3.207 \(\int \frac{x^{12}}{(b x^2+c x^4)^3} \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0324637, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {1584, 288, 321, 205} \[ -\frac{5 x^3}{8 c^2 \left (b+c x^2\right )}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 c^{7/2}}-\frac{x^5}{4 c \left (b+c x^2\right )^2}+\frac{15 x}{8 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^12/(b*x^2 + c*x^4)^3,x]

[Out]

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

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - 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^{12}}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{x^6}{\left (b+c x^2\right )^3} \, dx\\ &=-\frac{x^5}{4 c \left (b+c x^2\right )^2}+\frac{5 \int \frac{x^4}{\left (b+c x^2\right )^2} \, dx}{4 c}\\ &=-\frac{x^5}{4 c \left (b+c x^2\right )^2}-\frac{5 x^3}{8 c^2 \left (b+c x^2\right )}+\frac{15 \int \frac{x^2}{b+c x^2} \, dx}{8 c^2}\\ &=\frac{15 x}{8 c^3}-\frac{x^5}{4 c \left (b+c x^2\right )^2}-\frac{5 x^3}{8 c^2 \left (b+c x^2\right )}-\frac{(15 b) \int \frac{1}{b+c x^2} \, dx}{8 c^3}\\ &=\frac{15 x}{8 c^3}-\frac{x^5}{4 c \left (b+c x^2\right )^2}-\frac{5 x^3}{8 c^2 \left (b+c x^2\right )}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0460064, size = 66, normalized size = 0.89 \[ \frac{15 b^2 x+25 b c x^3+8 c^2 x^5}{8 c^3 \left (b+c x^2\right )^2}-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^12/(b*x^2 + c*x^4)^3,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.052, size = 63, normalized size = 0.9 \begin{align*}{\frac{x}{{c}^{3}}}+{\frac{9\,b{x}^{3}}{8\,{c}^{2} \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{7\,{b}^{2}x}{8\,{c}^{3} \left ( c{x}^{2}+b \right ) ^{2}}}-{\frac{15\,b}{8\,{c}^{3}}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12/(c*x^4+b*x^2)^3,x)

[Out]

x/c^3+9/8/c^2*b/(c*x^2+b)^2*x^3+7/8/c^3*b^2/(c*x^2+b)^2*x-15/8/c^3*b/(b*c)^(1/2)*arctan(x*c/(b*c)^(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^12/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.44713, size = 425, normalized size = 5.74 \begin{align*} \left [\frac{16 \, c^{2} x^{5} + 50 \, b c x^{3} + 30 \, b^{2} x + 15 \,{\left (c^{2} x^{4} + 2 \, b c x^{2} + b^{2}\right )} \sqrt{-\frac{b}{c}} \log \left (\frac{c x^{2} - 2 \, c x \sqrt{-\frac{b}{c}} - b}{c x^{2} + b}\right )}{16 \,{\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}}, \frac{8 \, c^{2} x^{5} + 25 \, b c x^{3} + 15 \, b^{2} x - 15 \,{\left (c^{2} x^{4} + 2 \, b c x^{2} + b^{2}\right )} \sqrt{\frac{b}{c}} \arctan \left (\frac{c x \sqrt{\frac{b}{c}}}{b}\right )}{8 \,{\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.601758, size = 107, normalized size = 1.45 \begin{align*} \frac{15 \sqrt{- \frac{b}{c^{7}}} \log{\left (- c^{3} \sqrt{- \frac{b}{c^{7}}} + x \right )}}{16} - \frac{15 \sqrt{- \frac{b}{c^{7}}} \log{\left (c^{3} \sqrt{- \frac{b}{c^{7}}} + x \right )}}{16} + \frac{7 b^{2} x + 9 b c x^{3}}{8 b^{2} c^{3} + 16 b c^{4} x^{2} + 8 c^{5} x^{4}} + \frac{x}{c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**12/(c*x**4+b*x**2)**3,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.35001, size = 73, normalized size = 0.99 \begin{align*} -\frac{15 \, b \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{8 \, \sqrt{b c} c^{3}} + \frac{x}{c^{3}} + \frac{9 \, b c x^{3} + 7 \, b^{2} x}{8 \,{\left (c x^{2} + b\right )}^{2} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

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