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

Optimal. Leaf size=72 \[ \frac{\cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{b^2 c^2}+\frac{\sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{b^2 c^2}-\frac{x}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.149846, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {4719, 4623, 3303, 3299, 3302} \[ \frac{\cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{b^2 c^2}+\frac{\sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{b^2 c^2}-\frac{x}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4719

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
((f*x)^m*(a + b*ArcSin[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x)^
(m - 1)*(a + b*ArcSin[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n,
-1] && GtQ[d, 0]

Rule 4623

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[1/(b*c), Subst[Int[x^n*Cos[a/b - x/b], x], x, a
+ b*ArcSin[c*x]], x] /; FreeQ[{a, b, c, n}, x]

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

Mathematica [A]  time = 0.106928, size = 59, normalized size = 0.82 \[ \frac{\cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )+\sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )-\frac{b c x}{a+b \sin ^{-1}(c x)}}{b^2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.042, size = 108, normalized size = 1.5 \begin{align*}{\frac{1}{{c}^{2} \left ( a+b\arcsin \left ( cx \right ) \right ){b}^{2}} \left ( \arcsin \left ( cx \right ){\it Si} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) b+\arcsin \left ( cx \right ){\it Ci} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) b+{\it Si} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) a+{\it Ci} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) a-xbc \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)*integrate(1/(b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt
(-c*x + 1)) + a*b*c), x) - x)/(b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)

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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^{2} c^{2} x^{2} +{\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arcsin \left (c x\right )^{2} - a^{2} + 2 \,{\left (a b c^{2} x^{2} - a b\right )} \arcsin \left (c x\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-c^2*x^2 + 1)*x/(a^2*c^2*x^2 + (b^2*c^2*x^2 - b^2)*arcsin(c*x)^2 - a^2 + 2*(a*b*c^2*x^2 - a*b)*
arcsin(c*x)), 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 )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.47322, size = 270, normalized size = 3.75 \begin{align*} \frac{b \arcsin \left (c x\right ) \cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{a}{b} + \arcsin \left (c x\right )\right )}{b^{3} c^{2} \arcsin \left (c x\right ) + a b^{2} c^{2}} + \frac{b \arcsin \left (c x\right ) \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{a}{b} + \arcsin \left (c x\right )\right )}{b^{3} c^{2} \arcsin \left (c x\right ) + a b^{2} c^{2}} - \frac{b c x}{b^{3} c^{2} \arcsin \left (c x\right ) + a b^{2} c^{2}} + \frac{a \cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{a}{b} + \arcsin \left (c x\right )\right )}{b^{3} c^{2} \arcsin \left (c x\right ) + a b^{2} c^{2}} + \frac{a \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{a}{b} + \arcsin \left (c x\right )\right )}{b^{3} c^{2} \arcsin \left (c x\right ) + a b^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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