3.31.49 \(\int \frac {(b+a x^4)^{3/4}}{b+2 a x^4+2 x^8} \, dx\)

Optimal. Leaf size=460 \[ \frac {\left (-1-\sqrt [4]{-1}\right ) \tan ^{-1}\left (\frac {(-1)^{7/8} \sqrt {2+\sqrt {2}} x \sqrt [8]{a^2-2 b} \sqrt [4]{a x^4+b}}{(-1)^{3/4} x^2 \sqrt [4]{a^2-2 b}+\sqrt {a x^4+b}}\right )}{8 \sqrt [8]{a^2-2 b}}+\frac {\left (\sqrt {2}+i \sqrt {2 \left (3-2 \sqrt {2}\right )}\right ) \tan ^{-1}\left (\frac {(-1)^{7/8} \left (\sqrt {2}-2\right ) x \sqrt [8]{a^2-2 b} \sqrt [4]{a x^4+b}}{(-1)^{3/4} \sqrt {2-\sqrt {2}} x^2 \sqrt [4]{a^2-2 b}+\sqrt {2-\sqrt {2}} \sqrt {a x^4+b}}\right )}{16 \sqrt [8]{a^2-2 b}}-\frac {i \left (\sqrt {2 \left (3-2 \sqrt {2}\right )}-i \sqrt {2}\right ) \tanh ^{-1}\left (\frac {(-1)^{7/8} x^2 \sqrt [4]{a^2-2 b}-\sqrt [8]{-1} \sqrt {a x^4+b}}{\sqrt {2-\sqrt {2}} x \sqrt [8]{a^2-2 b} \sqrt [4]{a x^4+b}}\right )}{16 \sqrt [8]{a^2-2 b}}+\frac {\left (-1-\sqrt [4]{-1}\right ) \tanh ^{-1}\left (\frac {(-1)^{7/8} x^2 \sqrt [4]{a^2-2 b}-\sqrt [8]{-1} \sqrt {a x^4+b}}{\sqrt {2+\sqrt {2}} x \sqrt [8]{a^2-2 b} \sqrt [4]{a x^4+b}}\right )}{8 \sqrt [8]{a^2-2 b}} \]

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Rubi [A]  time = 0.59, antiderivative size = 457, normalized size of antiderivative = 0.99, number of steps used = 19, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1428, 408, 240, 212, 206, 203, 377, 208, 205} \begin {gather*} -\frac {\left (-a \sqrt {a^2-2 b}+a^2-2 b\right )^{3/4} \tan ^{-1}\left (\frac {x \sqrt [4]{-a \sqrt {a^2-2 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2-2 b}} \sqrt [4]{a x^4+b}}\right )}{4 \left (a-\sqrt {a^2-2 b}\right )^{3/4} \sqrt {a^2-2 b}}+\frac {\left (a \sqrt {a^2-2 b}+a^2-2 b\right )^{3/4} \tan ^{-1}\left (\frac {x \sqrt [4]{a \sqrt {a^2-2 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2-2 b}+a} \sqrt [4]{a x^4+b}}\right )}{4 \left (\sqrt {a^2-2 b}+a\right )^{3/4} \sqrt {a^2-2 b}}-\frac {\left (-a \sqrt {a^2-2 b}+a^2-2 b\right )^{3/4} \tanh ^{-1}\left (\frac {x \sqrt [4]{-a \sqrt {a^2-2 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2-2 b}} \sqrt [4]{a x^4+b}}\right )}{4 \left (a-\sqrt {a^2-2 b}\right )^{3/4} \sqrt {a^2-2 b}}+\frac {\left (a \sqrt {a^2-2 b}+a^2-2 b\right )^{3/4} \tanh ^{-1}\left (\frac {x \sqrt [4]{a \sqrt {a^2-2 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2-2 b}+a} \sqrt [4]{a x^4+b}}\right )}{4 \left (\sqrt {a^2-2 b}+a\right )^{3/4} \sqrt {a^2-2 b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 240

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

Rule 377

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

Rule 408

Int[((a_) + (b_.)*(x_)^4)^(p_)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[b/d, Int[(a + b*x^4)^(p - 1), x], x] -
Dist[(b*c - a*d)/d, Int[(a + b*x^4)^(p - 1)/(c + d*x^4), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0
] && (EqQ[p, 3/4] || EqQ[p, 5/4])

Rule 1428

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[b^2 -
4*a*c, 2]}, Dist[(2*c)/r, Int[(d + e*x^n)^q/(b - r + 2*c*x^n), x], x] - Dist[(2*c)/r, Int[(d + e*x^n)^q/(b +
r + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] &&  !IntegerQ[q]

Rubi steps

\begin {align*} \int \frac {\left (b+a x^4\right )^{3/4}}{b+2 a x^4+2 x^8} \, dx &=\frac {2 \int \frac {\left (b+a x^4\right )^{3/4}}{2 a-2 \sqrt {a^2-2 b}+4 x^4} \, dx}{\sqrt {a^2-2 b}}-\frac {2 \int \frac {\left (b+a x^4\right )^{3/4}}{2 a+2 \sqrt {a^2-2 b}+4 x^4} \, dx}{\sqrt {a^2-2 b}}\\ &=\frac {\left (a \left (a+\sqrt {a^2-2 b}\right )-2 b\right ) \int \frac {1}{\left (2 a+2 \sqrt {a^2-2 b}+4 x^4\right ) \sqrt [4]{b+a x^4}} \, dx}{\sqrt {a^2-2 b}}-\frac {\left (a^2-a \sqrt {a^2-2 b}-2 b\right ) \int \frac {1}{\left (2 a-2 \sqrt {a^2-2 b}+4 x^4\right ) \sqrt [4]{b+a x^4}} \, dx}{\sqrt {a^2-2 b}}\\ &=\frac {\left (a \left (a+\sqrt {a^2-2 b}\right )-2 b\right ) \operatorname {Subst}\left (\int \frac {1}{2 a+2 \sqrt {a^2-2 b}-\left (a \left (2 a+2 \sqrt {a^2-2 b}\right )-4 b\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt {a^2-2 b}}-\frac {\left (a^2-a \sqrt {a^2-2 b}-2 b\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-2 \sqrt {a^2-2 b}-\left (a \left (2 a-2 \sqrt {a^2-2 b}\right )-4 b\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt {a^2-2 b}}\\ &=\frac {\left (a \left (a+\sqrt {a^2-2 b}\right )-2 b\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\sqrt {a^2-2 b}}-\sqrt {a^2+a \sqrt {a^2-2 b}-2 b} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{4 \sqrt {a+\sqrt {a^2-2 b}} \sqrt {a^2-2 b}}+\frac {\left (a \left (a+\sqrt {a^2-2 b}\right )-2 b\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\sqrt {a^2-2 b}}+\sqrt {a^2+a \sqrt {a^2-2 b}-2 b} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{4 \sqrt {a+\sqrt {a^2-2 b}} \sqrt {a^2-2 b}}-\frac {\left (a^2-a \sqrt {a^2-2 b}-2 b\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-\sqrt {a^2-2 b}}-\sqrt {a^2-a \sqrt {a^2-2 b}-2 b} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{4 \sqrt {a-\sqrt {a^2-2 b}} \sqrt {a^2-2 b}}-\frac {\left (a^2-a \sqrt {a^2-2 b}-2 b\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-\sqrt {a^2-2 b}}+\sqrt {a^2-a \sqrt {a^2-2 b}-2 b} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{4 \sqrt {a-\sqrt {a^2-2 b}} \sqrt {a^2-2 b}}\\ &=-\frac {\left (a^2-a \sqrt {a^2-2 b}-2 b\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a^2-a \sqrt {a^2-2 b}-2 b} x}{\sqrt [4]{a-\sqrt {a^2-2 b}} \sqrt [4]{b+a x^4}}\right )}{4 \left (a-\sqrt {a^2-2 b}\right )^{3/4} \sqrt {a^2-2 b}}+\frac {\left (a^2+a \sqrt {a^2-2 b}-2 b\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a^2+a \sqrt {a^2-2 b}-2 b} x}{\sqrt [4]{a+\sqrt {a^2-2 b}} \sqrt [4]{b+a x^4}}\right )}{4 \left (a+\sqrt {a^2-2 b}\right )^{3/4} \sqrt {a^2-2 b}}-\frac {\left (a^2-a \sqrt {a^2-2 b}-2 b\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a^2-a \sqrt {a^2-2 b}-2 b} x}{\sqrt [4]{a-\sqrt {a^2-2 b}} \sqrt [4]{b+a x^4}}\right )}{4 \left (a-\sqrt {a^2-2 b}\right )^{3/4} \sqrt {a^2-2 b}}+\frac {\left (a^2+a \sqrt {a^2-2 b}-2 b\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a^2+a \sqrt {a^2-2 b}-2 b} x}{\sqrt [4]{a+\sqrt {a^2-2 b}} \sqrt [4]{b+a x^4}}\right )}{4 \left (a+\sqrt {a^2-2 b}\right )^{3/4} \sqrt {a^2-2 b}}\\ \end {align*}

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Mathematica [F]  time = 0.11, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b+a x^4\right )^{3/4}}{b+2 a x^4+2 x^8} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 40.33, size = 510, normalized size = 1.11 \begin {gather*} \frac {\left (\sqrt {2}-i \sqrt {2 \left (3+2 \sqrt {2}\right )}\right ) \tan ^{-1}\left (\frac {\left ((-1+i)-(1+i) (-1)^{3/4}\right ) \sqrt [4]{a^2-2 b} x^2+(1+i) \sqrt {b+a x^4}+(1+i) (-1)^{3/4} \sqrt {b+a x^4}}{2 \sqrt [8]{a^2-2 b} x \sqrt [4]{b+a x^4}}\right )}{16 \sqrt [8]{a^2-2 b}}-\frac {i \left (-i \sqrt {2}+\sqrt {2 \left (3-2 \sqrt {2}\right )}\right ) \tan ^{-1}\left (\frac {2 \sqrt [8]{a^2-2 b} x \sqrt [4]{b+a x^4}}{\left ((1-i)+\sqrt {2}\right ) \sqrt [4]{a^2-2 b} x^2-(1+i) \sqrt {b+a x^4}-\sqrt {2} \sqrt {b+a x^4}}\right )}{16 \sqrt [8]{a^2-2 b}}+\frac {\left (-i-(-1)^{3/4}\right ) \tanh ^{-1}\left (\frac {\left ((-2+2 i)-(2+2 i) (-1)^{3/4}\right ) \sqrt [4]{a^2-2 b} x^2-(2+2 i) \sqrt {b+a x^4}-(2+2 i) (-1)^{3/4} \sqrt {b+a x^4}}{4 \sqrt [8]{a^2-2 b} x \sqrt [4]{b+a x^4}}\right )}{8 \sqrt [8]{a^2-2 b}}+\frac {\left (\sqrt {2}+i \sqrt {2 \left (3-2 \sqrt {2}\right )}\right ) \tanh ^{-1}\left (\frac {\left ((1-i)+\sqrt {2}\right ) \sqrt [4]{a^2-2 b} x^2+(1+i) \sqrt {b+a x^4}+\sqrt {2} \sqrt {b+a x^4}}{2 \sqrt [8]{a^2-2 b} x \sqrt [4]{b+a x^4}}\right )}{16 \sqrt [8]{a^2-2 b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b + a*x^4)^(3/4)/(b + 2*a*x^4 + 2*x^8),x]

[Out]

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

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fricas [F(-1)]  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^4+b)^(3/4)/(2*x^8+2*a*x^4+b),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a x^{4} + b\right )}^{\frac {3}{4}}}{2 \, x^{8} + 2 \, a x^{4} + b}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^4 + b)^(3/4)/(2*x^8 + 2*a*x^4 + b), x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{4}+b \right )^{\frac {3}{4}}}{2 x^{8}+2 a \,x^{4}+b}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4+b)^(3/4)/(2*x^8+2*a*x^4+b),x)

[Out]

int((a*x^4+b)^(3/4)/(2*x^8+2*a*x^4+b),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a x^{4} + b\right )}^{\frac {3}{4}}}{2 \, x^{8} + 2 \, a x^{4} + b}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^4 + b)^(3/4)/(2*x^8 + 2*a*x^4 + b), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b + a*x^4)^(3/4)/(b + 2*a*x^4 + 2*x^8),x)

[Out]

int((b + a*x^4)^(3/4)/(b + 2*a*x^4 + 2*x^8), x)

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sympy [F(-1)]  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**4+b)**(3/4)/(2*x**8+2*a*x**4+b),x)

[Out]

Timed out

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