3.184 \(\int \frac{x^6}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=74 \[ -\frac{5 x^3}{8 b^2 \left (a+b x^2\right )}-\frac{15 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 b^{7/2}}-\frac{x^5}{4 b \left (a+b x^2\right )^2}+\frac{15 x}{8 b^3} \]

[Out]

(15*x)/(8*b^3) - x^5/(4*b*(a + b*x^2)^2) - (5*x^3)/(8*b^2*(a + b*x^2)) - (15*Sqrt[a]*ArcTan[(Sqrt[b]*x)/Sqrt[a
]])/(8*b^(7/2))

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Rubi [A]  time = 0.0252003, antiderivative size = 74, 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, 321, 205} \[ -\frac{5 x^3}{8 b^2 \left (a+b x^2\right )}-\frac{15 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 b^{7/2}}-\frac{x^5}{4 b \left (a+b x^2\right )^2}+\frac{15 x}{8 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a + b*x^2)^3,x]

[Out]

(15*x)/(8*b^3) - x^5/(4*b*(a + b*x^2)^2) - (5*x^3)/(8*b^2*(a + b*x^2)) - (15*Sqrt[a]*ArcTan[(Sqrt[b]*x)/Sqrt[a
]])/(8*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 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^6}{\left (a+b x^2\right )^3} \, dx &=-\frac{x^5}{4 b \left (a+b x^2\right )^2}+\frac{5 \int \frac{x^4}{\left (a+b x^2\right )^2} \, dx}{4 b}\\ &=-\frac{x^5}{4 b \left (a+b x^2\right )^2}-\frac{5 x^3}{8 b^2 \left (a+b x^2\right )}+\frac{15 \int \frac{x^2}{a+b x^2} \, dx}{8 b^2}\\ &=\frac{15 x}{8 b^3}-\frac{x^5}{4 b \left (a+b x^2\right )^2}-\frac{5 x^3}{8 b^2 \left (a+b x^2\right )}-\frac{(15 a) \int \frac{1}{a+b x^2} \, dx}{8 b^3}\\ &=\frac{15 x}{8 b^3}-\frac{x^5}{4 b \left (a+b x^2\right )^2}-\frac{5 x^3}{8 b^2 \left (a+b x^2\right )}-\frac{15 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0459202, size = 66, normalized size = 0.89 \[ \frac{15 a^2 x+25 a b x^3+8 b^2 x^5}{8 b^3 \left (a+b x^2\right )^2}-\frac{15 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a + b*x^2)^3,x]

[Out]

(15*a^2*x + 25*a*b*x^3 + 8*b^2*x^5)/(8*b^3*(a + b*x^2)^2) - (15*Sqrt[a]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*b^(7/2
))

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Maple [A]  time = 0.008, size = 63, normalized size = 0.9 \begin{align*}{\frac{x}{{b}^{3}}}+{\frac{9\,a{x}^{3}}{8\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{7\,{a}^{2}x}{8\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{15\,a}{8\,{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)^3,x)

[Out]

x/b^3+9/8/b^2*a/(b*x^2+a)^2*x^3+7/8/b^3*a^2/(b*x^2+a)^2*x-15/8/b^3*a/(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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32053, size = 425, normalized size = 5.74 \begin{align*} \left [\frac{16 \, b^{2} x^{5} + 50 \, a b x^{3} + 30 \, a^{2} x + 15 \,{\left (b^{2} x^{4} + 2 \, 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 )}{16 \,{\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}, \frac{8 \, b^{2} x^{5} + 25 \, a b x^{3} + 15 \, a^{2} x - 15 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b x \sqrt{\frac{a}{b}}}{a}\right )}{8 \,{\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/16*(16*b^2*x^5 + 50*a*b*x^3 + 30*a^2*x + 15*(b^2*x^4 + 2*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^5*x^4 + 2*a*b^4*x^2 + a^2*b^3), 1/8*(8*b^2*x^5 + 25*a*b*x^3 + 15*a^2*x - 15*(b^2*x
^4 + 2*a*b*x^2 + a^2)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)]

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Sympy [A]  time = 0.545711, size = 107, normalized size = 1.45 \begin{align*} \frac{15 \sqrt{- \frac{a}{b^{7}}} \log{\left (- b^{3} \sqrt{- \frac{a}{b^{7}}} + x \right )}}{16} - \frac{15 \sqrt{- \frac{a}{b^{7}}} \log{\left (b^{3} \sqrt{- \frac{a}{b^{7}}} + x \right )}}{16} + \frac{7 a^{2} x + 9 a b x^{3}}{8 a^{2} b^{3} + 16 a b^{4} x^{2} + 8 b^{5} x^{4}} + \frac{x}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**2+a)**3,x)

[Out]

15*sqrt(-a/b**7)*log(-b**3*sqrt(-a/b**7) + x)/16 - 15*sqrt(-a/b**7)*log(b**3*sqrt(-a/b**7) + x)/16 + (7*a**2*x
+ 9*a*b*x**3)/(8*a**2*b**3 + 16*a*b**4*x**2 + 8*b**5*x**4) + x/b**3

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Giac [A]  time = 3.03909, size = 73, normalized size = 0.99 \begin{align*} -\frac{15 \, a \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} b^{3}} + \frac{x}{b^{3}} + \frac{9 \, a b x^{3} + 7 \, a^{2} x}{8 \,{\left (b x^{2} + a\right )}^{2} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a)^3,x, algorithm="giac")

[Out]

-15/8*a*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + x/b^3 + 1/8*(9*a*b*x^3 + 7*a^2*x)/((b*x^2 + a)^2*b^3)