3.375 \(\int \frac{(c-a^2 c x^2)^2}{\sin ^{-1}(a x)^2} \, dx\)

Optimal. Leaf size=78 \[ -\frac{c^2 \left (1-a^2 x^2\right )^{5/2}}{a \sin ^{-1}(a x)}-\frac{5 c^2 \text{Si}\left (\sin ^{-1}(a x)\right )}{8 a}-\frac{15 c^2 \text{Si}\left (3 \sin ^{-1}(a x)\right )}{16 a}-\frac{5 c^2 \text{Si}\left (5 \sin ^{-1}(a x)\right )}{16 a} \]

[Out]

-((c^2*(1 - a^2*x^2)^(5/2))/(a*ArcSin[a*x])) - (5*c^2*SinIntegral[ArcSin[a*x]])/(8*a) - (15*c^2*SinIntegral[3*
ArcSin[a*x]])/(16*a) - (5*c^2*SinIntegral[5*ArcSin[a*x]])/(16*a)

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Rubi [A]  time = 0.161396, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4659, 4723, 4406, 3299} \[ -\frac{c^2 \left (1-a^2 x^2\right )^{5/2}}{a \sin ^{-1}(a x)}-\frac{5 c^2 \text{Si}\left (\sin ^{-1}(a x)\right )}{8 a}-\frac{15 c^2 \text{Si}\left (3 \sin ^{-1}(a x)\right )}{16 a}-\frac{5 c^2 \text{Si}\left (5 \sin ^{-1}(a x)\right )}{16 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a^2*c*x^2)^2/ArcSin[a*x]^2,x]

[Out]

-((c^2*(1 - a^2*x^2)^(5/2))/(a*ArcSin[a*x])) - (5*c^2*SinIntegral[ArcSin[a*x]])/(8*a) - (15*c^2*SinIntegral[3*
ArcSin[a*x]])/(16*a) - (5*c^2*SinIntegral[5*ArcSin[a*x]])/(16*a)

Rule 4659

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(Sqrt[1 - c^2*x^2]*
(d + e*x^2)^p*(a + b*ArcSin[c*x])^(n + 1))/(b*c*(n + 1)), x] + Dist[(c*(2*p + 1)*d^IntPart[p]*(d + e*x^2)^Frac
Part[p])/(b*(n + 1)*(1 - c^2*x^2)^FracPart[p]), Int[x*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x],
x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (c-a^2 c x^2\right )^2}{\sin ^{-1}(a x)^2} \, dx &=-\frac{c^2 \left (1-a^2 x^2\right )^{5/2}}{a \sin ^{-1}(a x)}-\left (5 a c^2\right ) \int \frac{x \left (1-a^2 x^2\right )^{3/2}}{\sin ^{-1}(a x)} \, dx\\ &=-\frac{c^2 \left (1-a^2 x^2\right )^{5/2}}{a \sin ^{-1}(a x)}-\frac{\left (5 c^2\right ) \operatorname{Subst}\left (\int \frac{\cos ^4(x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=-\frac{c^2 \left (1-a^2 x^2\right )^{5/2}}{a \sin ^{-1}(a x)}-\frac{\left (5 c^2\right ) \operatorname{Subst}\left (\int \left (\frac{\sin (x)}{8 x}+\frac{3 \sin (3 x)}{16 x}+\frac{\sin (5 x)}{16 x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a}\\ &=-\frac{c^2 \left (1-a^2 x^2\right )^{5/2}}{a \sin ^{-1}(a x)}-\frac{\left (5 c^2\right ) \operatorname{Subst}\left (\int \frac{\sin (5 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{16 a}-\frac{\left (5 c^2\right ) \operatorname{Subst}\left (\int \frac{\sin (x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{8 a}-\frac{\left (15 c^2\right ) \operatorname{Subst}\left (\int \frac{\sin (3 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{16 a}\\ &=-\frac{c^2 \left (1-a^2 x^2\right )^{5/2}}{a \sin ^{-1}(a x)}-\frac{5 c^2 \text{Si}\left (\sin ^{-1}(a x)\right )}{8 a}-\frac{15 c^2 \text{Si}\left (3 \sin ^{-1}(a x)\right )}{16 a}-\frac{5 c^2 \text{Si}\left (5 \sin ^{-1}(a x)\right )}{16 a}\\ \end{align*}

Mathematica [A]  time = 0.487281, size = 70, normalized size = 0.9 \[ -\frac{c^2 \left (16 \left (1-a^2 x^2\right )^{5/2}+10 \sin ^{-1}(a x) \text{Si}\left (\sin ^{-1}(a x)\right )+15 \sin ^{-1}(a x) \text{Si}\left (3 \sin ^{-1}(a x)\right )+5 \sin ^{-1}(a x) \text{Si}\left (5 \sin ^{-1}(a x)\right )\right )}{16 a \sin ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - a^2*c*x^2)^2/ArcSin[a*x]^2,x]

[Out]

-(c^2*(16*(1 - a^2*x^2)^(5/2) + 10*ArcSin[a*x]*SinIntegral[ArcSin[a*x]] + 15*ArcSin[a*x]*SinIntegral[3*ArcSin[
a*x]] + 5*ArcSin[a*x]*SinIntegral[5*ArcSin[a*x]]))/(16*a*ArcSin[a*x])

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Maple [A]  time = 0.034, size = 83, normalized size = 1.1 \begin{align*} -{\frac{{c}^{2}}{16\,a\arcsin \left ( ax \right ) } \left ( 10\,{\it Si} \left ( \arcsin \left ( ax \right ) \right ) \arcsin \left ( ax \right ) +15\,{\it Si} \left ( 3\,\arcsin \left ( ax \right ) \right ) \arcsin \left ( ax \right ) +5\,{\it Si} \left ( 5\,\arcsin \left ( ax \right ) \right ) \arcsin \left ( ax \right ) +10\,\sqrt{-{a}^{2}{x}^{2}+1}+5\,\cos \left ( 3\,\arcsin \left ( ax \right ) \right ) +\cos \left ( 5\,\arcsin \left ( ax \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/a*c^2*(10*Si(arcsin(a*x))*arcsin(a*x)+15*Si(3*arcsin(a*x))*arcsin(a*x)+5*Si(5*arcsin(a*x))*arcsin(a*x)+1
0*(-a^2*x^2+1)^(1/2)+5*cos(3*arcsin(a*x))+cos(5*arcsin(a*x)))/arcsin(a*x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{5 \, a \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right ) \int \frac{{\left (a^{3} c^{2} x^{3} - a c^{2} x\right )} \sqrt{a x + 1} \sqrt{-a x + 1}}{\arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )}\,{d x} -{\left (a^{4} c^{2} x^{4} - 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \sqrt{a x + 1} \sqrt{-a x + 1}}{a \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))*integrate(5*(a^3*c^2*x^3 - a*c^2*x)*sqrt(a*x + 1)*sqrt(-a*x + 1)
/arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)), x) - (a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2)*sqrt(a*x + 1)*sqrt(-a*x
+ 1))/(a*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} c^{2} \left (\int - \frac{2 a^{2} x^{2}}{\operatorname{asin}^{2}{\left (a x \right )}}\, dx + \int \frac{a^{4} x^{4}}{\operatorname{asin}^{2}{\left (a x \right )}}\, dx + \int \frac{1}{\operatorname{asin}^{2}{\left (a x \right )}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**2*(Integral(-2*a**2*x**2/asin(a*x)**2, x) + Integral(a**4*x**4/asin(a*x)**2, x) + Integral(asin(a*x)**(-2),
x))

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Giac [A]  time = 1.43958, size = 109, normalized size = 1.4 \begin{align*} -\frac{{\left (a^{2} x^{2} - 1\right )}^{2} \sqrt{-a^{2} x^{2} + 1} c^{2}}{a \arcsin \left (a x\right )} - \frac{5 \, c^{2} \operatorname{Si}\left (5 \, \arcsin \left (a x\right )\right )}{16 \, a} - \frac{15 \, c^{2} \operatorname{Si}\left (3 \, \arcsin \left (a x\right )\right )}{16 \, a} - \frac{5 \, c^{2} \operatorname{Si}\left (\arcsin \left (a x\right )\right )}{8 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1)*c^2/(a*arcsin(a*x)) - 5/16*c^2*sin_integral(5*arcsin(a*x))/a - 15/16*c^2*s
in_integral(3*arcsin(a*x))/a - 5/8*c^2*sin_integral(arcsin(a*x))/a