3.152 \(\int \frac{x^6}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=66 \[ \frac{5 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{7/2}}-\frac{5 a x}{2 b^3}-\frac{x^5}{2 b \left (a+b x^2\right )}+\frac{5 x^3}{6 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0269408, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {288, 302, 205} \[ \frac{5 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{7/2}}-\frac{5 a x}{2 b^3}-\frac{x^5}{2 b \left (a+b x^2\right )}+\frac{5 x^3}{6 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a + b*x^2)^2,x]

[Out]

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

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

Mathematica [A]  time = 0.0435879, size = 60, normalized size = 0.91 \[ \frac{x \left (-\frac{3 a^2}{a+b x^2}-12 a+2 b x^2\right )}{6 b^3}+\frac{5 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a + b*x^2)^2,x]

[Out]

(x*(-12*a + 2*b*x^2 - (3*a^2)/(a + b*x^2)))/(6*b^3) + (5*a^(3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*b^(7/2))

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Maple [A]  time = 0.008, size = 57, normalized size = 0.9 \begin{align*}{\frac{{x}^{3}}{3\,{b}^{2}}}-2\,{\frac{ax}{{b}^{3}}}-{\frac{{a}^{2}x}{2\,{b}^{3} \left ( b{x}^{2}+a \right ) }}+{\frac{5\,{a}^{2}}{2\,{b}^{3}}\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^6/(b*x^2+a)^2,x)

[Out]

1/3*x^3/b^2-2*a*x/b^3-1/2/b^3*a^2*x/(b*x^2+a)+5/2/b^3*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^6/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.28089, size = 348, normalized size = 5.27 \begin{align*} \left [\frac{4 \, b^{2} x^{5} - 20 \, a b x^{3} - 30 \, a^{2} x + 15 \,{\left (a b x^{2} + a^{2}\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b x^{2} + 2 \, b x \sqrt{-\frac{a}{b}} - a}{b x^{2} + a}\right )}{12 \,{\left (b^{4} x^{2} + a b^{3}\right )}}, \frac{2 \, b^{2} x^{5} - 10 \, a b x^{3} - 15 \, a^{2} x + 15 \,{\left (a b x^{2} + a^{2}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b x \sqrt{\frac{a}{b}}}{a}\right )}{6 \,{\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

[1/12*(4*b^2*x^5 - 20*a*b*x^3 - 30*a^2*x + 15*(a*b*x^2 + a^2)*sqrt(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b
*x^2 + a)))/(b^4*x^2 + a*b^3), 1/6*(2*b^2*x^5 - 10*a*b*x^3 - 15*a^2*x + 15*(a*b*x^2 + a^2)*sqrt(a/b)*arctan(b*
x*sqrt(a/b)/a))/(b^4*x^2 + a*b^3)]

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Sympy [A]  time = 0.418157, size = 107, normalized size = 1.62 \begin{align*} - \frac{a^{2} x}{2 a b^{3} + 2 b^{4} x^{2}} - \frac{2 a x}{b^{3}} - \frac{5 \sqrt{- \frac{a^{3}}{b^{7}}} \log{\left (x - \frac{b^{3} \sqrt{- \frac{a^{3}}{b^{7}}}}{a} \right )}}{4} + \frac{5 \sqrt{- \frac{a^{3}}{b^{7}}} \log{\left (x + \frac{b^{3} \sqrt{- \frac{a^{3}}{b^{7}}}}{a} \right )}}{4} + \frac{x^{3}}{3 b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**2+a)**2,x)

[Out]

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

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Giac [A]  time = 2.57837, size = 82, normalized size = 1.24 \begin{align*} \frac{5 \, a^{2} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} b^{3}} - \frac{a^{2} x}{2 \,{\left (b x^{2} + a\right )} b^{3}} + \frac{b^{4} x^{3} - 6 \, a b^{3} x}{3 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a)^2,x, algorithm="giac")

[Out]

5/2*a^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/2*a^2*x/((b*x^2 + a)*b^3) + 1/3*(b^4*x^3 - 6*a*b^3*x)/b^6