3.192 \(\int \frac{x^{10}}{(b x^2+c x^4)^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0347578, antiderivative size = 66, 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} \[ \frac{5 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 c^{7/2}}-\frac{5 b x}{2 c^3}-\frac{x^5}{2 c \left (b+c x^2\right )}+\frac{5 x^3}{6 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x^10/(b*x^2 + c*x^4)^2,x]

[Out]

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

Mathematica [A]  time = 0.0414106, size = 60, normalized size = 0.91 \[ \frac{x \left (-\frac{3 b^2}{b+c x^2}-12 b+2 c x^2\right )}{6 c^3}+\frac{5 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/(b*x^2 + c*x^4)^2,x]

[Out]

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

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Maple [A]  time = 0.052, size = 57, normalized size = 0.9 \begin{align*}{\frac{{x}^{3}}{3\,{c}^{2}}}-2\,{\frac{bx}{{c}^{3}}}-{\frac{{b}^{2}x}{2\,{c}^{3} \left ( c{x}^{2}+b \right ) }}+{\frac{5\,{b}^{2}}{2\,{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^10/(c*x^4+b*x^2)^2,x)

[Out]

1/3*x^3/c^2-2*b*x/c^3-1/2/c^3*b^2*x/(c*x^2+b)+5/2/c^3*b^2/(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^10/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54293, size = 348, normalized size = 5.27 \begin{align*} \left [\frac{4 \, c^{2} x^{5} - 20 \, b c x^{3} - 30 \, b^{2} x + 15 \,{\left (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 )}{12 \,{\left (c^{4} x^{2} + b c^{3}\right )}}, \frac{2 \, c^{2} x^{5} - 10 \, b c x^{3} - 15 \, b^{2} x + 15 \,{\left (b c x^{2} + b^{2}\right )} \sqrt{\frac{b}{c}} \arctan \left (\frac{c x \sqrt{\frac{b}{c}}}{b}\right )}{6 \,{\left (c^{4} x^{2} + b c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.489909, size = 107, normalized size = 1.62 \begin{align*} - \frac{b^{2} x}{2 b c^{3} + 2 c^{4} x^{2}} - \frac{2 b x}{c^{3}} - \frac{5 \sqrt{- \frac{b^{3}}{c^{7}}} \log{\left (x - \frac{c^{3} \sqrt{- \frac{b^{3}}{c^{7}}}}{b} \right )}}{4} + \frac{5 \sqrt{- \frac{b^{3}}{c^{7}}} \log{\left (x + \frac{c^{3} \sqrt{- \frac{b^{3}}{c^{7}}}}{b} \right )}}{4} + \frac{x^{3}}{3 c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10/(c*x**4+b*x**2)**2,x)

[Out]

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

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Giac [A]  time = 1.2252, size = 82, normalized size = 1.24 \begin{align*} \frac{5 \, b^{2} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{2 \, \sqrt{b c} c^{3}} - \frac{b^{2} x}{2 \,{\left (c x^{2} + b\right )} c^{3}} + \frac{c^{4} x^{3} - 6 \, b c^{3} x}{3 \, c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

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