\(\int \frac {x}{\arccos (a x)^4} \, dx\) [70]

Optimal result

Integrand size = 8, antiderivative size = 97 \[ \int \frac {x}{\arccos (a x)^4} \, dx=\frac {x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^3}-\frac {1}{6 a^2 \arccos (a x)^2}+\frac {x^2}{3 \arccos (a x)^2}-\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)}+\frac {2 \operatorname {CosIntegral}(2 \arccos (a x))}{3 a^2} \]

[Out]

-1/6/a^2/arccos(a*x)^2+1/3*x^2/arccos(a*x)^2+2/3*Ci(2*arccos(a*x))/a^2+1/3*x*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)^
3-2/3*x*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4730, 4808, 4728, 3383, 4738} \[ \int \frac {x}{\arccos (a x)^4} \, dx=\frac {2 \operatorname {CosIntegral}(2 \arccos (a x))}{3 a^2}-\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)}+\frac {x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^3}-\frac {1}{6 a^2 \arccos (a x)^2}+\frac {x^2}{3 \arccos (a x)^2} \]

[In]

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

[Out]

(x*Sqrt[1 - a^2*x^2])/(3*a*ArcCos[a*x]^3) - 1/(6*a^2*ArcCos[a*x]^2) + x^2/(3*ArcCos[a*x]^2) - (2*x*Sqrt[1 - a^
2*x^2])/(3*a*ArcCos[a*x]) + (2*CosIntegral[2*ArcCos[a*x]])/(3*a^2)

Rule 3383

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

Rule 4728

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(-x^m)*Sqrt[1 - c^2*x^2]*((a + b*Arc
Cos[c*x])^(n + 1)/(b*c*(n + 1))), x] - Dist[1/(b^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), C
os[-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 4730

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

Rule 4738

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

Rule 4808

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] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^3}-\frac {\int \frac {1}{\sqrt {1-a^2 x^2} \arccos (a x)^3} \, dx}{3 a}+\frac {1}{3} (2 a) \int \frac {x^2}{\sqrt {1-a^2 x^2} \arccos (a x)^3} \, dx \\ & = \frac {x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^3}-\frac {1}{6 a^2 \arccos (a x)^2}+\frac {x^2}{3 \arccos (a x)^2}-\frac {2}{3} \int \frac {x}{\arccos (a x)^2} \, dx \\ & = \frac {x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^3}-\frac {1}{6 a^2 \arccos (a x)^2}+\frac {x^2}{3 \arccos (a x)^2}-\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)}+\frac {2 \text {Subst}\left (\int \frac {\cos (2 x)}{x} \, dx,x,\arccos (a x)\right )}{3 a^2} \\ & = \frac {x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^3}-\frac {1}{6 a^2 \arccos (a x)^2}+\frac {x^2}{3 \arccos (a x)^2}-\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)}+\frac {2 \operatorname {CosIntegral}(2 \arccos (a x))}{3 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.89 \[ \int \frac {x}{\arccos (a x)^4} \, dx=\frac {2 a x \sqrt {1-a^2 x^2}+\left (-1+2 a^2 x^2\right ) \arccos (a x)-4 a x \sqrt {1-a^2 x^2} \arccos (a x)^2+4 \arccos (a x)^3 \operatorname {CosIntegral}(2 \arccos (a x))}{6 a^2 \arccos (a x)^3} \]

[In]

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

[Out]

(2*a*x*Sqrt[1 - a^2*x^2] + (-1 + 2*a^2*x^2)*ArcCos[a*x] - 4*a*x*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^2 + 4*ArcCos[a*x
]^3*CosIntegral[2*ArcCos[a*x]])/(6*a^2*ArcCos[a*x]^3)

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.62

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (2 \arccos \left (a x \right )\right )}{6 \arccos \left (a x \right )^{3}}+\frac {\cos \left (2 \arccos \left (a x \right )\right )}{6 \arccos \left (a x \right )^{2}}-\frac {\sin \left (2 \arccos \left (a x \right )\right )}{3 \arccos \left (a x \right )}+\frac {2 \,\operatorname {Ci}\left (2 \arccos \left (a x \right )\right )}{3}}{a^{2}}\)	\(60\)
default	\(\frac {\frac {\sin \left (2 \arccos \left (a x \right )\right )}{6 \arccos \left (a x \right )^{3}}+\frac {\cos \left (2 \arccos \left (a x \right )\right )}{6 \arccos \left (a x \right )^{2}}-\frac {\sin \left (2 \arccos \left (a x \right )\right )}{3 \arccos \left (a x \right )}+\frac {2 \,\operatorname {Ci}\left (2 \arccos \left (a x \right )\right )}{3}}{a^{2}}\)	\(60\)

[In]

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

[Out]

1/a^2*(1/6/arccos(a*x)^3*sin(2*arccos(a*x))+1/6/arccos(a*x)^2*cos(2*arccos(a*x))-1/3/arccos(a*x)*sin(2*arccos(
a*x))+2/3*Ci(2*arccos(a*x)))

Fricas [F]

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

[In]

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

[Out]

integral(x/arccos(a*x)^4, x)

Sympy [F]

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

[In]

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

[Out]

Integral(x/acos(a*x)**4, x)

Maxima [F]

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

[In]

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

[Out]

1/6*(6*a^2*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^3*integrate(2/3*(2*a^2*x^2 - 1)*sqrt(a*x + 1)*sqrt(-a*x
+ 1)/((a^3*x^2 - a)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)), x) - 2*(2*a*x*arctan2(sqrt(a*x + 1)*sqrt(-a*x
+ 1), a*x)^2 - a*x)*sqrt(a*x + 1)*sqrt(-a*x + 1) + (2*a^2*x^2 - 1)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)
)/(a^2*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.86 \[ \int \frac {x}{\arccos (a x)^4} \, dx=\frac {x^{2}}{3 \, \arccos \left (a x\right )^{2}} - \frac {2 \, \sqrt {-a^{2} x^{2} + 1} x}{3 \, a \arccos \left (a x\right )} + \frac {2 \, \operatorname {Ci}\left (2 \, \arccos \left (a x\right )\right )}{3 \, a^{2}} + \frac {\sqrt {-a^{2} x^{2} + 1} x}{3 \, a \arccos \left (a x\right )^{3}} - \frac {1}{6 \, a^{2} \arccos \left (a x\right )^{2}} \]

[In]

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

[Out]

1/3*x^2/arccos(a*x)^2 - 2/3*sqrt(-a^2*x^2 + 1)*x/(a*arccos(a*x)) + 2/3*cos_integral(2*arccos(a*x))/a^2 + 1/3*s
qrt(-a^2*x^2 + 1)*x/(a*arccos(a*x)^3) - 1/6/(a^2*arccos(a*x)^2)

Mupad [F(-1)]

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

[In]

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

[Out]

int(x/acos(a*x)^4, x)