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

Optimal. Leaf size=60 \[ \frac {a^2 \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{b^3}+\frac {\text {Ci}\left (\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {\text {Ci}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {a \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.61, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {4805, 4747, 6741, 12, 6742, 3302, 4406, 3299} \[ \frac {a^2 \text {CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{b^3}+\frac {\text {CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {\text {CosIntegral}\left (3 \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {a \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/ArcSin[a + b*x],x]

[Out]

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

Rule 12

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

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 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 4406

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 4747

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

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 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 45, normalized size = 0.75 \[ -\frac {-\left (\left (4 a^2+1\right ) \text {Ci}\left (\sin ^{-1}(a+b x)\right )\right )+\text {Ci}\left (3 \sin ^{-1}(a+b x)\right )+4 a \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{4 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcSin[a + b*x],x]

[Out]

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

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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2}}{\arcsin \left (b x + a\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2/arcsin(b*x + a), x)

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giac [A]  time = 0.34, size = 56, normalized size = 0.93 \[ \frac {a^{2} \operatorname {Ci}\left (\arcsin \left (b x + a\right )\right )}{b^{3}} - \frac {a \operatorname {Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{b^{3}} - \frac {\operatorname {Ci}\left (3 \, \arcsin \left (b x + a\right )\right )}{4 \, b^{3}} + \frac {\operatorname {Ci}\left (\arcsin \left (b x + a\right )\right )}{4 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 49, normalized size = 0.82 \[ \frac {-a \Si \left (2 \arcsin \left (b x +a \right )\right )+\frac {\Ci \left (\arcsin \left (b x +a \right )\right )}{4}-\frac {\Ci \left (3 \arcsin \left (b x +a \right )\right )}{4}+a^{2} \Ci \left (\arcsin \left (b x +a \right )\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\arcsin \left (b x + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/arcsin(b*x + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^2}{\mathrm {asin}\left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/asin(a + b*x),x)

[Out]

int(x^2/asin(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\operatorname {asin}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/asin(a + b*x), x)

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