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

Optimal. Leaf size=54 \[ \frac{\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{b c^2}-\frac{\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{b c^2} \]

[Out]

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

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Rubi [A]  time = 0.153913, antiderivative size = 50, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4723, 3303, 3299, 3302} \[ \frac{\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{b c^2}-\frac{\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{b c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.104999, size = 45, normalized size = 0.83 \[ \frac{\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )-\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{b c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.039, size = 46, normalized size = 0.9 \begin{align*}{\frac{1}{{c}^{2}b} \left ({\it Si} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) -{\it Ci} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\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/(a+b*arcsin(c*x))/(-c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

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

[Out]

integral(-sqrt(-c^2*x^2 + 1)*x/(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}{\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/(a+b*asin(c*x))/(-c**2*x**2+1)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.42552, size = 68, normalized size = 1.26 \begin{align*} -\frac{\operatorname{Ci}\left (\frac{a}{b} + \arcsin \left (c x\right )\right ) \sin \left (\frac{a}{b}\right )}{b c^{2}} + \frac{\cos \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{a}{b} + \arcsin \left (c x\right )\right )}{b c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos_integral(a/b + arcsin(c*x))*sin(a/b)/(b*c^2) + cos(a/b)*sin_integral(a/b + arcsin(c*x))/(b*c^2)