3.282 \(\int \frac{x^2 (a+b x^2+c x^4)}{(d+e x^2)^2} \, dx\)

Optimal. Leaf size=106 \[ -\frac{x \left (a e^2-b d e+c d^2\right )}{2 e^3 \left (d+e x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (5 c d^2-e (3 b d-a e)\right )}{2 \sqrt{d} e^{7/2}}-\frac{x (2 c d-b e)}{e^3}+\frac{c x^3}{3 e^2} \]

[Out]

-(((2*c*d - b*e)*x)/e^3) + (c*x^3)/(3*e^2) - ((c*d^2 - b*d*e + a*e^2)*x)/(2*e^3*(d + e*x^2)) + ((5*c*d^2 - e*(
3*b*d - a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*Sqrt[d]*e^(7/2))

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Rubi [A]  time = 0.105909, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {1257, 1153, 205} \[ -\frac{x \left (a e^2-b d e+c d^2\right )}{2 e^3 \left (d+e x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (5 c d^2-e (3 b d-a e)\right )}{2 \sqrt{d} e^{7/2}}-\frac{x (2 c d-b e)}{e^3}+\frac{c x^3}{3 e^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*x^2 + c*x^4))/(d + e*x^2)^2,x]

[Out]

-(((2*c*d - b*e)*x)/e^3) + (c*x^3)/(3*e^2) - ((c*d^2 - b*d*e + a*e^2)*x)/(2*e^3*(d + e*x^2)) + ((5*c*d^2 - e*(
3*b*d - a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*Sqrt[d]*e^(7/2))

Rule 1257

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^
(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[1/(2*e^(2*p +
m/2)*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4
)^p - (-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x], x] /; FreeQ[{a, b
, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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 (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^2} \, dx &=-\frac{\left (c d^2-b d e+a e^2\right ) x}{2 e^3 \left (d+e x^2\right )}-\frac{\int \frac{-c d^2+b d e-a e^2+2 e (c d-b e) x^2-2 c e^2 x^4}{d+e x^2} \, dx}{2 e^3}\\ &=-\frac{\left (c d^2-b d e+a e^2\right ) x}{2 e^3 \left (d+e x^2\right )}-\frac{\int \left (2 (2 c d-b e)-2 c e x^2+\frac{-5 c d^2+3 b d e-a e^2}{d+e x^2}\right ) \, dx}{2 e^3}\\ &=-\frac{(2 c d-b e) x}{e^3}+\frac{c x^3}{3 e^2}-\frac{\left (c d^2-b d e+a e^2\right ) x}{2 e^3 \left (d+e x^2\right )}-\frac{\left (-5 c d^2+e (3 b d-a e)\right ) \int \frac{1}{d+e x^2} \, dx}{2 e^3}\\ &=-\frac{(2 c d-b e) x}{e^3}+\frac{c x^3}{3 e^2}-\frac{\left (c d^2-b d e+a e^2\right ) x}{2 e^3 \left (d+e x^2\right )}+\frac{\left (5 c d^2-e (3 b d-a e)\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{2 \sqrt{d} e^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0658701, size = 102, normalized size = 0.96 \[ -\frac{x \left (a e^2-b d e+c d^2\right )}{2 e^3 \left (d+e x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a e^2-3 b d e+5 c d^2\right )}{2 \sqrt{d} e^{7/2}}+\frac{x (b e-2 c d)}{e^3}+\frac{c x^3}{3 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*x^2 + c*x^4))/(d + e*x^2)^2,x]

[Out]

((-2*c*d + b*e)*x)/e^3 + (c*x^3)/(3*e^2) - ((c*d^2 - b*d*e + a*e^2)*x)/(2*e^3*(d + e*x^2)) + ((5*c*d^2 - 3*b*d
*e + a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*Sqrt[d]*e^(7/2))

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Maple [A]  time = 0.008, size = 141, normalized size = 1.3 \begin{align*}{\frac{c{x}^{3}}{3\,{e}^{2}}}+{\frac{bx}{{e}^{2}}}-2\,{\frac{cdx}{{e}^{3}}}-{\frac{xa}{2\,e \left ( e{x}^{2}+d \right ) }}+{\frac{dxb}{2\,{e}^{2} \left ( e{x}^{2}+d \right ) }}-{\frac{xc{d}^{2}}{2\,{e}^{3} \left ( e{x}^{2}+d \right ) }}+{\frac{a}{2\,e}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}-{\frac{3\,bd}{2\,{e}^{2}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{5\,c{d}^{2}}{2\,{e}^{3}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*x^4+b*x^2+a)/(e*x^2+d)^2,x)

[Out]

1/3*c*x^3/e^2+1/e^2*b*x-2/e^3*c*d*x-1/2/e*x/(e*x^2+d)*a+1/2/e^2*x/(e*x^2+d)*d*b-1/2/e^3*x/(e*x^2+d)*c*d^2+1/2/
e/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))*a-3/2/e^2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))*d*b+5/2/e^3/(d*e)^(1/2)*ar
ctan(e*x/(d*e)^(1/2))*c*d^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^2*(c*x^4+b*x^2+a)/(e*x^2+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85131, size = 648, normalized size = 6.11 \begin{align*} \left [\frac{4 \, c d e^{3} x^{5} - 4 \,{\left (5 \, c d^{2} e^{2} - 3 \, b d e^{3}\right )} x^{3} - 3 \,{\left (5 \, c d^{3} - 3 \, b d^{2} e + a d e^{2} +{\left (5 \, c d^{2} e - 3 \, b d e^{2} + a e^{3}\right )} x^{2}\right )} \sqrt{-d e} \log \left (\frac{e x^{2} - 2 \, \sqrt{-d e} x - d}{e x^{2} + d}\right ) - 6 \,{\left (5 \, c d^{3} e - 3 \, b d^{2} e^{2} + a d e^{3}\right )} x}{12 \,{\left (d e^{5} x^{2} + d^{2} e^{4}\right )}}, \frac{2 \, c d e^{3} x^{5} - 2 \,{\left (5 \, c d^{2} e^{2} - 3 \, b d e^{3}\right )} x^{3} + 3 \,{\left (5 \, c d^{3} - 3 \, b d^{2} e + a d e^{2} +{\left (5 \, c d^{2} e - 3 \, b d e^{2} + a e^{3}\right )} x^{2}\right )} \sqrt{d e} \arctan \left (\frac{\sqrt{d e} x}{d}\right ) - 3 \,{\left (5 \, c d^{3} e - 3 \, b d^{2} e^{2} + a d e^{3}\right )} x}{6 \,{\left (d e^{5} x^{2} + d^{2} e^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(4*c*d*e^3*x^5 - 4*(5*c*d^2*e^2 - 3*b*d*e^3)*x^3 - 3*(5*c*d^3 - 3*b*d^2*e + a*d*e^2 + (5*c*d^2*e - 3*b*d
*e^2 + a*e^3)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)) - 6*(5*c*d^3*e - 3*b*d^2*e^2 + a*d
*e^3)*x)/(d*e^5*x^2 + d^2*e^4), 1/6*(2*c*d*e^3*x^5 - 2*(5*c*d^2*e^2 - 3*b*d*e^3)*x^3 + 3*(5*c*d^3 - 3*b*d^2*e
+ a*d*e^2 + (5*c*d^2*e - 3*b*d*e^2 + a*e^3)*x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) - 3*(5*c*d^3*e - 3*b*d^2*e^2
+ a*d*e^3)*x)/(d*e^5*x^2 + d^2*e^4)]

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Sympy [A]  time = 1.21161, size = 160, normalized size = 1.51 \begin{align*} \frac{c x^{3}}{3 e^{2}} - \frac{x \left (a e^{2} - b d e + c d^{2}\right )}{2 d e^{3} + 2 e^{4} x^{2}} - \frac{\sqrt{- \frac{1}{d e^{7}}} \left (a e^{2} - 3 b d e + 5 c d^{2}\right ) \log{\left (- d e^{3} \sqrt{- \frac{1}{d e^{7}}} + x \right )}}{4} + \frac{\sqrt{- \frac{1}{d e^{7}}} \left (a e^{2} - 3 b d e + 5 c d^{2}\right ) \log{\left (d e^{3} \sqrt{- \frac{1}{d e^{7}}} + x \right )}}{4} + \frac{x \left (b e - 2 c d\right )}{e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*x**3/(3*e**2) - x*(a*e**2 - b*d*e + c*d**2)/(2*d*e**3 + 2*e**4*x**2) - sqrt(-1/(d*e**7))*(a*e**2 - 3*b*d*e +
5*c*d**2)*log(-d*e**3*sqrt(-1/(d*e**7)) + x)/4 + sqrt(-1/(d*e**7))*(a*e**2 - 3*b*d*e + 5*c*d**2)*log(d*e**3*s
qrt(-1/(d*e**7)) + x)/4 + x*(b*e - 2*c*d)/e**3

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Giac [A]  time = 1.10959, size = 123, normalized size = 1.16 \begin{align*} \frac{{\left (5 \, c d^{2} - 3 \, b d e + a e^{2}\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{7}{2}\right )}}{2 \, \sqrt{d}} + \frac{1}{3} \,{\left (c x^{3} e^{4} - 6 \, c d x e^{3} + 3 \, b x e^{4}\right )} e^{\left (-6\right )} - \frac{{\left (c d^{2} x - b d x e + a x e^{2}\right )} e^{\left (-3\right )}}{2 \,{\left (x^{2} e + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(5*c*d^2 - 3*b*d*e + a*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-7/2)/sqrt(d) + 1/3*(c*x^3*e^4 - 6*c*d*x*e^3 + 3*
b*x*e^4)*e^(-6) - 1/2*(c*d^2*x - b*d*x*e + a*x*e^2)*e^(-3)/(x^2*e + d)