\(\int \frac {x}{\arcsin (a x)^3} \, dx\) [63]

Optimal result

Integrand size = 8, antiderivative size = 64 \[ \int \frac {x}{\arcsin (a x)^3} \, dx=-\frac {x \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}-\frac {1}{2 a^2 \arcsin (a x)}+\frac {x^2}{\arcsin (a x)}-\frac {\text {Si}(2 \arcsin (a x))}{a^2} \]

[Out]

-1/2/a^2/arcsin(a*x)+x^2/arcsin(a*x)-Si(2*arcsin(a*x))/a^2-1/2*x*(-a^2*x^2+1)^(1/2)/a/arcsin(a*x)^2

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {4729, 4807, 4731, 4491, 12, 3380, 4737} \[ \int \frac {x}{\arcsin (a x)^3} \, dx=-\frac {\text {Si}(2 \arcsin (a x))}{a^2}-\frac {x \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}-\frac {1}{2 a^2 \arcsin (a x)}+\frac {x^2}{\arcsin (a x)} \]

[In]

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

[Out]

-1/2*(x*Sqrt[1 - a^2*x^2])/(a*ArcSin[a*x]^2) - 1/(2*a^2*ArcSin[a*x]) + x^2/ArcSin[a*x] - SinIntegral[2*ArcSin[
a*x]]/a^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3380

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

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(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]
&& IGtQ[p, 0]

Rule 4729

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[c*((m + 1)/(b*(n + 1))), Int[x^(m + 1)*((a + b*ArcSin[c*x])^(n + 1)/
Sqrt[1 - c^2*x^2]), x], x] - Dist[m/(b*c*(n + 1)), Int[x^(m - 1)*((a + b*ArcSin[c*x])^(n + 1)/Sqrt[1 - c^2*x^2
]), x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 4731

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

Rule 4737

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

Rule 4807

Int[(((a_.) + ArcSin[(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*ArcSin[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*ArcSin[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}}{2 a \arcsin (a x)^2}+\frac {\int \frac {1}{\sqrt {1-a^2 x^2} \arcsin (a x)^2} \, dx}{2 a}-a \int \frac {x^2}{\sqrt {1-a^2 x^2} \arcsin (a x)^2} \, dx \\ & = -\frac {x \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}-\frac {1}{2 a^2 \arcsin (a x)}+\frac {x^2}{\arcsin (a x)}-2 \int \frac {x}{\arcsin (a x)} \, dx \\ & = -\frac {x \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}-\frac {1}{2 a^2 \arcsin (a x)}+\frac {x^2}{\arcsin (a x)}-\frac {2 \text {Subst}\left (\int \frac {\cos (x) \sin (x)}{x} \, dx,x,\arcsin (a x)\right )}{a^2} \\ & = -\frac {x \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}-\frac {1}{2 a^2 \arcsin (a x)}+\frac {x^2}{\arcsin (a x)}-\frac {2 \text {Subst}\left (\int \frac {\sin (2 x)}{2 x} \, dx,x,\arcsin (a x)\right )}{a^2} \\ & = -\frac {x \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}-\frac {1}{2 a^2 \arcsin (a x)}+\frac {x^2}{\arcsin (a x)}-\frac {\text {Subst}\left (\int \frac {\sin (2 x)}{x} \, dx,x,\arcsin (a x)\right )}{a^2} \\ & = -\frac {x \sqrt {1-a^2 x^2}}{2 a \arcsin (a x)^2}-\frac {1}{2 a^2 \arcsin (a x)}+\frac {x^2}{\arcsin (a x)}-\frac {\text {Si}(2 \arcsin (a x))}{a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.95 \[ \int \frac {x}{\arcsin (a x)^3} \, dx=-\frac {a x \sqrt {1-a^2 x^2}+\left (1-2 a^2 x^2\right ) \arcsin (a x)+2 \arcsin (a x)^2 \text {Si}(2 \arcsin (a x))}{2 a^2 \arcsin (a x)^2} \]

[In]

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

[Out]

-1/2*(a*x*Sqrt[1 - a^2*x^2] + (1 - 2*a^2*x^2)*ArcSin[a*x] + 2*ArcSin[a*x]^2*SinIntegral[2*ArcSin[a*x]])/(a^2*A
rcSin[a*x]^2)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {-\frac {\sin \left (2 \arcsin \left (a x \right )\right )}{4 \arcsin \left (a x \right )^{2}}-\frac {\cos \left (2 \arcsin \left (a x \right )\right )}{2 \arcsin \left (a x \right )}-\operatorname {Si}\left (2 \arcsin \left (a x \right )\right )}{a^{2}}\)	\(45\)
default	\(\frac {-\frac {\sin \left (2 \arcsin \left (a x \right )\right )}{4 \arcsin \left (a x \right )^{2}}-\frac {\cos \left (2 \arcsin \left (a x \right )\right )}{2 \arcsin \left (a x \right )}-\operatorname {Si}\left (2 \arcsin \left (a x \right )\right )}{a^{2}}\)	\(45\)

[In]

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

[Out]

1/a^2*(-1/4/arcsin(a*x)^2*sin(2*arcsin(a*x))-1/2/arcsin(a*x)*cos(2*arcsin(a*x))-Si(2*arcsin(a*x)))

Fricas [F]

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

[In]

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

[Out]

integral(x/arcsin(a*x)^3, x)

Sympy [F]

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

[In]

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

[Out]

Integral(x/asin(a*x)**3, x)

Maxima [F]

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

[In]

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

[Out]

-1/2*(4*a^2*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^2*integrate(x/arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)
), x) + sqrt(a*x + 1)*sqrt(-a*x + 1)*a*x - (2*a^2*x^2 - 1)*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)))/(a^2*ar
ctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05 \[ \int \frac {x}{\arcsin (a x)^3} \, dx=-\frac {\operatorname {Si}\left (2 \, \arcsin \left (a x\right )\right )}{a^{2}} - \frac {\sqrt {-a^{2} x^{2} + 1} x}{2 \, a \arcsin \left (a x\right )^{2}} + \frac {a^{2} x^{2} - 1}{a^{2} \arcsin \left (a x\right )} + \frac {1}{2 \, a^{2} \arcsin \left (a x\right )} \]

[In]

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

[Out]

-sin_integral(2*arcsin(a*x))/a^2 - 1/2*sqrt(-a^2*x^2 + 1)*x/(a*arcsin(a*x)^2) + (a^2*x^2 - 1)/(a^2*arcsin(a*x)
) + 1/2/(a^2*arcsin(a*x))

Mupad [F(-1)]

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

[In]

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

[Out]

int(x/asin(a*x)^3, x)