3.211 \(\int \frac{x^8}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=65 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{3/2} c^{3/2}}+\frac{x}{8 b c \left (b+c x^2\right )}-\frac{x}{4 c \left (b+c x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0249015, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {1584, 288, 199, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{3/2} c^{3/2}}+\frac{x}{8 b c \left (b+c x^2\right )}-\frac{x}{4 c \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^8/(b*x^2 + c*x^4)^3,x]

[Out]

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

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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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

Mathematica [A]  time = 0.0297943, size = 58, normalized size = 0.89 \[ \frac{\frac{\sqrt{b} \sqrt{c} x \left (c x^2-b\right )}{\left (b+c x^2\right )^2}+\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{3/2} c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(b*x^2 + c*x^4)^3,x]

[Out]

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

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Maple [A]  time = 0.051, size = 49, normalized size = 0.8 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+b \right ) ^{2}} \left ({\frac{{x}^{3}}{8\,b}}-{\frac{x}{8\,c}} \right ) }+{\frac{1}{8\,bc}\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^8/(c*x^4+b*x^2)^3,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.49934, size = 394, normalized size = 6.06 \begin{align*} \left [\frac{2 \, b c^{2} x^{3} - 2 \, b^{2} c x -{\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^{2} c^{4} x^{4} + 2 \, b^{3} c^{3} x^{2} + b^{4} c^{2}\right )}}, \frac{b c^{2} x^{3} - b^{2} c x +{\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^{2} c^{4} x^{4} + 2 \, b^{3} c^{3} x^{2} + b^{4} c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 0.531297, size = 110, normalized size = 1.69 \begin{align*} - \frac{\sqrt{- \frac{1}{b^{3} c^{3}}} \log{\left (- b^{2} c \sqrt{- \frac{1}{b^{3} c^{3}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{b^{3} c^{3}}} \log{\left (b^{2} c \sqrt{- \frac{1}{b^{3} c^{3}}} + x \right )}}{16} + \frac{- b x + c x^{3}}{8 b^{3} c + 16 b^{2} c^{2} x^{2} + 8 b c^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(c*x**4+b*x**2)**3,x)

[Out]

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

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Giac [A]  time = 1.1987, size = 68, normalized size = 1.05 \begin{align*} \frac{\arctan \left (\frac{c x}{\sqrt{b c}}\right )}{8 \, \sqrt{b c} b c} + \frac{c x^{3} - b x}{8 \,{\left (c x^{2} + b\right )}^{2} b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

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