3.129 \(\int \frac{c+d x^2+e x^4+f x^6}{x^4 (a+b x^2)^2} \, dx\)

Optimal. Leaf size=121 \[ \frac{x \left (\frac{b^2 c}{a^2}-\frac{b d}{a}-\frac{a f}{b}+e\right )}{2 a \left (a+b x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ) \left (a^2 b e+a^3 f-3 a b^2 d+5 b^3 c\right )}{2 a^{7/2} b^{3/2}}+\frac{2 b c-a d}{a^3 x}-\frac{c}{3 a^2 x^3} \]

[Out]

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

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Rubi [A]  time = 0.157549, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {1805, 1261, 205} \[ \frac{x \left (\frac{b^2 c}{a^2}-\frac{b d}{a}-\frac{a f}{b}+e\right )}{2 a \left (a+b x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ) \left (a^2 b e+a^3 f-3 a b^2 d+5 b^3 c\right )}{2 a^{7/2} b^{3/2}}+\frac{2 b c-a d}{a^3 x}-\frac{c}{3 a^2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2 + e*x^4 + f*x^6)/(x^4*(a + b*x^2)^2),x]

[Out]

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 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{c+d x^2+e x^4+f x^6}{x^4 \left (a+b x^2\right )^2} \, dx &=\frac{\left (\frac{b^2 c}{a^2}-\frac{b d}{a}+e-\frac{a f}{b}\right ) x}{2 a \left (a+b x^2\right )}-\frac{\int \frac{-2 c+2 \left (\frac{b c}{a}-d\right ) x^2+\left (-\frac{b^2 c}{a^2}+\frac{b d}{a}-e-\frac{a f}{b}\right ) x^4}{x^4 \left (a+b x^2\right )} \, dx}{2 a}\\ &=\frac{\left (\frac{b^2 c}{a^2}-\frac{b d}{a}+e-\frac{a f}{b}\right ) x}{2 a \left (a+b x^2\right )}-\frac{\int \left (-\frac{2 c}{a x^4}-\frac{2 (-2 b c+a d)}{a^2 x^2}+\frac{-5 b^3 c+3 a b^2 d-a^2 b e-a^3 f}{a^2 b \left (a+b x^2\right )}\right ) \, dx}{2 a}\\ &=-\frac{c}{3 a^2 x^3}+\frac{2 b c-a d}{a^3 x}+\frac{\left (\frac{b^2 c}{a^2}-\frac{b d}{a}+e-\frac{a f}{b}\right ) x}{2 a \left (a+b x^2\right )}+\frac{\left (5 b^3 c-3 a b^2 d+a^2 b e+a^3 f\right ) \int \frac{1}{a+b x^2} \, dx}{2 a^3 b}\\ &=-\frac{c}{3 a^2 x^3}+\frac{2 b c-a d}{a^3 x}+\frac{\left (\frac{b^2 c}{a^2}-\frac{b d}{a}+e-\frac{a f}{b}\right ) x}{2 a \left (a+b x^2\right )}+\frac{\left (5 b^3 c-3 a b^2 d+a^2 b e+a^3 f\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2} b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0723972, size = 125, normalized size = 1.03 \[ -\frac{x \left (-a^2 b e+a^3 f+a b^2 d-b^3 c\right )}{2 a^3 b \left (a+b x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ) \left (a^2 b e+a^3 f-3 a b^2 d+5 b^3 c\right )}{2 a^{7/2} b^{3/2}}+\frac{2 b c-a d}{a^3 x}-\frac{c}{3 a^2 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2 + e*x^4 + f*x^6)/(x^4*(a + b*x^2)^2),x]

[Out]

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

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Maple [A]  time = 0.011, size = 182, normalized size = 1.5 \begin{align*} -{\frac{c}{3\,{x}^{3}{a}^{2}}}-{\frac{d}{{a}^{2}x}}+2\,{\frac{bc}{{a}^{3}x}}-{\frac{fx}{2\,b \left ( b{x}^{2}+a \right ) }}+{\frac{ex}{2\,a \left ( b{x}^{2}+a \right ) }}-{\frac{bdx}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) }}+{\frac{{b}^{2}xc}{2\,{a}^{3} \left ( b{x}^{2}+a \right ) }}+{\frac{f}{2\,b}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{e}{2\,a}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{3\,bd}{2\,{a}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{5\,{b}^{2}c}{2\,{a}^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^6+e*x^4+d*x^2+c)/x^4/(b*x^2+a)^2,x)

[Out]

-1/3*c/x^3/a^2-1/a^2/x*d+2/a^3/x*b*c-1/2/b*x/(b*x^2+a)*f+1/2/a*x/(b*x^2+a)*e-1/2/a^2*b*x/(b*x^2+a)*d+1/2/a^3*b
^2*x/(b*x^2+a)*c+1/2/b/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*f+1/2/a/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*e-3/2/a
^2*b/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*d+5/2/a^3*b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(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((f*x^6+e*x^4+d*x^2+c)/x^4/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.46541, size = 786, normalized size = 6.5 \begin{align*} \left [-\frac{4 \, a^{3} b^{2} c - 6 \,{\left (5 \, a b^{4} c - 3 \, a^{2} b^{3} d + a^{3} b^{2} e - a^{4} b f\right )} x^{4} - 4 \,{\left (5 \, a^{2} b^{3} c - 3 \, a^{3} b^{2} d\right )} x^{2} + 3 \,{\left ({\left (5 \, b^{4} c - 3 \, a b^{3} d + a^{2} b^{2} e + a^{3} b f\right )} x^{5} +{\left (5 \, a b^{3} c - 3 \, a^{2} b^{2} d + a^{3} b e + a^{4} f\right )} x^{3}\right )} \sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right )}{12 \,{\left (a^{4} b^{3} x^{5} + a^{5} b^{2} x^{3}\right )}}, -\frac{2 \, a^{3} b^{2} c - 3 \,{\left (5 \, a b^{4} c - 3 \, a^{2} b^{3} d + a^{3} b^{2} e - a^{4} b f\right )} x^{4} - 2 \,{\left (5 \, a^{2} b^{3} c - 3 \, a^{3} b^{2} d\right )} x^{2} - 3 \,{\left ({\left (5 \, b^{4} c - 3 \, a b^{3} d + a^{2} b^{2} e + a^{3} b f\right )} x^{5} +{\left (5 \, a b^{3} c - 3 \, a^{2} b^{2} d + a^{3} b e + a^{4} f\right )} x^{3}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right )}{6 \,{\left (a^{4} b^{3} x^{5} + a^{5} b^{2} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x^4/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 12.2098, size = 212, normalized size = 1.75 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{7} b^{3}}} \left (a^{3} f + a^{2} b e - 3 a b^{2} d + 5 b^{3} c\right ) \log{\left (- a^{4} b \sqrt{- \frac{1}{a^{7} b^{3}}} + x \right )}}{4} + \frac{\sqrt{- \frac{1}{a^{7} b^{3}}} \left (a^{3} f + a^{2} b e - 3 a b^{2} d + 5 b^{3} c\right ) \log{\left (a^{4} b \sqrt{- \frac{1}{a^{7} b^{3}}} + x \right )}}{4} - \frac{2 a^{2} b c + x^{4} \left (3 a^{3} f - 3 a^{2} b e + 9 a b^{2} d - 15 b^{3} c\right ) + x^{2} \left (6 a^{2} b d - 10 a b^{2} c\right )}{6 a^{4} b x^{3} + 6 a^{3} b^{2} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**6+e*x**4+d*x**2+c)/x**4/(b*x**2+a)**2,x)

[Out]

-sqrt(-1/(a**7*b**3))*(a**3*f + a**2*b*e - 3*a*b**2*d + 5*b**3*c)*log(-a**4*b*sqrt(-1/(a**7*b**3)) + x)/4 + sq
rt(-1/(a**7*b**3))*(a**3*f + a**2*b*e - 3*a*b**2*d + 5*b**3*c)*log(a**4*b*sqrt(-1/(a**7*b**3)) + x)/4 - (2*a**
2*b*c + x**4*(3*a**3*f - 3*a**2*b*e + 9*a*b**2*d - 15*b**3*c) + x**2*(6*a**2*b*d - 10*a*b**2*c))/(6*a**4*b*x**
3 + 6*a**3*b**2*x**5)

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Giac [A]  time = 1.25384, size = 166, normalized size = 1.37 \begin{align*} \frac{{\left (5 \, b^{3} c - 3 \, a b^{2} d + a^{3} f + a^{2} b e\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a^{3} b} + \frac{b^{3} c x - a b^{2} d x - a^{3} f x + a^{2} b x e}{2 \,{\left (b x^{2} + a\right )} a^{3} b} + \frac{6 \, b c x^{2} - 3 \, a d x^{2} - a c}{3 \, a^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x^4/(b*x^2+a)^2,x, algorithm="giac")

[Out]

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