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

Optimal. Leaf size=124 \[ -\frac {x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c^3 \sqrt {d-c^2 d x^2}}+\frac {b x^2 \sqrt {1-c^2 x^2}}{4 c \sqrt {d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 124, normalized size of antiderivative = 1.00, 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 = {4707, 4643, 4641, 30} \[ -\frac {x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c^3 \sqrt {d-c^2 d x^2}}+\frac {b x^2 \sqrt {1-c^2 x^2}}{4 c \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^2*Sqrt[1 - c^2*x^2])/(4*c*Sqrt[d - c^2*d*x^2]) - (x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(2*c^2*d) +
(Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(4*b*c^3*Sqrt[d - c^2*d*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 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 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 \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}} \, dx &=-\frac {x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d}+\frac {\int \frac {a+b \sin ^{-1}(c x)}{\sqrt {d-c^2 d x^2}} \, dx}{2 c^2}+\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int x \, dx}{2 c \sqrt {d-c^2 d x^2}}\\ &=\frac {b x^2 \sqrt {1-c^2 x^2}}{4 c \sqrt {d-c^2 d x^2}}-\frac {x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d}+\frac {\sqrt {1-c^2 x^2} \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{2 c^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {b x^2 \sqrt {1-c^2 x^2}}{4 c \sqrt {d-c^2 d x^2}}-\frac {x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c^3 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.41, size = 134, normalized size = 1.08 \[ -\frac {\frac {4 a c x \sqrt {d-c^2 d x^2}}{d}+\frac {4 a \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )}{\sqrt {d}}+\frac {b \sqrt {1-c^2 x^2} \left (-2 \sin ^{-1}(c x)^2+2 \sin \left (2 \sin ^{-1}(c x)\right ) \sin ^{-1}(c x)+\cos \left (2 \sin ^{-1}(c x)\right )\right )}{\sqrt {d-c^2 d x^2}}}{8 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b x^{2} \arcsin \left (c x\right ) + a x^{2}\right )}}{c^{2} d x^{2} - d}, x\right ) \]

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)^(1/2),x, algorithm="fricas")

[Out]

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

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

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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.28, size = 270, normalized size = 2.18 \[ -\frac {a x \sqrt {-c^{2} d \,x^{2}+d}}{2 c^{2} d}+\frac {a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 c^{2} \sqrt {c^{2} d}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{4 c^{3} d \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-c^{2} x^{2}+1}}{16 c^{3} \sqrt {-d \left (c^{2} x^{2}-1\right )}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{8 c^{2} d \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \cos \left (3 \arcsin \left (c x \right )\right )}{16 c^{3} d \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sin \left (3 \arcsin \left (c x \right )\right )}{8 c^{3} d \left (c^{2} x^{2}-1\right )} \]

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)^(1/2),x)

[Out]

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

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

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)^(1/2),x, algorithm="maxima")

[Out]

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

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

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 \[ \int \frac {x^{2} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \]

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)**(1/2),x)

[Out]

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

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