3.4.51 \(\int \frac {x^3}{\sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))} \, dx\) [351]

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

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4809, 3393, 3384, 3380, 3383} \begin {gather*} -\frac {3 \sin \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{4 b c^4}+\frac {\sin \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{4 b c^4}+\frac {3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \text {ArcSin}(c x)}{b}\right )}{4 b c^4}-\frac {\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \text {ArcSin}(c x))}{b}\right )}{4 b c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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

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Mathematica [A]
time = 0.14, size = 92, normalized size = 0.76 \begin {gather*} -\frac {3 \text {CosIntegral}\left (\frac {a}{b}+\text {ArcSin}(c x)\right ) \sin \left (\frac {a}{b}\right )-\text {CosIntegral}\left (3 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )-3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\text {ArcSin}(c x)\right )+\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )}{4 b c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 93, normalized size = 0.77

method	result	size
default	\(-\frac {\sinIntegral \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )-\cosineIntegral \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )-3 \sinIntegral \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )+3 \cosineIntegral \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )}{4 c^{4} b}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/c^4*(Si(3*arcsin(c*x)+3*a/b)*cos(3*a/b)-Ci(3*arcsin(c*x)+3*a/b)*sin(3*a/b)-3*Si(arcsin(c*x)+a/b)*cos(a/b)
+3*Ci(arcsin(c*x)+a/b)*sin(a/b))/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^3/(a+b*arcsin(c*x))/(-c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

integral(-sqrt(-c^2*x^2 + 1)*x^3/(a*c^2*x^2 + (b*c^2*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^{3}}{\sqrt {- \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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