3.122 \(\int \frac{x^2 (a+b \sin ^{-1}(c x))}{(d-c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=135 \[ -\frac{\sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt{d-c^2 d x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c^3 d \sqrt{d-c^2 d x^2}} \]

[Out]

(x*(a + b*ArcSin[c*x]))/(c^2*d*Sqrt[d - c^2*d*x^2]) - (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(2*b*c^3*d*Sqr
t[d - c^2*d*x^2]) + (b*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(2*c^3*d*Sqrt[d - c^2*d*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.159255, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {4703, 4643, 4641, 260} \[ -\frac{\sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt{d-c^2 d x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c^3 d \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*ArcSin[c*x]))/(c^2*d*Sqrt[d - c^2*d*x^2]) - (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(2*b*c^3*d*Sqr
t[d - c^2*d*x^2]) + (b*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(2*c^3*d*Sqrt[d - c^2*d*x^2])

Rule 4703

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

Rule 4643

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

Rule 4641

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{3/2}} \, dx &=\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt{d-c^2 d x^2}}-\frac{\int \frac{a+b \sin ^{-1}(c x)}{\sqrt{d-c^2 d x^2}} \, dx}{c^2 d}-\frac{\left (b \sqrt{1-c^2 x^2}\right ) \int \frac{x}{1-c^2 x^2} \, dx}{c d \sqrt{d-c^2 d x^2}}\\ &=\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c^3 d \sqrt{d-c^2 d x^2}}-\frac{\sqrt{1-c^2 x^2} \int \frac{a+b \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}} \, dx}{c^2 d \sqrt{d-c^2 d x^2}}\\ &=\frac{x \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt{d-c^2 d x^2}}-\frac{\sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c^3 d \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.196347, size = 160, normalized size = 1.19 \[ -\frac{a x \sqrt{-d \left (c^2 x^2-1\right )}}{c^2 d^2 \left (c^2 x^2-1\right )}+\frac{a \tan ^{-1}\left (\frac{c x \sqrt{-d \left (c^2 x^2-1\right )}}{\sqrt{d} \left (c^2 x^2-1\right )}\right )}{c^3 d^{3/2}}+\frac{b \left (2 c x \sin ^{-1}(c x)-\sqrt{1-c^2 x^2} \left (\sin ^{-1}(c x)^2-2 \log \left (\sqrt{1-c^2 x^2}\right )\right )\right )}{2 c^3 d \sqrt{d \left (1-c^2 x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*x*Sqrt[-(d*(-1 + c^2*x^2))])/(c^2*d^2*(-1 + c^2*x^2))) + (a*ArcTan[(c*x*Sqrt[-(d*(-1 + c^2*x^2))])/(Sqrt[
d]*(-1 + c^2*x^2))])/(c^3*d^(3/2)) + (b*(2*c*x*ArcSin[c*x] - Sqrt[1 - c^2*x^2]*(ArcSin[c*x]^2 - 2*Log[Sqrt[1 -
c^2*x^2]])))/(2*c^3*d*Sqrt[d*(1 - c^2*x^2)])

________________________________________________________________________________________

Maple [C]  time = 0.164, size = 274, normalized size = 2. \begin{align*}{\frac{ax}{{c}^{2}d}{\frac{1}{\sqrt{-{c}^{2}d{x}^{2}+d}}}}-{\frac{a}{{c}^{2}d}\arctan \left ({x\sqrt{{c}^{2}d}{\frac{1}{\sqrt{-{c}^{2}d{x}^{2}+d}}}} \right ){\frac{1}{\sqrt{{c}^{2}d}}}}+{\frac{b \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{2\,{d}^{2}{c}^{3} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{ib\arcsin \left ( cx \right ) }{{d}^{2}{c}^{3} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b\arcsin \left ( cx \right ) x}{{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{b}{{d}^{2}{c}^{3} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}\ln \left ( 1+ \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(3/2),x)

[Out]

a*x/c^2/d/(-c^2*d*x^2+d)^(1/2)-a/c^2/d/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+1/2*b*(-d*(c
^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/c^3/(c^2*x^2-1)*arcsin(c*x)^2+I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^
(1/2)/d^2/c^3/(c^2*x^2-1)*arcsin(c*x)-b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)/d^2/c^2/(c^2*x^2-1)*x-b*(-d*(c^2*x^
2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/c^3/(c^2*x^2-1)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \left (a + b \operatorname{asin}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsin(c*x) + a)*x^2/(-c^2*d*x^2 + d)^(3/2), x)