3.209 \(\int \frac{x^{10}}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=64 \[ -\frac{3 x}{8 c^2 \left (b+c x^2\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 \sqrt{b} c^{5/2}}-\frac{x^3}{4 c \left (b+c x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0255314, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {1584, 288, 205} \[ -\frac{3 x}{8 c^2 \left (b+c x^2\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 \sqrt{b} c^{5/2}}-\frac{x^3}{4 c \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^10/(b*x^2 + c*x^4)^3,x]

[Out]

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

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

Mathematica [A]  time = 0.0426801, size = 55, normalized size = 0.86 \[ \frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 \sqrt{b} c^{5/2}}-\frac{3 b x+5 c x^3}{8 c^2 \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/(b*x^2 + c*x^4)^3,x]

[Out]

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

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Maple [A]  time = 0.048, size = 47, normalized size = 0.7 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+b \right ) ^{2}} \left ( -{\frac{5\,{x}^{3}}{8\,c}}-{\frac{3\,bx}{8\,{c}^{2}}} \right ) }+{\frac{3}{8\,{c}^{2}}\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)^3,x)

[Out]

(-5/8*x^3/c-3/8*b*x/c^2)/(c*x^2+b)^2+3/8/c^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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59582, size = 404, normalized size = 6.31 \begin{align*} \left [-\frac{10 \, b c^{2} x^{3} + 6 \, b^{2} c x + 3 \,{\left (c^{2} x^{4} + 2 \, b c x^{2} + b^{2}\right )} \sqrt{-b c} \log \left (\frac{c x^{2} - 2 \, \sqrt{-b c} x - b}{c x^{2} + b}\right )}{16 \,{\left (b c^{5} x^{4} + 2 \, b^{2} c^{4} x^{2} + b^{3} c^{3}\right )}}, -\frac{5 \, b c^{2} x^{3} + 3 \, b^{2} c x - 3 \,{\left (c^{2} x^{4} + 2 \, b c x^{2} + b^{2}\right )} \sqrt{b c} \arctan \left (\frac{\sqrt{b c} x}{b}\right )}{8 \,{\left (b c^{5} x^{4} + 2 \, b^{2} c^{4} x^{2} + b^{3} 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)^3,x, algorithm="fricas")

[Out]

[-1/16*(10*b*c^2*x^3 + 6*b^2*c*x + 3*(c^2*x^4 + 2*b*c*x^2 + b^2)*sqrt(-b*c)*log((c*x^2 - 2*sqrt(-b*c)*x - b)/(
c*x^2 + b)))/(b*c^5*x^4 + 2*b^2*c^4*x^2 + b^3*c^3), -1/8*(5*b*c^2*x^3 + 3*b^2*c*x - 3*(c^2*x^4 + 2*b*c*x^2 + b
^2)*sqrt(b*c)*arctan(sqrt(b*c)*x/b))/(b*c^5*x^4 + 2*b^2*c^4*x^2 + b^3*c^3)]

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Sympy [A]  time = 0.532172, size = 109, normalized size = 1.7 \begin{align*} - \frac{3 \sqrt{- \frac{1}{b c^{5}}} \log{\left (- b c^{2} \sqrt{- \frac{1}{b c^{5}}} + x \right )}}{16} + \frac{3 \sqrt{- \frac{1}{b c^{5}}} \log{\left (b c^{2} \sqrt{- \frac{1}{b c^{5}}} + x \right )}}{16} - \frac{3 b x + 5 c x^{3}}{8 b^{2} c^{2} + 16 b c^{3} x^{2} + 8 c^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10/(c*x**4+b*x**2)**3,x)

[Out]

-3*sqrt(-1/(b*c**5))*log(-b*c**2*sqrt(-1/(b*c**5)) + x)/16 + 3*sqrt(-1/(b*c**5))*log(b*c**2*sqrt(-1/(b*c**5))
+ x)/16 - (3*b*x + 5*c*x**3)/(8*b**2*c**2 + 16*b*c**3*x**2 + 8*c**4*x**4)

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Giac [A]  time = 1.28926, size = 61, normalized size = 0.95 \begin{align*} \frac{3 \, \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{8 \, \sqrt{b c} c^{2}} - \frac{5 \, c x^{3} + 3 \, b x}{8 \,{\left (c x^{2} + b\right )}^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

3/8*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^2) - 1/8*(5*c*x^3 + 3*b*x)/((c*x^2 + b)^2*c^2)