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

Optimal. Leaf size=79 \[ \frac{7 a^2 x}{2 b^4}-\frac{7 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{9/2}}-\frac{7 a x^3}{6 b^3}-\frac{x^7}{2 b \left (a+b x^2\right )}+\frac{7 x^5}{10 b^2} \]

[Out]

(7*a^2*x)/(2*b^4) - (7*a*x^3)/(6*b^3) + (7*x^5)/(10*b^2) - x^7/(2*b*(a + b*x^2)) - (7*a^(5/2)*ArcTan[(Sqrt[b]*
x)/Sqrt[a]])/(2*b^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0452181, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {28, 288, 302, 205} \[ \frac{7 a^2 x}{2 b^4}-\frac{7 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{9/2}}-\frac{7 a x^3}{6 b^3}-\frac{x^7}{2 b \left (a+b x^2\right )}+\frac{7 x^5}{10 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^2*x)/(2*b^4) - (7*a*x^3)/(6*b^3) + (7*x^5)/(10*b^2) - x^7/(2*b*(a + b*x^2)) - (7*a^(5/2)*ArcTan[(Sqrt[b]*
x)/Sqrt[a]])/(2*b^(9/2))

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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]

Rubi steps

\begin{align*} \int \frac{x^8}{a^2+2 a b x^2+b^2 x^4} \, dx &=b^2 \int \frac{x^8}{\left (a b+b^2 x^2\right )^2} \, dx\\ &=-\frac{x^7}{2 b \left (a+b x^2\right )}+\frac{7}{2} \int \frac{x^6}{a b+b^2 x^2} \, dx\\ &=-\frac{x^7}{2 b \left (a+b x^2\right )}+\frac{7}{2} \int \left (\frac{a^2}{b^4}-\frac{a x^2}{b^3}+\frac{x^4}{b^2}-\frac{a^3}{b^3 \left (a b+b^2 x^2\right )}\right ) \, dx\\ &=\frac{7 a^2 x}{2 b^4}-\frac{7 a x^3}{6 b^3}+\frac{7 x^5}{10 b^2}-\frac{x^7}{2 b \left (a+b x^2\right )}-\frac{\left (7 a^3\right ) \int \frac{1}{a b+b^2 x^2} \, dx}{2 b^3}\\ &=\frac{7 a^2 x}{2 b^4}-\frac{7 a x^3}{6 b^3}+\frac{7 x^5}{10 b^2}-\frac{x^7}{2 b \left (a+b x^2\right )}-\frac{7 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.045999, size = 71, normalized size = 0.9 \[ \frac{x \left (\frac{15 a^3}{a+b x^2}+90 a^2-20 a b x^2+6 b^2 x^4\right )}{30 b^4}-\frac{7 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(90*a^2 - 20*a*b*x^2 + 6*b^2*x^4 + (15*a^3)/(a + b*x^2)))/(30*b^4) - (7*a^(5/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]]
)/(2*b^(9/2))

________________________________________________________________________________________

Maple [A]  time = 0.051, size = 68, normalized size = 0.9 \begin{align*}{\frac{{x}^{5}}{5\,{b}^{2}}}-{\frac{2\,a{x}^{3}}{3\,{b}^{3}}}+3\,{\frac{{a}^{2}x}{{b}^{4}}}+{\frac{x{a}^{3}}{2\,{b}^{4} \left ( b{x}^{2}+a \right ) }}-{\frac{7\,{a}^{3}}{2\,{b}^{4}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5/b^2-2/3*a*x^3/b^3+3*a^2*x/b^4+1/2/b^4*a^3*x/(b*x^2+a)-7/2/b^4*a^3/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.76173, size = 409, normalized size = 5.18 \begin{align*} \left [\frac{12 \, b^{3} x^{7} - 28 \, a b^{2} x^{5} + 140 \, a^{2} b x^{3} + 210 \, a^{3} x + 105 \,{\left (a^{2} b x^{2} + a^{3}\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b x^{2} - 2 \, b x \sqrt{-\frac{a}{b}} - a}{b x^{2} + a}\right )}{60 \,{\left (b^{5} x^{2} + a b^{4}\right )}}, \frac{6 \, b^{3} x^{7} - 14 \, a b^{2} x^{5} + 70 \, a^{2} b x^{3} + 105 \, a^{3} x - 105 \,{\left (a^{2} b x^{2} + a^{3}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b x \sqrt{\frac{a}{b}}}{a}\right )}{30 \,{\left (b^{5} x^{2} + a b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(12*b^3*x^7 - 28*a*b^2*x^5 + 140*a^2*b*x^3 + 210*a^3*x + 105*(a^2*b*x^2 + a^3)*sqrt(-a/b)*log((b*x^2 - 2
*b*x*sqrt(-a/b) - a)/(b*x^2 + a)))/(b^5*x^2 + a*b^4), 1/30*(6*b^3*x^7 - 14*a*b^2*x^5 + 70*a^2*b*x^3 + 105*a^3*
x - 105*(a^2*b*x^2 + a^3)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a))/(b^5*x^2 + a*b^4)]

________________________________________________________________________________________

Sympy [A]  time = 0.461368, size = 124, normalized size = 1.57 \begin{align*} \frac{a^{3} x}{2 a b^{4} + 2 b^{5} x^{2}} + \frac{3 a^{2} x}{b^{4}} - \frac{2 a x^{3}}{3 b^{3}} + \frac{7 \sqrt{- \frac{a^{5}}{b^{9}}} \log{\left (x - \frac{b^{4} \sqrt{- \frac{a^{5}}{b^{9}}}}{a^{2}} \right )}}{4} - \frac{7 \sqrt{- \frac{a^{5}}{b^{9}}} \log{\left (x + \frac{b^{4} \sqrt{- \frac{a^{5}}{b^{9}}}}{a^{2}} \right )}}{4} + \frac{x^{5}}{5 b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*x/(2*a*b**4 + 2*b**5*x**2) + 3*a**2*x/b**4 - 2*a*x**3/(3*b**3) + 7*sqrt(-a**5/b**9)*log(x - b**4*sqrt(-a*
*5/b**9)/a**2)/4 - 7*sqrt(-a**5/b**9)*log(x + b**4*sqrt(-a**5/b**9)/a**2)/4 + x**5/(5*b**2)

________________________________________________________________________________________

Giac [A]  time = 1.18509, size = 99, normalized size = 1.25 \begin{align*} -\frac{7 \, a^{3} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} b^{4}} + \frac{a^{3} x}{2 \,{\left (b x^{2} + a\right )} b^{4}} + \frac{3 \, b^{8} x^{5} - 10 \, a b^{7} x^{3} + 45 \, a^{2} b^{6} x}{15 \, b^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/2*a^3*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^4) + 1/2*a^3*x/((b*x^2 + a)*b^4) + 1/15*(3*b^8*x^5 - 10*a*b^7*x^3
+ 45*a^2*b^6*x)/b^10