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

Optimal. Leaf size=55 \[ \frac{\text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )}{b^2}+\frac{a \text{Si}\left (\sin ^{-1}(a+b x)\right )}{b^2}-\frac{x \sqrt{1-(a+b x)^2}}{b \sin ^{-1}(a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.140056, antiderivative size = 87, normalized size of antiderivative = 1.58, number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {4805, 4745, 4621, 4723, 3299, 4631, 3302} \[ \frac{\text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )}{b^2}+\frac{a \text{Si}\left (\sin ^{-1}(a+b x)\right )}{b^2}+\frac{a \sqrt{1-(a+b x)^2}}{b^2 \sin ^{-1}(a+b x)}-\frac{(a+b x) \sqrt{1-(a+b x)^2}}{b^2 \sin ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcSin[a + b*x]^2,x]

[Out]

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

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

Mathematica [A]  time = 0.182412, size = 63, normalized size = 1.15 \[ \frac{\sin ^{-1}(a+b x) \text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )+a \sin ^{-1}(a+b x) \text{Si}\left (\sin ^{-1}(a+b x)\right )-b x \sqrt{1-(a+b x)^2}}{b^2 \sin ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[x/ArcSin[a + b*x]^2,x]

[Out]

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

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Maple [A]  time = 0.038, size = 72, normalized size = 1.3 \begin{align*}{\frac{1}{{b}^{2}} \left ( -{\frac{\sin \left ( 2\,\arcsin \left ( bx+a \right ) \right ) }{2\,\arcsin \left ( bx+a \right ) }}+{\it Ci} \left ( 2\,\arcsin \left ( bx+a \right ) \right ) +{\frac{a}{\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/arcsin(b*x+a)^2,x)

[Out]

1/b^2*(-1/2/arcsin(b*x+a)*sin(2*arcsin(b*x+a))+Ci(2*arcsin(b*x+a))+a*(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/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}{\arcsin \left (b x + a\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x/arcsin(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/asin(a + b*x)**2, x)

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Giac [A]  time = 1.19313, size = 112, normalized size = 2.04 \begin{align*} \frac{a \operatorname{Si}\left (\arcsin \left (b x + a\right )\right )}{b^{2}} + \frac{\operatorname{Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{b^{2}} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )}}{b^{2} \arcsin \left (b x + a\right )} + \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1} a}{b^{2} \arcsin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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