3.350 \(\int \frac{x^4}{\sqrt{1-c^2 x^2} (a+b \sin ^{-1}(c x))} \, dx\)

Optimal. Leaf size=144 \[ -\frac{\cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{2 b c^5}+\frac{\cos \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{8 b c^5}-\frac{\sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{2 b c^5}+\frac{\sin \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{8 b c^5}+\frac{3 \log \left (a+b \sin ^{-1}(c x)\right )}{8 b c^5} \]

[Out]

-(Cos[(2*a)/b]*CosIntegral[(2*(a + b*ArcSin[c*x]))/b])/(2*b*c^5) + (Cos[(4*a)/b]*CosIntegral[(4*(a + b*ArcSin[
c*x]))/b])/(8*b*c^5) + (3*Log[a + b*ArcSin[c*x]])/(8*b*c^5) - (Sin[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c*x])
)/b])/(2*b*c^5) + (Sin[(4*a)/b]*SinIntegral[(4*(a + b*ArcSin[c*x]))/b])/(8*b*c^5)

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Rubi [A]  time = 0.330929, antiderivative size = 144, normalized size of antiderivative = 1., 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 = {4723, 3312, 3303, 3299, 3302} \[ -\frac{\cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{2 b c^5}+\frac{\cos \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 a}{b}+4 \sin ^{-1}(c x)\right )}{8 b c^5}-\frac{\sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{2 b c^5}+\frac{\sin \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 a}{b}+4 \sin ^{-1}(c x)\right )}{8 b c^5}+\frac{3 \log \left (a+b \sin ^{-1}(c x)\right )}{8 b c^5} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])),x]

[Out]

-(Cos[(2*a)/b]*CosIntegral[(2*a)/b + 2*ArcSin[c*x]])/(2*b*c^5) + (Cos[(4*a)/b]*CosIntegral[(4*a)/b + 4*ArcSin[
c*x]])/(8*b*c^5) + (3*Log[a + b*ArcSin[c*x]])/(8*b*c^5) - (Sin[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSin[c*x]])/
(2*b*c^5) + (Sin[(4*a)/b]*SinIntegral[(4*a)/b + 4*ArcSin[c*x]])/(8*b*c^5)

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])

Rule 3312

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.235287, size = 108, normalized size = 0.75 \[ \frac{-4 \cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+\cos \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-4 \sin \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+\sin \left (\frac{4 a}{b}\right ) \text{Si}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+3 \log \left (a+b \sin ^{-1}(c x)\right )}{8 b c^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])),x]

[Out]

(-4*Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcSin[c*x])] + Cos[(4*a)/b]*CosIntegral[4*(a/b + ArcSin[c*x])] + 3*Log[
a + b*ArcSin[c*x]] - 4*Sin[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] + Sin[(4*a)/b]*SinIntegral[4*(a/b + Arc
Sin[c*x])])/(8*b*c^5)

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Maple [A]  time = 0.046, size = 135, normalized size = 0.9 \begin{align*}{\frac{1}{8\,{c}^{5}b}{\it Si} \left ( 4\,\arcsin \left ( cx \right ) +4\,{\frac{a}{b}} \right ) \sin \left ( 4\,{\frac{a}{b}} \right ) }+{\frac{1}{8\,{c}^{5}b}{\it Ci} \left ( 4\,\arcsin \left ( cx \right ) +4\,{\frac{a}{b}} \right ) \cos \left ( 4\,{\frac{a}{b}} \right ) }-{\frac{1}{2\,{c}^{5}b}{\it Si} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) }-{\frac{1}{2\,{c}^{5}b}{\it Ci} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) }+{\frac{3\,\ln \left ( a+b\arcsin \left ( cx \right ) \right ) }{8\,{c}^{5}b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a+b*arcsin(c*x))/(-c^2*x^2+1)^(1/2),x)

[Out]

1/8/c^5/b*Si(4*arcsin(c*x)+4*a/b)*sin(4*a/b)+1/8/c^5/b*Ci(4*arcsin(c*x)+4*a/b)*cos(4*a/b)-1/2/c^5/b*Si(2*arcsi
n(c*x)+2*a/b)*sin(2*a/b)-1/2/c^5/b*Ci(2*arcsin(c*x)+2*a/b)*cos(2*a/b)+3/8*ln(a+b*arcsin(c*x))/b/c^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{-c^{2} x^{2} + 1}{\left (b \arcsin \left (c x\right ) + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b*arcsin(c*x))/(-c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/(sqrt(-c^2*x^2 + 1)*(b*arcsin(c*x) + a)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-c^{2} x^{2} + 1} x^{4}}{a c^{2} x^{2} +{\left (b c^{2} x^{2} - b\right )} \arcsin \left (c x\right ) - a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b*arcsin(c*x))/(-c^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*x^2 + 1)*x^4/(a*c^2*x^2 + (b*c^2*x^2 - b)*arcsin(c*x) - a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{- \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname{asin}{\left (c x \right )}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(a+b*asin(c*x))/(-c**2*x**2+1)**(1/2),x)

[Out]

Integral(x**4/(sqrt(-(c*x - 1)*(c*x + 1))*(a + b*asin(c*x))), x)

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Giac [A]  time = 1.3879, size = 343, normalized size = 2.38 \begin{align*} \frac{\cos \left (\frac{a}{b}\right )^{4} \operatorname{Ci}\left (\frac{4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{5}} + \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^{5}} - \frac{\cos \left (\frac{a}{b}\right )^{2} \operatorname{Ci}\left (\frac{4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{5}} - \frac{\cos \left (\frac{a}{b}\right )^{2} \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b c^{5}} - \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^{5}} - \frac{\cos \left (\frac{a}{b}\right ) \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b c^{5}} + \frac{\operatorname{Ci}\left (\frac{4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{8 \, b c^{5}} + \frac{\operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{2 \, b c^{5}} + \frac{3 \, \log \left (b \arcsin \left (c x\right ) + a\right )}{8 \, b c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b*arcsin(c*x))/(-c^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

cos(a/b)^4*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^5) + cos(a/b)^3*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x
))/(b*c^5) - cos(a/b)^2*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^5) - cos(a/b)^2*cos_integral(2*a/b + 2*arcsin
(c*x))/(b*c^5) - 1/2*cos(a/b)*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x))/(b*c^5) - cos(a/b)*sin(a/b)*sin_int
egral(2*a/b + 2*arcsin(c*x))/(b*c^5) + 1/8*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^5) + 1/2*cos_integral(2*a/
b + 2*arcsin(c*x))/(b*c^5) + 3/8*log(b*arcsin(c*x) + a)/(b*c^5)