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

Optimal. Leaf size=123 \[ -\frac {1}{6} c^2 d x^6 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} d x^4 \left (a+b \sin ^{-1}(c x)\right )-\frac {b d \sin ^{-1}(c x)}{24 c^4}-\frac {1}{36} b c d x^5 \sqrt {1-c^2 x^2}+\frac {b d x^3 \sqrt {1-c^2 x^2}}{36 c}+\frac {b d x \sqrt {1-c^2 x^2}}{24 c^3} \]

[Out]

-1/24*b*d*arcsin(c*x)/c^4+1/4*d*x^4*(a+b*arcsin(c*x))-1/6*c^2*d*x^6*(a+b*arcsin(c*x))+1/24*b*d*x*(-c^2*x^2+1)^
(1/2)/c^3+1/36*b*d*x^3*(-c^2*x^2+1)^(1/2)/c-1/36*b*c*d*x^5*(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {14, 4687, 12, 459, 321, 216} \[ -\frac {1}{6} c^2 d x^6 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} d x^4 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{36} b c d x^5 \sqrt {1-c^2 x^2}+\frac {b d x^3 \sqrt {1-c^2 x^2}}{36 c}+\frac {b d x \sqrt {1-c^2 x^2}}{24 c^3}-\frac {b d \sin ^{-1}(c x)}{24 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x*Sqrt[1 - c^2*x^2])/(24*c^3) + (b*d*x^3*Sqrt[1 - c^2*x^2])/(36*c) - (b*c*d*x^5*Sqrt[1 - c^2*x^2])/36 - (
b*d*ArcSin[c*x])/(24*c^4) + (d*x^4*(a + b*ArcSin[c*x]))/4 - (c^2*d*x^6*(a + b*ArcSin[c*x]))/6

Rule 12

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

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]]

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 459

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

Rule 4687

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = I
ntHide[(f*x)^m*(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]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x^3 \left (d-c^2 d x^2\right ) \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {1}{4} d x^4 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{6} c^2 d x^6 \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac {d x^4 \left (3-2 c^2 x^2\right )}{12 \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {1}{4} d x^4 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{6} c^2 d x^6 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{12} (b c d) \int \frac {x^4 \left (3-2 c^2 x^2\right )}{\sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {1}{36} b c d x^5 \sqrt {1-c^2 x^2}+\frac {1}{4} d x^4 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{6} c^2 d x^6 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{9} (b c d) \int \frac {x^4}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {b d x^3 \sqrt {1-c^2 x^2}}{36 c}-\frac {1}{36} b c d x^5 \sqrt {1-c^2 x^2}+\frac {1}{4} d x^4 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{6} c^2 d x^6 \left (a+b \sin ^{-1}(c x)\right )-\frac {(b d) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx}{12 c}\\ &=\frac {b d x \sqrt {1-c^2 x^2}}{24 c^3}+\frac {b d x^3 \sqrt {1-c^2 x^2}}{36 c}-\frac {1}{36} b c d x^5 \sqrt {1-c^2 x^2}+\frac {1}{4} d x^4 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{6} c^2 d x^6 \left (a+b \sin ^{-1}(c x)\right )-\frac {(b d) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{24 c^3}\\ &=\frac {b d x \sqrt {1-c^2 x^2}}{24 c^3}+\frac {b d x^3 \sqrt {1-c^2 x^2}}{36 c}-\frac {1}{36} b c d x^5 \sqrt {1-c^2 x^2}-\frac {b d \sin ^{-1}(c x)}{24 c^4}+\frac {1}{4} d x^4 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{6} c^2 d x^6 \left (a+b \sin ^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 89, normalized size = 0.72 \[ \frac {d \left (-6 a c^4 x^4 \left (2 c^2 x^2-3\right )-3 b \left (4 c^6 x^6-6 c^4 x^4+1\right ) \sin ^{-1}(c x)+b c x \sqrt {1-c^2 x^2} \left (-2 c^4 x^4+2 c^2 x^2+3\right )\right )}{72 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(-6*a*c^4*x^4*(-3 + 2*c^2*x^2) + b*c*x*Sqrt[1 - c^2*x^2]*(3 + 2*c^2*x^2 - 2*c^4*x^4) - 3*b*(1 - 6*c^4*x^4 +
4*c^6*x^6)*ArcSin[c*x]))/(72*c^4)

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fricas [A]  time = 0.55, size = 96, normalized size = 0.78 \[ -\frac {12 \, a c^{6} d x^{6} - 18 \, a c^{4} d x^{4} + 3 \, {\left (4 \, b c^{6} d x^{6} - 6 \, b c^{4} d x^{4} + b d\right )} \arcsin \left (c x\right ) + {\left (2 \, b c^{5} d x^{5} - 2 \, b c^{3} d x^{3} - 3 \, b c d x\right )} \sqrt {-c^{2} x^{2} + 1}}{72 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/72*(12*a*c^6*d*x^6 - 18*a*c^4*d*x^4 + 3*(4*b*c^6*d*x^6 - 6*b*c^4*d*x^4 + b*d)*arcsin(c*x) + (2*b*c^5*d*x^5
- 2*b*c^3*d*x^3 - 3*b*c*d*x)*sqrt(-c^2*x^2 + 1))/c^4

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giac [A]  time = 0.70, size = 144, normalized size = 1.17 \[ -\frac {1}{6} \, a c^{2} d x^{6} + \frac {1}{4} \, a d x^{4} - \frac {{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt {-c^{2} x^{2} + 1} b d x}{36 \, c^{3}} - \frac {{\left (c^{2} x^{2} - 1\right )}^{3} b d \arcsin \left (c x\right )}{6 \, c^{4}} + \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b d x}{36 \, c^{3}} - \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b d \arcsin \left (c x\right )}{4 \, c^{4}} + \frac {\sqrt {-c^{2} x^{2} + 1} b d x}{24 \, c^{3}} + \frac {b d \arcsin \left (c x\right )}{24 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a*c^2*d*x^6 + 1/4*a*d*x^4 - 1/36*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*d*x/c^3 - 1/6*(c^2*x^2 - 1)^3*b*d*a
rcsin(c*x)/c^4 + 1/36*(-c^2*x^2 + 1)^(3/2)*b*d*x/c^3 - 1/4*(c^2*x^2 - 1)^2*b*d*arcsin(c*x)/c^4 + 1/24*sqrt(-c^
2*x^2 + 1)*b*d*x/c^3 + 1/24*b*d*arcsin(c*x)/c^4

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maple [A]  time = 0.01, size = 118, normalized size = 0.96 \[ \frac {-d a \left (\frac {1}{6} c^{6} x^{6}-\frac {1}{4} c^{4} x^{4}\right )-d b \left (\frac {\arcsin \left (c x \right ) c^{6} x^{6}}{6}-\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}+\frac {c^{5} x^{5} \sqrt {-c^{2} x^{2}+1}}{36}-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{36}-\frac {c x \sqrt {-c^{2} x^{2}+1}}{24}+\frac {\arcsin \left (c x \right )}{24}\right )}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(-d*a*(1/6*c^6*x^6-1/4*c^4*x^4)-d*b*(1/6*arcsin(c*x)*c^6*x^6-1/4*c^4*x^4*arcsin(c*x)+1/36*c^5*x^5*(-c^2*
x^2+1)^(1/2)-1/36*c^3*x^3*(-c^2*x^2+1)^(1/2)-1/24*c*x*(-c^2*x^2+1)^(1/2)+1/24*arcsin(c*x)))

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maxima [A]  time = 0.45, size = 169, normalized size = 1.37 \[ -\frac {1}{6} \, a c^{2} d x^{6} + \frac {1}{4} \, a d x^{4} - \frac {1}{288} \, {\left (48 \, x^{6} \arcsin \left (c x\right ) + {\left (\frac {8 \, \sqrt {-c^{2} x^{2} + 1} x^{5}}{c^{2}} + \frac {10 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{4}} + \frac {15 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{6}} - \frac {15 \, \arcsin \left (c x\right )}{c^{7}}\right )} c\right )} b c^{2} d + \frac {1}{32} \, {\left (8 \, x^{4} \arcsin \left (c x\right ) + {\left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{4}} - \frac {3 \, \arcsin \left (c x\right )}{c^{5}}\right )} c\right )} b d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a*c^2*d*x^6 + 1/4*a*d*x^4 - 1/288*(48*x^6*arcsin(c*x) + (8*sqrt(-c^2*x^2 + 1)*x^5/c^2 + 10*sqrt(-c^2*x^2
+ 1)*x^3/c^4 + 15*sqrt(-c^2*x^2 + 1)*x/c^6 - 15*arcsin(c*x)/c^7)*c)*b*c^2*d + 1/32*(8*x^4*arcsin(c*x) + (2*sqr
t(-c^2*x^2 + 1)*x^3/c^2 + 3*sqrt(-c^2*x^2 + 1)*x/c^4 - 3*arcsin(c*x)/c^5)*c)*b*d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\left (d-c^2\,d\,x^2\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*asin(c*x))*(d - c^2*d*x^2),x)

[Out]

int(x^3*(a + b*asin(c*x))*(d - c^2*d*x^2), x)

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sympy [A]  time = 3.54, size = 138, normalized size = 1.12 \[ \begin {cases} - \frac {a c^{2} d x^{6}}{6} + \frac {a d x^{4}}{4} - \frac {b c^{2} d x^{6} \operatorname {asin}{\left (c x \right )}}{6} - \frac {b c d x^{5} \sqrt {- c^{2} x^{2} + 1}}{36} + \frac {b d x^{4} \operatorname {asin}{\left (c x \right )}}{4} + \frac {b d x^{3} \sqrt {- c^{2} x^{2} + 1}}{36 c} + \frac {b d x \sqrt {- c^{2} x^{2} + 1}}{24 c^{3}} - \frac {b d \operatorname {asin}{\left (c x \right )}}{24 c^{4}} & \text {for}\: c \neq 0 \\\frac {a d x^{4}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(-c**2*d*x**2+d)*(a+b*asin(c*x)),x)

[Out]

Piecewise((-a*c**2*d*x**6/6 + a*d*x**4/4 - b*c**2*d*x**6*asin(c*x)/6 - b*c*d*x**5*sqrt(-c**2*x**2 + 1)/36 + b*
d*x**4*asin(c*x)/4 + b*d*x**3*sqrt(-c**2*x**2 + 1)/(36*c) + b*d*x*sqrt(-c**2*x**2 + 1)/(24*c**3) - b*d*asin(c*
x)/(24*c**4), Ne(c, 0)), (a*d*x**4/4, True))

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