos(a*x)^2*sin(5*arccos(a*x)) 
+5/32*cos(5*arccos(a*x))/arccos(a*x)+25/32*Si(5*arccos(a*x)))
 

Fricas [F]

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

integrate(x^4/arccos(a*x)^3,x, algorithm="fricas")
 

Output:

integral(x^4/arccos(a*x)^3, x)
 

Sympy [F]

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

integrate(x**4/acos(a*x)**3,x)
 

Output:

Integral(x**4/acos(a*x)**3, x)
 

Maxima [F]

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

integrate(x^4/arccos(a*x)^3,x, algorithm="maxima")
 

Output:

1/2*(sqrt(a*x + 1)*sqrt(-a*x + 1)*a*x^4 - arctan2(sqrt(a*x + 1)*sqrt(-a*x 
+ 1), a*x)^2*integrate((25*a^2*x^4 - 12*x^2)/arctan2(sqrt(a*x + 1)*sqrt(-a 
*x + 1), a*x), x) + (5*a^2*x^5 - 4*x^3)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 
1), a*x))/(a^2*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.88 \[ \int \frac {x^4}{\arccos (a x)^3} \, dx=\frac {5 \, x^{5}}{2 \, \arccos \left (a x\right )} + \frac {\sqrt {-a^{2} x^{2} + 1} x^{4}}{2 \, a \arccos \left (a x\right )^{2}} - \frac {2 \, x^{3}}{a^{2} \arccos \left (a x\right )} + \frac {25 \, \operatorname {Si}\left (5 \, \arccos \left (a x\right )\right )}{32 \, a^{5}} + \frac {27 \, \operatorname {Si}\left (3 \, \arccos \left (a x\right )\right )}{32 \, a^{5}} + \frac {\operatorname {Si}\left (\arccos \left (a x\right )\right )}{16 \, a^{5}} \] Input:

integrate(x^4/arccos(a*x)^3,x, algorithm="giac")
 

Output:

5/2*x^5/arccos(a*x) + 1/2*sqrt(-a^2*x^2 + 1)*x^4/(a*arccos(a*x)^2) - 2*x^3 
/(a^2*arccos(a*x)) + 25/32*sin_integral(5*arccos(a*x))/a^5 + 27/32*sin_int 
egral(3*arccos(a*x))/a^5 + 1/16*sin_integral(arccos(a*x))/a^5
 

Mupad [F(-1)]

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

int(x^4/acos(a*x)^3,x)
 

Output:

int(x^4/acos(a*x)^3, x)
 

Reduce [F]

\[ \int \frac {x^4}{\arccos (a x)^3} \, dx=\int \frac {x^{4}}{\mathit {acos} \left (a x \right )^{3}}d x \] Input:

int(x^4/acos(a*x)^3,x)
 

Output:

int(x**4/acos(a*x)**3,x)