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

Optimal. Leaf size=55 \[ -\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 c^{5/2}}-\frac{x^3}{2 c \left (b+c x^2\right )}+\frac{3 x}{2 c^2} \]

[Out]

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

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Rubi [A]  time = 0.0244315, antiderivative size = 55, 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, 321, 205} \[ -\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 c^{5/2}}-\frac{x^3}{2 c \left (b+c x^2\right )}+\frac{3 x}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x^8/(b*x^2 + c*x^4)^2,x]

[Out]

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

Mathematica [A]  time = 0.0315579, size = 51, normalized size = 0.93 \[ \frac{b x}{2 c^2 \left (b+c x^2\right )}-\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 c^{5/2}}+\frac{x}{c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(b*x^2 + c*x^4)^2,x]

[Out]

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

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Maple [A]  time = 0.05, size = 43, normalized size = 0.8 \begin{align*}{\frac{x}{{c}^{2}}}+{\frac{bx}{2\,{c}^{2} \left ( c{x}^{2}+b \right ) }}-{\frac{3\,b}{2\,{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^8/(c*x^4+b*x^2)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53923, size = 285, normalized size = 5.18 \begin{align*} \left [\frac{4 \, c x^{3} + 3 \,{\left (c x^{2} + b\right )} \sqrt{-\frac{b}{c}} \log \left (\frac{c x^{2} - 2 \, c x \sqrt{-\frac{b}{c}} - b}{c x^{2} + b}\right ) + 6 \, b x}{4 \,{\left (c^{3} x^{2} + b c^{2}\right )}}, \frac{2 \, c x^{3} - 3 \,{\left (c x^{2} + b\right )} \sqrt{\frac{b}{c}} \arctan \left (\frac{c x \sqrt{\frac{b}{c}}}{b}\right ) + 3 \, b x}{2 \,{\left (c^{3} x^{2} + b 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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.464708, size = 83, normalized size = 1.51 \begin{align*} \frac{b x}{2 b c^{2} + 2 c^{3} x^{2}} + \frac{3 \sqrt{- \frac{b}{c^{5}}} \log{\left (- c^{2} \sqrt{- \frac{b}{c^{5}}} + x \right )}}{4} - \frac{3 \sqrt{- \frac{b}{c^{5}}} \log{\left (c^{2} \sqrt{- \frac{b}{c^{5}}} + x \right )}}{4} + \frac{x}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.2387, size = 57, normalized size = 1.04 \begin{align*} -\frac{3 \, b \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{2 \, \sqrt{b c} c^{2}} + \frac{b x}{2 \,{\left (c x^{2} + b\right )} c^{2}} + \frac{x}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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