3.147 \(\int \frac{x^2}{\sin ^{-1}(a+b x)^2} \, dx\)

Optimal. Leaf size=84 \[ -\frac{\left (4 a^2+1\right ) \text{Si}\left (\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac{2 a \text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}+\frac{3 \text{Si}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac{x^2 \sqrt{1-(a+b x)^2}}{b \sin ^{-1}(a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.227717, antiderivative size = 161, normalized size of antiderivative = 1.92, number of steps used = 12, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {4805, 4745, 4621, 4723, 3299, 4631, 3302} \[ -\frac{a^2 \text{Si}\left (\sin ^{-1}(a+b x)\right )}{b^3}-\frac{a^2 \sqrt{1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac{2 a \text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )}{b^3}-\frac{\text{Si}\left (\sin ^{-1}(a+b x)\right )}{4 b^3}+\frac{3 \text{Si}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3}+\frac{2 a (a+b x) \sqrt{1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)}-\frac{(a+b x)^2 \sqrt{1-(a+b x)^2}}{b^3 \sin ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[x^2/ArcSin[a + b*x]^2,x]

[Out]

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

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rule 4745

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d + e
*x)^m*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[m, 0] && LtQ[n, -1]

Rule 4621

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^(n + 1))
/(b*c*(n + 1)), x] + Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcSin[c*x])^(n + 1))/Sqrt[1 - c^2*x^2], x], x] /; Free
Q[{a, b, c}, x] && LtQ[n, -1]

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^(
m + 1), Subst[Int[(a + b*x)^n*Sin[x]^m*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 3299

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 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]

Rubi steps

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

Mathematica [A]  time = 0.553251, size = 86, normalized size = 1.02 \[ -\frac{\frac{4 b^2 x^2 \sqrt{-a^2-2 a b x-b^2 x^2+1}}{\sin ^{-1}(a+b x)}+\left (4 a^2+1\right ) \text{Si}\left (\sin ^{-1}(a+b x)\right )+8 a \text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )-3 \text{Si}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcSin[a + b*x]^2,x]

[Out]

-((4*b^2*x^2*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2])/ArcSin[a + b*x] + 8*a*CosIntegral[2*ArcSin[a + b*x]] + (1 + 4*
a^2)*SinIntegral[ArcSin[a + b*x]] - 3*SinIntegral[3*ArcSin[a + b*x]])/(4*b^3)

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Maple [A]  time = 0.047, size = 149, normalized size = 1.8 \begin{align*}{\frac{1}{{b}^{3}} \left ( -{\frac{a \left ( 2\,{\it Ci} \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \arcsin \left ( bx+a \right ) -\sin \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \right ) }{\arcsin \left ( bx+a \right ) }}-{\frac{1}{4\,\arcsin \left ( bx+a \right ) }\sqrt{1- \left ( bx+a \right ) ^{2}}}-{\frac{{\it Si} \left ( \arcsin \left ( bx+a \right ) \right ) }{4}}+{\frac{\cos \left ( 3\,\arcsin \left ( bx+a \right ) \right ) }{4\,\arcsin \left ( bx+a \right ) }}+{\frac{3\,{\it Si} \left ( 3\,\arcsin \left ( bx+a \right ) \right ) }{4}}-{\frac{{a}^{2}}{\arcsin \left ( bx+a \right ) } \left ({\it Si} \left ( \arcsin \left ( bx+a \right ) \right ) \arcsin \left ( bx+a \right ) +\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/arcsin(b*x+a)^2,x)

[Out]

1/b^3*(-a*(2*Ci(2*arcsin(b*x+a))*arcsin(b*x+a)-sin(2*arcsin(b*x+a)))/arcsin(b*x+a)-1/4/arcsin(b*x+a)*(1-(b*x+a
)^2)^(1/2)-1/4*Si(arcsin(b*x+a))+1/4/arcsin(b*x+a)*cos(3*arcsin(b*x+a))+3/4*Si(3*arcsin(b*x+a))-a^2*(Si(arcsin
(b*x+a))*arcsin(b*x+a)+(1-(b*x+a)^2)^(1/2))/arcsin(b*x+a))

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{2}}{\arcsin \left (b x + a\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2/arcsin(b*x + a)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\operatorname{asin}^{2}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/asin(b*x+a)**2,x)

[Out]

Integral(x**2/asin(a + b*x)**2, x)

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Giac [B]  time = 1.2137, size = 228, normalized size = 2.71 \begin{align*} -\frac{a^{2} \operatorname{Si}\left (\arcsin \left (b x + a\right )\right )}{b^{3}} - \frac{2 \, a \operatorname{Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{b^{3}} + \frac{2 \, \sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )} a}{b^{3} \arcsin \left (b x + a\right )} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1} a^{2}}{b^{3} \arcsin \left (b x + a\right )} + \frac{3 \, \operatorname{Si}\left (3 \, \arcsin \left (b x + a\right )\right )}{4 \, b^{3}} - \frac{\operatorname{Si}\left (\arcsin \left (b x + a\right )\right )}{4 \, b^{3}} + \frac{{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac{3}{2}}}{b^{3} \arcsin \left (b x + a\right )} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}}{b^{3} \arcsin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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