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

Optimal. Leaf size=189 \[ -\frac {x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c^2}+\frac {1}{4} x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c^3 \sqrt {1-c^2 x^2}}+\frac {b x^2 \sqrt {d-c^2 d x^2}}{16 c \sqrt {1-c^2 x^2}}-\frac {b c x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {4697, 4707, 4641, 30} \[ \frac {1}{4} x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c^2}+\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c^3 \sqrt {1-c^2 x^2}}-\frac {b c x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}+\frac {b x^2 \sqrt {d-c^2 d x^2}}{16 c \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

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

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 4697

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 140, normalized size = 0.74 \[ \frac {\sqrt {d-c^2 d x^2} \left (a^2+2 a b c x \sqrt {1-c^2 x^2} \left (2 c^2 x^2-1\right )+2 b \sin ^{-1}(c x) \left (a+b c x \sqrt {1-c^2 x^2} \left (2 c^2 x^2-1\right )\right )+b^2 c^2 x^2 \left (1-c^2 x^2\right )+b^2 \sin ^{-1}(c x)^2\right )}{16 b c^3 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-c^{2} d x^{2} + d} {\left (b x^{2} \arcsin \left (c x\right ) + a x^{2}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.29, size = 373, normalized size = 1.97 \[ -\frac {a x \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{4 c^{2} d}+\frac {a x \sqrt {-c^{2} d \,x^{2}+d}}{8 c^{2}}+\frac {a d \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{8 c^{2} \sqrt {c^{2} d}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \sqrt {-c^{2} x^{2}+1}\, x^{4}}{16 c^{2} x^{2}-16}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, x^{2}}{16 c \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{16 c^{3} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}}{128 c^{3} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{8 c^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} \arcsin \left (c x \right ) x^{5}}{4 c^{2} x^{2}-4}-\frac {3 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x^{3}}{8 \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a*x*(-c^2*d*x^2+d)^(3/2)/c^2/d+1/8*a/c^2*x*(-c^2*d*x^2+d)^(1/2)+1/8*a/c^2*d/(c^2*d)^(1/2)*arctan((c^2*d)^
(1/2)*x/(-c^2*d*x^2+d)^(1/2))+1/16*b*(-d*(c^2*x^2-1))^(1/2)*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^4-1/16*b*(-d*(c
^2*x^2-1))^(1/2)/c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^2-1/16*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^3/(c^
2*x^2-1)*arcsin(c*x)^2+1/128*b*(-d*(c^2*x^2-1))^(1/2)/c^3/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)+1/8*b*(-d*(c^2*x^2-1)
)^(1/2)/c^2/(c^2*x^2-1)*arcsin(c*x)*x+1/4*b*(-d*(c^2*x^2-1))^(1/2)*c^2/(c^2*x^2-1)*arcsin(c*x)*x^5-3/8*b*(-d*(
c^2*x^2-1))^(1/2)/(c^2*x^2-1)*arcsin(c*x)*x^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \sqrt {d} \int \sqrt {c x + 1} \sqrt {-c x + 1} x^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\,{d x} + \frac {1}{8} \, a {\left (\frac {\sqrt {-c^{2} d x^{2} + d} x}{c^{2}} - \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x}{c^{2} d} + \frac {\sqrt {d} \arcsin \left (c x\right )}{c^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*sqrt(d)*integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*x^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x) + 1/8*a*(s
qrt(-c^2*d*x^2 + d)*x/c^2 - 2*(-c^2*d*x^2 + d)^(3/2)*x/(c^2*d) + sqrt(d)*arcsin(c*x)/c^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {d-c^2\,d\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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