3.80 \(\int (a+b \cos ^{-1}(-1+d x^2))^4 \, dx\)

Optimal. Leaf size=127 \[ \frac{192 b^3 \sqrt{2 d x^2-d^2 x^4} \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )}{d x}-48 b^2 x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^2-\frac{8 b \sqrt{2 d x^2-d^2 x^4} \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^3}{d x}+x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^4+384 b^4 x \]

[Out]

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

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Rubi [A]  time = 0.0283064, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4815, 8} \[ \frac{192 b^3 \sqrt{2 d x^2-d^2 x^4} \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )}{d x}-48 b^2 x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^2-\frac{8 b \sqrt{2 d x^2-d^2 x^4} \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^3}{d x}+x \left (a+b \cos ^{-1}\left (d x^2-1\right )\right )^4+384 b^4 x \]

Antiderivative was successfully verified.

[In]

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

Rule 4815

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

Rule 8

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

Rubi steps

\begin{align*} \int \left (a+b \cos ^{-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 \cos ^{-1}\left (-1+d x^2\right )\right )^3}{d x}+x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^4-\left (48 b^2\right ) \int \left (a+b \cos ^{-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 \cos ^{-1}\left (-1+d x^2\right )\right )}{d x}-48 b^2 x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^2-\frac{8 b \sqrt{2 d x^2-d^2 x^4} \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^3}{d x}+x \left (a+b \cos ^{-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 \cos ^{-1}\left (-1+d x^2\right )\right )}{d x}-48 b^2 x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^2-\frac{8 b \sqrt{2 d x^2-d^2 x^4} \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^3}{d x}+x \left (a+b \cos ^{-1}\left (-1+d x^2\right )\right )^4\\ \end{align*}

Mathematica [A]  time = 0.239112, size = 249, normalized size = 1.96 \[ \frac{d x^2 \left (-48 a^2 b^2+a^4+384 b^4\right )-8 a b \left (a^2-24 b^2\right ) \sqrt{d x^2 \left (2-d x^2\right )}+6 b^2 \cos ^{-1}\left (d x^2-1\right )^2 \left (a^2 d x^2-4 a b \sqrt{-d x^2 \left (d x^2-2\right )}-8 b^2 d x^2\right )+4 b \cos ^{-1}\left (d x^2-1\right ) \left (-6 a^2 b \sqrt{-d x^2 \left (d x^2-2\right )}+a^3 d x^2-24 a b^2 d x^2+48 b^3 \sqrt{-d x^2 \left (d x^2-2\right )}\right )+4 b^3 \cos ^{-1}\left (d x^2-1\right )^3 \left (a d x^2-2 b \sqrt{-d x^2 \left (d x^2-2\right )}\right )+b^4 d x^2 \cos ^{-1}\left (d x^2-1\right )^4}{d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^4 - 48*a^2*b^2 + 384*b^4)*d*x^2 - 8*a*b*(a^2 - 24*b^2)*Sqrt[d*x^2*(2 - d*x^2)] + 4*b*(a^3*d*x^2 - 24*a*b^2
*d*x^2 - 6*a^2*b*Sqrt[-(d*x^2*(-2 + d*x^2))] + 48*b^3*Sqrt[-(d*x^2*(-2 + d*x^2))])*ArcCos[-1 + d*x^2] + 6*b^2*
(a^2*d*x^2 - 8*b^2*d*x^2 - 4*a*b*Sqrt[-(d*x^2*(-2 + d*x^2))])*ArcCos[-1 + d*x^2]^2 + 4*b^3*(a*d*x^2 - 2*b*Sqrt
[-(d*x^2*(-2 + d*x^2))])*ArcCos[-1 + d*x^2]^3 + b^4*d*x^2*ArcCos[-1 + d*x^2]^4)/(d*x)

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Maple [F]  time = 0.117, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arccos \left ( d{x}^{2}-1 \right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.65725, size = 474, normalized size = 3.73 \begin{align*} \frac{b^{4} d x^{2} \arccos \left (d x^{2} - 1\right )^{4} + 4 \, a b^{3} d x^{2} \arccos \left (d x^{2} - 1\right )^{3} + 6 \,{\left (a^{2} b^{2} - 8 \, b^{4}\right )} d x^{2} \arccos \left (d x^{2} - 1\right )^{2} + 4 \,{\left (a^{3} b - 24 \, a b^{3}\right )} d x^{2} \arccos \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} \arccos \left (d x^{2} - 1\right )^{3} + 3 \, a b^{3} \arccos \left (d x^{2} - 1\right )^{2} + a^{3} b - 24 \, a b^{3} + 3 \,{\left (a^{2} b^{2} - 8 \, b^{4}\right )} \arccos \left (d x^{2} - 1\right )\right )} \sqrt{-d^{2} x^{4} + 2 \, d x^{2}}}{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{acos}{\left (d x^{2} - 1 \right )}\right )^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arccos \left (d x^{2} - 1\right ) + a\right )}^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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