3.15.51 \(\int \frac {x^9}{a+b x^8} \, dx\) [1451]

Optimal. Leaf size=203 \[ \frac {x^2}{2 b}+\frac {\sqrt [4]{a} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{5/4}}-\frac {\sqrt [4]{a} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{5/4}}+\frac {\sqrt [4]{a} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} b^{5/4}}-\frac {\sqrt [4]{a} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} b^{5/4}} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {281, 327, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {\sqrt [4]{a} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{5/4}}-\frac {\sqrt [4]{a} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} b^{5/4}}+\frac {\sqrt [4]{a} \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {a}+\sqrt {b} x^4\right )}{8 \sqrt {2} b^{5/4}}-\frac {\sqrt [4]{a} \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {a}+\sqrt {b} x^4\right )}{8 \sqrt {2} b^{5/4}}+\frac {x^2}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9/(a + b*x^8),x]

[Out]

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

Rule 210

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

Rule 217

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 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 631

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 642

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 1176

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 1179

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^9}{a+b x^8} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x^4}{a+b x^4} \, dx,x,x^2\right )\\ &=\frac {x^2}{2 b}-\frac {a \text {Subst}\left (\int \frac {1}{a+b x^4} \, dx,x,x^2\right )}{2 b}\\ &=\frac {x^2}{2 b}-\frac {\sqrt {a} \text {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,x^2\right )}{4 b}-\frac {\sqrt {a} \text {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,x^2\right )}{4 b}\\ &=\frac {x^2}{2 b}-\frac {\sqrt {a} \text {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,x^2\right )}{8 b^{3/2}}-\frac {\sqrt {a} \text {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,x^2\right )}{8 b^{3/2}}+\frac {\sqrt [4]{a} \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,x^2\right )}{8 \sqrt {2} b^{5/4}}+\frac {\sqrt [4]{a} \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,x^2\right )}{8 \sqrt {2} b^{5/4}}\\ &=\frac {x^2}{2 b}+\frac {\sqrt [4]{a} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} b^{5/4}}-\frac {\sqrt [4]{a} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} b^{5/4}}-\frac {\sqrt [4]{a} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{5/4}}+\frac {\sqrt [4]{a} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{5/4}}\\ &=\frac {x^2}{2 b}+\frac {\sqrt [4]{a} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{5/4}}-\frac {\sqrt [4]{a} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt {2} b^{5/4}}+\frac {\sqrt [4]{a} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} b^{5/4}}-\frac {\sqrt [4]{a} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt {b} x^4\right )}{8 \sqrt {2} b^{5/4}}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 361, normalized size = 1.78 \begin {gather*} \frac {8 \sqrt [4]{b} x^2+2 \sqrt {2} \sqrt [4]{a} \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 \sqrt {2} \sqrt [4]{a} \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 \sqrt {2} \sqrt [4]{a} \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 \sqrt {2} \sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [8]{b} x \sec \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}+\tan \left (\frac {\pi }{8}\right )\right )+\sqrt {2} \sqrt [4]{a} \log \left (\sqrt [4]{a}+\sqrt [4]{b} x^2-2 \sqrt [8]{a} \sqrt [8]{b} x \cos \left (\frac {\pi }{8}\right )\right )+\sqrt {2} \sqrt [4]{a} \log \left (\sqrt [4]{a}+\sqrt [4]{b} x^2+2 \sqrt [8]{a} \sqrt [8]{b} x \cos \left (\frac {\pi }{8}\right )\right )-\sqrt {2} \sqrt [4]{a} \log \left (\sqrt [4]{a}+\sqrt [4]{b} x^2-2 \sqrt [8]{a} \sqrt [8]{b} x \sin \left (\frac {\pi }{8}\right )\right )-\sqrt {2} \sqrt [4]{a} \log \left (\sqrt [4]{a}+\sqrt [4]{b} x^2+2 \sqrt [8]{a} \sqrt [8]{b} x \sin \left (\frac {\pi }{8}\right )\right )}{16 b^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9/(a + b*x^8),x]

[Out]

(8*b^(1/4)*x^2 + 2*Sqrt[2]*a^(1/4)*ArcTan[Cot[Pi/8] - (b^(1/8)*x*Csc[Pi/8])/a^(1/8)] + 2*Sqrt[2]*a^(1/4)*ArcTa
n[Cot[Pi/8] + (b^(1/8)*x*Csc[Pi/8])/a^(1/8)] - 2*Sqrt[2]*a^(1/4)*ArcTan[(b^(1/8)*x*Sec[Pi/8])/a^(1/8) - Tan[Pi
/8]] + 2*Sqrt[2]*a^(1/4)*ArcTan[(b^(1/8)*x*Sec[Pi/8])/a^(1/8) + Tan[Pi/8]] + Sqrt[2]*a^(1/4)*Log[a^(1/4) + b^(
1/4)*x^2 - 2*a^(1/8)*b^(1/8)*x*Cos[Pi/8]] + Sqrt[2]*a^(1/4)*Log[a^(1/4) + b^(1/4)*x^2 + 2*a^(1/8)*b^(1/8)*x*Co
s[Pi/8]] - Sqrt[2]*a^(1/4)*Log[a^(1/4) + b^(1/4)*x^2 - 2*a^(1/8)*b^(1/8)*x*Sin[Pi/8]] - Sqrt[2]*a^(1/4)*Log[a^
(1/4) + b^(1/4)*x^2 + 2*a^(1/8)*b^(1/8)*x*Sin[Pi/8]])/(16*b^(5/4))

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Maple [A]
time = 0.18, size = 119, normalized size = 0.59

method	result	size
risch	\(\frac {x^{2}}{2 b}+\frac {\munderset {\textit {\_R} =\RootOf \left (b \,\textit {\_Z}^{4}+a \right )}{\sum }\textit {\_R} \ln \left (x^{2}-\textit {\_R} \right )}{8 b}\)	\(36\)
default	\(\frac {x^{2}}{2 b}-\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{4}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x^{2} \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{4}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x^{2} \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x^{2}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x^{2}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^8+a),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 190, normalized size = 0.94 \begin {gather*} \frac {x^{2}}{2 \, b} - \frac {\frac {2 \, \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} a^{\frac {1}{4}} \log \left (\sqrt {b} x^{4} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x^{2} + \sqrt {a}\right )}{b^{\frac {1}{4}}} - \frac {\sqrt {2} a^{\frac {1}{4}} \log \left (\sqrt {b} x^{4} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x^{2} + \sqrt {a}\right )}{b^{\frac {1}{4}}}}{16 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^8+a),x, algorithm="maxima")

[Out]

1/2*x^2/b - 1/16*(2*sqrt(2)*sqrt(a)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x^2 + sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*
sqrt(b)))/sqrt(sqrt(a)*sqrt(b)) + 2*sqrt(2)*sqrt(a)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4
))/sqrt(sqrt(a)*sqrt(b)))/sqrt(sqrt(a)*sqrt(b)) + sqrt(2)*a^(1/4)*log(sqrt(b)*x^4 + sqrt(2)*a^(1/4)*b^(1/4)*x^
2 + sqrt(a))/b^(1/4) - sqrt(2)*a^(1/4)*log(sqrt(b)*x^4 - sqrt(2)*a^(1/4)*b^(1/4)*x^2 + sqrt(a))/b^(1/4))/b

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Fricas [A]
time = 0.38, size = 127, normalized size = 0.63 \begin {gather*} -\frac {4 \, b \left (-\frac {a}{b^{5}}\right )^{\frac {1}{4}} \arctan \left (-\frac {b^{4} x^{2} \left (-\frac {a}{b^{5}}\right )^{\frac {3}{4}} - \sqrt {x^{4} + b^{2} \sqrt {-\frac {a}{b^{5}}}} b^{4} \left (-\frac {a}{b^{5}}\right )^{\frac {3}{4}}}{a}\right ) + b \left (-\frac {a}{b^{5}}\right )^{\frac {1}{4}} \log \left (x^{2} + b \left (-\frac {a}{b^{5}}\right )^{\frac {1}{4}}\right ) - b \left (-\frac {a}{b^{5}}\right )^{\frac {1}{4}} \log \left (x^{2} - b \left (-\frac {a}{b^{5}}\right )^{\frac {1}{4}}\right ) - 4 \, x^{2}}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^8+a),x, algorithm="fricas")

[Out]

-1/8*(4*b*(-a/b^5)^(1/4)*arctan(-(b^4*x^2*(-a/b^5)^(3/4) - sqrt(x^4 + b^2*sqrt(-a/b^5))*b^4*(-a/b^5)^(3/4))/a)
+ b*(-a/b^5)^(1/4)*log(x^2 + b*(-a/b^5)^(1/4)) - b*(-a/b^5)^(1/4)*log(x^2 - b*(-a/b^5)^(1/4)) - 4*x^2)/b

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Sympy [A]
time = 0.10, size = 27, normalized size = 0.13 \begin {gather*} \operatorname {RootSum} {\left (4096 t^{4} b^{5} + a, \left ( t \mapsto t \log {\left (- 8 t b + x^{2} \right )} \right )\right )} + \frac {x^{2}}{2 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b*x**8+a),x)

[Out]

RootSum(4096*_t**4*b**5 + a, Lambda(_t, _t*log(-8*_t*b + x**2))) + x**2/(2*b)

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Giac [A]
time = 1.45, size = 183, normalized size = 0.90 \begin {gather*} \frac {x^{2}}{2 \, b} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x^{2} + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, b^{2}} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (2 \, x^{2} - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, b^{2}} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \log \left (x^{4} + \sqrt {2} x^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{16 \, b^{2}} + \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {1}{4}} \log \left (x^{4} - \sqrt {2} x^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{16 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^8+a),x, algorithm="giac")

[Out]

1/2*x^2/b - 1/8*sqrt(2)*(a*b^3)^(1/4)*arctan(1/2*sqrt(2)*(2*x^2 + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/b^2 - 1/8*
sqrt(2)*(a*b^3)^(1/4)*arctan(1/2*sqrt(2)*(2*x^2 - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/b^2 - 1/16*sqrt(2)*(a*b^3)
^(1/4)*log(x^4 + sqrt(2)*x^2*(a/b)^(1/4) + sqrt(a/b))/b^2 + 1/16*sqrt(2)*(a*b^3)^(1/4)*log(x^4 - sqrt(2)*x^2*(
a/b)^(1/4) + sqrt(a/b))/b^2

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Mupad [B]
time = 1.14, size = 55, normalized size = 0.27 \begin {gather*} \frac {x^2}{2\,b}-\frac {{\left (-a\right )}^{1/4}\,\mathrm {atan}\left (\frac {b^{1/4}\,x^2}{{\left (-a\right )}^{1/4}}\right )}{4\,b^{5/4}}-\frac {{\left (-a\right )}^{1/4}\,\mathrm {atanh}\left (\frac {b^{1/4}\,x^2}{{\left (-a\right )}^{1/4}}\right )}{4\,b^{5/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(a + b*x^8),x)

[Out]

x^2/(2*b) - ((-a)^(1/4)*atan((b^(1/4)*x^2)/(-a)^(1/4)))/(4*b^(5/4)) - ((-a)^(1/4)*atanh((b^(1/4)*x^2)/(-a)^(1/
4)))/(4*b^(5/4))

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