3.4.18 \(\int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \text {ArcSin}(c x)} \, dx\) [318]

Optimal. Leaf size=82 \[ -\frac {\cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{8 b c^3}+\frac {\log (a+b \text {ArcSin}(c x))}{8 b c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{8 b c^3} \]

[Out]

-1/8*Ci(4*(a+b*arcsin(c*x))/b)*cos(4*a/b)/b/c^3+1/8*ln(a+b*arcsin(c*x))/b/c^3-1/8*Si(4*(a+b*arcsin(c*x))/b)*si
n(4*a/b)/b/c^3

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Rubi [A]
time = 0.16, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4809, 4491, 3384, 3380, 3383} \begin {gather*} -\frac {\cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{8 b c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{8 b c^3}+\frac {\log (a+b \text {ArcSin}(c x))}{8 b c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*Sqrt[1 - c^2*x^2])/(a + b*ArcSin[c*x]),x]

[Out]

-1/8*(Cos[(4*a)/b]*CosIntegral[(4*(a + b*ArcSin[c*x]))/b])/(b*c^3) + Log[a + b*ArcSin[c*x]]/(8*b*c^3) - (Sin[(
4*a)/b]*SinIntegral[(4*(a + b*ArcSin[c*x]))/b])/(8*b*c^3)

Rule 3380

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 3383

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 3384

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 4491

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 4809

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

Rubi steps

\begin {align*} \int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \sin ^{-1}(c x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\cos ^2(x) \sin ^2(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{c^3}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{8 (a+b x)}-\frac {\cos (4 x)}{8 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^3}\\ &=\frac {\log \left (a+b \sin ^{-1}(c x)\right )}{8 b c^3}-\frac {\text {Subst}\left (\int \frac {\cos (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 c^3}\\ &=\frac {\log \left (a+b \sin ^{-1}(c x)\right )}{8 b c^3}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 c^3}\\ &=-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Ci}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{8 b c^3}+\frac {\log \left (a+b \sin ^{-1}(c x)\right )}{8 b c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \sin ^{-1}(c x)\right )}{8 b c^3}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 66, normalized size = 0.80 \begin {gather*} -\frac {\cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )-\log (8 (a+b \text {ArcSin}(c x)))+\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )}{8 b c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*Sqrt[1 - c^2*x^2])/(a + b*ArcSin[c*x]),x]

[Out]

-1/8*(Cos[(4*a)/b]*CosIntegral[4*(a/b + ArcSin[c*x])] - Log[8*(a + b*ArcSin[c*x])] + Sin[(4*a)/b]*SinIntegral[
4*(a/b + ArcSin[c*x])])/(b*c^3)

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Maple [A]
time = 0.10, size = 65, normalized size = 0.79

method	result	size
default	\(-\frac {\sinIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )+\cosineIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )-\ln \left (a +b \arcsin \left (c x \right )\right )}{8 c^{3} b}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x)),x,method=_RETURNVERBOSE)

[Out]

-1/8/c^3*(Si(4*arcsin(c*x)+4*a/b)*sin(4*a/b)+Ci(4*arcsin(c*x)+4*a/b)*cos(4*a/b)-ln(a+b*arcsin(c*x)))/b

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

integrate(sqrt(-c^2*x^2 + 1)*x^2/(b*arcsin(c*x) + a), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

integral(sqrt(-c^2*x^2 + 1)*x^2/(b*arcsin(c*x) + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \sqrt {- \left (c x - 1\right ) \left (c x + 1\right )}}{a + b \operatorname {asin}{\left (c x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(-c**2*x**2+1)**(1/2)/(a+b*asin(c*x)),x)

[Out]

Integral(x**2*sqrt(-(c*x - 1)*(c*x + 1))/(a + b*asin(c*x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (76) = 152\).
time = 0.46, size = 169, normalized size = 2.06 \begin {gather*} -\frac {\cos \left (\frac {a}{b}\right )^{4} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{3}} - \frac {\cos \left (\frac {a}{b}\right )^{3} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{3}} + \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{3}} + \frac {\cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, b c^{3}} - \frac {\operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{8 \, b c^{3}} + \frac {\log \left (b \arcsin \left (c x\right ) + a\right )}{8 \, b c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

-cos(a/b)^4*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) - cos(a/b)^3*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*
x))/(b*c^3) + cos(a/b)^2*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) + 1/2*cos(a/b)*sin(a/b)*sin_integral(4*a/
b + 4*arcsin(c*x))/(b*c^3) - 1/8*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) + 1/8*log(b*arcsin(c*x) + a)/(b*c
^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(1 - c^2*x^2)^(1/2))/(a + b*asin(c*x)),x)

[Out]

int((x^2*(1 - c^2*x^2)^(1/2))/(a + b*asin(c*x)), x)

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