3.70 \(\int \frac{x}{\sin ^{-1}(a x)^4} \, dx\)

Optimal. Leaf size=97 \[ -\frac{2 \text{CosIntegral}\left (2 \sin ^{-1}(a x)\right )}{3 a^2}+\frac{2 x \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)}-\frac{x \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)^3}-\frac{1}{6 a^2 \sin ^{-1}(a x)^2}+\frac{x^2}{3 \sin ^{-1}(a x)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.163607, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {4633, 4719, 4631, 3302, 4641} \[ -\frac{2 \text{CosIntegral}\left (2 \sin ^{-1}(a x)\right )}{3 a^2}+\frac{2 x \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)}-\frac{x \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)^3}-\frac{1}{6 a^2 \sin ^{-1}(a x)^2}+\frac{x^2}{3 \sin ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4633

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 4719

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
((f*x)^m*(a + b*ArcSin[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), 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] && GtQ[d, 0]

Rule 4631

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

Rule 3302

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 4641

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

Rubi steps

\begin{align*} \int \frac{x}{\sin ^{-1}(a x)^4} \, dx &=-\frac{x \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)^3}+\frac{\int \frac{1}{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^3} \, dx}{3 a}-\frac{1}{3} (2 a) \int \frac{x^2}{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^3} \, dx\\ &=-\frac{x \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)^3}-\frac{1}{6 a^2 \sin ^{-1}(a x)^2}+\frac{x^2}{3 \sin ^{-1}(a x)^2}-\frac{2}{3} \int \frac{x}{\sin ^{-1}(a x)^2} \, dx\\ &=-\frac{x \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)^3}-\frac{1}{6 a^2 \sin ^{-1}(a x)^2}+\frac{x^2}{3 \sin ^{-1}(a x)^2}+\frac{2 x \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)}-\frac{2 \operatorname{Subst}\left (\int \frac{\cos (2 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{3 a^2}\\ &=-\frac{x \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)^3}-\frac{1}{6 a^2 \sin ^{-1}(a x)^2}+\frac{x^2}{3 \sin ^{-1}(a x)^2}+\frac{2 x \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)}-\frac{2 \text{Ci}\left (2 \sin ^{-1}(a x)\right )}{3 a^2}\\ \end{align*}

Mathematica [A]  time = 0.137627, size = 86, normalized size = 0.89 \[ \frac{-2 a x \sqrt{1-a^2 x^2}+4 a x \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2+\left (2 a^2 x^2-1\right ) \sin ^{-1}(a x)-4 \sin ^{-1}(a x)^3 \text{CosIntegral}\left (2 \sin ^{-1}(a x)\right )}{6 a^2 \sin ^{-1}(a x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.027, size = 60, normalized size = 0.6 \begin{align*}{\frac{1}{{a}^{2}} \left ( -{\frac{\sin \left ( 2\,\arcsin \left ( ax \right ) \right ) }{6\, \left ( \arcsin \left ( ax \right ) \right ) ^{3}}}-{\frac{\cos \left ( 2\,\arcsin \left ( ax \right ) \right ) }{6\, \left ( \arcsin \left ( ax \right ) \right ) ^{2}}}+{\frac{\sin \left ( 2\,\arcsin \left ( ax \right ) \right ) }{3\,\arcsin \left ( ax \right ) }}-{\frac{2\,{\it Ci} \left ( 2\,\arcsin \left ( ax \right ) \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{4 \, a^{2} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )^{3} \int \frac{{\left (2 \, a^{2} x^{2} - 1\right )} \sqrt{a x + 1} \sqrt{-a x + 1}}{{\left (a^{3} x^{2} - a\right )} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )}\,{d x} - 2 \,{\left (2 \, a x \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )^{2} - a x\right )} \sqrt{a x + 1} \sqrt{-a x + 1} -{\left (2 \, a^{2} x^{2} - 1\right )} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )}{6 \, a^{2} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x}{\arcsin \left (a x\right )^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x/arcsin(a*x)^4, x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\operatorname{asin}^{4}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/asin(a*x)**4, x)

________________________________________________________________________________________

Giac [A]  time = 1.2573, size = 124, normalized size = 1.28 \begin{align*} \frac{2 \, \sqrt{-a^{2} x^{2} + 1} x}{3 \, a \arcsin \left (a x\right )} - \frac{2 \, \operatorname{Ci}\left (2 \, \arcsin \left (a x\right )\right )}{3 \, a^{2}} - \frac{\sqrt{-a^{2} x^{2} + 1} x}{3 \, a \arcsin \left (a x\right )^{3}} + \frac{a^{2} x^{2} - 1}{3 \, a^{2} \arcsin \left (a x\right )^{2}} + \frac{1}{6 \, a^{2} \arcsin \left (a x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(-a^2*x^2 + 1)*x/(a*arcsin(a*x)) - 2/3*cos_integral(2*arcsin(a*x))/a^2 - 1/3*sqrt(-a^2*x^2 + 1)*x/(a*a
rcsin(a*x)^3) + 1/3*(a^2*x^2 - 1)/(a^2*arcsin(a*x)^2) + 1/6/(a^2*arcsin(a*x)^2)