3.19 \(\int \frac{(A+B x+C x^2) (a+b x^2+c x^4)^2}{x^6} \, dx\)

Optimal. Leaf size=143 \[ -\frac{a^2 A}{5 x^5}-\frac{a^2 B}{4 x^4}+x \left (C \left (2 a c+b^2\right )+2 A b c\right )-\frac{A \left (2 a c+b^2\right )+2 a b C}{x}-\frac{a (a C+2 A b)}{3 x^3}+B \log (x) \left (2 a c+b^2\right )-\frac{a b B}{x^2}+\frac{1}{3} c x^3 (A c+2 b C)+b B c x^2+\frac{1}{4} B c^2 x^4+\frac{1}{5} c^2 C x^5 \]

[Out]

-(a^2*A)/(5*x^5) - (a^2*B)/(4*x^4) - (a*(2*A*b + a*C))/(3*x^3) - (a*b*B)/x^2 - (A*(b^2 + 2*a*c) + 2*a*b*C)/x +
(2*A*b*c + (b^2 + 2*a*c)*C)*x + b*B*c*x^2 + (c*(A*c + 2*b*C)*x^3)/3 + (B*c^2*x^4)/4 + (c^2*C*x^5)/5 + B*(b^2
+ 2*a*c)*Log[x]

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Rubi [A]  time = 0.146662, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.036, Rules used = {1628} \[ -\frac{a^2 A}{5 x^5}-\frac{a^2 B}{4 x^4}+x \left (C \left (2 a c+b^2\right )+2 A b c\right )-\frac{A \left (2 a c+b^2\right )+2 a b C}{x}-\frac{a (a C+2 A b)}{3 x^3}+B \log (x) \left (2 a c+b^2\right )-\frac{a b B}{x^2}+\frac{1}{3} c x^3 (A c+2 b C)+b B c x^2+\frac{1}{4} B c^2 x^4+\frac{1}{5} c^2 C x^5 \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x + C*x^2)*(a + b*x^2 + c*x^4)^2)/x^6,x]

[Out]

-(a^2*A)/(5*x^5) - (a^2*B)/(4*x^4) - (a*(2*A*b + a*C))/(3*x^3) - (a*b*B)/x^2 - (A*(b^2 + 2*a*c) + 2*a*b*C)/x +
(2*A*b*c + (b^2 + 2*a*c)*C)*x + b*B*c*x^2 + (c*(A*c + 2*b*C)*x^3)/3 + (B*c^2*x^4)/4 + (c^2*C*x^5)/5 + B*(b^2
+ 2*a*c)*Log[x]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin{align*} \int \frac{\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^6} \, dx &=\int \left (2 A b c \left (1+\frac{b \left (1+\frac{2 a c}{b^2}\right ) C}{2 A c}\right )+\frac{a^2 A}{x^6}+\frac{a^2 B}{x^5}+\frac{a (2 A b+a C)}{x^4}+\frac{2 a b B}{x^3}+\frac{A \left (b^2+2 a c\right )+2 a b C}{x^2}+\frac{B \left (b^2+2 a c\right )}{x}+2 b B c x+c (A c+2 b C) x^2+B c^2 x^3+c^2 C x^4\right ) \, dx\\ &=-\frac{a^2 A}{5 x^5}-\frac{a^2 B}{4 x^4}-\frac{a (2 A b+a C)}{3 x^3}-\frac{a b B}{x^2}-\frac{A \left (b^2+2 a c\right )+2 a b C}{x}+\left (2 A b c+\left (b^2+2 a c\right ) C\right ) x+b B c x^2+\frac{1}{3} c (A c+2 b C) x^3+\frac{1}{4} B c^2 x^4+\frac{1}{5} c^2 C x^5+B \left (b^2+2 a c\right ) \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0823089, size = 142, normalized size = 0.99 \[ -\frac{a^2 A}{5 x^5}-\frac{a^2 B}{4 x^4}-\frac{2 a A c+2 a b C+A b^2}{x}-\frac{a (a C+2 A b)}{3 x^3}+B \log (x) \left (2 a c+b^2\right )+C x \left (2 a c+b^2\right )-\frac{a b B}{x^2}+\frac{1}{3} c x^3 (A c+2 b C)+2 A b c x+b B c x^2+\frac{1}{4} B c^2 x^4+\frac{1}{5} c^2 C x^5 \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x + C*x^2)*(a + b*x^2 + c*x^4)^2)/x^6,x]

[Out]

-(a^2*A)/(5*x^5) - (a^2*B)/(4*x^4) - (a*(2*A*b + a*C))/(3*x^3) - (a*b*B)/x^2 - (A*b^2 + 2*a*A*c + 2*a*b*C)/x +
2*A*b*c*x + (b^2 + 2*a*c)*C*x + b*B*c*x^2 + (c*(A*c + 2*b*C)*x^3)/3 + (B*c^2*x^4)/4 + (c^2*C*x^5)/5 + B*(b^2
+ 2*a*c)*Log[x]

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Maple [A]  time = 0.007, size = 144, normalized size = 1. \begin{align*}{\frac{{c}^{2}C{x}^{5}}{5}}+{\frac{B{c}^{2}{x}^{4}}{4}}+{\frac{A{x}^{3}{c}^{2}}{3}}+{\frac{2\,C{x}^{3}bc}{3}}+bBc{x}^{2}+2\,Abcx+2\,acCx+{b}^{2}Cx-2\,{\frac{aAc}{x}}-{\frac{A{b}^{2}}{x}}-2\,{\frac{abC}{x}}-{\frac{A{a}^{2}}{5\,{x}^{5}}}-{\frac{Bab}{{x}^{2}}}-{\frac{B{a}^{2}}{4\,{x}^{4}}}-{\frac{2\,Aab}{3\,{x}^{3}}}-{\frac{C{a}^{2}}{3\,{x}^{3}}}+2\,B\ln \left ( x \right ) ac+B\ln \left ( x \right ){b}^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)*(c*x^4+b*x^2+a)^2/x^6,x)

[Out]

1/5*c^2*C*x^5+1/4*B*c^2*x^4+1/3*A*x^3*c^2+2/3*C*x^3*b*c+b*B*c*x^2+2*A*b*c*x+2*a*c*C*x+b^2*C*x-2/x*a*A*c-1/x*A*
b^2-2/x*a*b*C-1/5*a^2*A/x^5-a*b*B/x^2-1/4*a^2*B/x^4-2/3*a/x^3*A*b-1/3*a^2/x^3*C+2*B*ln(x)*a*c+B*ln(x)*b^2

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Maxima [A]  time = 0.970014, size = 186, normalized size = 1.3 \begin{align*} \frac{1}{5} \, C c^{2} x^{5} + \frac{1}{4} \, B c^{2} x^{4} + B b c x^{2} + \frac{1}{3} \,{\left (2 \, C b c + A c^{2}\right )} x^{3} +{\left (C b^{2} + 2 \,{\left (C a + A b\right )} c\right )} x +{\left (B b^{2} + 2 \, B a c\right )} \log \left (x\right ) - \frac{60 \, B a b x^{3} + 60 \,{\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} + 15 \, B a^{2} x + 12 \, A a^{2} + 20 \,{\left (C a^{2} + 2 \, A a b\right )} x^{2}}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(c*x^4+b*x^2+a)^2/x^6,x, algorithm="maxima")

[Out]

1/5*C*c^2*x^5 + 1/4*B*c^2*x^4 + B*b*c*x^2 + 1/3*(2*C*b*c + A*c^2)*x^3 + (C*b^2 + 2*(C*a + A*b)*c)*x + (B*b^2 +
2*B*a*c)*log(x) - 1/60*(60*B*a*b*x^3 + 60*(2*C*a*b + A*b^2 + 2*A*a*c)*x^4 + 15*B*a^2*x + 12*A*a^2 + 20*(C*a^2
+ 2*A*a*b)*x^2)/x^5

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Fricas [A]  time = 1.20422, size = 344, normalized size = 2.41 \begin{align*} \frac{12 \, C c^{2} x^{10} + 15 \, B c^{2} x^{9} + 60 \, B b c x^{7} + 20 \,{\left (2 \, C b c + A c^{2}\right )} x^{8} + 60 \,{\left (C b^{2} + 2 \,{\left (C a + A b\right )} c\right )} x^{6} + 60 \,{\left (B b^{2} + 2 \, B a c\right )} x^{5} \log \left (x\right ) - 60 \, B a b x^{3} - 60 \,{\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} - 15 \, B a^{2} x - 12 \, A a^{2} - 20 \,{\left (C a^{2} + 2 \, A a b\right )} x^{2}}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(c*x^4+b*x^2+a)^2/x^6,x, algorithm="fricas")

[Out]

1/60*(12*C*c^2*x^10 + 15*B*c^2*x^9 + 60*B*b*c*x^7 + 20*(2*C*b*c + A*c^2)*x^8 + 60*(C*b^2 + 2*(C*a + A*b)*c)*x^
6 + 60*(B*b^2 + 2*B*a*c)*x^5*log(x) - 60*B*a*b*x^3 - 60*(2*C*a*b + A*b^2 + 2*A*a*c)*x^4 - 15*B*a^2*x - 12*A*a^
2 - 20*(C*a^2 + 2*A*a*b)*x^2)/x^5

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Sympy [A]  time = 8.27827, size = 151, normalized size = 1.06 \begin{align*} B b c x^{2} + \frac{B c^{2} x^{4}}{4} + B \left (2 a c + b^{2}\right ) \log{\left (x \right )} + \frac{C c^{2} x^{5}}{5} + x^{3} \left (\frac{A c^{2}}{3} + \frac{2 C b c}{3}\right ) + x \left (2 A b c + 2 C a c + C b^{2}\right ) - \frac{12 A a^{2} + 15 B a^{2} x + 60 B a b x^{3} + x^{4} \left (120 A a c + 60 A b^{2} + 120 C a b\right ) + x^{2} \left (40 A a b + 20 C a^{2}\right )}{60 x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)*(c*x**4+b*x**2+a)**2/x**6,x)

[Out]

B*b*c*x**2 + B*c**2*x**4/4 + B*(2*a*c + b**2)*log(x) + C*c**2*x**5/5 + x**3*(A*c**2/3 + 2*C*b*c/3) + x*(2*A*b*
c + 2*C*a*c + C*b**2) - (12*A*a**2 + 15*B*a**2*x + 60*B*a*b*x**3 + x**4*(120*A*a*c + 60*A*b**2 + 120*C*a*b) +
x**2*(40*A*a*b + 20*C*a**2))/(60*x**5)

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Giac [A]  time = 1.12074, size = 189, normalized size = 1.32 \begin{align*} \frac{1}{5} \, C c^{2} x^{5} + \frac{1}{4} \, B c^{2} x^{4} + \frac{2}{3} \, C b c x^{3} + \frac{1}{3} \, A c^{2} x^{3} + B b c x^{2} + C b^{2} x + 2 \, C a c x + 2 \, A b c x +{\left (B b^{2} + 2 \, B a c\right )} \log \left ({\left | x \right |}\right ) - \frac{60 \, B a b x^{3} + 60 \,{\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} + 15 \, B a^{2} x + 12 \, A a^{2} + 20 \,{\left (C a^{2} + 2 \, A a b\right )} x^{2}}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(c*x^4+b*x^2+a)^2/x^6,x, algorithm="giac")

[Out]

1/5*C*c^2*x^5 + 1/4*B*c^2*x^4 + 2/3*C*b*c*x^3 + 1/3*A*c^2*x^3 + B*b*c*x^2 + C*b^2*x + 2*C*a*c*x + 2*A*b*c*x +
(B*b^2 + 2*B*a*c)*log(abs(x)) - 1/60*(60*B*a*b*x^3 + 60*(2*C*a*b + A*b^2 + 2*A*a*c)*x^4 + 15*B*a^2*x + 12*A*a^
2 + 20*(C*a^2 + 2*A*a*b)*x^2)/x^5