3.70 \(\int (d-c^2 d x^2)^{3/2} (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=188 \[ \frac{1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{3}{8} d x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{3 d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt{1-c^2 x^2}}+\frac{b c^3 d x^4 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}-\frac{5 b c d x^2 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.105156, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {4649, 4647, 4641, 30, 14} \[ \frac{1}{4} x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac{3}{8} d x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{3 d \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt{1-c^2 x^2}}+\frac{b c^3 d x^4 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}}-\frac{5 b c d x^2 \sqrt{d-c^2 d x^2}}{16 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]),x]

[Out]

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

Rule 4649

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

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]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.549964, size = 210, normalized size = 1.12 \[ \frac{d \sqrt{d-c^2 d x^2} \left (16 a c x \sqrt{1-c^2 x^2} \left (5-2 c^2 x^2\right )+16 b \cos \left (2 \sin ^{-1}(c x)\right )+b \cos \left (4 \sin ^{-1}(c x)\right )\right )-48 a d^{3/2} \sqrt{1-c^2 x^2} \tan ^{-1}\left (\frac{c x \sqrt{d-c^2 d x^2}}{\sqrt{d} \left (c^2 x^2-1\right )}\right )+24 b d \sqrt{d-c^2 d x^2} \sin ^{-1}(c x)^2+4 b d \sqrt{d-c^2 d x^2} \left (8 \sin \left (2 \sin ^{-1}(c x)\right )+\sin \left (4 \sin ^{-1}(c x)\right )\right ) \sin ^{-1}(c x)}{128 c \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.14, size = 371, normalized size = 2. \begin{align*}{\frac{ax}{4} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{3\,adx}{8}\sqrt{-{c}^{2}d{x}^{2}+d}}+{\frac{3\,a{d}^{2}}{8}\arctan \left ({x\sqrt{{c}^{2}d}{\frac{1}{\sqrt{-{c}^{2}d{x}^{2}+d}}}} \right ){\frac{1}{\sqrt{{c}^{2}d}}}}-{\frac{3\,b \left ( \arcsin \left ( cx \right ) \right ) ^{2}d}{16\,c \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{bd{c}^{4}\arcsin \left ( cx \right ){x}^{5}}{4\,{c}^{2}{x}^{2}-4}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{7\,b{c}^{2}d\arcsin \left ( cx \right ){x}^{3}}{8\,{c}^{2}{x}^{2}-8}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{17\,bd}{128\,c \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{5\,bd\arcsin \left ( cx \right ) x}{8\,{c}^{2}{x}^{2}-8}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{bd{c}^{3}{x}^{4}}{16\,{c}^{2}{x}^{2}-16}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{5\,bdc{x}^{2}}{16\,{c}^{2}{x}^{2}-16}\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)^(3/2)*(a+b*arcsin(c*x)),x)

[Out]

1/4*a*x*(-c^2*d*x^2+d)^(3/2)+3/8*a*d*x*(-c^2*d*x^2+d)^(1/2)+3/8*a*d^2/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c
^2*d*x^2+d)^(1/2))-3/16*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*d-1/4*b*(-d*(c
^2*x^2-1))^(1/2)*d*c^4/(c^2*x^2-1)*arcsin(c*x)*x^5+7/8*b*(-d*(c^2*x^2-1))^(1/2)*d*c^2/(c^2*x^2-1)*arcsin(c*x)*
x^3-17/128*b*(-d*(c^2*x^2-1))^(1/2)*d/c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)-5/8*b*(-d*(c^2*x^2-1))^(1/2)*d/(c^2*x^2
-1)*arcsin(c*x)*x-1/16*b*(-d*(c^2*x^2-1))^(1/2)*d*c^3/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^4+5/16*b*(-d*(c^2*x^2-1
))^(1/2)*d*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^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)^(3/2)*(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 (-{\left (a c^{2} d x^{2} - a d +{\left (b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} d x^{2} + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac{3}{2}} \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)**(3/2)*(a+b*asin(c*x)),x)

[Out]

Integral((-d*(c*x - 1)*(c*x + 1))**(3/2)*(a + b*asin(c*x)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}}{\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)^(3/2)*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

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