\(\int \frac {x^3}{\arcsin (a x)^2} \, dx\) [54]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 10, antiderivative size = 57 \[ \int \frac {x^3}{\arcsin (a x)^2} \, dx=-\frac {x^3 \sqrt {1-a^2 x^2}}{a \arcsin (a x)}+\frac {\operatorname {CosIntegral}(2 \arcsin (a x))}{2 a^4}-\frac {\operatorname {CosIntegral}(4 \arcsin (a x))}{2 a^4} \]

[Out]

1/2*Ci(2*arcsin(a*x))/a^4-1/2*Ci(4*arcsin(a*x))/a^4-x^3*(-a^2*x^2+1)^(1/2)/a/arcsin(a*x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4727, 3383} \[ \int \frac {x^3}{\arcsin (a x)^2} \, dx=\frac {\operatorname {CosIntegral}(2 \arcsin (a x))}{2 a^4}-\frac {\operatorname {CosIntegral}(4 \arcsin (a x))}{2 a^4}-\frac {x^3 \sqrt {1-a^2 x^2}}{a \arcsin (a x)} \]

[In]

Int[x^3/ArcSin[a*x]^2,x]

[Out]

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

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 4727

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

Rubi steps \begin{align*} \text {integral}& = -\frac {x^3 \sqrt {1-a^2 x^2}}{a \arcsin (a x)}+\frac {\text {Subst}\left (\int \left (\frac {\cos (2 x)}{2 x}-\frac {\cos (4 x)}{2 x}\right ) \, dx,x,\arcsin (a x)\right )}{a^4} \\ & = -\frac {x^3 \sqrt {1-a^2 x^2}}{a \arcsin (a x)}+\frac {\text {Subst}\left (\int \frac {\cos (2 x)}{x} \, dx,x,\arcsin (a x)\right )}{2 a^4}-\frac {\text {Subst}\left (\int \frac {\cos (4 x)}{x} \, dx,x,\arcsin (a x)\right )}{2 a^4} \\ & = -\frac {x^3 \sqrt {1-a^2 x^2}}{a \arcsin (a x)}+\frac {\operatorname {CosIntegral}(2 \arcsin (a x))}{2 a^4}-\frac {\operatorname {CosIntegral}(4 \arcsin (a x))}{2 a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98 \[ \int \frac {x^3}{\arcsin (a x)^2} \, dx=\frac {4 \arcsin (a x) \operatorname {CosIntegral}(2 \arcsin (a x))-4 \arcsin (a x) \operatorname {CosIntegral}(4 \arcsin (a x))-2 \sin (2 \arcsin (a x))+\sin (4 \arcsin (a x))}{8 a^4 \arcsin (a x)} \]

[In]

Integrate[x^3/ArcSin[a*x]^2,x]

[Out]

(4*ArcSin[a*x]*CosIntegral[2*ArcSin[a*x]] - 4*ArcSin[a*x]*CosIntegral[4*ArcSin[a*x]] - 2*Sin[2*ArcSin[a*x]] +
Sin[4*ArcSin[a*x]])/(8*a^4*ArcSin[a*x])

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.95

method result size
derivativedivides \(\frac {-\frac {\sin \left (2 \arcsin \left (a x \right )\right )}{4 \arcsin \left (a x \right )}+\frac {\operatorname {Ci}\left (2 \arcsin \left (a x \right )\right )}{2}+\frac {\sin \left (4 \arcsin \left (a x \right )\right )}{8 \arcsin \left (a x \right )}-\frac {\operatorname {Ci}\left (4 \arcsin \left (a x \right )\right )}{2}}{a^{4}}\) \(54\)
default \(\frac {-\frac {\sin \left (2 \arcsin \left (a x \right )\right )}{4 \arcsin \left (a x \right )}+\frac {\operatorname {Ci}\left (2 \arcsin \left (a x \right )\right )}{2}+\frac {\sin \left (4 \arcsin \left (a x \right )\right )}{8 \arcsin \left (a x \right )}-\frac {\operatorname {Ci}\left (4 \arcsin \left (a x \right )\right )}{2}}{a^{4}}\) \(54\)

[In]

int(x^3/arcsin(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/a^4*(-1/4/arcsin(a*x)*sin(2*arcsin(a*x))+1/2*Ci(2*arcsin(a*x))+1/8/arcsin(a*x)*sin(4*arcsin(a*x))-1/2*Ci(4*a
rcsin(a*x)))

Fricas [F]

\[ \int \frac {x^3}{\arcsin (a x)^2} \, dx=\int { \frac {x^{3}}{\arcsin \left (a x\right )^{2}} \,d x } \]

[In]

integrate(x^3/arcsin(a*x)^2,x, algorithm="fricas")

[Out]

integral(x^3/arcsin(a*x)^2, x)

Sympy [F]

\[ \int \frac {x^3}{\arcsin (a x)^2} \, dx=\int \frac {x^{3}}{\operatorname {asin}^{2}{\left (a x \right )}}\, dx \]

[In]

integrate(x**3/asin(a*x)**2,x)

[Out]

Integral(x**3/asin(a*x)**2, x)

Maxima [F]

\[ \int \frac {x^3}{\arcsin (a x)^2} \, dx=\int { \frac {x^{3}}{\arcsin \left (a x\right )^{2}} \,d x } \]

[In]

integrate(x^3/arcsin(a*x)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.26 \[ \int \frac {x^3}{\arcsin (a x)^2} \, dx=\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{a^{3} \arcsin \left (a x\right )} - \frac {\sqrt {-a^{2} x^{2} + 1} x}{a^{3} \arcsin \left (a x\right )} - \frac {\operatorname {Ci}\left (4 \, \arcsin \left (a x\right )\right )}{2 \, a^{4}} + \frac {\operatorname {Ci}\left (2 \, \arcsin \left (a x\right )\right )}{2 \, a^{4}} \]

[In]

integrate(x^3/arcsin(a*x)^2,x, algorithm="giac")

[Out]

(-a^2*x^2 + 1)^(3/2)*x/(a^3*arcsin(a*x)) - sqrt(-a^2*x^2 + 1)*x/(a^3*arcsin(a*x)) - 1/2*cos_integral(4*arcsin(
a*x))/a^4 + 1/2*cos_integral(2*arcsin(a*x))/a^4

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\arcsin (a x)^2} \, dx=\int \frac {x^3}{{\mathrm {asin}\left (a\,x\right )}^2} \,d x \]

[In]

int(x^3/asin(a*x)^2,x)

[Out]

int(x^3/asin(a*x)^2, x)