3.532 \(\int \frac{x^2 (3 e+2 f x^2)}{e^2+4 e f x^2+4 f^2 x^4+4 d f x^6} \, dx\)

Optimal. Leaf size=40 \[ \frac{\tan ^{-1}\left (\frac{2 \sqrt{d} \sqrt{f} x^3}{e+2 f x^2}\right )}{2 \sqrt{d} \sqrt{f}} \]

[Out]

ArcTan[(2*Sqrt[d]*Sqrt[f]*x^3)/(e + 2*f*x^2)]/(2*Sqrt[d]*Sqrt[f])

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Rubi [A]  time = 0.129959, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {2094, 205} \[ \frac{\tan ^{-1}\left (\frac{2 \sqrt{d} \sqrt{f} x^3}{e+2 f x^2}\right )}{2 \sqrt{d} \sqrt{f}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(3*e + 2*f*x^2))/(e^2 + 4*e*f*x^2 + 4*f^2*x^4 + 4*d*f*x^6),x]

[Out]

ArcTan[(2*Sqrt[d]*Sqrt[f]*x^3)/(e + 2*f*x^2)]/(2*Sqrt[d]*Sqrt[f])

Rule 2094

Int[((x_)^(m_.)*((A_) + (B_.)*(x_)^(n_.)))/((a_) + (b_.)*(x_)^(k_.) + (c_.)*(x_)^(n_.) + (d_.)*(x_)^(n2_)), x_
Symbol] :> Dist[(A^2*(m - n + 1))/(m + 1), Subst[Int[1/(a + A^2*b*(m - n + 1)^2*x^2), x], x, x^(m + 1)/(A*(m -
n + 1) + B*(m + 1)*x^n)], x] /; FreeQ[{a, b, c, d, A, B, m, n}, x] && EqQ[n2, 2*n] && EqQ[k, 2*(m + 1)] && Eq
Q[a*B^2*(m + 1)^2 - A^2*d*(m - n + 1)^2, 0] && EqQ[B*c*(m + 1) - 2*A*d*(m - n + 1), 0]

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^2 \left (3 e+2 f x^2\right )}{e^2+4 e f x^2+4 f^2 x^4+4 d f x^6} \, dx &=\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{e^2+36 d e^2 f x^2} \, dx,x,\frac{x^3}{3 e+6 f x^2}\right )\\ &=\frac{\tan ^{-1}\left (\frac{2 \sqrt{d} \sqrt{f} x^3}{e+2 f x^2}\right )}{2 \sqrt{d} \sqrt{f}}\\ \end{align*}

Mathematica [C]  time = 0.0507576, size = 85, normalized size = 2.12 \[ \frac{\text{RootSum}\left [4 \text{$\#$1}^6 d f+4 \text{$\#$1}^2 e f+4 \text{$\#$1}^4 f^2+e^2\& ,\frac{2 \text{$\#$1}^3 f \log (x-\text{$\#$1})+3 \text{$\#$1} e \log (x-\text{$\#$1})}{3 \text{$\#$1}^4 d+2 \text{$\#$1}^2 f+e}\& \right ]}{8 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(3*e + 2*f*x^2))/(e^2 + 4*e*f*x^2 + 4*f^2*x^4 + 4*d*f*x^6),x]

[Out]

RootSum[e^2 + 4*e*f*#1^2 + 4*f^2*#1^4 + 4*d*f*#1^6 & , (3*e*Log[x - #1]*#1 + 2*f*Log[x - #1]*#1^3)/(e + 2*f*#1
^2 + 3*d*#1^4) & ]/(8*f)

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Maple [C]  time = 0.112, size = 74, normalized size = 1.9 \begin{align*}{\frac{1}{8\,f}\sum _{{\it \_R}={\it RootOf} \left ( 4\,df{{\it \_Z}}^{6}+4\,{f}^{2}{{\it \_Z}}^{4}+4\,fe{{\it \_Z}}^{2}+{e}^{2} \right ) }{\frac{ \left ( 2\,{{\it \_R}}^{4}f+3\,{{\it \_R}}^{2}e \right ) \ln \left ( x-{\it \_R} \right ) }{3\,d{{\it \_R}}^{5}+2\,f{{\it \_R}}^{3}+e{\it \_R}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(2*f*x^2+3*e)/(4*d*f*x^6+4*f^2*x^4+4*e*f*x^2+e^2),x)

[Out]

1/8/f*sum((2*_R^4*f+3*_R^2*e)/(3*_R^5*d+2*_R^3*f+_R*e)*ln(x-_R),_R=RootOf(4*_Z^6*d*f+4*_Z^4*f^2+4*_Z^2*e*f+e^2
))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, f x^{2} + 3 \, e\right )} x^{2}}{4 \, d f x^{6} + 4 \, f^{2} x^{4} + 4 \, e f x^{2} + e^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(2*f*x^2+3*e)/(4*d*f*x^6+4*f^2*x^4+4*e*f*x^2+e^2),x, algorithm="maxima")

[Out]

integrate((2*f*x^2 + 3*e)*x^2/(4*d*f*x^6 + 4*f^2*x^4 + 4*e*f*x^2 + e^2), x)

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Fricas [B]  time = 1.26836, size = 462, normalized size = 11.55 \begin{align*} \left [-\frac{\sqrt{-d f} \log \left (\frac{4 \, d f x^{6} - 4 \, f^{2} x^{4} - 4 \, e f x^{2} - e^{2} - 4 \,{\left (2 \, f x^{5} + e x^{3}\right )} \sqrt{-d f}}{4 \, d f x^{6} + 4 \, f^{2} x^{4} + 4 \, e f x^{2} + e^{2}}\right )}{4 \, d f}, \frac{\sqrt{d f} \arctan \left (\frac{\sqrt{d f} x}{f}\right ) - \sqrt{d f} \arctan \left (\frac{2 \,{\left (2 \, d f x^{5} -{\left (d e - 2 \, f^{2}\right )} x^{3} + e f x\right )} \sqrt{d f}}{d e^{2}}\right ) + \sqrt{d f} \arctan \left (\frac{{\left (2 \, d f x^{3} -{\left (d e - 2 \, f^{2}\right )} x\right )} \sqrt{d f}}{d e f}\right )}{2 \, d f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(2*f*x^2+3*e)/(4*d*f*x^6+4*f^2*x^4+4*e*f*x^2+e^2),x, algorithm="fricas")

[Out]

[-1/4*sqrt(-d*f)*log((4*d*f*x^6 - 4*f^2*x^4 - 4*e*f*x^2 - e^2 - 4*(2*f*x^5 + e*x^3)*sqrt(-d*f))/(4*d*f*x^6 + 4
*f^2*x^4 + 4*e*f*x^2 + e^2))/(d*f), 1/2*(sqrt(d*f)*arctan(sqrt(d*f)*x/f) - sqrt(d*f)*arctan(2*(2*d*f*x^5 - (d*
e - 2*f^2)*x^3 + e*f*x)*sqrt(d*f)/(d*e^2)) + sqrt(d*f)*arctan((2*d*f*x^3 - (d*e - 2*f^2)*x)*sqrt(d*f)/(d*e*f))
)/(d*f)]

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Sympy [B]  time = 1.08516, size = 90, normalized size = 2.25 \begin{align*} - \frac{\sqrt{- \frac{1}{d f}} \log{\left (- \frac{e \sqrt{- \frac{1}{d f}}}{2} - f x^{2} \sqrt{- \frac{1}{d f}} + x^{3} \right )}}{4} + \frac{\sqrt{- \frac{1}{d f}} \log{\left (\frac{e \sqrt{- \frac{1}{d f}}}{2} + f x^{2} \sqrt{- \frac{1}{d f}} + x^{3} \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(2*f*x**2+3*e)/(4*d*f*x**6+4*f**2*x**4+4*e*f*x**2+e**2),x)

[Out]

-sqrt(-1/(d*f))*log(-e*sqrt(-1/(d*f))/2 - f*x**2*sqrt(-1/(d*f)) + x**3)/4 + sqrt(-1/(d*f))*log(e*sqrt(-1/(d*f)
)/2 + f*x**2*sqrt(-1/(d*f)) + x**3)/4

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, f x^{2} + 3 \, e\right )} x^{2}}{4 \, d f x^{6} + 4 \, f^{2} x^{4} + 4 \, e f x^{2} + e^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(2*f*x^2+3*e)/(4*d*f*x^6+4*f^2*x^4+4*e*f*x^2+e^2),x, algorithm="giac")

[Out]

integrate((2*f*x^2 + 3*e)*x^2/(4*d*f*x^6 + 4*f^2*x^4 + 4*e*f*x^2 + e^2), x)