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]

1/2*arctan(2*x^3*d^(1/2)*f^(1/2)/(2*f*x^2+e))/d^(1/2)/f^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 40, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.05, size = 85, normalized size = 2.12 \[ \frac {\text {RootSum}\left [4 \text {$\#$1}^6 d f+4 \text {$\#$1}^4 f^2+4 \text {$\#$1}^2 e f+e^2\& ,\frac {2 \text {$\#$1}^3 f \log (x-\text {$\#$1})+3 \text {$\#$1} e \log (x-\text {$\#$1})}{3 \text {$\#$1}^4 d+2 \text {$\#$1}^2 f+e}\& \right ]}{8 f} \]

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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fricas [B]  time = 0.42, size = 208, normalized size = 5.20 \[ \left [-\frac {\sqrt {-d f} \log \left (\frac {4 \, d f x^{6} - 4 \, f^{2} x^{4} - 4 \, e f x^{2} - e^{2} - 4 \, {\left (2 \, f x^{5} + e x^{3}\right )} \sqrt {-d f}}{4 \, d f x^{6} + 4 \, f^{2} x^{4} + 4 \, e f x^{2} + e^{2}}\right )}{4 \, d f}, \frac {\sqrt {d f} \arctan \left (\frac {\sqrt {d f} x}{f}\right ) - \sqrt {d f} \arctan \left (\frac {2 \, {\left (2 \, d f x^{5} - {\left (d e - 2 \, f^{2}\right )} x^{3} + e f x\right )} \sqrt {d f}}{d e^{2}}\right ) + \sqrt {d f} \arctan \left (\frac {{\left (2 \, d f x^{3} - {\left (d e - 2 \, f^{2}\right )} x\right )} \sqrt {d f}}{d e f}\right )}{2 \, d f}\right ] \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, f x^{2} + 3 \, e\right )} x^{2}}{4 \, d f x^{6} + 4 \, f^{2} x^{4} + 4 \, e f x^{2} + e^{2}}\,{d x} \]

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)

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maple [C]  time = 0.15, size = 74, normalized size = 1.85 \[ \frac {\left (2 \RootOf \left (4 d f \,\textit {\_Z}^{6}+4 f^{2} \textit {\_Z}^{4}+4 e f \,\textit {\_Z}^{2}+e^{2}\right )^{4} f +3 \RootOf \left (4 d f \,\textit {\_Z}^{6}+4 f^{2} \textit {\_Z}^{4}+4 e f \,\textit {\_Z}^{2}+e^{2}\right )^{2} e \right ) \ln \left (-\RootOf \left (4 d f \,\textit {\_Z}^{6}+4 f^{2} \textit {\_Z}^{4}+4 e f \,\textit {\_Z}^{2}+e^{2}\right )+x \right )}{8 f \left (3 d \RootOf \left (4 d f \,\textit {\_Z}^{6}+4 f^{2} \textit {\_Z}^{4}+4 e f \,\textit {\_Z}^{2}+e^{2}\right )^{5}+2 f \RootOf \left (4 d f \,\textit {\_Z}^{6}+4 f^{2} \textit {\_Z}^{4}+4 e f \,\textit {\_Z}^{2}+e^{2}\right )^{3}+e \RootOf \left (4 d f \,\textit {\_Z}^{6}+4 f^{2} \textit {\_Z}^{4}+4 e f \,\textit {\_Z}^{2}+e^{2}\right )\right )} \]

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(-_R+x),_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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (2 \, f x^{2} + 3 \, e\right )} x^{2}}{4 \, d f x^{6} + 4 \, f^{2} x^{4} + 4 \, e f x^{2} + e^{2}}\,{d x} \]

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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mupad [B]  time = 3.21, size = 278, normalized size = 6.95 \[ \frac {\mathrm {atan}\left (\frac {2\,f^2\,x+2\,d\,f\,x^3-d\,e\,x}{\sqrt {d}\,e\,\sqrt {f}}\right )-\mathrm {atan}\left (\frac {1984\,d^{3/2}\,f^{9/2}\,x^3}{432\,d^2\,e^2\,f^2-128\,d\,e\,f^4}+\frac {1728\,d^{5/2}\,f^{7/2}\,x^5}{432\,d^2\,e^2\,f^2-128\,d\,e\,f^4}+\frac {512\,\sqrt {d}\,f^{13/2}\,x^3}{128\,d\,e^2\,f^4-432\,d^2\,e^3\,f^2}+\frac {512\,d^{3/2}\,f^{11/2}\,x^5}{128\,d\,e^2\,f^4-432\,d^2\,e^3\,f^2}-\frac {256\,\sqrt {d}\,f^{11/2}\,x}{432\,d^2\,e^2\,f^2-128\,d\,e\,f^4}+\frac {864\,d^{3/2}\,e\,f^{7/2}\,x}{432\,d^2\,e^2\,f^2-128\,d\,e\,f^4}-\frac {864\,d^{5/2}\,e\,f^{5/2}\,x^3}{432\,d^2\,e^2\,f^2-128\,d\,e\,f^4}\right )+\mathrm {atan}\left (\frac {\sqrt {d}\,x}{\sqrt {f}}\right )}{2\,\sqrt {d}\,\sqrt {f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((2*f^2*x + 2*d*f*x^3 - d*e*x)/(d^(1/2)*e*f^(1/2))) - atan((1984*d^(3/2)*f^(9/2)*x^3)/(432*d^2*e^2*f^2 -
128*d*e*f^4) + (1728*d^(5/2)*f^(7/2)*x^5)/(432*d^2*e^2*f^2 - 128*d*e*f^4) + (512*d^(1/2)*f^(13/2)*x^3)/(128*d*
e^2*f^4 - 432*d^2*e^3*f^2) + (512*d^(3/2)*f^(11/2)*x^5)/(128*d*e^2*f^4 - 432*d^2*e^3*f^2) - (256*d^(1/2)*f^(11
/2)*x)/(432*d^2*e^2*f^2 - 128*d*e*f^4) + (864*d^(3/2)*e*f^(7/2)*x)/(432*d^2*e^2*f^2 - 128*d*e*f^4) - (864*d^(5
/2)*e*f^(5/2)*x^3)/(432*d^2*e^2*f^2 - 128*d*e*f^4)) + atan((d^(1/2)*x)/f^(1/2)))/(2*d^(1/2)*f^(1/2))

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sympy [B]  time = 1.13, size = 90, normalized size = 2.25 \[ - \frac {\sqrt {- \frac {1}{d f}} \log {\left (- \frac {e \sqrt {- \frac {1}{d f}}}{2} - f x^{2} \sqrt {- \frac {1}{d f}} + x^{3} \right )}}{4} + \frac {\sqrt {- \frac {1}{d f}} \log {\left (\frac {e \sqrt {- \frac {1}{d f}}}{2} + f x^{2} \sqrt {- \frac {1}{d f}} + x^{3} \right )}}{4} \]

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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