3.133 \(\int \frac{x^2 (a+b \sin ^{-1}(c x))}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=125 \[ \frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{b}{6 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}-\frac{b \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{6 c^3 d^2 \sqrt{d-c^2 d x^2}} \]

[Out]

-b/(6*c^3*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (x^3*(a + b*ArcSin[c*x]))/(3*d*(d - c^2*d*x^2)^(3/2)) -
(b*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(6*c^3*d^2*Sqrt[d - c^2*d*x^2])

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Rubi [A]  time = 0.130668, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4681, 266, 43} \[ \frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{b}{6 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}-\frac{b \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{6 c^3 d^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

-b/(6*c^3*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (x^3*(a + b*ArcSin[c*x]))/(3*d*(d - c^2*d*x^2)^(3/2)) -
(b*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(6*c^3*d^2*Sqrt[d - c^2*d*x^2])

Rule 4681

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n)/(d*f*(m + 1)), x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x
^2)^FracPart[p])/(f*(m + 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, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{\left (b c \sqrt{1-c^2 x^2}\right ) \int \frac{x^3}{\left (1-c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{\left (b c \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{x}{\left (1-c^2 x\right )^2} \, dx,x,x^2\right )}{6 d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{\left (b c \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{c^2 \left (-1+c^2 x\right )^2}+\frac{1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{6 d^2 \sqrt{d-c^2 d x^2}}\\ &=-\frac{b}{6 c^3 d^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac{b \sqrt{1-c^2 x^2} \log \left (1-c^2 x^2\right )}{6 c^3 d^2 \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.193043, size = 103, normalized size = 0.82 \[ \frac{\sqrt{d-c^2 d x^2} \left (2 a c^3 x^3-b \sqrt{1-c^2 x^2}-b \left (1-c^2 x^2\right )^{3/2} \log \left (c^2 x^2-1\right )+2 b c^3 x^3 \sin ^{-1}(c x)\right )}{6 c^3 d^3 \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

(Sqrt[d - c^2*d*x^2]*(2*a*c^3*x^3 - b*Sqrt[1 - c^2*x^2] + 2*b*c^3*x^3*ArcSin[c*x] - b*(1 - c^2*x^2)^(3/2)*Log[
-1 + c^2*x^2]))/(6*c^3*d^3*(-1 + c^2*x^2)^2)

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Maple [C]  time = 0.204, size = 1219, normalized size = 9.8 \begin{align*} \text{result too large to display} \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)^(5/2),x)

[Out]

1/3*a/c^2/d*x/(-c^2*d*x^2+d)^(3/2)-1/3*a/c^2/d^2*x/(-c^2*d*x^2+d)^(1/2)+I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^8*
x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*c^3*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^6+1/6*I*b*(-d*(c^2*x^2-1))^(1/2)/d^
3/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*(-c^2*x^2+1)*x^3-1/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^8*x^8-
9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/c^3*arcsin(c*x)*(-c^2*x^2+1)^(1/2)+b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^8*x^8-9
*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*c^4*arcsin(c*x)*x^7-2*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^8*x^8-9*c^6*x^6+10*
c^4*x^4-5*c^2*x^2+1)*c*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^4-1/2*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^8*x^8-9*c^6*x^
6+10*c^4*x^4-5*c^2*x^2+1)*c*(-c^2*x^2+1)^(1/2)*x^4+1/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^8*x^8-9*c^6*x^6+10*
c^4*x^4-5*c^2*x^2+1)*c^2*x^5+4/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/c
*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^2-b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*
c^2*arcsin(c*x)*x^5-1/6*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*c^2*(-c^2*
x^2+1)*x^5+1/2*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/c*(-c^2*x^2+1)^(1/2)*
x^2-1/6*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*x^3-1/6*I*b*(-d*(c^2*x^2-1
))^(1/2)/d^3/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*c^4*x^7+1/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^8*x^8-
9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*arcsin(c*x)*x^3-1/6*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^8*x^8-9*c^6*x^6+10*c^4
*x^4-5*c^2*x^2+1)/c^3*(-c^2*x^2+1)^(1/2)-2/3*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^3/d^3/(c^2*x^2-1)
*arcsin(c*x)+1/3*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^3/d^3/(c^2*x^2-1)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2
))^2)

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Maxima [A]  time = 1.67374, size = 207, normalized size = 1.66 \begin{align*} \frac{1}{6} \, b c{\left (\frac{1}{c^{6} d^{\frac{5}{2}} x^{2} - c^{4} d^{\frac{5}{2}}} - \frac{\log \left (c x + 1\right )}{c^{4} d^{\frac{5}{2}}} - \frac{\log \left (c x - 1\right )}{c^{4} d^{\frac{5}{2}}}\right )} - \frac{1}{3} \, b{\left (\frac{x}{\sqrt{-c^{2} d x^{2} + d} c^{2} d^{2}} - \frac{x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{2} d}\right )} \arcsin \left (c x\right ) - \frac{1}{3} \, a{\left (\frac{x}{\sqrt{-c^{2} d x^{2} + d} c^{2} d^{2}} - \frac{x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{2} d}\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)^(5/2),x, algorithm="maxima")

[Out]

1/6*b*c*(1/(c^6*d^(5/2)*x^2 - c^4*d^(5/2)) - log(c*x + 1)/(c^4*d^(5/2)) - log(c*x - 1)/(c^4*d^(5/2))) - 1/3*b*
(x/(sqrt(-c^2*d*x^2 + d)*c^2*d^2) - x/((-c^2*d*x^2 + d)^(3/2)*c^2*d))*arcsin(c*x) - 1/3*a*(x/(sqrt(-c^2*d*x^2
+ d)*c^2*d^2) - x/((-c^2*d*x^2 + d)^(3/2)*c^2*d))

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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^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, 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)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*d*x^2 + d)*(b*x^2*arcsin(c*x) + a*x^2)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3)
, x)

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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{5}{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)**(5/2),x)

[Out]

Integral(x**2*(a + b*asin(c*x))/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)

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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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)*x^2/(-c^2*d*x^2 + d)^(5/2), x)