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

Optimal. Leaf size=124 \[ -\frac {d^2 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{6 c^2}+\frac {b d^2 x \left (1-c^2 x^2\right )^{5/2}}{36 c}+\frac {5 b d^2 x \left (1-c^2 x^2\right )^{3/2}}{144 c}+\frac {5 b d^2 x \sqrt {1-c^2 x^2}}{96 c}+\frac {5 b d^2 \sin ^{-1}(c x)}{96 c^2} \]

[Out]

5/144*b*d^2*x*(-c^2*x^2+1)^(3/2)/c+1/36*b*d^2*x*(-c^2*x^2+1)^(5/2)/c+5/96*b*d^2*arcsin(c*x)/c^2-1/6*d^2*(-c^2*
x^2+1)^3*(a+b*arcsin(c*x))/c^2+5/96*b*d^2*x*(-c^2*x^2+1)^(1/2)/c

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Rubi [A]  time = 0.07, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4677, 195, 216} \[ -\frac {d^2 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{6 c^2}+\frac {b d^2 x \left (1-c^2 x^2\right )^{5/2}}{36 c}+\frac {5 b d^2 x \left (1-c^2 x^2\right )^{3/2}}{144 c}+\frac {5 b d^2 x \sqrt {1-c^2 x^2}}{96 c}+\frac {5 b d^2 \sin ^{-1}(c x)}{96 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*d^2*x*Sqrt[1 - c^2*x^2])/(96*c) + (5*b*d^2*x*(1 - c^2*x^2)^(3/2))/(144*c) + (b*d^2*x*(1 - c^2*x^2)^(5/2))
/(36*c) + (5*b*d^2*ArcSin[c*x])/(96*c^2) - (d^2*(1 - c^2*x^2)^3*(a + b*ArcSin[c*x]))/(6*c^2)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 4677

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

Rubi steps

\begin {align*} \int x \left (d-c^2 d x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right ) \, dx &=-\frac {d^2 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{6 c^2}+\frac {\left (b d^2\right ) \int \left (1-c^2 x^2\right )^{5/2} \, dx}{6 c}\\ &=\frac {b d^2 x \left (1-c^2 x^2\right )^{5/2}}{36 c}-\frac {d^2 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{6 c^2}+\frac {\left (5 b d^2\right ) \int \left (1-c^2 x^2\right )^{3/2} \, dx}{36 c}\\ &=\frac {5 b d^2 x \left (1-c^2 x^2\right )^{3/2}}{144 c}+\frac {b d^2 x \left (1-c^2 x^2\right )^{5/2}}{36 c}-\frac {d^2 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{6 c^2}+\frac {\left (5 b d^2\right ) \int \sqrt {1-c^2 x^2} \, dx}{48 c}\\ &=\frac {5 b d^2 x \sqrt {1-c^2 x^2}}{96 c}+\frac {5 b d^2 x \left (1-c^2 x^2\right )^{3/2}}{144 c}+\frac {b d^2 x \left (1-c^2 x^2\right )^{5/2}}{36 c}-\frac {d^2 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{6 c^2}+\frac {\left (5 b d^2\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{96 c}\\ &=\frac {5 b d^2 x \sqrt {1-c^2 x^2}}{96 c}+\frac {5 b d^2 x \left (1-c^2 x^2\right )^{3/2}}{144 c}+\frac {b d^2 x \left (1-c^2 x^2\right )^{5/2}}{36 c}+\frac {5 b d^2 \sin ^{-1}(c x)}{96 c^2}-\frac {d^2 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{6 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 94, normalized size = 0.76 \[ \frac {d^2 \left (48 a \left (c^2 x^2-1\right )^3+b c x \sqrt {1-c^2 x^2} \left (8 c^4 x^4-26 c^2 x^2+33\right )+3 b \left (16 c^6 x^6-48 c^4 x^4+48 c^2 x^2-11\right ) \sin ^{-1}(c x)\right )}{288 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(48*a*(-1 + c^2*x^2)^3 + b*c*x*Sqrt[1 - c^2*x^2]*(33 - 26*c^2*x^2 + 8*c^4*x^4) + 3*b*(-11 + 48*c^2*x^2 -
48*c^4*x^4 + 16*c^6*x^6)*ArcSin[c*x]))/(288*c^2)

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fricas [A]  time = 1.15, size = 137, normalized size = 1.10 \[ \frac {48 \, a c^{6} d^{2} x^{6} - 144 \, a c^{4} d^{2} x^{4} + 144 \, a c^{2} d^{2} x^{2} + 3 \, {\left (16 \, b c^{6} d^{2} x^{6} - 48 \, b c^{4} d^{2} x^{4} + 48 \, b c^{2} d^{2} x^{2} - 11 \, b d^{2}\right )} \arcsin \left (c x\right ) + {\left (8 \, b c^{5} d^{2} x^{5} - 26 \, b c^{3} d^{2} x^{3} + 33 \, b c d^{2} x\right )} \sqrt {-c^{2} x^{2} + 1}}{288 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/288*(48*a*c^6*d^2*x^6 - 144*a*c^4*d^2*x^4 + 144*a*c^2*d^2*x^2 + 3*(16*b*c^6*d^2*x^6 - 48*b*c^4*d^2*x^4 + 48*
b*c^2*d^2*x^2 - 11*b*d^2)*arcsin(c*x) + (8*b*c^5*d^2*x^5 - 26*b*c^3*d^2*x^3 + 33*b*c*d^2*x)*sqrt(-c^2*x^2 + 1)
)/c^2

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giac [A]  time = 0.65, size = 157, normalized size = 1.27 \[ \frac {1}{6} \, a c^{4} d^{2} x^{6} - \frac {1}{2} \, a c^{2} d^{2} x^{4} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt {-c^{2} x^{2} + 1} b d^{2} x}{36 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )}^{3} b d^{2} \arcsin \left (c x\right )}{6 \, c^{2}} + \frac {5 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b d^{2} x}{144 \, c} + \frac {5 \, \sqrt {-c^{2} x^{2} + 1} b d^{2} x}{96 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )} a d^{2}}{2 \, c^{2}} + \frac {5 \, b d^{2} \arcsin \left (c x\right )}{96 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*c^4*d^2*x^6 - 1/2*a*c^2*d^2*x^4 + 1/36*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*d^2*x/c + 1/6*(c^2*x^2 - 1)^
3*b*d^2*arcsin(c*x)/c^2 + 5/144*(-c^2*x^2 + 1)^(3/2)*b*d^2*x/c + 5/96*sqrt(-c^2*x^2 + 1)*b*d^2*x/c + 1/2*(c^2*
x^2 - 1)*a*d^2/c^2 + 5/96*b*d^2*arcsin(c*x)/c^2

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maple [A]  time = 0.01, size = 140, normalized size = 1.13 \[ \frac {d^{2} a \left (\frac {1}{6} c^{6} x^{6}-\frac {1}{2} c^{4} x^{4}+\frac {1}{2} c^{2} x^{2}\right )+d^{2} b \left (\frac {\arcsin \left (c x \right ) c^{6} x^{6}}{6}-\frac {c^{4} x^{4} \arcsin \left (c x \right )}{2}+\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {c^{5} x^{5} \sqrt {-c^{2} x^{2}+1}}{36}-\frac {13 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{144}+\frac {11 c x \sqrt {-c^{2} x^{2}+1}}{96}-\frac {11 \arcsin \left (c x \right )}{96}\right )}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(d^2*a*(1/6*c^6*x^6-1/2*c^4*x^4+1/2*c^2*x^2)+d^2*b*(1/6*arcsin(c*x)*c^6*x^6-1/2*c^4*x^4*arcsin(c*x)+1/2*
c^2*x^2*arcsin(c*x)+1/36*c^5*x^5*(-c^2*x^2+1)^(1/2)-13/144*c^3*x^3*(-c^2*x^2+1)^(1/2)+11/96*c*x*(-c^2*x^2+1)^(
1/2)-11/96*arcsin(c*x)))

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maxima [B]  time = 0.54, size = 237, normalized size = 1.91 \[ \frac {1}{6} \, a c^{4} d^{2} x^{6} - \frac {1}{2} \, a c^{2} d^{2} x^{4} + \frac {1}{288} \, {\left (48 \, x^{6} \arcsin \left (c x\right ) + {\left (\frac {8 \, \sqrt {-c^{2} x^{2} + 1} x^{5}}{c^{2}} + \frac {10 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{4}} + \frac {15 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{6}} - \frac {15 \, \arcsin \left (c x\right )}{c^{7}}\right )} c\right )} b c^{4} d^{2} - \frac {1}{16} \, {\left (8 \, x^{4} \arcsin \left (c x\right ) + {\left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{4}} - \frac {3 \, \arcsin \left (c x\right )}{c^{5}}\right )} c\right )} b c^{2} d^{2} + \frac {1}{2} \, a d^{2} x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x}{c^{2}} - \frac {\arcsin \left (c x\right )}{c^{3}}\right )}\right )} b d^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*c^4*d^2*x^6 - 1/2*a*c^2*d^2*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^4*d^2 - 1/16*(8*x^4*arcsin(c*x)
+ (2*sqrt(-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*c^2*d^2 + 1/2*a*d^2*x^
2 + 1/4*(2*x^2*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x/c^2 - arcsin(c*x)/c^3))*b*d^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.93, size = 190, normalized size = 1.53 \[ \begin {cases} \frac {a c^{4} d^{2} x^{6}}{6} - \frac {a c^{2} d^{2} x^{4}}{2} + \frac {a d^{2} x^{2}}{2} + \frac {b c^{4} d^{2} x^{6} \operatorname {asin}{\left (c x \right )}}{6} + \frac {b c^{3} d^{2} x^{5} \sqrt {- c^{2} x^{2} + 1}}{36} - \frac {b c^{2} d^{2} x^{4} \operatorname {asin}{\left (c x \right )}}{2} - \frac {13 b c d^{2} x^{3} \sqrt {- c^{2} x^{2} + 1}}{144} + \frac {b d^{2} x^{2} \operatorname {asin}{\left (c x \right )}}{2} + \frac {11 b d^{2} x \sqrt {- c^{2} x^{2} + 1}}{96 c} - \frac {11 b d^{2} \operatorname {asin}{\left (c x \right )}}{96 c^{2}} & \text {for}\: c \neq 0 \\\frac {a d^{2} x^{2}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**4*d**2*x**6/6 - a*c**2*d**2*x**4/2 + a*d**2*x**2/2 + b*c**4*d**2*x**6*asin(c*x)/6 + b*c**3*d**
2*x**5*sqrt(-c**2*x**2 + 1)/36 - b*c**2*d**2*x**4*asin(c*x)/2 - 13*b*c*d**2*x**3*sqrt(-c**2*x**2 + 1)/144 + b*
d**2*x**2*asin(c*x)/2 + 11*b*d**2*x*sqrt(-c**2*x**2 + 1)/(96*c) - 11*b*d**2*asin(c*x)/(96*c**2), Ne(c, 0)), (a
*d**2*x**2/2, True))

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