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

Optimal. Leaf size=79 \[ \frac{7 b^2 x}{2 c^4}-\frac{7 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 c^{9/2}}-\frac{7 b x^3}{6 c^3}-\frac{x^7}{2 c \left (b+c x^2\right )}+\frac{7 x^5}{10 c^2} \]

[Out]

(7*b^2*x)/(2*c^4) - (7*b*x^3)/(6*c^3) + (7*x^5)/(10*c^2) - x^7/(2*c*(b + c*x^2)) - (7*b^(5/2)*ArcTan[(Sqrt[c]*
x)/Sqrt[b]])/(2*c^(9/2))

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Rubi [A]  time = 0.0387618, antiderivative size = 79, 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, 302, 205} \[ \frac{7 b^2 x}{2 c^4}-\frac{7 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 c^{9/2}}-\frac{7 b x^3}{6 c^3}-\frac{x^7}{2 c \left (b+c x^2\right )}+\frac{7 x^5}{10 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x^12/(b*x^2 + c*x^4)^2,x]

[Out]

(7*b^2*x)/(2*c^4) - (7*b*x^3)/(6*c^3) + (7*x^5)/(10*c^2) - x^7/(2*c*(b + c*x^2)) - (7*b^(5/2)*ArcTan[(Sqrt[c]*
x)/Sqrt[b]])/(2*c^(9/2))

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 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^{12}}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac{x^8}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac{x^7}{2 c \left (b+c x^2\right )}+\frac{7 \int \frac{x^6}{b+c x^2} \, dx}{2 c}\\ &=-\frac{x^7}{2 c \left (b+c x^2\right )}+\frac{7 \int \left (\frac{b^2}{c^3}-\frac{b x^2}{c^2}+\frac{x^4}{c}-\frac{b^3}{c^3 \left (b+c x^2\right )}\right ) \, dx}{2 c}\\ &=\frac{7 b^2 x}{2 c^4}-\frac{7 b x^3}{6 c^3}+\frac{7 x^5}{10 c^2}-\frac{x^7}{2 c \left (b+c x^2\right )}-\frac{\left (7 b^3\right ) \int \frac{1}{b+c x^2} \, dx}{2 c^4}\\ &=\frac{7 b^2 x}{2 c^4}-\frac{7 b x^3}{6 c^3}+\frac{7 x^5}{10 c^2}-\frac{x^7}{2 c \left (b+c x^2\right )}-\frac{7 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0524963, size = 71, normalized size = 0.9 \[ \frac{x \left (\frac{15 b^3}{b+c x^2}+90 b^2-20 b c x^2+6 c^2 x^4\right )}{30 c^4}-\frac{7 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 c^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^12/(b*x^2 + c*x^4)^2,x]

[Out]

(x*(90*b^2 - 20*b*c*x^2 + 6*c^2*x^4 + (15*b^3)/(b + c*x^2)))/(30*c^4) - (7*b^(5/2)*ArcTan[(Sqrt[c]*x)/Sqrt[b]]
)/(2*c^(9/2))

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Maple [A]  time = 0.052, size = 68, normalized size = 0.9 \begin{align*}{\frac{{x}^{5}}{5\,{c}^{2}}}-{\frac{2\,b{x}^{3}}{3\,{c}^{3}}}+3\,{\frac{{b}^{2}x}{{c}^{4}}}+{\frac{{b}^{3}x}{2\,{c}^{4} \left ( c{x}^{2}+b \right ) }}-{\frac{7\,{b}^{3}}{2\,{c}^{4}}\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)^2,x)

[Out]

1/5*x^5/c^2-2/3*b*x^3/c^3+3*b^2*x/c^4+1/2/c^4*b^3*x/(c*x^2+b)-7/2/c^4*b^3/(b*c)^(1/2)*arctan(x*c/(b*c)^(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^12/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.47675, size = 409, normalized size = 5.18 \begin{align*} \left [\frac{12 \, c^{3} x^{7} - 28 \, b c^{2} x^{5} + 140 \, b^{2} c x^{3} + 210 \, b^{3} x + 105 \,{\left (b^{2} c x^{2} + b^{3}\right )} \sqrt{-\frac{b}{c}} \log \left (\frac{c x^{2} - 2 \, c x \sqrt{-\frac{b}{c}} - b}{c x^{2} + b}\right )}{60 \,{\left (c^{5} x^{2} + b c^{4}\right )}}, \frac{6 \, c^{3} x^{7} - 14 \, b c^{2} x^{5} + 70 \, b^{2} c x^{3} + 105 \, b^{3} x - 105 \,{\left (b^{2} c x^{2} + b^{3}\right )} \sqrt{\frac{b}{c}} \arctan \left (\frac{c x \sqrt{\frac{b}{c}}}{b}\right )}{30 \,{\left (c^{5} x^{2} + b c^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/60*(12*c^3*x^7 - 28*b*c^2*x^5 + 140*b^2*c*x^3 + 210*b^3*x + 105*(b^2*c*x^2 + b^3)*sqrt(-b/c)*log((c*x^2 - 2
*c*x*sqrt(-b/c) - b)/(c*x^2 + b)))/(c^5*x^2 + b*c^4), 1/30*(6*c^3*x^7 - 14*b*c^2*x^5 + 70*b^2*c*x^3 + 105*b^3*
x - 105*(b^2*c*x^2 + b^3)*sqrt(b/c)*arctan(c*x*sqrt(b/c)/b))/(c^5*x^2 + b*c^4)]

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Sympy [A]  time = 0.482897, size = 124, normalized size = 1.57 \begin{align*} \frac{b^{3} x}{2 b c^{4} + 2 c^{5} x^{2}} + \frac{3 b^{2} x}{c^{4}} - \frac{2 b x^{3}}{3 c^{3}} + \frac{7 \sqrt{- \frac{b^{5}}{c^{9}}} \log{\left (x - \frac{c^{4} \sqrt{- \frac{b^{5}}{c^{9}}}}{b^{2}} \right )}}{4} - \frac{7 \sqrt{- \frac{b^{5}}{c^{9}}} \log{\left (x + \frac{c^{4} \sqrt{- \frac{b^{5}}{c^{9}}}}{b^{2}} \right )}}{4} + \frac{x^{5}}{5 c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21212, size = 99, normalized size = 1.25 \begin{align*} -\frac{7 \, b^{3} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{2 \, \sqrt{b c} c^{4}} + \frac{b^{3} x}{2 \,{\left (c x^{2} + b\right )} c^{4}} + \frac{3 \, c^{8} x^{5} - 10 \, b c^{7} x^{3} + 45 \, b^{2} c^{6} x}{15 \, c^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/2*b^3*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^4) + 1/2*b^3*x/((c*x^2 + b)*c^4) + 1/15*(3*c^8*x^5 - 10*b*c^7*x^3
+ 45*b^2*c^6*x)/c^10