3.291 \(\int \frac{a+b x^2+c x^4}{(d+e x^2)^3} \, dx\)

Optimal. Leaf size=115 \[ -\frac{x \left (5 c d^2-e (3 a e+b d)\right )}{8 d^2 e^2 \left (d+e x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (e (3 a e+b d)+3 c d^2\right )}{8 d^{5/2} e^{5/2}}+\frac{x \left (a+\frac{d (c d-b e)}{e^2}\right )}{4 d \left (d+e x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.116052, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {1157, 385, 205} \[ -\frac{x \left (5 c d^2-e (3 a e+b d)\right )}{8 d^2 e^2 \left (d+e x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (e (3 a e+b d)+3 c d^2\right )}{8 d^{5/2} e^{5/2}}+\frac{x \left (a+\frac{d (c d-b e)}{e^2}\right )}{4 d \left (d+e x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)/(d + e*x^2)^3,x]

[Out]

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

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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{a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx &=\frac{\left (a+\frac{d (c d-b e)}{e^2}\right ) x}{4 d \left (d+e x^2\right )^2}-\frac{\int \frac{-3 a+\frac{d (c d-b e)}{e^2}-\frac{4 c d x^2}{e}}{\left (d+e x^2\right )^2} \, dx}{4 d}\\ &=\frac{\left (a+\frac{d (c d-b e)}{e^2}\right ) x}{4 d \left (d+e x^2\right )^2}-\frac{\left (5 c d^2-e (b d+3 a e)\right ) x}{8 d^2 e^2 \left (d+e x^2\right )}-\frac{\left (-\frac{4 c d^2}{e}+e \left (-3 a+\frac{d (c d-b e)}{e^2}\right )\right ) \int \frac{1}{d+e x^2} \, dx}{8 d^2 e}\\ &=\frac{\left (a+\frac{d (c d-b e)}{e^2}\right ) x}{4 d \left (d+e x^2\right )^2}-\frac{\left (5 c d^2-e (b d+3 a e)\right ) x}{8 d^2 e^2 \left (d+e x^2\right )}+\frac{\left (3 c d^2+e (b d+3 a e)\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{8 d^{5/2} e^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0970242, size = 110, normalized size = 0.96 \[ \frac{x \left (e \left (a e \left (5 d+3 e x^2\right )+b d \left (e x^2-d\right )\right )-c d^2 \left (3 d+5 e x^2\right )\right )}{8 d^2 e^2 \left (d+e x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (e (3 a e+b d)+3 c d^2\right )}{8 d^{5/2} e^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)/(d + e*x^2)^3,x]

[Out]

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

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Maple [A]  time = 0.008, size = 131, normalized size = 1.1 \begin{align*}{\frac{1}{ \left ( e{x}^{2}+d \right ) ^{2}} \left ({\frac{ \left ( 3\,a{e}^{2}+deb-5\,c{d}^{2} \right ){x}^{3}}{8\,{d}^{2}e}}+{\frac{ \left ( 5\,a{e}^{2}-deb-3\,c{d}^{2} \right ) x}{8\,{e}^{2}d}} \right ) }+{\frac{3\,a}{8\,{d}^{2}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{b}{8\,de}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{3\,c}{8\,{e}^{2}}\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((c*x^4+b*x^2+a)/(e*x^2+d)^3,x)

[Out]

(1/8*(3*a*e^2+b*d*e-5*c*d^2)/d^2/e*x^3+1/8*(5*a*e^2-b*d*e-3*c*d^2)/e^2/d*x)/(e*x^2+d)^2+3/8/d^2/(d*e)^(1/2)*ar
ctan(e*x/(d*e)^(1/2))*a+1/8/d/e/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))*b+3/8/e^2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/
2))*c

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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((c*x^4+b*x^2+a)/(e*x^2+d)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.83935, size = 813, normalized size = 7.07 \begin{align*} \left [-\frac{2 \,{\left (5 \, c d^{3} e^{2} - b d^{2} e^{3} - 3 \, a d e^{4}\right )} x^{3} +{\left (3 \, c d^{4} + b d^{3} e + 3 \, a d^{2} e^{2} +{\left (3 \, c d^{2} e^{2} + b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \,{\left (3 \, c d^{3} e + b d^{2} e^{2} + 3 \, a d 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 ) + 2 \,{\left (3 \, c d^{4} e + b d^{3} e^{2} - 5 \, a d^{2} e^{3}\right )} x}{16 \,{\left (d^{3} e^{5} x^{4} + 2 \, d^{4} e^{4} x^{2} + d^{5} e^{3}\right )}}, -\frac{{\left (5 \, c d^{3} e^{2} - b d^{2} e^{3} - 3 \, a d e^{4}\right )} x^{3} -{\left (3 \, c d^{4} + b d^{3} e + 3 \, a d^{2} e^{2} +{\left (3 \, c d^{2} e^{2} + b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \,{\left (3 \, c d^{3} e + b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2}\right )} \sqrt{d e} \arctan \left (\frac{\sqrt{d e} x}{d}\right ) +{\left (3 \, c d^{4} e + b d^{3} e^{2} - 5 \, a d^{2} e^{3}\right )} x}{8 \,{\left (d^{3} e^{5} x^{4} + 2 \, d^{4} e^{4} x^{2} + d^{5} e^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(2*(5*c*d^3*e^2 - b*d^2*e^3 - 3*a*d*e^4)*x^3 + (3*c*d^4 + b*d^3*e + 3*a*d^2*e^2 + (3*c*d^2*e^2 + b*d*e^
3 + 3*a*e^4)*x^4 + 2*(3*c*d^3*e + b*d^2*e^2 + 3*a*d*e^3)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x
^2 + d)) + 2*(3*c*d^4*e + b*d^3*e^2 - 5*a*d^2*e^3)*x)/(d^3*e^5*x^4 + 2*d^4*e^4*x^2 + d^5*e^3), -1/8*((5*c*d^3*
e^2 - b*d^2*e^3 - 3*a*d*e^4)*x^3 - (3*c*d^4 + b*d^3*e + 3*a*d^2*e^2 + (3*c*d^2*e^2 + b*d*e^3 + 3*a*e^4)*x^4 +
2*(3*c*d^3*e + b*d^2*e^2 + 3*a*d*e^3)*x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) + (3*c*d^4*e + b*d^3*e^2 - 5*a*d^2*
e^3)*x)/(d^3*e^5*x^4 + 2*d^4*e^4*x^2 + d^5*e^3)]

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Sympy [A]  time = 1.77191, size = 196, normalized size = 1.7 \begin{align*} - \frac{\sqrt{- \frac{1}{d^{5} e^{5}}} \left (3 a e^{2} + b d e + 3 c d^{2}\right ) \log{\left (- d^{3} e^{2} \sqrt{- \frac{1}{d^{5} e^{5}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{d^{5} e^{5}}} \left (3 a e^{2} + b d e + 3 c d^{2}\right ) \log{\left (d^{3} e^{2} \sqrt{- \frac{1}{d^{5} e^{5}}} + x \right )}}{16} + \frac{x^{3} \left (3 a e^{3} + b d e^{2} - 5 c d^{2} e\right ) + x \left (5 a d e^{2} - b d^{2} e - 3 c d^{3}\right )}{8 d^{4} e^{2} + 16 d^{3} e^{3} x^{2} + 8 d^{2} e^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)/(e*x**2+d)**3,x)

[Out]

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

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Giac [A]  time = 1.10543, size = 136, normalized size = 1.18 \begin{align*} \frac{{\left (3 \, c d^{2} + b d e + 3 \, a e^{2}\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{5}{2}\right )}}{8 \, d^{\frac{5}{2}}} - \frac{{\left (5 \, c d^{2} x^{3} e - b d x^{3} e^{2} + 3 \, c d^{3} x - 3 \, a x^{3} e^{3} + b d^{2} x e - 5 \, a d x e^{2}\right )} e^{\left (-2\right )}}{8 \,{\left (x^{2} e + d\right )}^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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