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

Optimal. Leaf size=131 \[ \frac{1}{5} c^4 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )+d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{b d^2 \left (1-c^2 x^2\right )^{5/2}}{25 c}+\frac{4 b d^2 \left (1-c^2 x^2\right )^{3/2}}{45 c}+\frac{8 b d^2 \sqrt{1-c^2 x^2}}{15 c} \]

[Out]

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

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Rubi [A]  time = 0.104451, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {194, 4645, 12, 1247, 698} \[ \frac{1}{5} c^4 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )+d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{b d^2 \left (1-c^2 x^2\right )^{5/2}}{25 c}+\frac{4 b d^2 \left (1-c^2 x^2\right )^{3/2}}{45 c}+\frac{8 b d^2 \sqrt{1-c^2 x^2}}{15 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 4645

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

Rule 12

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

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int \left (d-c^2 d x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right ) \, dx &=d^2 x \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac{d^2 x \left (15-10 c^2 x^2+3 c^4 x^4\right )}{15 \sqrt{1-c^2 x^2}} \, dx\\ &=d^2 x \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{15} \left (b c d^2\right ) \int \frac{x \left (15-10 c^2 x^2+3 c^4 x^4\right )}{\sqrt{1-c^2 x^2}} \, dx\\ &=d^2 x \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{30} \left (b c d^2\right ) \operatorname{Subst}\left (\int \frac{15-10 c^2 x+3 c^4 x^2}{\sqrt{1-c^2 x}} \, dx,x,x^2\right )\\ &=d^2 x \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{30} \left (b c d^2\right ) \operatorname{Subst}\left (\int \left (\frac{8}{\sqrt{1-c^2 x}}+4 \sqrt{1-c^2 x}+3 \left (1-c^2 x\right )^{3/2}\right ) \, dx,x,x^2\right )\\ &=\frac{8 b d^2 \sqrt{1-c^2 x^2}}{15 c}+\frac{4 b d^2 \left (1-c^2 x^2\right )^{3/2}}{45 c}+\frac{b d^2 \left (1-c^2 x^2\right )^{5/2}}{25 c}+d^2 x \left (a+b \sin ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0915752, size = 95, normalized size = 0.73 \[ \frac{d^2 \left (15 a c x \left (3 c^4 x^4-10 c^2 x^2+15\right )+b \sqrt{1-c^2 x^2} \left (9 c^4 x^4-38 c^2 x^2+149\right )+15 b c x \left (3 c^4 x^4-10 c^2 x^2+15\right ) \sin ^{-1}(c x)\right )}{225 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(15*a*c*x*(15 - 10*c^2*x^2 + 3*c^4*x^4) + b*Sqrt[1 - c^2*x^2]*(149 - 38*c^2*x^2 + 9*c^4*x^4) + 15*b*c*x*(
15 - 10*c^2*x^2 + 3*c^4*x^4)*ArcSin[c*x]))/(225*c)

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Maple [A]  time = 0.004, size = 122, normalized size = 0.9 \begin{align*}{\frac{1}{c} \left ({d}^{2}a \left ({\frac{{c}^{5}{x}^{5}}{5}}-{\frac{2\,{c}^{3}{x}^{3}}{3}}+cx \right ) +{d}^{2}b \left ({\frac{\arcsin \left ( cx \right ){c}^{5}{x}^{5}}{5}}-{\frac{2\,{c}^{3}{x}^{3}\arcsin \left ( cx \right ) }{3}}+cx\arcsin \left ( cx \right ) +{\frac{{c}^{4}{x}^{4}}{25}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{38\,{c}^{2}{x}^{2}}{225}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{149}{225}\sqrt{-{c}^{2}{x}^{2}+1}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(d^2*a*(1/5*c^5*x^5-2/3*c^3*x^3+c*x)+d^2*b*(1/5*arcsin(c*x)*c^5*x^5-2/3*c^3*x^3*arcsin(c*x)+c*x*arcsin(c*x
)+1/25*c^4*x^4*(-c^2*x^2+1)^(1/2)-38/225*c^2*x^2*(-c^2*x^2+1)^(1/2)+149/225*(-c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.70968, size = 265, normalized size = 2.02 \begin{align*} \frac{1}{5} \, a c^{4} d^{2} x^{5} + \frac{1}{75} \,{\left (15 \, x^{5} \arcsin \left (c x\right ) +{\left (\frac{3 \, \sqrt{-c^{2} x^{2} + 1} x^{4}}{c^{2}} + \frac{4 \, \sqrt{-c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac{8 \, \sqrt{-c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b c^{4} d^{2} - \frac{2}{3} \, a c^{2} d^{2} x^{3} - \frac{2}{9} \,{\left (3 \, x^{3} \arcsin \left (c x\right ) + c{\left (\frac{\sqrt{-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac{2 \, \sqrt{-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b c^{2} d^{2} + a d^{2} x + \frac{{\left (c x \arcsin \left (c x\right ) + \sqrt{-c^{2} x^{2} + 1}\right )} b d^{2}}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*c^4*d^2*x^5 + 1/75*(15*x^5*arcsin(c*x) + (3*sqrt(-c^2*x^2 + 1)*x^4/c^2 + 4*sqrt(-c^2*x^2 + 1)*x^2/c^4 +
8*sqrt(-c^2*x^2 + 1)/c^6)*c)*b*c^4*d^2 - 2/3*a*c^2*d^2*x^3 - 2/9*(3*x^3*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x^
2/c^2 + 2*sqrt(-c^2*x^2 + 1)/c^4))*b*c^2*d^2 + a*d^2*x + (c*x*arcsin(c*x) + sqrt(-c^2*x^2 + 1))*b*d^2/c

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Fricas [A]  time = 2.09164, size = 274, normalized size = 2.09 \begin{align*} \frac{45 \, a c^{5} d^{2} x^{5} - 150 \, a c^{3} d^{2} x^{3} + 225 \, a c d^{2} x + 15 \,{\left (3 \, b c^{5} d^{2} x^{5} - 10 \, b c^{3} d^{2} x^{3} + 15 \, b c d^{2} x\right )} \arcsin \left (c x\right ) +{\left (9 \, b c^{4} d^{2} x^{4} - 38 \, b c^{2} d^{2} x^{2} + 149 \, b d^{2}\right )} \sqrt{-c^{2} x^{2} + 1}}{225 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/225*(45*a*c^5*d^2*x^5 - 150*a*c^3*d^2*x^3 + 225*a*c*d^2*x + 15*(3*b*c^5*d^2*x^5 - 10*b*c^3*d^2*x^3 + 15*b*c*
d^2*x)*arcsin(c*x) + (9*b*c^4*d^2*x^4 - 38*b*c^2*d^2*x^2 + 149*b*d^2)*sqrt(-c^2*x^2 + 1))/c

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Sympy [A]  time = 4.83471, size = 165, normalized size = 1.26 \begin{align*} \begin{cases} \frac{a c^{4} d^{2} x^{5}}{5} - \frac{2 a c^{2} d^{2} x^{3}}{3} + a d^{2} x + \frac{b c^{4} d^{2} x^{5} \operatorname{asin}{\left (c x \right )}}{5} + \frac{b c^{3} d^{2} x^{4} \sqrt{- c^{2} x^{2} + 1}}{25} - \frac{2 b c^{2} d^{2} x^{3} \operatorname{asin}{\left (c x \right )}}{3} - \frac{38 b c d^{2} x^{2} \sqrt{- c^{2} x^{2} + 1}}{225} + b d^{2} x \operatorname{asin}{\left (c x \right )} + \frac{149 b d^{2} \sqrt{- c^{2} x^{2} + 1}}{225 c} & \text{for}\: c \neq 0 \\a d^{2} x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**4*d**2*x**5/5 - 2*a*c**2*d**2*x**3/3 + a*d**2*x + b*c**4*d**2*x**5*asin(c*x)/5 + b*c**3*d**2*x
**4*sqrt(-c**2*x**2 + 1)/25 - 2*b*c**2*d**2*x**3*asin(c*x)/3 - 38*b*c*d**2*x**2*sqrt(-c**2*x**2 + 1)/225 + b*d
**2*x*asin(c*x) + 149*b*d**2*sqrt(-c**2*x**2 + 1)/(225*c), Ne(c, 0)), (a*d**2*x, True))

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Giac [A]  time = 1.21212, size = 213, normalized size = 1.63 \begin{align*} \frac{1}{5} \, a c^{4} d^{2} x^{5} - \frac{2}{3} \, a c^{2} d^{2} x^{3} + \frac{1}{5} \,{\left (c^{2} x^{2} - 1\right )}^{2} b d^{2} x \arcsin \left (c x\right ) - \frac{4}{15} \,{\left (c^{2} x^{2} - 1\right )} b d^{2} x \arcsin \left (c x\right ) + \frac{8}{15} \, b d^{2} x \arcsin \left (c x\right ) + \frac{{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt{-c^{2} x^{2} + 1} b d^{2}}{25 \, c} + a d^{2} x + \frac{4 \,{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b d^{2}}{45 \, c} + \frac{8 \, \sqrt{-c^{2} x^{2} + 1} b d^{2}}{15 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*c^4*d^2*x^5 - 2/3*a*c^2*d^2*x^3 + 1/5*(c^2*x^2 - 1)^2*b*d^2*x*arcsin(c*x) - 4/15*(c^2*x^2 - 1)*b*d^2*x*a
rcsin(c*x) + 8/15*b*d^2*x*arcsin(c*x) + 1/25*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*d^2/c + a*d^2*x + 4/45*(-c^2
*x^2 + 1)^(3/2)*b*d^2/c + 8/15*sqrt(-c^2*x^2 + 1)*b*d^2/c