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

Optimal. Leaf size=45 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{3/2} \sqrt{c}}+\frac{x}{2 b \left (b+c x^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0173041, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {1584, 199, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{3/2} \sqrt{c}}+\frac{x}{2 b \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(b*x^2 + c*x^4)^2,x]

[Out]

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

Mathematica [A]  time = 0.0247956, size = 45, normalized size = 1. \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{2 b^{3/2} \sqrt{c}}+\frac{x}{2 b \left (b+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(b*x^2 + c*x^4)^2,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.048, size = 36, normalized size = 0.8 \begin{align*}{\frac{x}{2\,b \left ( c{x}^{2}+b \right ) }}+{\frac{1}{2\,b}\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^4/(c*x^4+b*x^2)^2,x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.48401, size = 261, normalized size = 5.8 \begin{align*} \left [\frac{2 \, b c x -{\left (c x^{2} + b\right )} \sqrt{-b c} \log \left (\frac{c x^{2} - 2 \, \sqrt{-b c} x - b}{c x^{2} + b}\right )}{4 \,{\left (b^{2} c^{2} x^{2} + b^{3} c\right )}}, \frac{b c x +{\left (c x^{2} + b\right )} \sqrt{b c} \arctan \left (\frac{\sqrt{b c} x}{b}\right )}{2 \,{\left (b^{2} c^{2} x^{2} + b^{3} c\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [B]  time = 0.390433, size = 78, normalized size = 1.73 \begin{align*} \frac{x}{2 b^{2} + 2 b c x^{2}} - \frac{\sqrt{- \frac{1}{b^{3} c}} \log{\left (- b^{2} \sqrt{- \frac{1}{b^{3} c}} + x \right )}}{4} + \frac{\sqrt{- \frac{1}{b^{3} c}} \log{\left (b^{2} \sqrt{- \frac{1}{b^{3} c}} + x \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(2*b**2 + 2*b*c*x**2) - sqrt(-1/(b**3*c))*log(-b**2*sqrt(-1/(b**3*c)) + x)/4 + sqrt(-1/(b**3*c))*log(b**2*sq
rt(-1/(b**3*c)) + x)/4

________________________________________________________________________________________

Giac [A]  time = 1.26283, size = 47, normalized size = 1.04 \begin{align*} \frac{\arctan \left (\frac{c x}{\sqrt{b c}}\right )}{2 \, \sqrt{b c} b} + \frac{x}{2 \,{\left (c x^{2} + b\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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