3.140 \(\int \frac{c+d x^2+e x^4+f x^6}{x^6 (a+b x^2)^3} \, dx\)

Optimal. Leaf size=196 \[ -\frac{x \left (7 a^2 b e-3 a^3 f-11 a b^2 d+15 b^3 c\right )}{8 a^5 \left (a+b x^2\right )}-\frac{x \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{4 a^4 \left (a+b x^2\right )^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ) \left (15 a^2 b e-3 a^3 f-35 a b^2 d+63 b^3 c\right )}{8 a^{11/2} \sqrt{b}}-\frac{a^2 e-3 a b d+6 b^2 c}{a^5 x}+\frac{3 b c-a d}{3 a^4 x^3}-\frac{c}{5 a^3 x^5} \]

[Out]

-c/(5*a^3*x^5) + (3*b*c - a*d)/(3*a^4*x^3) - (6*b^2*c - 3*a*b*d + a^2*e)/(a^5*x) - ((b^3*c - a*b^2*d + a^2*b*e
- a^3*f)*x)/(4*a^4*(a + b*x^2)^2) - ((15*b^3*c - 11*a*b^2*d + 7*a^2*b*e - 3*a^3*f)*x)/(8*a^5*(a + b*x^2)) - (
(63*b^3*c - 35*a*b^2*d + 15*a^2*b*e - 3*a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(11/2)*Sqrt[b])

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Rubi [A]  time = 0.350203, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {1805, 1802, 205} \[ -\frac{x \left (7 a^2 b e-3 a^3 f-11 a b^2 d+15 b^3 c\right )}{8 a^5 \left (a+b x^2\right )}-\frac{x \left (a^2 b e+a^3 (-f)-a b^2 d+b^3 c\right )}{4 a^4 \left (a+b x^2\right )^2}-\frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ) \left (15 a^2 b e-3 a^3 f-35 a b^2 d+63 b^3 c\right )}{8 a^{11/2} \sqrt{b}}-\frac{a^2 e-3 a b d+6 b^2 c}{a^5 x}+\frac{3 b c-a d}{3 a^4 x^3}-\frac{c}{5 a^3 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2 + e*x^4 + f*x^6)/(x^6*(a + b*x^2)^3),x]

[Out]

-c/(5*a^3*x^5) + (3*b*c - a*d)/(3*a^4*x^3) - (6*b^2*c - 3*a*b*d + a^2*e)/(a^5*x) - ((b^3*c - a*b^2*d + a^2*b*e
- a^3*f)*x)/(4*a^4*(a + b*x^2)^2) - ((15*b^3*c - 11*a*b^2*d + 7*a^2*b*e - 3*a^3*f)*x)/(8*a^5*(a + b*x^2)) - (
(63*b^3*c - 35*a*b^2*d + 15*a^2*b*e - 3*a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(11/2)*Sqrt[b])

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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -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^6 \left (a+b x^2\right )^3} \, dx &=-\frac{\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{4 a^4 \left (a+b x^2\right )^2}-\frac{\int \frac{-4 c+4 \left (\frac{b c}{a}-d\right ) x^2-\frac{4 \left (b^2 c-a b d+a^2 e\right ) x^4}{a^2}+\frac{3 \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x^6}{a^3}}{x^6 \left (a+b x^2\right )^2} \, dx}{4 a}\\ &=-\frac{\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{4 a^4 \left (a+b x^2\right )^2}-\frac{\left (15 b^3 c-11 a b^2 d+7 a^2 b e-3 a^3 f\right ) x}{8 a^5 \left (a+b x^2\right )}+\frac{\int \frac{8 c-8 \left (\frac{2 b c}{a}-d\right ) x^2+8 \left (\frac{3 b^2 c}{a^2}-\frac{2 b d}{a}+e\right ) x^4-\frac{\left (15 b^3 c-11 a b^2 d+7 a^2 b e-3 a^3 f\right ) x^6}{a^3}}{x^6 \left (a+b x^2\right )} \, dx}{8 a^2}\\ &=-\frac{\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{4 a^4 \left (a+b x^2\right )^2}-\frac{\left (15 b^3 c-11 a b^2 d+7 a^2 b e-3 a^3 f\right ) x}{8 a^5 \left (a+b x^2\right )}+\frac{\int \left (\frac{8 c}{a x^6}+\frac{8 (-3 b c+a d)}{a^2 x^4}+\frac{8 \left (6 b^2 c-3 a b d+a^2 e\right )}{a^3 x^2}+\frac{-63 b^3 c+35 a b^2 d-15 a^2 b e+3 a^3 f}{a^3 \left (a+b x^2\right )}\right ) \, dx}{8 a^2}\\ &=-\frac{c}{5 a^3 x^5}+\frac{3 b c-a d}{3 a^4 x^3}-\frac{6 b^2 c-3 a b d+a^2 e}{a^5 x}-\frac{\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{4 a^4 \left (a+b x^2\right )^2}-\frac{\left (15 b^3 c-11 a b^2 d+7 a^2 b e-3 a^3 f\right ) x}{8 a^5 \left (a+b x^2\right )}-\frac{\left (63 b^3 c-35 a b^2 d+15 a^2 b e-3 a^3 f\right ) \int \frac{1}{a+b x^2} \, dx}{8 a^5}\\ &=-\frac{c}{5 a^3 x^5}+\frac{3 b c-a d}{3 a^4 x^3}-\frac{6 b^2 c-3 a b d+a^2 e}{a^5 x}-\frac{\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{4 a^4 \left (a+b x^2\right )^2}-\frac{\left (15 b^3 c-11 a b^2 d+7 a^2 b e-3 a^3 f\right ) x}{8 a^5 \left (a+b x^2\right )}-\frac{\left (63 b^3 c-35 a b^2 d+15 a^2 b e-3 a^3 f\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{11/2} \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.110882, size = 196, normalized size = 1. \[ \frac{x \left (-7 a^2 b e+3 a^3 f+11 a b^2 d-15 b^3 c\right )}{8 a^5 \left (a+b x^2\right )}+\frac{x \left (-a^2 b e+a^3 f+a b^2 d-b^3 c\right )}{4 a^4 \left (a+b x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ) \left (-15 a^2 b e+3 a^3 f+35 a b^2 d-63 b^3 c\right )}{8 a^{11/2} \sqrt{b}}+\frac{a^2 (-e)+3 a b d-6 b^2 c}{a^5 x}+\frac{3 b c-a d}{3 a^4 x^3}-\frac{c}{5 a^3 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2 + e*x^4 + f*x^6)/(x^6*(a + b*x^2)^3),x]

[Out]

-c/(5*a^3*x^5) + (3*b*c - a*d)/(3*a^4*x^3) + (-6*b^2*c + 3*a*b*d - a^2*e)/(a^5*x) + ((-(b^3*c) + a*b^2*d - a^2
*b*e + a^3*f)*x)/(4*a^4*(a + b*x^2)^2) + ((-15*b^3*c + 11*a*b^2*d - 7*a^2*b*e + 3*a^3*f)*x)/(8*a^5*(a + b*x^2)
) + ((-63*b^3*c + 35*a*b^2*d - 15*a^2*b*e + 3*a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(11/2)*Sqrt[b])

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Maple [A]  time = 0.016, size = 300, normalized size = 1.5 \begin{align*} -{\frac{c}{5\,{a}^{3}{x}^{5}}}-{\frac{d}{3\,{a}^{3}{x}^{3}}}+{\frac{bc}{{a}^{4}{x}^{3}}}-{\frac{e}{{a}^{3}x}}+3\,{\frac{bd}{{a}^{4}x}}-6\,{\frac{{b}^{2}c}{{a}^{5}x}}+{\frac{3\,{x}^{3}bf}{8\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{7\,{x}^{3}{b}^{2}e}{8\,{a}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{11\,{x}^{3}{b}^{3}d}{8\,{a}^{4} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{15\,{x}^{3}{b}^{4}c}{8\,{a}^{5} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{5\,fx}{8\,a \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{9\,bxe}{8\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{13\,{b}^{2}dx}{8\,{a}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{17\,{b}^{3}cx}{8\,{a}^{4} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{3\,f}{8\,{a}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{15\,be}{8\,{a}^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{35\,{b}^{2}d}{8\,{a}^{4}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{63\,{b}^{3}c}{8\,{a}^{5}}\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^6/(b*x^2+a)^3,x)

[Out]

-1/5*c/a^3/x^5-1/3/a^3/x^3*d+1/a^4/x^3*b*c-1/a^3/x*e+3/a^4/x*b*d-6/a^5/x*b^2*c+3/8/a^2/(b*x^2+a)^2*x^3*b*f-7/8
/a^3/(b*x^2+a)^2*x^3*b^2*e+11/8/a^4/(b*x^2+a)^2*x^3*b^3*d-15/8/a^5/(b*x^2+a)^2*x^3*b^4*c+5/8/a/(b*x^2+a)^2*f*x
-9/8/a^2/(b*x^2+a)^2*b*e*x+13/8/a^3/(b*x^2+a)^2*b^2*d*x-17/8/a^4/(b*x^2+a)^2*b^3*c*x+3/8/a^2/(a*b)^(1/2)*arcta
n(b*x/(a*b)^(1/2))*f-15/8/a^3/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*b*e+35/8/a^4/(a*b)^(1/2)*arctan(b*x/(a*b)^(1
/2))*b^2*d-63/8/a^5/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*b^3*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^6/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54559, size = 1386, normalized size = 7.07 \begin{align*} \left [-\frac{30 \,{\left (63 \, a b^{5} c - 35 \, a^{2} b^{4} d + 15 \, a^{3} b^{3} e - 3 \, a^{4} b^{2} f\right )} x^{8} + 48 \, a^{5} b c + 50 \,{\left (63 \, a^{2} b^{4} c - 35 \, a^{3} b^{3} d + 15 \, a^{4} b^{2} e - 3 \, a^{5} b f\right )} x^{6} + 16 \,{\left (63 \, a^{3} b^{3} c - 35 \, a^{4} b^{2} d + 15 \, a^{5} b e\right )} x^{4} - 16 \,{\left (9 \, a^{4} b^{2} c - 5 \, a^{5} b d\right )} x^{2} - 15 \,{\left ({\left (63 \, b^{5} c - 35 \, a b^{4} d + 15 \, a^{2} b^{3} e - 3 \, a^{3} b^{2} f\right )} x^{9} + 2 \,{\left (63 \, a b^{4} c - 35 \, a^{2} b^{3} d + 15 \, a^{3} b^{2} e - 3 \, a^{4} b f\right )} x^{7} +{\left (63 \, a^{2} b^{3} c - 35 \, a^{3} b^{2} d + 15 \, a^{4} b e - 3 \, a^{5} f\right )} x^{5}\right )} \sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right )}{240 \,{\left (a^{6} b^{3} x^{9} + 2 \, a^{7} b^{2} x^{7} + a^{8} b x^{5}\right )}}, -\frac{15 \,{\left (63 \, a b^{5} c - 35 \, a^{2} b^{4} d + 15 \, a^{3} b^{3} e - 3 \, a^{4} b^{2} f\right )} x^{8} + 24 \, a^{5} b c + 25 \,{\left (63 \, a^{2} b^{4} c - 35 \, a^{3} b^{3} d + 15 \, a^{4} b^{2} e - 3 \, a^{5} b f\right )} x^{6} + 8 \,{\left (63 \, a^{3} b^{3} c - 35 \, a^{4} b^{2} d + 15 \, a^{5} b e\right )} x^{4} - 8 \,{\left (9 \, a^{4} b^{2} c - 5 \, a^{5} b d\right )} x^{2} + 15 \,{\left ({\left (63 \, b^{5} c - 35 \, a b^{4} d + 15 \, a^{2} b^{3} e - 3 \, a^{3} b^{2} f\right )} x^{9} + 2 \,{\left (63 \, a b^{4} c - 35 \, a^{2} b^{3} d + 15 \, a^{3} b^{2} e - 3 \, a^{4} b f\right )} x^{7} +{\left (63 \, a^{2} b^{3} c - 35 \, a^{3} b^{2} d + 15 \, a^{4} b e - 3 \, a^{5} f\right )} x^{5}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right )}{120 \,{\left (a^{6} b^{3} x^{9} + 2 \, a^{7} b^{2} x^{7} + a^{8} b x^{5}\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^6/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[-1/240*(30*(63*a*b^5*c - 35*a^2*b^4*d + 15*a^3*b^3*e - 3*a^4*b^2*f)*x^8 + 48*a^5*b*c + 50*(63*a^2*b^4*c - 35*
a^3*b^3*d + 15*a^4*b^2*e - 3*a^5*b*f)*x^6 + 16*(63*a^3*b^3*c - 35*a^4*b^2*d + 15*a^5*b*e)*x^4 - 16*(9*a^4*b^2*
c - 5*a^5*b*d)*x^2 - 15*((63*b^5*c - 35*a*b^4*d + 15*a^2*b^3*e - 3*a^3*b^2*f)*x^9 + 2*(63*a*b^4*c - 35*a^2*b^3
*d + 15*a^3*b^2*e - 3*a^4*b*f)*x^7 + (63*a^2*b^3*c - 35*a^3*b^2*d + 15*a^4*b*e - 3*a^5*f)*x^5)*sqrt(-a*b)*log(
(b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^6*b^3*x^9 + 2*a^7*b^2*x^7 + a^8*b*x^5), -1/120*(15*(63*a*b^5*c -
35*a^2*b^4*d + 15*a^3*b^3*e - 3*a^4*b^2*f)*x^8 + 24*a^5*b*c + 25*(63*a^2*b^4*c - 35*a^3*b^3*d + 15*a^4*b^2*e
- 3*a^5*b*f)*x^6 + 8*(63*a^3*b^3*c - 35*a^4*b^2*d + 15*a^5*b*e)*x^4 - 8*(9*a^4*b^2*c - 5*a^5*b*d)*x^2 + 15*((6
3*b^5*c - 35*a*b^4*d + 15*a^2*b^3*e - 3*a^3*b^2*f)*x^9 + 2*(63*a*b^4*c - 35*a^2*b^3*d + 15*a^3*b^2*e - 3*a^4*b
*f)*x^7 + (63*a^2*b^3*c - 35*a^3*b^2*d + 15*a^4*b*e - 3*a^5*f)*x^5)*sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a^6*b^3*
x^9 + 2*a^7*b^2*x^7 + a^8*b*x^5)]

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Sympy [A]  time = 135.007, size = 284, normalized size = 1.45 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{11} b}} \left (3 a^{3} f - 15 a^{2} b e + 35 a b^{2} d - 63 b^{3} c\right ) \log{\left (- a^{6} \sqrt{- \frac{1}{a^{11} b}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{a^{11} b}} \left (3 a^{3} f - 15 a^{2} b e + 35 a b^{2} d - 63 b^{3} c\right ) \log{\left (a^{6} \sqrt{- \frac{1}{a^{11} b}} + x \right )}}{16} + \frac{- 24 a^{4} c + x^{8} \left (45 a^{3} b f - 225 a^{2} b^{2} e + 525 a b^{3} d - 945 b^{4} c\right ) + x^{6} \left (75 a^{4} f - 375 a^{3} b e + 875 a^{2} b^{2} d - 1575 a b^{3} c\right ) + x^{4} \left (- 120 a^{4} e + 280 a^{3} b d - 504 a^{2} b^{2} c\right ) + x^{2} \left (- 40 a^{4} d + 72 a^{3} b c\right )}{120 a^{7} x^{5} + 240 a^{6} b x^{7} + 120 a^{5} b^{2} x^{9}} \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**6/(b*x**2+a)**3,x)

[Out]

-sqrt(-1/(a**11*b))*(3*a**3*f - 15*a**2*b*e + 35*a*b**2*d - 63*b**3*c)*log(-a**6*sqrt(-1/(a**11*b)) + x)/16 +
sqrt(-1/(a**11*b))*(3*a**3*f - 15*a**2*b*e + 35*a*b**2*d - 63*b**3*c)*log(a**6*sqrt(-1/(a**11*b)) + x)/16 + (-
24*a**4*c + x**8*(45*a**3*b*f - 225*a**2*b**2*e + 525*a*b**3*d - 945*b**4*c) + x**6*(75*a**4*f - 375*a**3*b*e
+ 875*a**2*b**2*d - 1575*a*b**3*c) + x**4*(-120*a**4*e + 280*a**3*b*d - 504*a**2*b**2*c) + x**2*(-40*a**4*d +
72*a**3*b*c))/(120*a**7*x**5 + 240*a**6*b*x**7 + 120*a**5*b**2*x**9)

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Giac [A]  time = 1.23475, size = 267, normalized size = 1.36 \begin{align*} -\frac{{\left (63 \, b^{3} c - 35 \, a b^{2} d - 3 \, a^{3} f + 15 \, a^{2} b e\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{5}} - \frac{15 \, b^{4} c x^{3} - 11 \, a b^{3} d x^{3} - 3 \, a^{3} b f x^{3} + 7 \, a^{2} b^{2} x^{3} e + 17 \, a b^{3} c x - 13 \, a^{2} b^{2} d x - 5 \, a^{4} f x + 9 \, a^{3} b x e}{8 \,{\left (b x^{2} + a\right )}^{2} a^{5}} - \frac{90 \, b^{2} c x^{4} - 45 \, a b d x^{4} + 15 \, a^{2} x^{4} e - 15 \, a b c x^{2} + 5 \, a^{2} d x^{2} + 3 \, a^{2} c}{15 \, a^{5} 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^6/(b*x^2+a)^3,x, algorithm="giac")

[Out]

-1/8*(63*b^3*c - 35*a*b^2*d - 3*a^3*f + 15*a^2*b*e)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^5) - 1/8*(15*b^4*c*x^3
- 11*a*b^3*d*x^3 - 3*a^3*b*f*x^3 + 7*a^2*b^2*x^3*e + 17*a*b^3*c*x - 13*a^2*b^2*d*x - 5*a^4*f*x + 9*a^3*b*x*e)/
((b*x^2 + a)^2*a^5) - 1/15*(90*b^2*c*x^4 - 45*a*b*d*x^4 + 15*a^2*x^4*e - 15*a*b*c*x^2 + 5*a^2*d*x^2 + 3*a^2*c)
/(a^5*x^5)