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

Optimal. Leaf size=192 \[ \frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c \sqrt {1-c^2 x^2}}+\frac {1}{2} x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2-\frac {b c x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}-\frac {1}{4} b^2 x \sqrt {d-c^2 d x^2}+\frac {b^2 \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{4 c \sqrt {1-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {4647, 4641, 4627, 321, 216} \[ \frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c \sqrt {1-c^2 x^2}}+\frac {1}{2} x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2-\frac {b c x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}-\frac {1}{4} b^2 x \sqrt {d-c^2 d x^2}+\frac {b^2 \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{4 c \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*x*Sqrt[d - c^2*d*x^2])/4 + (b^2*Sqrt[d - c^2*d*x^2]*ArcSin[c*x])/(4*c*Sqrt[1 - c^2*x^2]) - (b*c*x^2*Sqrt
[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(2*Sqrt[1 - c^2*x^2]) + (x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/2 +
(Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^3)/(6*b*c*Sqrt[1 - c^2*x^2])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && 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 4647

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

Rubi steps

\begin {align*} \int \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \, dx &=\frac {1}{2} x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {\sqrt {d-c^2 d x^2} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}} \, dx}{2 \sqrt {1-c^2 x^2}}-\frac {\left (b c \sqrt {d-c^2 d x^2}\right ) \int x \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}\\ &=-\frac {b c x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}+\frac {1}{2} x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c \sqrt {1-c^2 x^2}}+\frac {\left (b^2 c^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx}{2 \sqrt {1-c^2 x^2}}\\ &=-\frac {1}{4} b^2 x \sqrt {d-c^2 d x^2}-\frac {b c x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}+\frac {1}{2} x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c \sqrt {1-c^2 x^2}}+\frac {\left (b^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{4 \sqrt {1-c^2 x^2}}\\ &=-\frac {1}{4} b^2 x \sqrt {d-c^2 d x^2}+\frac {b^2 \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{4 c \sqrt {1-c^2 x^2}}-\frac {b c x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}+\frac {1}{2} x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 128, normalized size = 0.67 \[ \frac {1}{6} \sqrt {d-c^2 d x^2} \left (\frac {\left (a+b \sin ^{-1}(c x)\right )^3}{b c \sqrt {1-c^2 x^2}}-\frac {3 b \left (c x \left (2 a c x+b \sqrt {1-c^2 x^2}\right )+b \left (2 c^2 x^2-1\right ) \sin ^{-1}(c x)\right )}{2 c \sqrt {1-c^2 x^2}}+3 x \left (a+b \sin ^{-1}(c x)\right )^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-c^{2} d x^{2} + d} {\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.20, size = 564, normalized size = 2.94 \[ \frac {x \sqrt {-c^{2} d \,x^{2}+d}\, a^{2}}{2}+\frac {a^{2} d \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 \sqrt {c^{2} d}}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{3}}{6 c \left (c^{2} x^{2}-1\right )}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{4 c \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} \arcsin \left (c x \right )^{2} x^{3}}{2 c^{2} x^{2}-2}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2} x}{2 \left (c^{2} x^{2}-1\right )}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} x^{3}}{4 \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x}{4 c^{2} x^{2}-4}+\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, x^{2}}{2 c^{2} x^{2}-2}-\frac {a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{2 c \left (c^{2} x^{2}-1\right )}+\frac {a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \sqrt {-c^{2} x^{2}+1}\, x^{2}}{2 c^{2} x^{2}-2}+\frac {a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} \arcsin \left (c x \right ) x^{3}}{c^{2} x^{2}-1}-\frac {a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}}{4 c \left (c^{2} x^{2}-1\right )}-\frac {a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{c^{2} x^{2}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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