3.317 \(\int \frac {x^3 \sqrt {1-c^2 x^2}}{a+b \sin ^{-1}(c x)} \, dx\)

Optimal. Leaf size=183 \[ -\frac {\sin \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{8 b c^4}-\frac {\sin \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b c^4}+\frac {\sin \left (\frac {5 a}{b}\right ) \text {Ci}\left (\frac {5 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b c^4}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{8 b c^4}+\frac {\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b c^4}-\frac {\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b c^4} \]

[Out]

1/8*cos(a/b)*Si((a+b*arcsin(c*x))/b)/b/c^4+1/16*cos(3*a/b)*Si(3*(a+b*arcsin(c*x))/b)/b/c^4-1/16*cos(5*a/b)*Si(
5*(a+b*arcsin(c*x))/b)/b/c^4-1/8*Ci((a+b*arcsin(c*x))/b)*sin(a/b)/b/c^4-1/16*Ci(3*(a+b*arcsin(c*x))/b)*sin(3*a
/b)/b/c^4+1/16*Ci(5*(a+b*arcsin(c*x))/b)*sin(5*a/b)/b/c^4

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Rubi [A]  time = 0.43, antiderivative size = 179, normalized size of antiderivative = 0.98, number of steps used = 12, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4723, 4406, 3303, 3299, 3302} \[ -\frac {\sin \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{8 b c^4}-\frac {\sin \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b c^4}+\frac {\sin \left (\frac {5 a}{b}\right ) \text {CosIntegral}\left (\frac {5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b c^4}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{8 b c^4}+\frac {\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b c^4}-\frac {\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b c^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*Sqrt[1 - c^2*x^2])/(a + b*ArcSin[c*x]),x]

[Out]

-(CosIntegral[a/b + ArcSin[c*x]]*Sin[a/b])/(8*b*c^4) - (CosIntegral[(3*a)/b + 3*ArcSin[c*x]]*Sin[(3*a)/b])/(16
*b*c^4) + (CosIntegral[(5*a)/b + 5*ArcSin[c*x]]*Sin[(5*a)/b])/(16*b*c^4) + (Cos[a/b]*SinIntegral[a/b + ArcSin[
c*x]])/(8*b*c^4) + (Cos[(3*a)/b]*SinIntegral[(3*a)/b + 3*ArcSin[c*x]])/(16*b*c^4) - (Cos[(5*a)/b]*SinIntegral[
(5*a)/b + 5*ArcSin[c*x]])/(16*b*c^4)

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 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])

Rubi steps

\begin {align*} \int \frac {x^3 \sqrt {1-c^2 x^2}}{a+b \sin ^{-1}(c x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\cos ^2(x) \sin ^3(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{c^4}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {\sin (x)}{8 (a+b x)}+\frac {\sin (3 x)}{16 (a+b x)}-\frac {\sin (5 x)}{16 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^4}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\sin (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 c^4}-\frac {\operatorname {Subst}\left (\int \frac {\sin (5 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 c^4}+\frac {\operatorname {Subst}\left (\int \frac {\sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 c^4}\\ &=\frac {\cos \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 )}{8 c^4}+\frac {\cos \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 )}{16 c^4}-\frac {\cos \left (\frac {5 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 c^4}-\frac {\sin \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 )}{8 c^4}-\frac {\sin \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 )}{16 c^4}+\frac {\sin \left (\frac {5 a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 c^4}\\ &=-\frac {\text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right ) \sin \left (\frac {a}{b}\right )}{8 b c^4}-\frac {\text {Ci}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right ) \sin \left (\frac {3 a}{b}\right )}{16 b c^4}+\frac {\text {Ci}\left (\frac {5 a}{b}+5 \sin ^{-1}(c x)\right ) \sin \left (\frac {5 a}{b}\right )}{16 b c^4}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{8 b c^4}+\frac {\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b c^4}-\frac {\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b c^4}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 135, normalized size = 0.74 \[ \frac {-2 \sin \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )-\sin \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+\sin \left (\frac {5 a}{b}\right ) \text {Ci}\left (5 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+2 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )+\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )}{16 b c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*Sqrt[1 - c^2*x^2])/(a + b*ArcSin[c*x]),x]

[Out]

(-2*CosIntegral[a/b + ArcSin[c*x]]*Sin[a/b] - CosIntegral[3*(a/b + ArcSin[c*x])]*Sin[(3*a)/b] + CosIntegral[5*
(a/b + ArcSin[c*x])]*Sin[(5*a)/b] + 2*Cos[a/b]*SinIntegral[a/b + ArcSin[c*x]] + Cos[(3*a)/b]*SinIntegral[3*(a/
b + ArcSin[c*x])] - Cos[(5*a)/b]*SinIntegral[5*(a/b + ArcSin[c*x])])/(16*b*c^4)

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{3}}{b \arcsin \left (c x\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

integral(sqrt(-c^2*x^2 + 1)*x^3/(b*arcsin(c*x) + a), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.08, size = 138, normalized size = 0.75 \[ \frac {\Ci \left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right )-2 \Ci \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+2 \Si \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )+\Si \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )-\Ci \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )-\Si \left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right )}{16 c^{4} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x)),x)

[Out]

1/16/c^4*(Ci(5*arcsin(c*x)+5*a/b)*sin(5*a/b)-2*Ci(arcsin(c*x)+a/b)*sin(a/b)+2*Si(arcsin(c*x)+a/b)*cos(a/b)+Si(
3*arcsin(c*x)+3*a/b)*cos(3*a/b)-Ci(3*arcsin(c*x)+3*a/b)*sin(3*a/b)-Si(5*arcsin(c*x)+5*a/b)*cos(5*a/b))/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-c^{2} x^{2} + 1} x^{3}}{b \arcsin \left (c x\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

integrate(sqrt(-c^2*x^2 + 1)*x^3/(b*arcsin(c*x) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\sqrt {1-c^2\,x^2}}{a+b\,\mathrm {asin}\left (c\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(1 - c^2*x^2)^(1/2))/(a + b*asin(c*x)),x)

[Out]

int((x^3*(1 - c^2*x^2)^(1/2))/(a + b*asin(c*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \sqrt {- \left (c x - 1\right ) \left (c x + 1\right )}}{a + b \operatorname {asin}{\left (c x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(-c**2*x**2+1)**(1/2)/(a+b*asin(c*x)),x)

[Out]

Integral(x**3*sqrt(-(c*x - 1)*(c*x + 1))/(a + b*asin(c*x)), x)

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