3.110 \(\int \frac{x^4 (a+b \sin ^{-1}(c x))}{\sqrt{d-c^2 d x^2}} \, dx\)

Optimal. Leaf size=200 \[ -\frac{x^3 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d}-\frac{3 x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d}+\frac{3 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c^5 \sqrt{d-c^2 d x^2}}+\frac{b x^4 \sqrt{1-c^2 x^2}}{16 c \sqrt{d-c^2 d x^2}}+\frac{3 b x^2 \sqrt{1-c^2 x^2}}{16 c^3 \sqrt{d-c^2 d x^2}} \]

[Out]

(3*b*x^2*Sqrt[1 - c^2*x^2])/(16*c^3*Sqrt[d - c^2*d*x^2]) + (b*x^4*Sqrt[1 - c^2*x^2])/(16*c*Sqrt[d - c^2*d*x^2]
) - (3*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(8*c^4*d) - (x^3*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(4
*c^2*d) + (3*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(16*b*c^5*Sqrt[d - c^2*d*x^2])

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Rubi [A]  time = 0.249813, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {4707, 4643, 4641, 30} \[ -\frac{x^3 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d}-\frac{3 x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d}+\frac{3 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c^5 \sqrt{d-c^2 d x^2}}+\frac{b x^4 \sqrt{1-c^2 x^2}}{16 c \sqrt{d-c^2 d x^2}}+\frac{3 b x^2 \sqrt{1-c^2 x^2}}{16 c^3 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSin[c*x]))/Sqrt[d - c^2*d*x^2],x]

[Out]

(3*b*x^2*Sqrt[1 - c^2*x^2])/(16*c^3*Sqrt[d - c^2*d*x^2]) + (b*x^4*Sqrt[1 - c^2*x^2])/(16*c*Sqrt[d - c^2*d*x^2]
) - (3*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(8*c^4*d) - (x^3*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(4
*c^2*d) + (3*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(16*b*c^5*Sqrt[d - c^2*d*x^2])

Rule 4707

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

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{x^4 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}} \, dx &=-\frac{x^3 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d}+\frac{3 \int \frac{x^2 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}} \, dx}{4 c^2}+\frac{\left (b \sqrt{1-c^2 x^2}\right ) \int x^3 \, dx}{4 c \sqrt{d-c^2 d x^2}}\\ &=\frac{b x^4 \sqrt{1-c^2 x^2}}{16 c \sqrt{d-c^2 d x^2}}-\frac{3 x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d}-\frac{x^3 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d}+\frac{3 \int \frac{a+b \sin ^{-1}(c x)}{\sqrt{d-c^2 d x^2}} \, dx}{8 c^4}+\frac{\left (3 b \sqrt{1-c^2 x^2}\right ) \int x \, dx}{8 c^3 \sqrt{d-c^2 d x^2}}\\ &=\frac{3 b x^2 \sqrt{1-c^2 x^2}}{16 c^3 \sqrt{d-c^2 d x^2}}+\frac{b x^4 \sqrt{1-c^2 x^2}}{16 c \sqrt{d-c^2 d x^2}}-\frac{3 x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d}-\frac{x^3 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d}+\frac{\left (3 \sqrt{1-c^2 x^2}\right ) \int \frac{a+b \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}} \, dx}{8 c^4 \sqrt{d-c^2 d x^2}}\\ &=\frac{3 b x^2 \sqrt{1-c^2 x^2}}{16 c^3 \sqrt{d-c^2 d x^2}}+\frac{b x^4 \sqrt{1-c^2 x^2}}{16 c \sqrt{d-c^2 d x^2}}-\frac{3 x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c^4 d}-\frac{x^3 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d}+\frac{3 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c^5 \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.804621, size = 161, normalized size = 0.8 \[ \frac{-\frac{16 a c x \left (2 c^2 x^2+3\right ) \sqrt{d-c^2 d x^2}}{d}-\frac{48 a \tan ^{-1}\left (\frac{c x \sqrt{d-c^2 d x^2}}{\sqrt{d} \left (c^2 x^2-1\right )}\right )}{\sqrt{d}}+\frac{b \sqrt{1-c^2 x^2} \left (4 \sin ^{-1}(c x) \left (6 \sin ^{-1}(c x)-8 \sin \left (2 \sin ^{-1}(c x)\right )+\sin \left (4 \sin ^{-1}(c x)\right )\right )-16 \cos \left (2 \sin ^{-1}(c x)\right )+\cos \left (4 \sin ^{-1}(c x)\right )\right )}{\sqrt{d-c^2 d x^2}}}{128 c^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*ArcSin[c*x]))/Sqrt[d - c^2*d*x^2],x]

[Out]

((-16*a*c*x*(3 + 2*c^2*x^2)*Sqrt[d - c^2*d*x^2])/d - (48*a*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^2
*x^2))])/Sqrt[d] + (b*Sqrt[1 - c^2*x^2]*(-16*Cos[2*ArcSin[c*x]] + Cos[4*ArcSin[c*x]] + 4*ArcSin[c*x]*(6*ArcSin
[c*x] - 8*Sin[2*ArcSin[c*x]] + Sin[4*ArcSin[c*x]])))/Sqrt[d - c^2*d*x^2])/(128*c^5)

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Maple [B]  time = 0.332, size = 400, normalized size = 2. \begin{align*} -{\frac{a{x}^{3}}{4\,{c}^{2}d}\sqrt{-{c}^{2}d{x}^{2}+d}}-{\frac{3\,ax}{8\,{c}^{4}d}\sqrt{-{c}^{2}d{x}^{2}+d}}+{\frac{3\,a}{8\,{c}^{4}}\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\arcsin \left ( cx \right ){x}^{5}}{4\,d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{b\arcsin \left ( cx \right ){x}^{3}}{8\,{c}^{2}d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{3\,b\arcsin \left ( cx \right ) x}{8\,{c}^{4}d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{15\,b}{128\,d{c}^{5} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{3\,b \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{16\,d{c}^{5} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b{x}^{4}}{16\,dc \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{3\,b{x}^{2}}{16\,{c}^{3}d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x)

[Out]

-1/4*a*x^3/c^2/d*(-c^2*d*x^2+d)^(1/2)-3/8*a/c^4*x/d*(-c^2*d*x^2+d)^(1/2)+3/8*a/c^4/(c^2*d)^(1/2)*arctan((c^2*d
)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))-1/4*b*(-d*(c^2*x^2-1))^(1/2)/d/(c^2*x^2-1)*arcsin(c*x)*x^5-1/8*b*(-d*(c^2*x^2-
1))^(1/2)/c^2/d/(c^2*x^2-1)*arcsin(c*x)*x^3+3/8*b*(-d*(c^2*x^2-1))^(1/2)/c^4/d/(c^2*x^2-1)*arcsin(c*x)*x+15/12
8*b*(-d*(c^2*x^2-1))^(1/2)/c^5/d/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)-3/16*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/
2)/c^5/d/(c^2*x^2-1)*arcsin(c*x)^2-1/16*b*(-d*(c^2*x^2-1))^(1/2)/c/d/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^4-3/16*b
*(-d*(c^2*x^2-1))^(1/2)/c^3/d/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^2

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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^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(1/2),x)

[Out]

Integral(x**4*(a + b*asin(c*x))/sqrt(-d*(c*x - 1)*(c*x + 1)), 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^{4}}{\sqrt{-c^{2} d x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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