3.1451 \(\int \frac{x^9}{a+b x^8} \, dx\)

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

[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))

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Rubi [A]  time = 0.193976, antiderivative size = 203, normalized size of antiderivative = 1., 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 = {275, 321, 211, 1165, 628, 1162, 617, 204} \[ \frac{\sqrt [4]{a} \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt{a}+\sqrt{b} x^4\right )}{8 \sqrt{2} b^{5/4}}-\frac{\sqrt [4]{a} \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x^2+\sqrt{a}+\sqrt{b} x^4\right )}{8 \sqrt{2} b^{5/4}}+\frac{\sqrt [4]{a} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}\right )}{4 \sqrt{2} b^{5/4}}-\frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x^2}{\sqrt [4]{a}}+1\right )}{4 \sqrt{2} b^{5/4}}+\frac{x^2}{2 b} \]

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 275

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 321

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 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 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]

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 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 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])

Rubi steps

\begin{align*} \int \frac{x^9}{a+b x^8} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^4}{a+b x^4} \, dx,x,x^2\right )\\ &=\frac{x^2}{2 b}-\frac{a \operatorname{Subst}\left (\int \frac{1}{a+b x^4} \, dx,x,x^2\right )}{2 b}\\ &=\frac{x^2}{2 b}-\frac{\sqrt{a} \operatorname{Subst}\left (\int \frac{\sqrt{a}-\sqrt{b} x^2}{a+b x^4} \, dx,x,x^2\right )}{4 b}-\frac{\sqrt{a} \operatorname{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} \operatorname{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} \operatorname{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} \operatorname{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} \operatorname{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} \operatorname{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} \operatorname{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*}

Mathematica [A]  time = 0.205206, size = 361, normalized size = 1.78 \[ \frac{-\sqrt{2} \sqrt [4]{a} \log \left (-2 \sqrt [8]{a} \sqrt [8]{b} x \sin \left (\frac{\pi }{8}\right )+\sqrt [4]{a}+\sqrt [4]{b} x^2\right )-\sqrt{2} \sqrt [4]{a} \log \left (2 \sqrt [8]{a} \sqrt [8]{b} x \sin \left (\frac{\pi }{8}\right )+\sqrt [4]{a}+\sqrt [4]{b} x^2\right )+\sqrt{2} \sqrt [4]{a} \log \left (-2 \sqrt [8]{a} \sqrt [8]{b} x \cos \left (\frac{\pi }{8}\right )+\sqrt [4]{a}+\sqrt [4]{b} x^2\right )+\sqrt{2} \sqrt [4]{a} \log \left (2 \sqrt [8]{a} \sqrt [8]{b} x \cos \left (\frac{\pi }{8}\right )+\sqrt [4]{a}+\sqrt [4]{b} x^2\right )-2 \sqrt{2} \sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt [8]{b} x \sec \left (\frac{\pi }{8}\right )}{\sqrt [8]{a}}-\tan \left (\frac{\pi }{8}\right )\right )+2 \sqrt{2} \sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt [8]{b} x \sec \left (\frac{\pi }{8}\right )}{\sqrt [8]{a}}+\tan \left (\frac{\pi }{8}\right )\right )+2 \sqrt{2} \sqrt [4]{a} \tan ^{-1}\left (\cot \left (\frac{\pi }{8}\right )-\frac{\sqrt [8]{b} x \csc \left (\frac{\pi }{8}\right )}{\sqrt [8]{a}}\right )+2 \sqrt{2} \sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt [8]{b} x \csc \left (\frac{\pi }{8}\right )}{\sqrt [8]{a}}+\cot \left (\frac{\pi }{8}\right )\right )+8 \sqrt [4]{b} x^2}{16 b^{5/4}} \]

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 = 0.006, size = 144, normalized size = 0.7 \begin{align*}{\frac{{x}^{2}}{2\,b}}-{\frac{\sqrt{2}}{16\,b}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ({x}^{4}+\sqrt [4]{{\frac{a}{b}}}{x}^{2}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ({x}^{4}-\sqrt [4]{{\frac{a}{b}}}{x}^{2}\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ) }-{\frac{\sqrt{2}}{8\,b}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({{x}^{2}\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }-{\frac{\sqrt{2}}{8\,b}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({{x}^{2}\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(x^9/(b*x^8+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.36145, size = 293, normalized size = 1.44 \begin{align*} -\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{align*}

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 = 0.370487, size = 27, normalized size = 0.13 \begin{align*} \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{align*}

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 = 1.15054, size = 247, normalized size = 1.22 \begin{align*} \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{align*}

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