3.183 \(\int \frac{x^8}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=85 \[ \frac{35 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 b^{9/2}}-\frac{7 x^5}{8 b^2 \left (a+b x^2\right )}-\frac{35 a x}{8 b^4}-\frac{x^7}{4 b \left (a+b x^2\right )^2}+\frac{35 x^3}{24 b^3} \]

[Out]

(-35*a*x)/(8*b^4) + (35*x^3)/(24*b^3) - x^7/(4*b*(a + b*x^2)^2) - (7*x^5)/(8*b^2*(a + b*x^2)) + (35*a^(3/2)*Ar
cTan[(Sqrt[b]*x)/Sqrt[a]])/(8*b^(9/2))

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Rubi [A]  time = 0.0351929, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {288, 302, 205} \[ \frac{35 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 b^{9/2}}-\frac{7 x^5}{8 b^2 \left (a+b x^2\right )}-\frac{35 a x}{8 b^4}-\frac{x^7}{4 b \left (a+b x^2\right )^2}+\frac{35 x^3}{24 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^8/(a + b*x^2)^3,x]

[Out]

(-35*a*x)/(8*b^4) + (35*x^3)/(24*b^3) - x^7/(4*b*(a + b*x^2)^2) - (7*x^5)/(8*b^2*(a + b*x^2)) + (35*a^(3/2)*Ar
cTan[(Sqrt[b]*x)/Sqrt[a]])/(8*b^(9/2))

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

Mathematica [A]  time = 0.0447759, size = 77, normalized size = 0.91 \[ \frac{35 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 b^{9/2}}-\frac{175 a^2 b x^3+105 a^3 x+56 a b^2 x^5-8 b^3 x^7}{24 b^4 \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(a + b*x^2)^3,x]

[Out]

-(105*a^3*x + 175*a^2*b*x^3 + 56*a*b^2*x^5 - 8*b^3*x^7)/(24*b^4*(a + b*x^2)^2) + (35*a^(3/2)*ArcTan[(Sqrt[b]*x
)/Sqrt[a]])/(8*b^(9/2))

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Maple [A]  time = 0.01, size = 77, normalized size = 0.9 \begin{align*}{\frac{{x}^{3}}{3\,{b}^{3}}}-3\,{\frac{ax}{{b}^{4}}}-{\frac{13\,{a}^{2}{x}^{3}}{8\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{11\,{a}^{3}x}{8\,{b}^{4} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{35\,{a}^{2}}{8\,{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*x^2+a)^3,x)

[Out]

1/3*x^3/b^3-3*a*x/b^4-13/8/b^3*a^2/(b*x^2+a)^2*x^3-11/8/b^4*a^3/(b*x^2+a)^2*x+35/8/b^4*a^2/(a*b)^(1/2)*arctan(
b*x/(a*b)^(1/2))

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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^8/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.29079, size = 493, normalized size = 5.8 \begin{align*} \left [\frac{16 \, b^{3} x^{7} - 112 \, a b^{2} x^{5} - 350 \, a^{2} b x^{3} - 210 \, a^{3} x + 105 \,{\left (a b^{2} x^{4} + 2 \, 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 )}{48 \,{\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}, \frac{8 \, b^{3} x^{7} - 56 \, a b^{2} x^{5} - 175 \, a^{2} b x^{3} - 105 \, a^{3} x + 105 \,{\left (a b^{2} x^{4} + 2 \, a^{2} b x^{2} + a^{3}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b x \sqrt{\frac{a}{b}}}{a}\right )}{24 \,{\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/48*(16*b^3*x^7 - 112*a*b^2*x^5 - 350*a^2*b*x^3 - 210*a^3*x + 105*(a*b^2*x^4 + 2*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^6*x^4 + 2*a*b^5*x^2 + a^2*b^4), 1/24*(8*b^3*x^7 - 56*a*b^
2*x^5 - 175*a^2*b*x^3 - 105*a^3*x + 105*(a*b^2*x^4 + 2*a^2*b*x^2 + a^3)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a))/(b^
6*x^4 + 2*a*b^5*x^2 + a^2*b^4)]

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Sympy [A]  time = 0.579039, size = 131, normalized size = 1.54 \begin{align*} - \frac{3 a x}{b^{4}} - \frac{35 \sqrt{- \frac{a^{3}}{b^{9}}} \log{\left (x - \frac{b^{4} \sqrt{- \frac{a^{3}}{b^{9}}}}{a} \right )}}{16} + \frac{35 \sqrt{- \frac{a^{3}}{b^{9}}} \log{\left (x + \frac{b^{4} \sqrt{- \frac{a^{3}}{b^{9}}}}{a} \right )}}{16} - \frac{11 a^{3} x + 13 a^{2} b x^{3}}{8 a^{2} b^{4} + 16 a b^{5} x^{2} + 8 b^{6} x^{4}} + \frac{x^{3}}{3 b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(b*x**2+a)**3,x)

[Out]

-3*a*x/b**4 - 35*sqrt(-a**3/b**9)*log(x - b**4*sqrt(-a**3/b**9)/a)/16 + 35*sqrt(-a**3/b**9)*log(x + b**4*sqrt(
-a**3/b**9)/a)/16 - (11*a**3*x + 13*a**2*b*x**3)/(8*a**2*b**4 + 16*a*b**5*x**2 + 8*b**6*x**4) + x**3/(3*b**3)

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Giac [A]  time = 1.53496, size = 99, normalized size = 1.16 \begin{align*} \frac{35 \, a^{2} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} b^{4}} - \frac{13 \, a^{2} b x^{3} + 11 \, a^{3} x}{8 \,{\left (b x^{2} + a\right )}^{2} b^{4}} + \frac{b^{6} x^{3} - 9 \, a b^{5} x}{3 \, b^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^2+a)^3,x, algorithm="giac")

[Out]

35/8*a^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^4) - 1/8*(13*a^2*b*x^3 + 11*a^3*x)/((b*x^2 + a)^2*b^4) + 1/3*(b^6*
x^3 - 9*a*b^5*x)/b^9