3.182 \(\int \frac{x^{10}}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=98 \[ \frac{63 a^2 x}{8 b^5}-\frac{63 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 b^{11/2}}-\frac{9 x^7}{8 b^2 \left (a+b x^2\right )}-\frac{21 a x^3}{8 b^4}-\frac{x^9}{4 b \left (a+b x^2\right )^2}+\frac{63 x^5}{40 b^3} \]

[Out]

(63*a^2*x)/(8*b^5) - (21*a*x^3)/(8*b^4) + (63*x^5)/(40*b^3) - x^9/(4*b*(a + b*x^2)^2) - (9*x^7)/(8*b^2*(a + b*
x^2)) - (63*a^(5/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*b^(11/2))

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Rubi [A]  time = 0.0407871, antiderivative size = 98, 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} \[ \frac{63 a^2 x}{8 b^5}-\frac{63 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 b^{11/2}}-\frac{9 x^7}{8 b^2 \left (a+b x^2\right )}-\frac{21 a x^3}{8 b^4}-\frac{x^9}{4 b \left (a+b x^2\right )^2}+\frac{63 x^5}{40 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^10/(a + b*x^2)^3,x]

[Out]

(63*a^2*x)/(8*b^5) - (21*a*x^3)/(8*b^4) + (63*x^5)/(40*b^3) - x^9/(4*b*(a + b*x^2)^2) - (9*x^7)/(8*b^2*(a + b*
x^2)) - (63*a^(5/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*b^(11/2))

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^{10}}{\left (a+b x^2\right )^3} \, dx &=-\frac{x^9}{4 b \left (a+b x^2\right )^2}+\frac{9 \int \frac{x^8}{\left (a+b x^2\right )^2} \, dx}{4 b}\\ &=-\frac{x^9}{4 b \left (a+b x^2\right )^2}-\frac{9 x^7}{8 b^2 \left (a+b x^2\right )}+\frac{63 \int \frac{x^6}{a+b x^2} \, dx}{8 b^2}\\ &=-\frac{x^9}{4 b \left (a+b x^2\right )^2}-\frac{9 x^7}{8 b^2 \left (a+b x^2\right )}+\frac{63 \int \left (\frac{a^2}{b^3}-\frac{a x^2}{b^2}+\frac{x^4}{b}-\frac{a^3}{b^3 \left (a+b x^2\right )}\right ) \, dx}{8 b^2}\\ &=\frac{63 a^2 x}{8 b^5}-\frac{21 a x^3}{8 b^4}+\frac{63 x^5}{40 b^3}-\frac{x^9}{4 b \left (a+b x^2\right )^2}-\frac{9 x^7}{8 b^2 \left (a+b x^2\right )}-\frac{\left (63 a^3\right ) \int \frac{1}{a+b x^2} \, dx}{8 b^5}\\ &=\frac{63 a^2 x}{8 b^5}-\frac{21 a x^3}{8 b^4}+\frac{63 x^5}{40 b^3}-\frac{x^9}{4 b \left (a+b x^2\right )^2}-\frac{9 x^7}{8 b^2 \left (a+b x^2\right )}-\frac{63 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 b^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.0472025, size = 88, normalized size = 0.9 \[ \frac{168 a^2 b^2 x^5+525 a^3 b x^3+315 a^4 x-24 a b^3 x^7+8 b^4 x^9}{40 b^5 \left (a+b x^2\right )^2}-\frac{63 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 b^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/(a + b*x^2)^3,x]

[Out]

(315*a^4*x + 525*a^3*b*x^3 + 168*a^2*b^2*x^5 - 24*a*b^3*x^7 + 8*b^4*x^9)/(40*b^5*(a + b*x^2)^2) - (63*a^(5/2)*
ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*b^(11/2))

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Maple [A]  time = 0.008, size = 88, normalized size = 0.9 \begin{align*}{\frac{{x}^{5}}{5\,{b}^{3}}}-{\frac{a{x}^{3}}{{b}^{4}}}+6\,{\frac{{a}^{2}x}{{b}^{5}}}+{\frac{17\,{a}^{3}{x}^{3}}{8\,{b}^{4} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{15\,{a}^{4}x}{8\,{b}^{5} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{63\,{a}^{3}}{8\,{b}^{5}}\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^10/(b*x^2+a)^3,x)

[Out]

1/5*x^5/b^3-a*x^3/b^4+6*a^2*x/b^5+17/8/b^4*a^3/(b*x^2+a)^2*x^3+15/8/b^5*a^4/(b*x^2+a)^2*x-63/8/b^5*a^3/(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^10/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.2842, size = 547, normalized size = 5.58 \begin{align*} \left [\frac{16 \, b^{4} x^{9} - 48 \, a b^{3} x^{7} + 336 \, a^{2} b^{2} x^{5} + 1050 \, a^{3} b x^{3} + 630 \, a^{4} x + 315 \,{\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b x^{2} - 2 \, b x \sqrt{-\frac{a}{b}} - a}{b x^{2} + a}\right )}{80 \,{\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}}, \frac{8 \, b^{4} x^{9} - 24 \, a b^{3} x^{7} + 168 \, a^{2} b^{2} x^{5} + 525 \, a^{3} b x^{3} + 315 \, a^{4} x - 315 \,{\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b x \sqrt{\frac{a}{b}}}{a}\right )}{40 \,{\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/80*(16*b^4*x^9 - 48*a*b^3*x^7 + 336*a^2*b^2*x^5 + 1050*a^3*b*x^3 + 630*a^4*x + 315*(a^2*b^2*x^4 + 2*a^3*b*x
^2 + a^4)*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)))/(b^7*x^4 + 2*a*b^6*x^2 + a^2*b^5), 1/40*
(8*b^4*x^9 - 24*a*b^3*x^7 + 168*a^2*b^2*x^5 + 525*a^3*b*x^3 + 315*a^4*x - 315*(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4
)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a))/(b^7*x^4 + 2*a*b^6*x^2 + a^2*b^5)]

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Sympy [A]  time = 0.610995, size = 144, normalized size = 1.47 \begin{align*} \frac{6 a^{2} x}{b^{5}} - \frac{a x^{3}}{b^{4}} + \frac{63 \sqrt{- \frac{a^{5}}{b^{11}}} \log{\left (x - \frac{b^{5} \sqrt{- \frac{a^{5}}{b^{11}}}}{a^{2}} \right )}}{16} - \frac{63 \sqrt{- \frac{a^{5}}{b^{11}}} \log{\left (x + \frac{b^{5} \sqrt{- \frac{a^{5}}{b^{11}}}}{a^{2}} \right )}}{16} + \frac{15 a^{4} x + 17 a^{3} b x^{3}}{8 a^{2} b^{5} + 16 a b^{6} x^{2} + 8 b^{7} x^{4}} + \frac{x^{5}}{5 b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10/(b*x**2+a)**3,x)

[Out]

6*a**2*x/b**5 - a*x**3/b**4 + 63*sqrt(-a**5/b**11)*log(x - b**5*sqrt(-a**5/b**11)/a**2)/16 - 63*sqrt(-a**5/b**
11)*log(x + b**5*sqrt(-a**5/b**11)/a**2)/16 + (15*a**4*x + 17*a**3*b*x**3)/(8*a**2*b**5 + 16*a*b**6*x**2 + 8*b
**7*x**4) + x**5/(5*b**3)

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Giac [A]  time = 2.73309, size = 113, normalized size = 1.15 \begin{align*} -\frac{63 \, a^{3} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} b^{5}} + \frac{17 \, a^{3} b x^{3} + 15 \, a^{4} x}{8 \,{\left (b x^{2} + a\right )}^{2} b^{5}} + \frac{b^{12} x^{5} - 5 \, a b^{11} x^{3} + 30 \, a^{2} b^{10} x}{5 \, b^{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(b*x^2+a)^3,x, algorithm="giac")

[Out]

-63/8*a^3*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5) + 1/8*(17*a^3*b*x^3 + 15*a^4*x)/((b*x^2 + a)^2*b^5) + 1/5*(b^1
2*x^5 - 5*a*b^11*x^3 + 30*a^2*b^10*x)/b^15