3.31.28 \(\int \frac {(-1+a x^8) (1+a x^8)^{3/4}}{1+x^8+a^2 x^{16}} \, dx\) [3028]

Optimal. Leaf size=432 \[ \frac {\left (1+\sqrt [4]{-1}\right ) \text {ArcTan}\left (\frac {(-1)^{7/8} \sqrt {2+\sqrt {2}} \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}{(-1)^{3/4} \sqrt [4]{-1+2 a} x^2+\sqrt {1+a x^8}}\right )}{8 \sqrt [8]{-1+2 a}}-\frac {i \left (-i \sqrt {2}+\sqrt {2 \left (3-2 \sqrt {2}\right )}\right ) \text {ArcTan}\left (\frac {(-1)^{7/8} \left (-2+\sqrt {2}\right ) \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}{(-1)^{3/4} \sqrt {2-\sqrt {2}} \sqrt [4]{-1+2 a} x^2+\sqrt {2-\sqrt {2}} \sqrt {1+a x^8}}\right )}{16 \sqrt [8]{-1+2 a}}+\frac {\left (\sqrt {2}+i \sqrt {2 \left (3-2 \sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {(-1)^{7/8} \sqrt [4]{-1+2 a} x^2-\sqrt [8]{-1} \sqrt {1+a x^8}}{\sqrt {2-\sqrt {2}} \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}\right )}{16 \sqrt [8]{-1+2 a}}+\frac {\left (1+\sqrt [4]{-1}\right ) \tanh ^{-1}\left (\frac {(-1)^{7/8} \sqrt [4]{-1+2 a} x^2-\sqrt [8]{-1} \sqrt {1+a x^8}}{\sqrt {2+\sqrt {2}} \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}\right )}{8 \sqrt [8]{-1+2 a}} \]

[Out]

1/8*(1+(-1)^(1/4))*arctan((-1)^(7/8)*(2+2^(1/2))^(1/2)*(-1+2*a)^(1/8)*x*(a*x^8+1)^(1/4)/((-1)^(3/4)*(-1+2*a)^(
1/4)*x^2+(a*x^8+1)^(1/2)))/(-1+2*a)^(1/8)-1/16*I*(-I*2^(1/2)+2-2^(1/2))*arctan((-1)^(7/8)*(-2+2^(1/2))*(-1+2*a
)^(1/8)*x*(a*x^8+1)^(1/4)/((-1)^(3/4)*(2-2^(1/2))^(1/2)*(-1+2*a)^(1/4)*x^2+(2-2^(1/2))^(1/2)*(a*x^8+1)^(1/2)))
/(-1+2*a)^(1/8)+1/16*(2^(1/2)+I*(2-2^(1/2)))*arctanh(((-1)^(7/8)*(-1+2*a)^(1/4)*x^2-(-1)^(1/8)*(a*x^8+1)^(1/2)
)/(2-2^(1/2))^(1/2)/(-1+2*a)^(1/8)/x/(a*x^8+1)^(1/4))/(-1+2*a)^(1/8)+1/8*(1+(-1)^(1/4))*arctanh(((-1)^(7/8)*(-
1+2*a)^(1/4)*x^2-(-1)^(1/8)*(a*x^8+1)^(1/2))/(2+2^(1/2))^(1/2)/(-1+2*a)^(1/8)/x/(a*x^8+1)^(1/4))/(-1+2*a)^(1/8
)

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Rubi [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 0.27, antiderivative size = 160, normalized size of antiderivative = 0.37, number of steps used = 4, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {6860, 440} \begin {gather*} \frac {a \left (1-\frac {2 a+1}{\sqrt {1-4 a^2}}\right ) x F_1\left (\frac {1}{8};-\frac {3}{4},1;\frac {9}{8};-a x^8,-\frac {2 a^2 x^8}{1-\sqrt {1-4 a^2}}\right )}{1-\sqrt {1-4 a^2}}+\frac {a \left (\frac {2 a+1}{\sqrt {1-4 a^2}}+1\right ) x F_1\left (\frac {1}{8};-\frac {3}{4},1;\frac {9}{8};-a x^8,-\frac {2 a^2 x^8}{\sqrt {1-4 a^2}+1}\right )}{\sqrt {1-4 a^2}+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 + a*x^8)*(1 + a*x^8)^(3/4))/(1 + x^8 + a^2*x^16),x]

[Out]

(a*(1 - (1 + 2*a)/Sqrt[1 - 4*a^2])*x*AppellF1[1/8, -3/4, 1, 9/8, -(a*x^8), (-2*a^2*x^8)/(1 - Sqrt[1 - 4*a^2])]
)/(1 - Sqrt[1 - 4*a^2]) + (a*(1 + (1 + 2*a)/Sqrt[1 - 4*a^2])*x*AppellF1[1/8, -3/4, 1, 9/8, -(a*x^8), (-2*a^2*x
^8)/(1 + Sqrt[1 - 4*a^2])])/(1 + Sqrt[1 - 4*a^2])

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (-1+a x^8\right ) \left (1+a x^8\right )^{3/4}}{1+x^8+a^2 x^{16}} \, dx &=\int \left (\frac {\left (a-\frac {a (1+2 a)}{\sqrt {1-4 a^2}}\right ) \left (1+a x^8\right )^{3/4}}{1-\sqrt {1-4 a^2}+2 a^2 x^8}+\frac {\left (a+\frac {a (1+2 a)}{\sqrt {1-4 a^2}}\right ) \left (1+a x^8\right )^{3/4}}{1+\sqrt {1-4 a^2}+2 a^2 x^8}\right ) \, dx\\ &=\left (a \left (1-\frac {1+2 a}{\sqrt {1-4 a^2}}\right )\right ) \int \frac {\left (1+a x^8\right )^{3/4}}{1-\sqrt {1-4 a^2}+2 a^2 x^8} \, dx+\left (a \left (1+\frac {1+2 a}{\sqrt {1-4 a^2}}\right )\right ) \int \frac {\left (1+a x^8\right )^{3/4}}{1+\sqrt {1-4 a^2}+2 a^2 x^8} \, dx\\ &=\frac {a \left (1-\frac {1+2 a}{\sqrt {1-4 a^2}}\right ) x F_1\left (\frac {1}{8};-\frac {3}{4},1;\frac {9}{8};-a x^8,-\frac {2 a^2 x^8}{1-\sqrt {1-4 a^2}}\right )}{1-\sqrt {1-4 a^2}}+\frac {a \left (1+\frac {1+2 a}{\sqrt {1-4 a^2}}\right ) x F_1\left (\frac {1}{8};-\frac {3}{4},1;\frac {9}{8};-a x^8,-\frac {2 a^2 x^8}{1+\sqrt {1-4 a^2}}\right )}{1+\sqrt {1-4 a^2}}\\ \end {align*}

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Mathematica [A]
time = 8.74, size = 385, normalized size = 0.89 \begin {gather*} \frac {-2 \left (i+(-1)^{3/4}\right ) \tanh ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (\left (-i+(-1)^{3/4}\right ) \sqrt [4]{-1+2 a} x^2+\left (1+(-1)^{3/4}\right ) \sqrt {1+a x^8}\right )}{\sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}\right )+\sqrt {2} \left (i \left (i+\sqrt {3+2 \sqrt {2}}\right ) \text {ArcTan}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (-\left (\left (-i+(-1)^{3/4}\right ) \sqrt [4]{-1+2 a} x^2\right )+\left (1+(-1)^{3/4}\right ) \sqrt {1+a x^8}\right )}{\sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}\right )+\left (1+i \sqrt {3-2 \sqrt {2}}\right ) \text {ArcTan}\left (\frac {2 \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}{\left ((1-i)+\sqrt {2}\right ) \sqrt [4]{-1+2 a} x^2-\left ((1+i)+\sqrt {2}\right ) \sqrt {1+a x^8}}\right )+\left (-1-i \sqrt {3-2 \sqrt {2}}\right ) \tanh ^{-1}\left (\frac {\left ((1-i)+\sqrt {2}\right ) \sqrt [4]{-1+2 a} x^2+\left ((1+i)+\sqrt {2}\right ) \sqrt {1+a x^8}}{2 \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}\right )\right )}{16 \sqrt [8]{-1+2 a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1 + a*x^8)*(1 + a*x^8)^(3/4))/(1 + x^8 + a^2*x^16),x]

[Out]

(-2*(I + (-1)^(3/4))*ArcTanh[((1/2 + I/2)*((-I + (-1)^(3/4))*(-1 + 2*a)^(1/4)*x^2 + (1 + (-1)^(3/4))*Sqrt[1 +
a*x^8]))/((-1 + 2*a)^(1/8)*x*(1 + a*x^8)^(1/4))] + Sqrt[2]*(I*(I + Sqrt[3 + 2*Sqrt[2]])*ArcTan[((1/2 + I/2)*(-
((-I + (-1)^(3/4))*(-1 + 2*a)^(1/4)*x^2) + (1 + (-1)^(3/4))*Sqrt[1 + a*x^8]))/((-1 + 2*a)^(1/8)*x*(1 + a*x^8)^
(1/4))] + (1 + I*Sqrt[3 - 2*Sqrt[2]])*ArcTan[(2*(-1 + 2*a)^(1/8)*x*(1 + a*x^8)^(1/4))/(((1 - I) + Sqrt[2])*(-1
+ 2*a)^(1/4)*x^2 - ((1 + I) + Sqrt[2])*Sqrt[1 + a*x^8])] + (-1 - I*Sqrt[3 - 2*Sqrt[2]])*ArcTanh[(((1 - I) + S
qrt[2])*(-1 + 2*a)^(1/4)*x^2 + ((1 + I) + Sqrt[2])*Sqrt[1 + a*x^8])/(2*(-1 + 2*a)^(1/8)*x*(1 + a*x^8)^(1/4))])
)/(16*(-1 + 2*a)^(1/8))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{8}-1\right ) \left (a \,x^{8}+1\right )^{\frac {3}{4}}}{a^{2} x^{16}+x^{8}+1}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^8-1)*(a*x^8+1)^(3/4)/(a^2*x^16+x^8+1),x)

[Out]

int((a*x^8-1)*(a*x^8+1)^(3/4)/(a^2*x^16+x^8+1),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^8-1)*(a*x^8+1)^(3/4)/(a^2*x^16+x^8+1),x, algorithm="maxima")

[Out]

integrate((a*x^8 + 1)^(3/4)*(a*x^8 - 1)/(a^2*x^16 + x^8 + 1), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^8-1)*(a*x^8+1)^(3/4)/(a^2*x^16+x^8+1),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**8-1)*(a*x**8+1)**(3/4)/(a**2*x**16+x**8+1),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^8-1)*(a*x^8+1)^(3/4)/(a^2*x^16+x^8+1),x, algorithm="giac")

[Out]

integrate((a*x^8 + 1)^(3/4)*(a*x^8 - 1)/(a^2*x^16 + x^8 + 1), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a\,x^8-1\right )\,{\left (a\,x^8+1\right )}^{3/4}}{a^2\,x^{16}+x^8+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x^8 - 1)*(a*x^8 + 1)^(3/4))/(x^8 + a^2*x^16 + 1),x)

[Out]

int(((a*x^8 - 1)*(a*x^8 + 1)^(3/4))/(x^8 + a^2*x^16 + 1), x)

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