3.1.68 \(\int \frac {x^3}{\arccos (a x)^4} \, dx\) [68]

3.1.68.1 Optimal result
3.1.68.2 Mathematica [A] (verified)
3.1.68.3 Rubi [A] (verified)
3.1.68.4 Maple [A] (verified)
3.1.68.5 Fricas [F]
3.1.68.6 Sympy [F]
3.1.68.7 Maxima [F]
3.1.68.8 Giac [A] (verification not implemented)
3.1.68.9 Mupad [F(-1)]

3.1.68.1 Optimal result

Integrand size = 10, antiderivative size = 143 \[ \int \frac {x^3}{\arccos (a x)^4} \, dx=\frac {x^3 \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^3}-\frac {x^2}{2 a^2 \arccos (a x)^2}+\frac {2 x^4}{3 \arccos (a x)^2}+\frac {x \sqrt {1-a^2 x^2}}{a^3 \arccos (a x)}-\frac {8 x^3 \sqrt {1-a^2 x^2}}{3 a \arccos (a x)}+\frac {\operatorname {CosIntegral}(2 \arccos (a x))}{3 a^4}+\frac {4 \operatorname {CosIntegral}(4 \arccos (a x))}{3 a^4} \]

output
-1/2*x^2/a^2/arccos(a*x)^2+2/3*x^4/arccos(a*x)^2+1/3*Ci(2*arccos(a*x))/a^4 
+4/3*Ci(4*arccos(a*x))/a^4+1/3*x^3*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)^3+x*(- 
a^2*x^2+1)^(1/2)/a^3/arccos(a*x)-8/3*x^3*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)
 
3.1.68.2 Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{\arccos (a x)^4} \, dx=\frac {\frac {a x \left (2 a^2 x^2 \sqrt {1-a^2 x^2}+a x \left (-3+4 a^2 x^2\right ) \arccos (a x)-2 \sqrt {1-a^2 x^2} \left (-3+8 a^2 x^2\right ) \arccos (a x)^2\right )}{\arccos (a x)^3}+2 \operatorname {CosIntegral}(2 \arccos (a x))+8 \operatorname {CosIntegral}(4 \arccos (a x))}{6 a^4} \]

input
Integrate[x^3/ArcCos[a*x]^4,x]
 
output
((a*x*(2*a^2*x^2*Sqrt[1 - a^2*x^2] + a*x*(-3 + 4*a^2*x^2)*ArcCos[a*x] - 2* 
Sqrt[1 - a^2*x^2]*(-3 + 8*a^2*x^2)*ArcCos[a*x]^2))/ArcCos[a*x]^3 + 2*CosIn 
tegral[2*ArcCos[a*x]] + 8*CosIntegral[4*ArcCos[a*x]])/(6*a^4)
 
3.1.68.3 Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.24, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5145, 5223, 5143, 25, 2009, 3042, 3783}

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.

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

\(\Big \downarrow \) 5145

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

\(\Big \downarrow \) 5223

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

\(\Big \downarrow \) 5143

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {x^2}{2 a \arccos (a x)^2}-\frac {\frac {x \sqrt {1-a^2 x^2}}{a \arccos (a x)}-\frac {\int \frac {\sin \left (2 \arccos (a x)+\frac {\pi }{2}\right )}{\arccos (a x)}d\arccos (a x)}{a^2}}{a}}{a}+\frac {x^3 \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^3}+\frac {4}{3} a \left (\frac {x^4}{2 a \arccos (a x)^2}-\frac {2 \left (\frac {-\frac {1}{2} \operatorname {CosIntegral}(2 \arccos (a x))-\frac {1}{2} \operatorname {CosIntegral}(4 \arccos (a x))}{a^4}+\frac {x^3 \sqrt {1-a^2 x^2}}{a \arccos (a x)}\right )}{a}\right )\)

\(\Big \downarrow \) 3783

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

input
Int[x^3/ArcCos[a*x]^4,x]
 
output
(x^3*Sqrt[1 - a^2*x^2])/(3*a*ArcCos[a*x]^3) - (x^2/(2*a*ArcCos[a*x]^2) - ( 
(x*Sqrt[1 - a^2*x^2])/(a*ArcCos[a*x]) - CosIntegral[2*ArcCos[a*x]]/a^2)/a) 
/a + (4*a*(x^4/(2*a*ArcCos[a*x]^2) - (2*((x^3*Sqrt[1 - a^2*x^2])/(a*ArcCos 
[a*x]) + (-1/2*CosIntegral[2*ArcCos[a*x]] - CosIntegral[4*ArcCos[a*x]]/2)/ 
a^4))/a))/3
 

3.1.68.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3783
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte 
gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - 
c*f, 0]
 

rule 5143
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] - S 
imp[1/(b^2*c^(m + 1)*(n + 1))   Subst[Int[ExpandTrigReduce[x^(n + 1), Cos[- 
a/b + x/b]^(m - 1)*(m - (m + 1)*Cos[-a/b + x/b]^2), x], x], x, a + b*ArcCos 
[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]
 

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 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]
 
3.1.68.4 Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.80

method result size
derivativedivides \(\frac {\frac {\sin \left (2 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{3}}+\frac {\cos \left (2 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{2}}-\frac {\sin \left (2 \arccos \left (a x \right )\right )}{6 \arccos \left (a x \right )}+\frac {\operatorname {Ci}\left (2 \arccos \left (a x \right )\right )}{3}+\frac {\sin \left (4 \arccos \left (a x \right )\right )}{24 \arccos \left (a x \right )^{3}}+\frac {\cos \left (4 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{2}}-\frac {\sin \left (4 \arccos \left (a x \right )\right )}{3 \arccos \left (a x \right )}+\frac {4 \,\operatorname {Ci}\left (4 \arccos \left (a x \right )\right )}{3}}{a^{4}}\) \(114\)
default \(\frac {\frac {\sin \left (2 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{3}}+\frac {\cos \left (2 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{2}}-\frac {\sin \left (2 \arccos \left (a x \right )\right )}{6 \arccos \left (a x \right )}+\frac {\operatorname {Ci}\left (2 \arccos \left (a x \right )\right )}{3}+\frac {\sin \left (4 \arccos \left (a x \right )\right )}{24 \arccos \left (a x \right )^{3}}+\frac {\cos \left (4 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{2}}-\frac {\sin \left (4 \arccos \left (a x \right )\right )}{3 \arccos \left (a x \right )}+\frac {4 \,\operatorname {Ci}\left (4 \arccos \left (a x \right )\right )}{3}}{a^{4}}\) \(114\)

input
int(x^3/arccos(a*x)^4,x,method=_RETURNVERBOSE)
 
output
1/a^4*(1/12/arccos(a*x)^3*sin(2*arccos(a*x))+1/12/arccos(a*x)^2*cos(2*arcc 
os(a*x))-1/6/arccos(a*x)*sin(2*arccos(a*x))+1/3*Ci(2*arccos(a*x))+1/24/arc 
cos(a*x)^3*sin(4*arccos(a*x))+1/12/arccos(a*x)^2*cos(4*arccos(a*x))-1/3/ar 
ccos(a*x)*sin(4*arccos(a*x))+4/3*Ci(4*arccos(a*x)))
 
3.1.68.5 Fricas [F]

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

input
integrate(x^3/arccos(a*x)^4,x, algorithm="fricas")
 
output
integral(x^3/arccos(a*x)^4, x)
 
3.1.68.6 Sympy [F]

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

input
integrate(x**3/acos(a*x)**4,x)
 
output
Integral(x**3/acos(a*x)**4, x)
 
3.1.68.7 Maxima [F]

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

input
integrate(x^3/arccos(a*x)^4,x, algorithm="maxima")
 
output
1/6*(6*a^3*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^3*integrate(1/3*(32* 
a^4*x^4 - 30*a^2*x^2 + 3)*sqrt(a*x + 1)*sqrt(-a*x + 1)/((a^5*x^2 - a^3)*ar 
ctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)), x) + 2*(a^2*x^3 - (8*a^2*x^3 - 
3*x)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2)*sqrt(a*x + 1)*sqrt(-a*x 
 + 1) + (4*a^3*x^4 - 3*a*x^2)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x))/ 
(a^3*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^3)
 
3.1.68.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.87 \[ \int \frac {x^3}{\arccos (a x)^4} \, dx=\frac {2 \, x^{4}}{3 \, \arccos \left (a x\right )^{2}} - \frac {8 \, \sqrt {-a^{2} x^{2} + 1} x^{3}}{3 \, a \arccos \left (a x\right )} + \frac {\sqrt {-a^{2} x^{2} + 1} x^{3}}{3 \, a \arccos \left (a x\right )^{3}} - \frac {x^{2}}{2 \, a^{2} \arccos \left (a x\right )^{2}} + \frac {\sqrt {-a^{2} x^{2} + 1} x}{a^{3} \arccos \left (a x\right )} + \frac {4 \, \operatorname {Ci}\left (4 \, \arccos \left (a x\right )\right )}{3 \, a^{4}} + \frac {\operatorname {Ci}\left (2 \, \arccos \left (a x\right )\right )}{3 \, a^{4}} \]

input
integrate(x^3/arccos(a*x)^4,x, algorithm="giac")
 
output
2/3*x^4/arccos(a*x)^2 - 8/3*sqrt(-a^2*x^2 + 1)*x^3/(a*arccos(a*x)) + 1/3*s 
qrt(-a^2*x^2 + 1)*x^3/(a*arccos(a*x)^3) - 1/2*x^2/(a^2*arccos(a*x)^2) + sq 
rt(-a^2*x^2 + 1)*x/(a^3*arccos(a*x)) + 4/3*cos_integral(4*arccos(a*x))/a^4 
 + 1/3*cos_integral(2*arccos(a*x))/a^4
 
3.1.68.9 Mupad [F(-1)]

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

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