3.213 \(\int \frac{x^6}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=62 \[ \frac{3 x}{8 b^2 \left (b+c x^2\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{5/2} \sqrt{c}}+\frac{x}{4 b \left (b+c x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0234265, antiderivative size = 62, 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, 199, 205} \[ \frac{3 x}{8 b^2 \left (b+c x^2\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{5/2} \sqrt{c}}+\frac{x}{4 b \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(b*x^2 + c*x^4)^3,x]

[Out]

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

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

Mathematica [A]  time = 0.0338992, size = 55, normalized size = 0.89 \[ \frac{5 b x+3 c x^3}{8 b^2 \left (b+c x^2\right )^2}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{8 b^{5/2} \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(b*x^2 + c*x^4)^3,x]

[Out]

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

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Maple [A]  time = 0.046, size = 51, normalized size = 0.8 \begin{align*}{\frac{x}{4\,b \left ( c{x}^{2}+b \right ) ^{2}}}+{\frac{3\,x}{8\,{b}^{2} \left ( c{x}^{2}+b \right ) }}+{\frac{3}{8\,{b}^{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^6/(c*x^4+b*x^2)^3,x)

[Out]

1/4*x/b/(c*x^2+b)^2+3/8*x/b^2/(c*x^2+b)+3/8/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^6/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.518, size = 401, normalized size = 6.47 \begin{align*} \left [\frac{6 \, b c^{2} x^{3} + 10 \, 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^{3} c^{3} x^{4} + 2 \, b^{4} c^{2} x^{2} + b^{5} c\right )}}, \frac{3 \, b c^{2} x^{3} + 5 \, 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^{3} c^{3} x^{4} + 2 \, b^{4} c^{2} x^{2} + b^{5} c\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.517811, size = 105, normalized size = 1.69 \begin{align*} - \frac{3 \sqrt{- \frac{1}{b^{5} c}} \log{\left (- b^{3} \sqrt{- \frac{1}{b^{5} c}} + x \right )}}{16} + \frac{3 \sqrt{- \frac{1}{b^{5} c}} \log{\left (b^{3} \sqrt{- \frac{1}{b^{5} c}} + x \right )}}{16} + \frac{5 b x + 3 c x^{3}}{8 b^{4} + 16 b^{3} c x^{2} + 8 b^{2} c^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(c*x**4+b*x**2)**3,x)

[Out]

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

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Giac [A]  time = 1.30779, size = 61, normalized size = 0.98 \begin{align*} \frac{3 \, \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{8 \, \sqrt{b c} b^{2}} + \frac{3 \, c x^{3} + 5 \, b x}{8 \,{\left (c x^{2} + b\right )}^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

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