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

Optimal. Leaf size=89 \[ -\frac{x \left (a e^2-b d e+c d^2\right )}{2 d^2 e \left (d+e x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (e (b d-3 a e)+c d^2\right )}{2 d^{5/2} e^{3/2}}-\frac{a}{d^2 x} \]

[Out]

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

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Rubi [A]  time = 0.118452, antiderivative size = 86, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {1259, 453, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (e (b d-3 a e)+c d^2\right )}{2 d^{5/2} e^{3/2}}-\frac{x \left (\frac{c}{e}-\frac{b d-a e}{d^2}\right )}{2 \left (d+e x^2\right )}-\frac{a}{d^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1259

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[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(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] && ILtQ[m/2, 0]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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

Mathematica [A]  time = 0.061662, size = 89, normalized size = 1. \[ -\frac{x \left (a e^2-b d e+c d^2\right )}{2 d^2 e \left (d+e x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (-3 a e^2+b d e+c d^2\right )}{2 d^{5/2} e^{3/2}}-\frac{a}{d^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 121, normalized size = 1.4 \begin{align*} -{\frac{a}{{d}^{2}x}}-{\frac{exa}{2\,{d}^{2} \left ( e{x}^{2}+d \right ) }}+{\frac{xb}{2\,d \left ( e{x}^{2}+d \right ) }}-{\frac{xc}{2\,e \left ( e{x}^{2}+d \right ) }}-{\frac{3\,ae}{2\,{d}^{2}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{b}{2\,d}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{c}{2\,e}\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)/x^2/(e*x^2+d)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.89304, size = 555, normalized size = 6.24 \begin{align*} \left [-\frac{4 \, a d^{2} e^{2} + 2 \,{\left (c d^{3} e - b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2} -{\left ({\left (c d^{2} e + b d e^{2} - 3 \, a e^{3}\right )} x^{3} +{\left (c d^{3} + b d^{2} e - 3 \, a d e^{2}\right )} x\right )} \sqrt{-d e} \log \left (\frac{e x^{2} + 2 \, \sqrt{-d e} x - d}{e x^{2} + d}\right )}{4 \,{\left (d^{3} e^{3} x^{3} + d^{4} e^{2} x\right )}}, -\frac{2 \, a d^{2} e^{2} +{\left (c d^{3} e - b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2} -{\left ({\left (c d^{2} e + b d e^{2} - 3 \, a e^{3}\right )} x^{3} +{\left (c d^{3} + b d^{2} e - 3 \, a d e^{2}\right )} x\right )} \sqrt{d e} \arctan \left (\frac{\sqrt{d e} x}{d}\right )}{2 \,{\left (d^{3} e^{3} x^{3} + d^{4} e^{2} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.33698, size = 155, normalized size = 1.74 \begin{align*} \frac{\sqrt{- \frac{1}{d^{5} e^{3}}} \left (3 a e^{2} - b d e - c d^{2}\right ) \log{\left (- d^{3} e \sqrt{- \frac{1}{d^{5} e^{3}}} + x \right )}}{4} - \frac{\sqrt{- \frac{1}{d^{5} e^{3}}} \left (3 a e^{2} - b d e - c d^{2}\right ) \log{\left (d^{3} e \sqrt{- \frac{1}{d^{5} e^{3}}} + x \right )}}{4} - \frac{2 a d e + x^{2} \left (3 a e^{2} - b d e + c d^{2}\right )}{2 d^{3} e x + 2 d^{2} e^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.09488, size = 112, normalized size = 1.26 \begin{align*} \frac{{\left (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{3}{2}\right )}}{2 \, d^{\frac{5}{2}}} - \frac{{\left (c d^{2} x^{2} - b d x^{2} e + 3 \, a x^{2} e^{2} + 2 \, a d e\right )} e^{\left (-1\right )}}{2 \,{\left (x^{3} e + d x\right )} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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