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

Optimal. Leaf size=93 \[ -\frac{7 x^5}{24 b^2 \left (a+b x^2\right )^2}-\frac{35 x^3}{48 b^3 \left (a+b x^2\right )}-\frac{35 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 b^{9/2}}-\frac{x^7}{6 b \left (a+b x^2\right )^3}+\frac{35 x}{16 b^4} \]

[Out]

(35*x)/(16*b^4) - x^7/(6*b*(a + b*x^2)^3) - (7*x^5)/(24*b^2*(a + b*x^2)^2) - (35*x^3)/(48*b^3*(a + b*x^2)) - (
35*Sqrt[a]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(16*b^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0466813, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {28, 288, 321, 205} \[ -\frac{7 x^5}{24 b^2 \left (a+b x^2\right )^2}-\frac{35 x^3}{48 b^3 \left (a+b x^2\right )}-\frac{35 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 b^{9/2}}-\frac{x^7}{6 b \left (a+b x^2\right )^3}+\frac{35 x}{16 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*x)/(16*b^4) - x^7/(6*b*(a + b*x^2)^3) - (7*x^5)/(24*b^2*(a + b*x^2)^2) - (35*x^3)/(48*b^3*(a + b*x^2)) - (
35*Sqrt[a]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(16*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 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 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}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac{x^8}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=-\frac{x^7}{6 b \left (a+b x^2\right )^3}+\frac{1}{6} \left (7 b^2\right ) \int \frac{x^6}{\left (a b+b^2 x^2\right )^3} \, dx\\ &=-\frac{x^7}{6 b \left (a+b x^2\right )^3}-\frac{7 x^5}{24 b^2 \left (a+b x^2\right )^2}+\frac{35}{24} \int \frac{x^4}{\left (a b+b^2 x^2\right )^2} \, dx\\ &=-\frac{x^7}{6 b \left (a+b x^2\right )^3}-\frac{7 x^5}{24 b^2 \left (a+b x^2\right )^2}-\frac{35 x^3}{48 b^3 \left (a+b x^2\right )}+\frac{35 \int \frac{x^2}{a b+b^2 x^2} \, dx}{16 b^2}\\ &=\frac{35 x}{16 b^4}-\frac{x^7}{6 b \left (a+b x^2\right )^3}-\frac{7 x^5}{24 b^2 \left (a+b x^2\right )^2}-\frac{35 x^3}{48 b^3 \left (a+b x^2\right )}-\frac{(35 a) \int \frac{1}{a b+b^2 x^2} \, dx}{16 b^3}\\ &=\frac{35 x}{16 b^4}-\frac{x^7}{6 b \left (a+b x^2\right )^3}-\frac{7 x^5}{24 b^2 \left (a+b x^2\right )^2}-\frac{35 x^3}{48 b^3 \left (a+b x^2\right )}-\frac{35 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0438906, size = 77, normalized size = 0.83 \[ \frac{280 a^2 b x^3+105 a^3 x+231 a b^2 x^5+48 b^3 x^7}{48 b^4 \left (a+b x^2\right )^3}-\frac{35 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 b^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(105*a^3*x + 280*a^2*b*x^3 + 231*a*b^2*x^5 + 48*b^3*x^7)/(48*b^4*(a + b*x^2)^3) - (35*Sqrt[a]*ArcTan[(Sqrt[b]*
x)/Sqrt[a]])/(16*b^(9/2))

________________________________________________________________________________________

Maple [A]  time = 0.054, size = 83, normalized size = 0.9 \begin{align*}{\frac{x}{{b}^{4}}}+{\frac{29\,a{x}^{5}}{16\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{3}}}+{\frac{17\,{a}^{2}{x}^{3}}{6\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{3}}}+{\frac{19\,x{a}^{3}}{16\,{b}^{4} \left ( b{x}^{2}+a \right ) ^{3}}}-{\frac{35\,a}{16\,{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)^2,x)

[Out]

x/b^4+29/16/b^2*a/(b*x^2+a)^3*x^5+17/6/b^3*a^2/(b*x^2+a)^3*x^3+19/16/b^4*a^3/(b*x^2+a)^3*x-35/16/b^4*a/(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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.73532, size = 571, normalized size = 6.14 \begin{align*} \left [\frac{96 \, b^{3} x^{7} + 462 \, a b^{2} x^{5} + 560 \, a^{2} b x^{3} + 210 \, a^{3} x + 105 \,{\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, 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 )}{96 \,{\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}, \frac{48 \, b^{3} x^{7} + 231 \, a b^{2} x^{5} + 280 \, a^{2} b x^{3} + 105 \, a^{3} x - 105 \,{\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b x \sqrt{\frac{a}{b}}}{a}\right )}{48 \,{\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} 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)^2,x, algorithm="fricas")

[Out]

[1/96*(96*b^3*x^7 + 462*a*b^2*x^5 + 560*a^2*b*x^3 + 210*a^3*x + 105*(b^3*x^6 + 3*a*b^2*x^4 + 3*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^7*x^6 + 3*a*b^6*x^4 + 3*a^2*b^5*x^2 + a^3*b^4
), 1/48*(48*b^3*x^7 + 231*a*b^2*x^5 + 280*a^2*b*x^3 + 105*a^3*x - 105*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a
^3)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a))/(b^7*x^6 + 3*a*b^6*x^4 + 3*a^2*b^5*x^2 + a^3*b^4)]

________________________________________________________________________________________

Sympy [A]  time = 0.744338, size = 131, normalized size = 1.41 \begin{align*} \frac{35 \sqrt{- \frac{a}{b^{9}}} \log{\left (- b^{4} \sqrt{- \frac{a}{b^{9}}} + x \right )}}{32} - \frac{35 \sqrt{- \frac{a}{b^{9}}} \log{\left (b^{4} \sqrt{- \frac{a}{b^{9}}} + x \right )}}{32} + \frac{57 a^{3} x + 136 a^{2} b x^{3} + 87 a b^{2} x^{5}}{48 a^{3} b^{4} + 144 a^{2} b^{5} x^{2} + 144 a b^{6} x^{4} + 48 b^{7} x^{6}} + \frac{x}{b^{4}} \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)**2,x)

[Out]

35*sqrt(-a/b**9)*log(-b**4*sqrt(-a/b**9) + x)/32 - 35*sqrt(-a/b**9)*log(b**4*sqrt(-a/b**9) + x)/32 + (57*a**3*
x + 136*a**2*b*x**3 + 87*a*b**2*x**5)/(48*a**3*b**4 + 144*a**2*b**5*x**2 + 144*a*b**6*x**4 + 48*b**7*x**6) + x
/b**4

________________________________________________________________________________________

Giac [A]  time = 1.1114, size = 88, normalized size = 0.95 \begin{align*} -\frac{35 \, a \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{16 \, \sqrt{a b} b^{4}} + \frac{x}{b^{4}} + \frac{87 \, a b^{2} x^{5} + 136 \, a^{2} b x^{3} + 57 \, a^{3} x}{48 \,{\left (b x^{2} + a\right )}^{3} b^{4}} \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)^2,x, algorithm="giac")

[Out]

-35/16*a*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^4) + x/b^4 + 1/48*(87*a*b^2*x^5 + 136*a^2*b*x^3 + 57*a^3*x)/((b*x^
2 + a)^3*b^4)