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

Optimal. Leaf size=156 \[ \frac {2 b d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c}+\frac {2 b e x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+\frac {4 b e \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3}+d x \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{3} e x^3 \left (a+b \sin ^{-1}(c x)\right )^2-\frac {4 b^2 e x}{9 c^2}-2 b^2 d x-\frac {2}{27} b^2 e x^3 \]

[Out]

-2*b^2*d*x-4/9*b^2*e*x/c^2-2/27*b^2*e*x^3+d*x*(a+b*arcsin(c*x))^2+1/3*e*x^3*(a+b*arcsin(c*x))^2+2*b*d*(a+b*arc
sin(c*x))*(-c^2*x^2+1)^(1/2)/c+4/9*b*e*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/c^3+2/9*b*e*x^2*(a+b*arcsin(c*x))*
(-c^2*x^2+1)^(1/2)/c

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Rubi [A]  time = 0.26, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {4667, 4619, 4677, 8, 4627, 4707, 30} \[ \frac {2 b d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c}+\frac {2 b e x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+\frac {4 b e \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3}+d x \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{3} e x^3 \left (a+b \sin ^{-1}(c x)\right )^2-\frac {4 b^2 e x}{9 c^2}-2 b^2 d x-\frac {2}{27} b^2 e x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*b^2*d*x - (4*b^2*e*x)/(9*c^2) - (2*b^2*e*x^3)/27 + (2*b*d*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/c + (4*b*e
*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(9*c^3) + (2*b*e*x^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(9*c) + d*
x*(a + b*ArcSin[c*x])^2 + (e*x^3*(a + b*ArcSin[c*x])^2)/3

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4619

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSin[c*x])^n, x] - Dist[b*c*n, Int[
(x*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4667

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

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]

Rule 4707

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSin[c*x])^n)/(e*m), x] + (Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)^(m
- 2)*(a + b*ArcSin[c*x])^n)/Sqrt[d + e*x^2], x], x] + Dist[(b*f*n*Sqrt[1 - c^2*x^2])/(c*m*Sqrt[d + e*x^2]), I
nt[(f*x)^(m - 1)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] &&
GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

Rubi steps

\begin {align*} \int \left (d+e x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2 \, dx &=\int \left (d \left (a+b \sin ^{-1}(c x)\right )^2+e x^2 \left (a+b \sin ^{-1}(c x)\right )^2\right ) \, dx\\ &=d \int \left (a+b \sin ^{-1}(c x)\right )^2 \, dx+e \int x^2 \left (a+b \sin ^{-1}(c x)\right )^2 \, dx\\ &=d x \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{3} e x^3 \left (a+b \sin ^{-1}(c x)\right )^2-(2 b c d) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx-\frac {1}{3} (2 b c e) \int \frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {2 b d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c}+\frac {2 b e x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+d x \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{3} e x^3 \left (a+b \sin ^{-1}(c x)\right )^2-\left (2 b^2 d\right ) \int 1 \, dx-\frac {1}{9} \left (2 b^2 e\right ) \int x^2 \, dx-\frac {(4 b e) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{9 c}\\ &=-2 b^2 d x-\frac {2}{27} b^2 e x^3+\frac {2 b d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c}+\frac {4 b e \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3}+\frac {2 b e x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+d x \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{3} e x^3 \left (a+b \sin ^{-1}(c x)\right )^2-\frac {\left (4 b^2 e\right ) \int 1 \, dx}{9 c^2}\\ &=-2 b^2 d x-\frac {4 b^2 e x}{9 c^2}-\frac {2}{27} b^2 e x^3+\frac {2 b d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c}+\frac {4 b e \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c^3}+\frac {2 b e x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+d x \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{3} e x^3 \left (a+b \sin ^{-1}(c x)\right )^2\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 148, normalized size = 0.95 \[ -2 b d \left (b x-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c}\right )-\frac {2}{27} b e \left (-\frac {3 x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c}+\frac {6 \left (\frac {b x}{c}-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^2}\right )}{c}+b x^3\right )+d x \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{3} e x^3 \left (a+b \sin ^{-1}(c x)\right )^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

d*x*(a + b*ArcSin[c*x])^2 + (e*x^3*(a + b*ArcSin[c*x])^2)/3 - 2*b*d*(b*x - (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*
x]))/c) - (2*b*e*(b*x^3 - (3*x^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/c + (6*((b*x)/c - (Sqrt[1 - c^2*x^2]*(
a + b*ArcSin[c*x]))/c^2))/c))/27

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fricas [A]  time = 0.88, size = 177, normalized size = 1.13 \[ \frac {{\left (9 \, a^{2} - 2 \, b^{2}\right )} c^{3} e x^{3} + 9 \, {\left (b^{2} c^{3} e x^{3} + 3 \, b^{2} c^{3} d x\right )} \arcsin \left (c x\right )^{2} + 3 \, {\left (9 \, {\left (a^{2} - 2 \, b^{2}\right )} c^{3} d - 4 \, b^{2} c e\right )} x + 18 \, {\left (a b c^{3} e x^{3} + 3 \, a b c^{3} d x\right )} \arcsin \left (c x\right ) + 6 \, {\left (a b c^{2} e x^{2} + 9 \, a b c^{2} d + 2 \, a b e + {\left (b^{2} c^{2} e x^{2} + 9 \, b^{2} c^{2} d + 2 \, b^{2} e\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} x^{2} + 1}}{27 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*((9*a^2 - 2*b^2)*c^3*e*x^3 + 9*(b^2*c^3*e*x^3 + 3*b^2*c^3*d*x)*arcsin(c*x)^2 + 3*(9*(a^2 - 2*b^2)*c^3*d -
4*b^2*c*e)*x + 18*(a*b*c^3*e*x^3 + 3*a*b*c^3*d*x)*arcsin(c*x) + 6*(a*b*c^2*e*x^2 + 9*a*b*c^2*d + 2*a*b*e + (b
^2*c^2*e*x^2 + 9*b^2*c^2*d + 2*b^2*e)*arcsin(c*x))*sqrt(-c^2*x^2 + 1))/c^3

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giac [B]  time = 0.88, size = 296, normalized size = 1.90 \[ b^{2} d x \arcsin \left (c x\right )^{2} + \frac {1}{3} \, a^{2} x^{3} e + 2 \, a b d x \arcsin \left (c x\right ) + \frac {{\left (c^{2} x^{2} - 1\right )} b^{2} x \arcsin \left (c x\right )^{2} e}{3 \, c^{2}} + a^{2} d x - 2 \, b^{2} d x + \frac {2 \, {\left (c^{2} x^{2} - 1\right )} a b x \arcsin \left (c x\right ) e}{3 \, c^{2}} + \frac {b^{2} x \arcsin \left (c x\right )^{2} e}{3 \, c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} b^{2} d \arcsin \left (c x\right )}{c} - \frac {2 \, {\left (c^{2} x^{2} - 1\right )} b^{2} x e}{27 \, c^{2}} + \frac {2 \, a b x \arcsin \left (c x\right ) e}{3 \, c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} a b d}{c} - \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b^{2} \arcsin \left (c x\right ) e}{9 \, c^{3}} - \frac {14 \, b^{2} x e}{27 \, c^{2}} - \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} a b e}{9 \, c^{3}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} b^{2} \arcsin \left (c x\right ) e}{3 \, c^{3}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} a b e}{3 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*d*x*arcsin(c*x)^2 + 1/3*a^2*x^3*e + 2*a*b*d*x*arcsin(c*x) + 1/3*(c^2*x^2 - 1)*b^2*x*arcsin(c*x)^2*e/c^2 +
a^2*d*x - 2*b^2*d*x + 2/3*(c^2*x^2 - 1)*a*b*x*arcsin(c*x)*e/c^2 + 1/3*b^2*x*arcsin(c*x)^2*e/c^2 + 2*sqrt(-c^2*
x^2 + 1)*b^2*d*arcsin(c*x)/c - 2/27*(c^2*x^2 - 1)*b^2*x*e/c^2 + 2/3*a*b*x*arcsin(c*x)*e/c^2 + 2*sqrt(-c^2*x^2
+ 1)*a*b*d/c - 2/9*(-c^2*x^2 + 1)^(3/2)*b^2*arcsin(c*x)*e/c^3 - 14/27*b^2*x*e/c^2 - 2/9*(-c^2*x^2 + 1)^(3/2)*a
*b*e/c^3 + 2/3*sqrt(-c^2*x^2 + 1)*b^2*arcsin(c*x)*e/c^3 + 2/3*sqrt(-c^2*x^2 + 1)*a*b*e/c^3

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maple [A]  time = 0.10, size = 276, normalized size = 1.77 \[ \frac {\frac {a^{2} \left (\frac {1}{3} c^{3} x^{3} e +c^{3} d x \right )}{c^{2}}+\frac {b^{2} \left (\frac {e \left (9 \arcsin \left (c x \right )^{2} c^{3} x^{3}+6 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}-27 c x \arcsin \left (c x \right )^{2}-2 c^{3} x^{3}-42 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}+42 c x \right )}{27}+c^{2} d \left (c x \arcsin \left (c x \right )^{2}-2 c x +2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\right )+e \left (c x \arcsin \left (c x \right )^{2}-2 c x +2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{2}}+\frac {2 a b \left (\frac {\arcsin \left (c x \right ) c^{3} x^{3} e}{3}+\arcsin \left (c x \right ) c^{3} d x -\frac {e \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}+c^{2} d \sqrt {-c^{2} x^{2}+1}\right )}{c^{2}}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(a^2/c^2*(1/3*c^3*x^3*e+c^3*d*x)+b^2/c^2*(1/27*e*(9*arcsin(c*x)^2*c^3*x^3+6*arcsin(c*x)*(-c^2*x^2+1)^(1/2)
*c^2*x^2-27*c*x*arcsin(c*x)^2-2*c^3*x^3-42*arcsin(c*x)*(-c^2*x^2+1)^(1/2)+42*c*x)+c^2*d*(c*x*arcsin(c*x)^2-2*c
*x+2*arcsin(c*x)*(-c^2*x^2+1)^(1/2))+e*(c*x*arcsin(c*x)^2-2*c*x+2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)))+2*a*b/c^2*(
1/3*arcsin(c*x)*c^3*x^3*e+arcsin(c*x)*c^3*d*x-1/3*e*(-1/3*c^2*x^2*(-c^2*x^2+1)^(1/2)-2/3*(-c^2*x^2+1)^(1/2))+c
^2*d*(-c^2*x^2+1)^(1/2)))

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maxima [A]  time = 0.44, size = 221, normalized size = 1.42 \[ \frac {1}{3} \, b^{2} e x^{3} \arcsin \left (c x\right )^{2} + \frac {1}{3} \, a^{2} e x^{3} + b^{2} d x \arcsin \left (c x\right )^{2} + \frac {2}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} a b e + \frac {2}{27} \, {\left (3 \, c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )} \arcsin \left (c x\right ) - \frac {c^{2} x^{3} + 6 \, x}{c^{2}}\right )} b^{2} e - 2 \, b^{2} d {\left (x - \frac {\sqrt {-c^{2} x^{2} + 1} \arcsin \left (c x\right )}{c}\right )} + a^{2} d x + \frac {2 \, {\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} a b d}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*e*x^3*arcsin(c*x)^2 + 1/3*a^2*e*x^3 + b^2*d*x*arcsin(c*x)^2 + 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))*a*b*e + 2/27*(3*c*(sqrt(-c^2*x^2 + 1)*x^2/c^2 + 2*sqrt(-c^2*x^2 +
1)/c^4)*arcsin(c*x) - (c^2*x^3 + 6*x)/c^2)*b^2*e - 2*b^2*d*(x - sqrt(-c^2*x^2 + 1)*arcsin(c*x)/c) + a^2*d*x +
2*(c*x*arcsin(c*x) + sqrt(-c^2*x^2 + 1))*a*b*d/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.27, size = 279, normalized size = 1.79 \[ \begin {cases} a^{2} d x + \frac {a^{2} e x^{3}}{3} + 2 a b d x \operatorname {asin}{\left (c x \right )} + \frac {2 a b e x^{3} \operatorname {asin}{\left (c x \right )}}{3} + \frac {2 a b d \sqrt {- c^{2} x^{2} + 1}}{c} + \frac {2 a b e x^{2} \sqrt {- c^{2} x^{2} + 1}}{9 c} + \frac {4 a b e \sqrt {- c^{2} x^{2} + 1}}{9 c^{3}} + b^{2} d x \operatorname {asin}^{2}{\left (c x \right )} - 2 b^{2} d x + \frac {b^{2} e x^{3} \operatorname {asin}^{2}{\left (c x \right )}}{3} - \frac {2 b^{2} e x^{3}}{27} + \frac {2 b^{2} d \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{c} + \frac {2 b^{2} e x^{2} \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{9 c} - \frac {4 b^{2} e x}{9 c^{2}} + \frac {4 b^{2} e \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{9 c^{3}} & \text {for}\: c \neq 0 \\a^{2} \left (d x + \frac {e x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*d*x + a**2*e*x**3/3 + 2*a*b*d*x*asin(c*x) + 2*a*b*e*x**3*asin(c*x)/3 + 2*a*b*d*sqrt(-c**2*x**2
+ 1)/c + 2*a*b*e*x**2*sqrt(-c**2*x**2 + 1)/(9*c) + 4*a*b*e*sqrt(-c**2*x**2 + 1)/(9*c**3) + b**2*d*x*asin(c*x)
**2 - 2*b**2*d*x + b**2*e*x**3*asin(c*x)**2/3 - 2*b**2*e*x**3/27 + 2*b**2*d*sqrt(-c**2*x**2 + 1)*asin(c*x)/c +
2*b**2*e*x**2*sqrt(-c**2*x**2 + 1)*asin(c*x)/(9*c) - 4*b**2*e*x/(9*c**2) + 4*b**2*e*sqrt(-c**2*x**2 + 1)*asin
(c*x)/(9*c**3), Ne(c, 0)), (a**2*(d*x + e*x**3/3), True))

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