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

Optimal. Leaf size=183 \[ \frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4 d^2}-\frac{\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d}-\frac{b c x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}+\frac{b x^3 \sqrt{d-c^2 d x^2}}{45 c \sqrt{1-c^2 x^2}}+\frac{2 b x \sqrt{d-c^2 d x^2}}{15 c^3 \sqrt{1-c^2 x^2}} \]

[Out]

(2*b*x*Sqrt[d - c^2*d*x^2])/(15*c^3*Sqrt[1 - c^2*x^2]) + (b*x^3*Sqrt[d - c^2*d*x^2])/(45*c*Sqrt[1 - c^2*x^2])
- (b*c*x^5*Sqrt[d - c^2*d*x^2])/(25*Sqrt[1 - c^2*x^2]) - ((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(3*c^4*d)
+ ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(5*c^4*d^2)

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Rubi [A]  time = 0.168046, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {266, 43, 4691, 12} \[ \frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4 d^2}-\frac{\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d}-\frac{b c x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}+\frac{b x^3 \sqrt{d-c^2 d x^2}}{45 c \sqrt{1-c^2 x^2}}+\frac{2 b x \sqrt{d-c^2 d x^2}}{15 c^3 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*x*Sqrt[d - c^2*d*x^2])/(15*c^3*Sqrt[1 - c^2*x^2]) + (b*x^3*Sqrt[d - c^2*d*x^2])/(45*c*Sqrt[1 - c^2*x^2])
- (b*c*x^5*Sqrt[d - c^2*d*x^2])/(25*Sqrt[1 - c^2*x^2]) - ((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(3*c^4*d)
+ ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(5*c^4*d^2)

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 4691

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x^
m*(1 - c^2*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], Int[x^m*(d + e*x^2)^p, x], x] - Dist[(b*c*d^(p - 1/2)*Sqrt[d +
e*x^2])/Sqrt[1 - c^2*x^2], Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x
] && EqQ[c^2*d + e, 0] && IGtQ[p + 1/2, 0] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin{align*} \int x^3 \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=-\frac{\left (b c \sqrt{d-c^2 d x^2}\right ) \int \frac{-2-c^2 x^2+3 c^4 x^4}{15 c^4} \, dx}{\sqrt{1-c^2 x^2}}+\left (a+b \sin ^{-1}(c x)\right ) \int x^3 \sqrt{d-c^2 d x^2} \, dx\\ &=-\frac{\left (b \sqrt{d-c^2 d x^2}\right ) \int \left (-2-c^2 x^2+3 c^4 x^4\right ) \, dx}{15 c^3 \sqrt{1-c^2 x^2}}+\frac{1}{2} \left (a+b \sin ^{-1}(c x)\right ) \operatorname{Subst}\left (\int x \sqrt{d-c^2 d x} \, dx,x,x^2\right )\\ &=\frac{2 b x \sqrt{d-c^2 d x^2}}{15 c^3 \sqrt{1-c^2 x^2}}+\frac{b x^3 \sqrt{d-c^2 d x^2}}{45 c \sqrt{1-c^2 x^2}}-\frac{b c x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}+\frac{1}{2} \left (a+b \sin ^{-1}(c x)\right ) \operatorname{Subst}\left (\int \left (\frac{\sqrt{d-c^2 d x}}{c^2}-\frac{\left (d-c^2 d x\right )^{3/2}}{c^2 d}\right ) \, dx,x,x^2\right )\\ &=\frac{2 b x \sqrt{d-c^2 d x^2}}{15 c^3 \sqrt{1-c^2 x^2}}+\frac{b x^3 \sqrt{d-c^2 d x^2}}{45 c \sqrt{1-c^2 x^2}}-\frac{b c x^5 \sqrt{d-c^2 d x^2}}{25 \sqrt{1-c^2 x^2}}-\frac{\left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d}+\frac{\left (d-c^2 d x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4 d^2}\\ \end{align*}

Mathematica [A]  time = 0.0853771, size = 134, normalized size = 0.73 \[ \frac{\sqrt{d-c^2 d x^2} \left (15 a \sqrt{1-c^2 x^2} \left (3 c^4 x^4-c^2 x^2-2\right )+b \left (-9 c^5 x^5+5 c^3 x^3+30 c x\right )+15 b \sqrt{1-c^2 x^2} \left (3 c^4 x^4-c^2 x^2-2\right ) \sin ^{-1}(c x)\right )}{225 c^4 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d - c^2*d*x^2]*(15*a*Sqrt[1 - c^2*x^2]*(-2 - c^2*x^2 + 3*c^4*x^4) + b*(30*c*x + 5*c^3*x^3 - 9*c^5*x^5) +
15*b*Sqrt[1 - c^2*x^2]*(-2 - c^2*x^2 + 3*c^4*x^4)*ArcSin[c*x]))/(225*c^4*Sqrt[1 - c^2*x^2])

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Maple [C]  time = 0.277, size = 617, normalized size = 3.4 \begin{align*} a \left ( -{\frac{{x}^{2}}{5\,{c}^{2}d} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{2}{15\,d{c}^{4}} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}} \right ) +b \left ({\frac{i+5\,\arcsin \left ( cx \right ) }{800\,{c}^{4} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) } \left ( 16\,{c}^{6}{x}^{6}-28\,{c}^{4}{x}^{4}-16\,i\sqrt{-{c}^{2}{x}^{2}+1}{x}^{5}{c}^{5}+13\,{c}^{2}{x}^{2}+20\,i\sqrt{-{c}^{2}{x}^{2}+1}{x}^{3}{c}^{3}-5\,i\sqrt{-{c}^{2}{x}^{2}+1}xc-1 \right ) }+{\frac{i+3\,\arcsin \left ( cx \right ) }{288\,{c}^{4} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) } \left ( 4\,{c}^{4}{x}^{4}-5\,{c}^{2}{x}^{2}-4\,i\sqrt{-{c}^{2}{x}^{2}+1}{x}^{3}{c}^{3}+3\,i\sqrt{-{c}^{2}{x}^{2}+1}xc+1 \right ) }-{\frac{\arcsin \left ( cx \right ) +i}{16\,{c}^{4} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) } \left ({c}^{2}{x}^{2}-i\sqrt{-{c}^{2}{x}^{2}+1}xc-1 \right ) }-{\frac{\arcsin \left ( cx \right ) -i}{16\,{c}^{4} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) } \left ( i\sqrt{-{c}^{2}{x}^{2}+1}xc+{c}^{2}{x}^{2}-1 \right ) }+{\frac{-i+3\,\arcsin \left ( cx \right ) }{288\,{c}^{4} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) } \left ( 4\,i\sqrt{-{c}^{2}{x}^{2}+1}{x}^{3}{c}^{3}+4\,{c}^{4}{x}^{4}-3\,i\sqrt{-{c}^{2}{x}^{2}+1}xc-5\,{c}^{2}{x}^{2}+1 \right ) }+{\frac{-i+5\,\arcsin \left ( cx \right ) }{800\,{c}^{4} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) } \left ( 16\,i\sqrt{-{c}^{2}{x}^{2}+1}{x}^{5}{c}^{5}+16\,{c}^{6}{x}^{6}-20\,i\sqrt{-{c}^{2}{x}^{2}+1}{x}^{3}{c}^{3}-28\,{c}^{4}{x}^{4}+5\,i\sqrt{-{c}^{2}{x}^{2}+1}xc+13\,{c}^{2}{x}^{2}-1 \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(-1/5*x^2*(-c^2*d*x^2+d)^(3/2)/c^2/d-2/15/d/c^4*(-c^2*d*x^2+d)^(3/2))+b*(1/800*(-d*(c^2*x^2-1))^(1/2)*(16*c^
6*x^6-28*c^4*x^4-16*I*(-c^2*x^2+1)^(1/2)*x^5*c^5+13*c^2*x^2+20*I*(-c^2*x^2+1)^(1/2)*x^3*c^3-5*I*(-c^2*x^2+1)^(
1/2)*x*c-1)*(I+5*arcsin(c*x))/c^4/(c^2*x^2-1)+1/288*(-d*(c^2*x^2-1))^(1/2)*(4*c^4*x^4-5*c^2*x^2-4*I*(-c^2*x^2+
1)^(1/2)*x^3*c^3+3*I*(-c^2*x^2+1)^(1/2)*x*c+1)*(I+3*arcsin(c*x))/c^4/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*(
c^2*x^2-I*(-c^2*x^2+1)^(1/2)*x*c-1)*(arcsin(c*x)+I)/c^4/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1
)^(1/2)*x*c+c^2*x^2-1)*(arcsin(c*x)-I)/c^4/(c^2*x^2-1)+1/288*(-d*(c^2*x^2-1))^(1/2)*(4*I*(-c^2*x^2+1)^(1/2)*x^
3*c^3+4*c^4*x^4-3*I*(-c^2*x^2+1)^(1/2)*x*c-5*c^2*x^2+1)*(-I+3*arcsin(c*x))/c^4/(c^2*x^2-1)+1/800*(-d*(c^2*x^2-
1))^(1/2)*(16*I*(-c^2*x^2+1)^(1/2)*x^5*c^5+16*c^6*x^6-20*I*(-c^2*x^2+1)^(1/2)*x^3*c^3-28*c^4*x^4+5*I*(-c^2*x^2
+1)^(1/2)*x*c+13*c^2*x^2-1)*(-I+5*arcsin(c*x))/c^4/(c^2*x^2-1))

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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(x^3*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.44425, size = 319, normalized size = 1.74 \begin{align*} \frac{{\left (9 \, b c^{5} x^{5} - 5 \, b c^{3} x^{3} - 30 \, b c x\right )} \sqrt{-c^{2} d x^{2} + d} \sqrt{-c^{2} x^{2} + 1} + 15 \,{\left (3 \, a c^{6} x^{6} - 4 \, a c^{4} x^{4} - a c^{2} x^{2} +{\left (3 \, b c^{6} x^{6} - 4 \, b c^{4} x^{4} - b c^{2} x^{2} + 2 \, b\right )} \arcsin \left (c x\right ) + 2 \, a\right )} \sqrt{-c^{2} d x^{2} + d}}{225 \,{\left (c^{6} x^{2} - c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

1/225*((9*b*c^5*x^5 - 5*b*c^3*x^3 - 30*b*c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) + 15*(3*a*c^6*x^6 - 4*a*
c^4*x^4 - a*c^2*x^2 + (3*b*c^6*x^6 - 4*b*c^4*x^4 - b*c^2*x^2 + 2*b)*arcsin(c*x) + 2*a)*sqrt(-c^2*d*x^2 + d))/(
c^6*x^2 - c^4)

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError