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

Optimal. Leaf size=116 \[ \frac{1}{2} x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt{1-c^2 x^2}}-\frac{b c x^2 \sqrt{d-c^2 d x^2}}{4 \sqrt{1-c^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0548086, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4647, 4641, 30} \[ \frac{1}{2} x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{\sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt{1-c^2 x^2}}-\frac{b c x^2 \sqrt{d-c^2 d x^2}}{4 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

Mathematica [A]  time = 0.0504091, size = 111, normalized size = 0.96 \[ \frac{\sqrt{d-c^2 d x^2} \left (a^2+2 a b c x \sqrt{1-c^2 x^2}+2 b \sin ^{-1}(c x) \left (a+b c x \sqrt{1-c^2 x^2}\right )-b^2 c^2 x^2+b^2 \sin ^{-1}(c x)^2\right )}{4 b c \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[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 + 2*a*b*c*x*Sqrt[1 - c^2*x^2] + 2*b*(a + b*c*x*Sqrt[1 - c^2*x^2])*ArcS
in[c*x] + b^2*ArcSin[c*x]^2))/(4*b*c*Sqrt[1 - c^2*x^2])

________________________________________________________________________________________

Maple [B]  time = 0.112, size = 260, normalized size = 2.2 \begin{align*}{\frac{ax}{2}\sqrt{-{c}^{2}d{x}^{2}+d}}+{\frac{ad}{2}\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 \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{4\,c \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{b{c}^{2}\arcsin \left ( cx \right ){x}^{3}}{2\,{c}^{2}{x}^{2}-2}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{bc{x}^{2}}{4\,{c}^{2}{x}^{2}-4}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b\arcsin \left ( cx \right ) x}{2\,{c}^{2}{x}^{2}-2}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{b}{8\,c \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((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x)),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

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)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

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)),x)

[Out]

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

________________________________________________________________________________________

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

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)),x, algorithm="giac")

[Out]

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