3.1463 \(\int \frac {x^4}{a+b x^8} \, dx\)

Optimal. Leaf size=267 \[ -\frac {\log \left (-\sqrt {2} \sqrt [8]{-a} \sqrt [8]{b} x+\sqrt [4]{-a}+\sqrt [4]{b} x^2\right )}{8 \sqrt {2} (-a)^{3/8} b^{5/8}}+\frac {\log \left (\sqrt {2} \sqrt [8]{-a} \sqrt [8]{b} x+\sqrt [4]{-a}+\sqrt [4]{b} x^2\right )}{8 \sqrt {2} (-a)^{3/8} b^{5/8}}-\frac {\tan ^{-1}\left (\frac {\sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 (-a)^{3/8} b^{5/8}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 \sqrt {2} (-a)^{3/8} b^{5/8}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [8]{b} x}{\sqrt [8]{-a}}+1\right )}{4 \sqrt {2} (-a)^{3/8} b^{5/8}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 (-a)^{3/8} b^{5/8}} \]

[Out]

-1/4*arctan(b^(1/8)*x/(-a)^(1/8))/(-a)^(3/8)/b^(5/8)-1/4*arctanh(b^(1/8)*x/(-a)^(1/8))/(-a)^(3/8)/b^(5/8)+1/8*
arctan(-1+b^(1/8)*x*2^(1/2)/(-a)^(1/8))/(-a)^(3/8)/b^(5/8)*2^(1/2)+1/8*arctan(1+b^(1/8)*x*2^(1/2)/(-a)^(1/8))/
(-a)^(3/8)/b^(5/8)*2^(1/2)-1/16*ln((-a)^(1/4)+b^(1/4)*x^2-(-a)^(1/8)*b^(1/8)*x*2^(1/2))/(-a)^(3/8)/b^(5/8)*2^(
1/2)+1/16*ln((-a)^(1/4)+b^(1/4)*x^2+(-a)^(1/8)*b^(1/8)*x*2^(1/2))/(-a)^(3/8)/b^(5/8)*2^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {301, 211, 1165, 628, 1162, 617, 204, 212, 208, 205} \[ -\frac {\log \left (-\sqrt {2} \sqrt [8]{-a} \sqrt [8]{b} x+\sqrt [4]{-a}+\sqrt [4]{b} x^2\right )}{8 \sqrt {2} (-a)^{3/8} b^{5/8}}+\frac {\log \left (\sqrt {2} \sqrt [8]{-a} \sqrt [8]{b} x+\sqrt [4]{-a}+\sqrt [4]{b} x^2\right )}{8 \sqrt {2} (-a)^{3/8} b^{5/8}}-\frac {\tan ^{-1}\left (\frac {\sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 (-a)^{3/8} b^{5/8}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 \sqrt {2} (-a)^{3/8} b^{5/8}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [8]{b} x}{\sqrt [8]{-a}}+1\right )}{4 \sqrt {2} (-a)^{3/8} b^{5/8}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 (-a)^{3/8} b^{5/8}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a + b*x^8),x]

[Out]

-ArcTan[(b^(1/8)*x)/(-a)^(1/8)]/(4*(-a)^(3/8)*b^(5/8)) - ArcTan[1 - (Sqrt[2]*b^(1/8)*x)/(-a)^(1/8)]/(4*Sqrt[2]
*(-a)^(3/8)*b^(5/8)) + ArcTan[1 + (Sqrt[2]*b^(1/8)*x)/(-a)^(1/8)]/(4*Sqrt[2]*(-a)^(3/8)*b^(5/8)) - ArcTanh[(b^
(1/8)*x)/(-a)^(1/8)]/(4*(-a)^(3/8)*b^(5/8)) - Log[(-a)^(1/4) - Sqrt[2]*(-a)^(1/8)*b^(1/8)*x + b^(1/4)*x^2]/(8*
Sqrt[2]*(-a)^(3/8)*b^(5/8)) + Log[(-a)^(1/4) + Sqrt[2]*(-a)^(1/8)*b^(1/8)*x + b^(1/4)*x^2]/(8*Sqrt[2]*(-a)^(3/
8)*b^(5/8))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[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 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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, 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 301

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {x^4}{a+b x^8} \, dx &=-\frac {\int \frac {1}{\sqrt {-a}-\sqrt {b} x^4} \, dx}{2 \sqrt {b}}+\frac {\int \frac {1}{\sqrt {-a}+\sqrt {b} x^4} \, dx}{2 \sqrt {b}}\\ &=-\frac {\int \frac {1}{\sqrt [4]{-a}-\sqrt [4]{b} x^2} \, dx}{4 \sqrt [4]{-a} \sqrt {b}}-\frac {\int \frac {1}{\sqrt [4]{-a}+\sqrt [4]{b} x^2} \, dx}{4 \sqrt [4]{-a} \sqrt {b}}+\frac {\int \frac {\sqrt [4]{-a}-\sqrt [4]{b} x^2}{\sqrt {-a}+\sqrt {b} x^4} \, dx}{4 \sqrt [4]{-a} \sqrt {b}}+\frac {\int \frac {\sqrt [4]{-a}+\sqrt [4]{b} x^2}{\sqrt {-a}+\sqrt {b} x^4} \, dx}{4 \sqrt [4]{-a} \sqrt {b}}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 (-a)^{3/8} b^{5/8}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 (-a)^{3/8} b^{5/8}}+\frac {\int \frac {1}{\frac {\sqrt [4]{-a}}{\sqrt [4]{b}}-\frac {\sqrt {2} \sqrt [8]{-a} x}{\sqrt [8]{b}}+x^2} \, dx}{8 \sqrt [4]{-a} b^{3/4}}+\frac {\int \frac {1}{\frac {\sqrt [4]{-a}}{\sqrt [4]{b}}+\frac {\sqrt {2} \sqrt [8]{-a} x}{\sqrt [8]{b}}+x^2} \, dx}{8 \sqrt [4]{-a} b^{3/4}}-\frac {\int \frac {\frac {\sqrt {2} \sqrt [8]{-a}}{\sqrt [8]{b}}+2 x}{-\frac {\sqrt [4]{-a}}{\sqrt [4]{b}}-\frac {\sqrt {2} \sqrt [8]{-a} x}{\sqrt [8]{b}}-x^2} \, dx}{8 \sqrt {2} (-a)^{3/8} b^{5/8}}-\frac {\int \frac {\frac {\sqrt {2} \sqrt [8]{-a}}{\sqrt [8]{b}}-2 x}{-\frac {\sqrt [4]{-a}}{\sqrt [4]{b}}+\frac {\sqrt {2} \sqrt [8]{-a} x}{\sqrt [8]{b}}-x^2} \, dx}{8 \sqrt {2} (-a)^{3/8} b^{5/8}}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 (-a)^{3/8} b^{5/8}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 (-a)^{3/8} b^{5/8}}-\frac {\log \left (\sqrt [4]{-a}-\sqrt {2} \sqrt [8]{-a} \sqrt [8]{b} x+\sqrt [4]{b} x^2\right )}{8 \sqrt {2} (-a)^{3/8} b^{5/8}}+\frac {\log \left (\sqrt [4]{-a}+\sqrt {2} \sqrt [8]{-a} \sqrt [8]{b} x+\sqrt [4]{b} x^2\right )}{8 \sqrt {2} (-a)^{3/8} b^{5/8}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 \sqrt {2} (-a)^{3/8} b^{5/8}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 \sqrt {2} (-a)^{3/8} b^{5/8}}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 (-a)^{3/8} b^{5/8}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 \sqrt {2} (-a)^{3/8} b^{5/8}}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 \sqrt {2} (-a)^{3/8} b^{5/8}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [8]{b} x}{\sqrt [8]{-a}}\right )}{4 (-a)^{3/8} b^{5/8}}-\frac {\log \left (\sqrt [4]{-a}-\sqrt {2} \sqrt [8]{-a} \sqrt [8]{b} x+\sqrt [4]{b} x^2\right )}{8 \sqrt {2} (-a)^{3/8} b^{5/8}}+\frac {\log \left (\sqrt [4]{-a}+\sqrt {2} \sqrt [8]{-a} \sqrt [8]{b} x+\sqrt [4]{b} x^2\right )}{8 \sqrt {2} (-a)^{3/8} b^{5/8}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 324, normalized size = 1.21 \[ -\frac {\cos \left (\frac {\pi }{8}\right ) \log \left (-2 \sqrt [8]{a} \sqrt [8]{b} x \sin \left (\frac {\pi }{8}\right )+\sqrt [4]{a}+\sqrt [4]{b} x^2\right )-\cos \left (\frac {\pi }{8}\right ) \log \left (2 \sqrt [8]{a} \sqrt [8]{b} x \sin \left (\frac {\pi }{8}\right )+\sqrt [4]{a}+\sqrt [4]{b} x^2\right )-\sin \left (\frac {\pi }{8}\right ) \log \left (-2 \sqrt [8]{a} \sqrt [8]{b} x \cos \left (\frac {\pi }{8}\right )+\sqrt [4]{a}+\sqrt [4]{b} x^2\right )+\sin \left (\frac {\pi }{8}\right ) \log \left (2 \sqrt [8]{a} \sqrt [8]{b} x \cos \left (\frac {\pi }{8}\right )+\sqrt [4]{a}+\sqrt [4]{b} x^2\right )+2 \sin \left (\frac {\pi }{8}\right ) \tan ^{-1}\left (\frac {\sqrt [8]{b} x \sec \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}-\tan \left (\frac {\pi }{8}\right )\right )+2 \sin \left (\frac {\pi }{8}\right ) \tan ^{-1}\left (\frac {\sqrt [8]{b} x \sec \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}+\tan \left (\frac {\pi }{8}\right )\right )+2 \cos \left (\frac {\pi }{8}\right ) \tan ^{-1}\left (\cot \left (\frac {\pi }{8}\right )-\frac {\sqrt [8]{b} x \csc \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}\right )-2 \cos \left (\frac {\pi }{8}\right ) \tan ^{-1}\left (\frac {\sqrt [8]{b} x \csc \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}+\cot \left (\frac {\pi }{8}\right )\right )}{8 a^{3/8} b^{5/8}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a + b*x^8),x]

[Out]

-1/8*(2*ArcTan[Cot[Pi/8] - (b^(1/8)*x*Csc[Pi/8])/a^(1/8)]*Cos[Pi/8] - 2*ArcTan[Cot[Pi/8] + (b^(1/8)*x*Csc[Pi/8
])/a^(1/8)]*Cos[Pi/8] + Cos[Pi/8]*Log[a^(1/4) + b^(1/4)*x^2 - 2*a^(1/8)*b^(1/8)*x*Sin[Pi/8]] - Cos[Pi/8]*Log[a
^(1/4) + b^(1/4)*x^2 + 2*a^(1/8)*b^(1/8)*x*Sin[Pi/8]] + 2*ArcTan[(b^(1/8)*x*Sec[Pi/8])/a^(1/8) - Tan[Pi/8]]*Si
n[Pi/8] + 2*ArcTan[(b^(1/8)*x*Sec[Pi/8])/a^(1/8) + Tan[Pi/8]]*Sin[Pi/8] - Log[a^(1/4) + b^(1/4)*x^2 - 2*a^(1/8
)*b^(1/8)*x*Cos[Pi/8]]*Sin[Pi/8] + Log[a^(1/4) + b^(1/4)*x^2 + 2*a^(1/8)*b^(1/8)*x*Cos[Pi/8]]*Sin[Pi/8])/(a^(3
/8)*b^(5/8))

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fricas [B]  time = 0.70, size = 438, normalized size = 1.64 \[ \frac {1}{4} \, \sqrt {2} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{8}} \arctan \left (-\sqrt {2} a b^{2} x \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {3}{8}} + \sqrt {2} \sqrt {\sqrt {2} a^{2} b^{3} x \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {5}{8}} - a b \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{4}} + x^{2}} a b^{2} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {3}{8}} + 1\right ) + \frac {1}{4} \, \sqrt {2} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{8}} \arctan \left (-\sqrt {2} a b^{2} x \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {3}{8}} + \sqrt {2} \sqrt {-\sqrt {2} a^{2} b^{3} x \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {5}{8}} - a b \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{4}} + x^{2}} a b^{2} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {3}{8}} - 1\right ) + \frac {1}{16} \, \sqrt {2} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{8}} \log \left (\sqrt {2} a^{2} b^{3} x \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {5}{8}} - a b \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{4}} + x^{2}\right ) - \frac {1}{16} \, \sqrt {2} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{8}} \log \left (-\sqrt {2} a^{2} b^{3} x \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {5}{8}} - a b \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{4}} + x^{2}\right ) - \frac {1}{2} \, \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{8}} \arctan \left (-a b^{2} x \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {3}{8}} + \sqrt {-a b \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{4}} + x^{2}} a b^{2} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {3}{8}}\right ) - \frac {1}{8} \, \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{8}} \log \left (a^{2} b^{3} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {5}{8}} + x\right ) + \frac {1}{8} \, \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {1}{8}} \log \left (-a^{2} b^{3} \left (-\frac {1}{a^{3} b^{5}}\right )^{\frac {5}{8}} + x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*(-1/(a^3*b^5))^(1/8)*arctan(-sqrt(2)*a*b^2*x*(-1/(a^3*b^5))^(3/8) + sqrt(2)*sqrt(sqrt(2)*a^2*b^3*x
*(-1/(a^3*b^5))^(5/8) - a*b*(-1/(a^3*b^5))^(1/4) + x^2)*a*b^2*(-1/(a^3*b^5))^(3/8) + 1) + 1/4*sqrt(2)*(-1/(a^3
*b^5))^(1/8)*arctan(-sqrt(2)*a*b^2*x*(-1/(a^3*b^5))^(3/8) + sqrt(2)*sqrt(-sqrt(2)*a^2*b^3*x*(-1/(a^3*b^5))^(5/
8) - a*b*(-1/(a^3*b^5))^(1/4) + x^2)*a*b^2*(-1/(a^3*b^5))^(3/8) - 1) + 1/16*sqrt(2)*(-1/(a^3*b^5))^(1/8)*log(s
qrt(2)*a^2*b^3*x*(-1/(a^3*b^5))^(5/8) - a*b*(-1/(a^3*b^5))^(1/4) + x^2) - 1/16*sqrt(2)*(-1/(a^3*b^5))^(1/8)*lo
g(-sqrt(2)*a^2*b^3*x*(-1/(a^3*b^5))^(5/8) - a*b*(-1/(a^3*b^5))^(1/4) + x^2) - 1/2*(-1/(a^3*b^5))^(1/8)*arctan(
-a*b^2*x*(-1/(a^3*b^5))^(3/8) + sqrt(-a*b*(-1/(a^3*b^5))^(1/4) + x^2)*a*b^2*(-1/(a^3*b^5))^(3/8)) - 1/8*(-1/(a
^3*b^5))^(1/8)*log(a^2*b^3*(-1/(a^3*b^5))^(5/8) + x) + 1/8*(-1/(a^3*b^5))^(1/8)*log(-a^2*b^3*(-1/(a^3*b^5))^(5
/8) + x)

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giac [B]  time = 0.37, size = 437, normalized size = 1.64 \[ -\frac {\left (\frac {a}{b}\right )^{\frac {5}{8}} \arctan \left (\frac {2 \, x + \sqrt {-\sqrt {2} + 2} \left (\frac {a}{b}\right )^{\frac {1}{8}}}{\sqrt {\sqrt {2} + 2} \left (\frac {a}{b}\right )^{\frac {1}{8}}}\right )}{4 \, a \sqrt {2 \, \sqrt {2} + 4}} - \frac {\left (\frac {a}{b}\right )^{\frac {5}{8}} \arctan \left (\frac {2 \, x - \sqrt {-\sqrt {2} + 2} \left (\frac {a}{b}\right )^{\frac {1}{8}}}{\sqrt {\sqrt {2} + 2} \left (\frac {a}{b}\right )^{\frac {1}{8}}}\right )}{4 \, a \sqrt {2 \, \sqrt {2} + 4}} + \frac {\left (\frac {a}{b}\right )^{\frac {5}{8}} \arctan \left (\frac {2 \, x + \sqrt {\sqrt {2} + 2} \left (\frac {a}{b}\right )^{\frac {1}{8}}}{\sqrt {-\sqrt {2} + 2} \left (\frac {a}{b}\right )^{\frac {1}{8}}}\right )}{4 \, a \sqrt {-2 \, \sqrt {2} + 4}} + \frac {\left (\frac {a}{b}\right )^{\frac {5}{8}} \arctan \left (\frac {2 \, x - \sqrt {\sqrt {2} + 2} \left (\frac {a}{b}\right )^{\frac {1}{8}}}{\sqrt {-\sqrt {2} + 2} \left (\frac {a}{b}\right )^{\frac {1}{8}}}\right )}{4 \, a \sqrt {-2 \, \sqrt {2} + 4}} - \frac {\left (\frac {a}{b}\right )^{\frac {5}{8}} \log \left (x^{2} + x \sqrt {\sqrt {2} + 2} \left (\frac {a}{b}\right )^{\frac {1}{8}} + \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}{8 \, a \sqrt {2 \, \sqrt {2} + 4}} + \frac {\left (\frac {a}{b}\right )^{\frac {5}{8}} \log \left (x^{2} - x \sqrt {\sqrt {2} + 2} \left (\frac {a}{b}\right )^{\frac {1}{8}} + \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}{8 \, a \sqrt {2 \, \sqrt {2} + 4}} + \frac {\left (\frac {a}{b}\right )^{\frac {5}{8}} \log \left (x^{2} + x \sqrt {-\sqrt {2} + 2} \left (\frac {a}{b}\right )^{\frac {1}{8}} + \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}{8 \, a \sqrt {-2 \, \sqrt {2} + 4}} - \frac {\left (\frac {a}{b}\right )^{\frac {5}{8}} \log \left (x^{2} - x \sqrt {-\sqrt {2} + 2} \left (\frac {a}{b}\right )^{\frac {1}{8}} + \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}{8 \, a \sqrt {-2 \, \sqrt {2} + 4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a/b)^(5/8)*arctan((2*x + sqrt(-sqrt(2) + 2)*(a/b)^(1/8))/(sqrt(sqrt(2) + 2)*(a/b)^(1/8)))/(a*sqrt(2*sqrt
(2) + 4)) - 1/4*(a/b)^(5/8)*arctan((2*x - sqrt(-sqrt(2) + 2)*(a/b)^(1/8))/(sqrt(sqrt(2) + 2)*(a/b)^(1/8)))/(a*
sqrt(2*sqrt(2) + 4)) + 1/4*(a/b)^(5/8)*arctan((2*x + sqrt(sqrt(2) + 2)*(a/b)^(1/8))/(sqrt(-sqrt(2) + 2)*(a/b)^
(1/8)))/(a*sqrt(-2*sqrt(2) + 4)) + 1/4*(a/b)^(5/8)*arctan((2*x - sqrt(sqrt(2) + 2)*(a/b)^(1/8))/(sqrt(-sqrt(2)
+ 2)*(a/b)^(1/8)))/(a*sqrt(-2*sqrt(2) + 4)) - 1/8*(a/b)^(5/8)*log(x^2 + x*sqrt(sqrt(2) + 2)*(a/b)^(1/8) + (a/
b)^(1/4))/(a*sqrt(2*sqrt(2) + 4)) + 1/8*(a/b)^(5/8)*log(x^2 - x*sqrt(sqrt(2) + 2)*(a/b)^(1/8) + (a/b)^(1/4))/(
a*sqrt(2*sqrt(2) + 4)) + 1/8*(a/b)^(5/8)*log(x^2 + x*sqrt(-sqrt(2) + 2)*(a/b)^(1/8) + (a/b)^(1/4))/(a*sqrt(-2*
sqrt(2) + 4)) - 1/8*(a/b)^(5/8)*log(x^2 - x*sqrt(-sqrt(2) + 2)*(a/b)^(1/8) + (a/b)^(1/4))/(a*sqrt(-2*sqrt(2) +
4))

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maple [C]  time = 0.01, size = 27, normalized size = 0.10 \[ \frac {\ln \left (-\RootOf \left (b \,\textit {\_Z}^{8}+a \right )+x \right )}{8 b \RootOf \left (b \,\textit {\_Z}^{8}+a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/b*sum(1/_R^3*ln(-_R+x),_R=RootOf(_Z^8*b+a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{b x^{8} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/(b*x^8 + a), x)

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mupad [B]  time = 1.17, size = 110, normalized size = 0.41 \[ -\frac {\mathrm {atan}\left (\frac {b^{1/8}\,x}{{\left (-a\right )}^{1/8}}\right )}{4\,{\left (-a\right )}^{3/8}\,b^{5/8}}+\frac {\mathrm {atan}\left (\frac {b^{1/8}\,x\,1{}\mathrm {i}}{{\left (-a\right )}^{1/8}}\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{3/8}\,b^{5/8}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,b^{1/8}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )}{{\left (-a\right )}^{1/8}}\right )\,\left (\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )}{{\left (-a\right )}^{3/8}\,b^{5/8}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,b^{1/8}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )}{{\left (-a\right )}^{1/8}}\right )\,\left (\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )}{{\left (-a\right )}^{3/8}\,b^{5/8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + b*x^8),x)

[Out]

(atan((b^(1/8)*x*1i)/(-a)^(1/8))*1i)/(4*(-a)^(3/8)*b^(5/8)) - atan((b^(1/8)*x)/(-a)^(1/8))/(4*(-a)^(3/8)*b^(5/
8)) + (2^(1/2)*atan((2^(1/2)*b^(1/8)*x*(1/2 - 1i/2))/(-a)^(1/8))*(1/8 + 1i/8))/((-a)^(3/8)*b^(5/8)) + (2^(1/2)
*atan((2^(1/2)*b^(1/8)*x*(1/2 + 1i/2))/(-a)^(1/8))*(1/8 - 1i/8))/((-a)^(3/8)*b^(5/8))

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sympy [A]  time = 0.34, size = 29, normalized size = 0.11 \[ \operatorname {RootSum} {\left (16777216 t^{8} a^{3} b^{5} + 1, \left (t \mapsto t \log {\left (- 32768 t^{5} a^{2} b^{3} + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(16777216*_t**8*a**3*b**5 + 1, Lambda(_t, _t*log(-32768*_t**5*a**2*b**3 + x)))

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