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

Optimal. Leaf size=83 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (3 c d^2-e (a e+b d)\right )}{2 d^{3/2} e^{5/2}}+\frac{x \left (a+\frac{d (c d-b e)}{e^2}\right )}{2 d \left (d+e x^2\right )}+\frac{c x}{e^2} \]

[Out]

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

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Rubi [A]  time = 0.0933651, antiderivative size = 83, 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, 388, 205} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (3 c d^2-e (a e+b d)\right )}{2 d^{3/2} e^{5/2}}+\frac{x \left (a+\frac{d (c d-b e)}{e^2}\right )}{2 d \left (d+e x^2\right )}+\frac{c x}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*x)/e^2 + ((a + (d*(c*d - b*e))/e^2)*x)/(2*d*(d + e*x^2)) - ((3*c*d^2 - e*(b*d + a*e))*ArcTan[(Sqrt[e]*x)/Sq
rt[d]])/(2*d^(3/2)*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 388

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

Mathematica [A]  time = 0.0472368, size = 88, normalized size = 1.06 \[ \frac{x \left (a e^2-b d e+c d^2\right )}{2 d e^2 \left (d+e x^2\right )}-\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (-a e^2-b d e+3 c d^2\right )}{2 d^{3/2} e^{5/2}}+\frac{c x}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0., size = 118, normalized size = 1.4 \begin{align*}{\frac{cx}{{e}^{2}}}+{\frac{xa}{2\,d \left ( e{x}^{2}+d \right ) }}-{\frac{bx}{2\,e \left ( e{x}^{2}+d \right ) }}+{\frac{dxc}{2\,{e}^{2} \left ( e{x}^{2}+d \right ) }}+{\frac{a}{2\,d}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{b}{2\,e}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}-{\frac{3\,cd}{2\,{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)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65278, size = 541, normalized size = 6.52 \begin{align*} \left [\frac{4 \, c d^{2} e^{2} x^{3} +{\left (3 \, c d^{3} - b d^{2} e - a d e^{2} +{\left (3 \, c d^{2} e - 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 ) + 2 \,{\left (3 \, c d^{3} e - b d^{2} e^{2} + a d e^{3}\right )} x}{4 \,{\left (d^{2} e^{4} x^{2} + d^{3} e^{3}\right )}}, \frac{2 \, c d^{2} e^{2} x^{3} -{\left (3 \, c d^{3} - b d^{2} e - a d e^{2} +{\left (3 \, c d^{2} e - b d e^{2} - a e^{3}\right )} x^{2}\right )} \sqrt{d e} \arctan \left (\frac{\sqrt{d e} x}{d}\right ) +{\left (3 \, c d^{3} e - b d^{2} e^{2} + a d e^{3}\right )} x}{2 \,{\left (d^{2} e^{4} x^{2} + d^{3} 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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 1.03622, size = 153, normalized size = 1.84 \begin{align*} \frac{c x}{e^{2}} + \frac{x \left (a e^{2} - b d e + c d^{2}\right )}{2 d^{2} e^{2} + 2 d e^{3} x^{2}} - \frac{\sqrt{- \frac{1}{d^{3} e^{5}}} \left (a e^{2} + b d e - 3 c d^{2}\right ) \log{\left (- d^{2} e^{2} \sqrt{- \frac{1}{d^{3} e^{5}}} + x \right )}}{4} + \frac{\sqrt{- \frac{1}{d^{3} e^{5}}} \left (a e^{2} + b d e - 3 c d^{2}\right ) \log{\left (d^{2} e^{2} \sqrt{- \frac{1}{d^{3} e^{5}}} + x \right )}}{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)**2,x)

[Out]

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

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Giac [A]  time = 1.15544, size = 101, normalized size = 1.22 \begin{align*} c x e^{\left (-2\right )} - \frac{{\left (3 \, c d^{2} - b d e - a e^{2}\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{5}{2}\right )}}{2 \, d^{\frac{3}{2}}} + \frac{{\left (c d^{2} x - b d x e + a x e^{2}\right )} e^{\left (-2\right )}}{2 \,{\left (x^{2} e + d\right )} d} \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)^2,x, algorithm="giac")

[Out]

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