3.408 \(\int (a-b \sin ^{-1}(1-d x^2))^4 \, dx\)

Optimal. Leaf size=135 \[ -\frac {192 b^3 \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )}{d x}-48 b^2 x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2+\frac {8 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^4+384 b^4 x \]

[Out]

384*b^4*x-48*b^2*x*(a+b*arcsin(d*x^2-1))^2+x*(a+b*arcsin(d*x^2-1))^4-192*b^3*(a+b*arcsin(d*x^2-1))*(-d^2*x^4+2
*d*x^2)^(1/2)/d/x+8*b*(a+b*arcsin(d*x^2-1))^3*(-d^2*x^4+2*d*x^2)^(1/2)/d/x

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Rubi [A]  time = 0.03, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4814, 8} \[ -\frac {192 b^3 \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )}{d x}-48 b^2 x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2+\frac {8 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^4+384 b^4 x \]

Antiderivative was successfully verified.

[In]

Int[(a - b*ArcSin[1 - d*x^2])^4,x]

[Out]

384*b^4*x - (192*b^3*Sqrt[2*d*x^2 - d^2*x^4]*(a - b*ArcSin[1 - d*x^2]))/(d*x) - 48*b^2*x*(a - b*ArcSin[1 - d*x
^2])^2 + (8*b*Sqrt[2*d*x^2 - d^2*x^4]*(a - b*ArcSin[1 - d*x^2])^3)/(d*x) + x*(a - b*ArcSin[1 - d*x^2])^4

Rule 8

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

Rule 4814

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*(a + b*ArcSin[c + d*x^2])^n, x] + (-
Dist[4*b^2*n*(n - 1), Int[(a + b*ArcSin[c + d*x^2])^(n - 2), x], x] + Simp[(2*b*n*Sqrt[-2*c*d*x^2 - d^2*x^4]*(
a + b*ArcSin[c + d*x^2])^(n - 1))/(d*x), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1] && GtQ[n, 1]

Rubi steps

\begin {align*} \int \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^4 \, dx &=\frac {8 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^4-\left (48 b^2\right ) \int \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2 \, dx\\ &=-\frac {192 b^3 \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )}{d x}-48 b^2 x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2+\frac {8 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^4+\left (384 b^4\right ) \int 1 \, dx\\ &=384 b^4 x-\frac {192 b^3 \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )}{d x}-48 b^2 x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2+\frac {8 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^4\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 131, normalized size = 0.97 \[ -48 b^2 \left (\frac {4 b \sqrt {-d x^2 \left (d x^2-2\right )} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )}{d x}+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2-8 b^2 x\right )+x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^4+\frac {8 b \sqrt {-d x^2 \left (d x^2-2\right )} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^3}{d x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*ArcSin[1 - d*x^2])^4,x]

[Out]

(8*b*Sqrt[-(d*x^2*(-2 + d*x^2))]*(a - b*ArcSin[1 - d*x^2])^3)/(d*x) + x*(a - b*ArcSin[1 - d*x^2])^4 - 48*b^2*(
-8*b^2*x + (4*b*Sqrt[-(d*x^2*(-2 + d*x^2))]*(a - b*ArcSin[1 - d*x^2]))/(d*x) + x*(a - b*ArcSin[1 - d*x^2])^2)

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fricas [A]  time = 1.31, size = 207, normalized size = 1.53 \[ \frac {b^{4} d x^{2} \arcsin \left (d x^{2} - 1\right )^{4} + 4 \, a b^{3} d x^{2} \arcsin \left (d x^{2} - 1\right )^{3} + 6 \, {\left (a^{2} b^{2} - 8 \, b^{4}\right )} d x^{2} \arcsin \left (d x^{2} - 1\right )^{2} + 4 \, {\left (a^{3} b - 24 \, a b^{3}\right )} d x^{2} \arcsin \left (d x^{2} - 1\right ) + {\left (a^{4} - 48 \, a^{2} b^{2} + 384 \, b^{4}\right )} d x^{2} + 8 \, {\left (b^{4} \arcsin \left (d x^{2} - 1\right )^{3} + 3 \, a b^{3} \arcsin \left (d x^{2} - 1\right )^{2} + a^{3} b - 24 \, a b^{3} + 3 \, {\left (a^{2} b^{2} - 8 \, b^{4}\right )} \arcsin \left (d x^{2} - 1\right )\right )} \sqrt {-d^{2} x^{4} + 2 \, d x^{2}}}{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2-1))^4,x, algorithm="fricas")

[Out]

(b^4*d*x^2*arcsin(d*x^2 - 1)^4 + 4*a*b^3*d*x^2*arcsin(d*x^2 - 1)^3 + 6*(a^2*b^2 - 8*b^4)*d*x^2*arcsin(d*x^2 -
1)^2 + 4*(a^3*b - 24*a*b^3)*d*x^2*arcsin(d*x^2 - 1) + (a^4 - 48*a^2*b^2 + 384*b^4)*d*x^2 + 8*(b^4*arcsin(d*x^2
- 1)^3 + 3*a*b^3*arcsin(d*x^2 - 1)^2 + a^3*b - 24*a*b^3 + 3*(a^2*b^2 - 8*b^4)*arcsin(d*x^2 - 1))*sqrt(-d^2*x^
4 + 2*d*x^2))/(d*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2-1))^4,x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x^2 - 1) + a)^4, x)

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \left (a +b \arcsin \left (d \,x^{2}-1\right )\right )^{4}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x^2-1))^4,x)

[Out]

int((a+b*arcsin(d*x^2-1))^4,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b^{4} x \arctan \left (d x^{2} - 1, \sqrt {-d x^{2} + 2} \sqrt {d} x\right )^{4} + 4 \, {\left (x \arcsin \left (d x^{2} - 1\right ) - \frac {2 \, {\left (d^{\frac {3}{2}} x^{2} - 2 \, \sqrt {d}\right )}}{\sqrt {-d x^{2} + 2} d}\right )} a^{3} b + a^{4} x + \int \frac {2 \, {\left (4 \, \sqrt {-d x^{2} + 2} b^{4} \sqrt {d} x \arctan \left (d x^{2} - 1, \sqrt {-d x^{2} + 2} \sqrt {d} x\right )^{3} + 2 \, {\left (a b^{3} d x^{2} - 2 \, a b^{3}\right )} \arctan \left (d x^{2} - 1, \sqrt {-d x^{2} + 2} \sqrt {d} x\right )^{3} + 3 \, {\left (a^{2} b^{2} d x^{2} - 2 \, a^{2} b^{2}\right )} \arctan \left (d x^{2} - 1, \sqrt {-d x^{2} + 2} \sqrt {d} x\right )^{2}\right )}}{d x^{2} - 2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x^2-1))^4,x, algorithm="maxima")

[Out]

b^4*x*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x)^4 + 4*(x*arcsin(d*x^2 - 1) - 2*(d^(3/2)*x^2 - 2*sqrt(d))/
(sqrt(-d*x^2 + 2)*d))*a^3*b + a^4*x + integrate(2*(4*sqrt(-d*x^2 + 2)*b^4*sqrt(d)*x*arctan2(d*x^2 - 1, sqrt(-d
*x^2 + 2)*sqrt(d)*x)^3 + 2*(a*b^3*d*x^2 - 2*a*b^3)*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x)^3 + 3*(a^2*b
^2*d*x^2 - 2*a^2*b^2)*arctan2(d*x^2 - 1, sqrt(-d*x^2 + 2)*sqrt(d)*x)^2)/(d*x^2 - 2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\mathrm {asin}\left (d\,x^2-1\right )\right )}^4 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(d*x^2 - 1))^4,x)

[Out]

int((a + b*asin(d*x^2 - 1))^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {asin}{\left (d x^{2} - 1 \right )}\right )^{4}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x**2-1))**4,x)

[Out]

Integral((a + b*asin(d*x**2 - 1))**4, x)

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