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

Optimal. Leaf size=121 \[ \frac {\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{4 b c^3}-\frac {\cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b c^3}+\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{4 b c^3}-\frac {\sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b c^3} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 117, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4635, 4406, 3303, 3299, 3302} \[ \frac {\cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{4 b c^3}-\frac {\cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b c^3}+\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{4 b c^3}-\frac {\sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*ArcSin[c*x]),x]

[Out]

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

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 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - 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 4635

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 91, normalized size = 0.75 \[ \frac {\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )-\cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )-\sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )}{4 b c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*ArcSin[c*x]),x]

[Out]

(Cos[a/b]*CosIntegral[a/b + ArcSin[c*x]] - Cos[(3*a)/b]*CosIntegral[3*(a/b + ArcSin[c*x])] + Sin[a/b]*SinInteg
ral[a/b + ArcSin[c*x]] - Sin[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c*x])])/(4*b*c^3)

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{2}}{b \arcsin \left (c x\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2/(b*arcsin(c*x) + a), x)

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giac [A]  time = 0.45, size = 173, normalized size = 1.43 \[ -\frac {\cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right )}{b c^{3}} - \frac {\cos \left (\frac {a}{b}\right )^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right )}{b c^{3}} + \frac {3 \, \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right )}{4 \, b c^{3}} + \frac {\cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{4 \, b c^{3}} + \frac {\sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (c x\right )\right )}{4 \, b c^{3}} + \frac {\sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{4 \, b c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(a/b)^3*cos_integral(3*a/b + 3*arcsin(c*x))/(b*c^3) - cos(a/b)^2*sin(a/b)*sin_integral(3*a/b + 3*arcsin(c*
x))/(b*c^3) + 3/4*cos(a/b)*cos_integral(3*a/b + 3*arcsin(c*x))/(b*c^3) + 1/4*cos(a/b)*cos_integral(a/b + arcsi
n(c*x))/(b*c^3) + 1/4*sin(a/b)*sin_integral(3*a/b + 3*arcsin(c*x))/(b*c^3) + 1/4*sin(a/b)*sin_integral(a/b + a
rcsin(c*x))/(b*c^3)

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maple [A]  time = 0.03, size = 102, normalized size = 0.84 \[ \frac {\frac {\Si \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )}{4 b}+\frac {\Ci \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )}{4 b}-\frac {\Si \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{4 b}-\frac {\Ci \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )}{4 b}}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(b*arcsin(c*x) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{a+b\,\mathrm {asin}\left (c\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*asin(c*x)),x)

[Out]

int(x^2/(a + b*asin(c*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/(a + b*asin(c*x)), x)

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