### 3.20 $$\int \frac{(A+B x+C x^2) (a+b x^2+c x^4)^2}{x^7} \, dx$$

Optimal. Leaf size=149 $-\frac{a^2 A}{6 x^6}-\frac{a^2 B}{5 x^5}-\frac{A \left (2 a c+b^2\right )+2 a b C}{2 x^2}+\log (x) \left (C \left (2 a c+b^2\right )+2 A b c\right )-\frac{a (a C+2 A b)}{4 x^4}-\frac{B \left (2 a c+b^2\right )}{x}-\frac{2 a b B}{3 x^3}+\frac{1}{2} c x^2 (A c+2 b C)+2 b B c x+\frac{1}{3} B c^2 x^3+\frac{1}{4} c^2 C x^4$

[Out]

-(a^2*A)/(6*x^6) - (a^2*B)/(5*x^5) - (a*(2*A*b + a*C))/(4*x^4) - (2*a*b*B)/(3*x^3) - (A*(b^2 + 2*a*c) + 2*a*b*
C)/(2*x^2) - (B*(b^2 + 2*a*c))/x + 2*b*B*c*x + (c*(A*c + 2*b*C)*x^2)/2 + (B*c^2*x^3)/3 + (c^2*C*x^4)/4 + (2*A*
b*c + (b^2 + 2*a*c)*C)*Log[x]

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Rubi [A]  time = 0.143141, antiderivative size = 149, 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} $-\frac{a^2 A}{6 x^6}-\frac{a^2 B}{5 x^5}-\frac{A \left (2 a c+b^2\right )+2 a b C}{2 x^2}+\log (x) \left (C \left (2 a c+b^2\right )+2 A b c\right )-\frac{a (a C+2 A b)}{4 x^4}-\frac{B \left (2 a c+b^2\right )}{x}-\frac{2 a b B}{3 x^3}+\frac{1}{2} c x^2 (A c+2 b C)+2 b B c x+\frac{1}{3} B c^2 x^3+\frac{1}{4} c^2 C x^4$

Antiderivative was successfully veriﬁed.

[In]

Int[((A + B*x + C*x^2)*(a + b*x^2 + c*x^4)^2)/x^7,x]

[Out]

-(a^2*A)/(6*x^6) - (a^2*B)/(5*x^5) - (a*(2*A*b + a*C))/(4*x^4) - (2*a*b*B)/(3*x^3) - (A*(b^2 + 2*a*c) + 2*a*b*
C)/(2*x^2) - (B*(b^2 + 2*a*c))/x + 2*b*B*c*x + (c*(A*c + 2*b*C)*x^2)/2 + (B*c^2*x^3)/3 + (c^2*C*x^4)/4 + (2*A*
b*c + (b^2 + 2*a*c)*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^7} \, dx &=\int \left (2 b B c+\frac{a^2 A}{x^7}+\frac{a^2 B}{x^6}+\frac{a (2 A b+a C)}{x^5}+\frac{2 a b B}{x^4}+\frac{A \left (b^2+2 a c\right )+2 a b C}{x^3}+\frac{B \left (b^2+2 a c\right )}{x^2}+\frac{2 A b c+\left (b^2+2 a c\right ) C}{x}+c (A c+2 b C) x+B c^2 x^2+c^2 C x^3\right ) \, dx\\ &=-\frac{a^2 A}{6 x^6}-\frac{a^2 B}{5 x^5}-\frac{a (2 A b+a C)}{4 x^4}-\frac{2 a b B}{3 x^3}-\frac{A \left (b^2+2 a c\right )+2 a b C}{2 x^2}-\frac{B \left (b^2+2 a c\right )}{x}+2 b B c x+\frac{1}{2} c (A c+2 b C) x^2+\frac{1}{3} B c^2 x^3+\frac{1}{4} c^2 C x^4+\left (2 A b c+\left (b^2+2 a c\right ) C\right ) \log (x)\\ \end{align*}

Mathematica [A]  time = 0.10065, size = 144, normalized size = 0.97 $-\frac{a^2 (10 A+3 x (4 B+5 C x))}{60 x^6}+\log (x) \left (C \left (2 a c+b^2\right )+2 A b c\right )-\frac{a \left (3 A \left (b+2 c x^2\right )+2 x \left (2 b B+3 b C x+6 B c x^2\right )\right )}{6 x^4}+\frac{A \left (c^2 x^4-b^2\right )}{2 x^2}-\frac{b^2 B}{x}+b c x (2 B+C x)+\frac{1}{12} c^2 x^3 (4 B+3 C x)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[((A + B*x + C*x^2)*(a + b*x^2 + c*x^4)^2)/x^7,x]

[Out]

-((b^2*B)/x) + b*c*x*(2*B + C*x) + (c^2*x^3*(4*B + 3*C*x))/12 + (A*(-b^2 + c^2*x^4))/(2*x^2) - (a^2*(10*A + 3*
x*(4*B + 5*C*x)))/(60*x^6) - (a*(3*A*(b + 2*c*x^2) + 2*x*(2*b*B + 3*b*C*x + 6*B*c*x^2)))/(6*x^4) + (2*A*b*c +
(b^2 + 2*a*c)*C)*Log[x]

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Maple [A]  time = 0.009, size = 148, normalized size = 1. \begin{align*}{\frac{{c}^{2}C{x}^{4}}{4}}+{\frac{B{c}^{2}{x}^{3}}{3}}+{\frac{A{x}^{2}{c}^{2}}{2}}+C{x}^{2}bc+2\,bBcx-{\frac{Aab}{2\,{x}^{4}}}-{\frac{C{a}^{2}}{4\,{x}^{4}}}-2\,{\frac{Bac}{x}}-{\frac{B{b}^{2}}{x}}-{\frac{aAc}{{x}^{2}}}-{\frac{A{b}^{2}}{2\,{x}^{2}}}-{\frac{abC}{{x}^{2}}}-{\frac{A{a}^{2}}{6\,{x}^{6}}}-{\frac{B{a}^{2}}{5\,{x}^{5}}}-{\frac{2\,Bab}{3\,{x}^{3}}}+2\,A\ln \left ( x \right ) bc+2\,C\ln \left ( x \right ) ac+C\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^7,x)

[Out]

1/4*c^2*C*x^4+1/3*B*c^2*x^3+1/2*A*x^2*c^2+C*x^2*b*c+2*b*B*c*x-1/2*a/x^4*A*b-1/4*a^2/x^4*C-2*B/x*a*c-B/x*b^2-1/
x^2*a*A*c-1/2/x^2*A*b^2-1/x^2*a*b*C-1/6*a^2*A/x^6-1/5*a^2*B/x^5-2/3*a*b*B/x^3+2*A*ln(x)*b*c+2*C*ln(x)*a*c+C*ln
(x)*b^2

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Maxima [A]  time = 0.952514, size = 189, normalized size = 1.27 \begin{align*} \frac{1}{4} \, C c^{2} x^{4} + \frac{1}{3} \, B c^{2} x^{3} + 2 \, B b c x + \frac{1}{2} \,{\left (2 \, C b c + A c^{2}\right )} x^{2} +{\left (C b^{2} + 2 \,{\left (C a + A b\right )} c\right )} \log \left (x\right ) - \frac{40 \, B a b x^{3} + 60 \,{\left (B b^{2} + 2 \, B a c\right )} x^{5} + 30 \,{\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} + 12 \, B a^{2} x + 10 \, A a^{2} + 15 \,{\left (C a^{2} + 2 \, A a b\right )} x^{2}}{60 \, x^{6}} \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^7,x, algorithm="maxima")

[Out]

1/4*C*c^2*x^4 + 1/3*B*c^2*x^3 + 2*B*b*c*x + 1/2*(2*C*b*c + A*c^2)*x^2 + (C*b^2 + 2*(C*a + A*b)*c)*log(x) - 1/6
0*(40*B*a*b*x^3 + 60*(B*b^2 + 2*B*a*c)*x^5 + 30*(2*C*a*b + A*b^2 + 2*A*a*c)*x^4 + 12*B*a^2*x + 10*A*a^2 + 15*(
C*a^2 + 2*A*a*b)*x^2)/x^6

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Fricas [A]  time = 1.21679, size = 346, normalized size = 2.32 \begin{align*} \frac{15 \, C c^{2} x^{10} + 20 \, B c^{2} x^{9} + 120 \, B b c x^{7} + 30 \,{\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} \log \left (x\right ) - 40 \, B a b x^{3} - 60 \,{\left (B b^{2} + 2 \, B a c\right )} x^{5} - 30 \,{\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} - 12 \, B a^{2} x - 10 \, A a^{2} - 15 \,{\left (C a^{2} + 2 \, A a b\right )} x^{2}}{60 \, x^{6}} \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^7,x, algorithm="fricas")

[Out]

1/60*(15*C*c^2*x^10 + 20*B*c^2*x^9 + 120*B*b*c*x^7 + 30*(2*C*b*c + A*c^2)*x^8 + 60*(C*b^2 + 2*(C*a + A*b)*c)*x
^6*log(x) - 40*B*a*b*x^3 - 60*(B*b^2 + 2*B*a*c)*x^5 - 30*(2*C*a*b + A*b^2 + 2*A*a*c)*x^4 - 12*B*a^2*x - 10*A*a
^2 - 15*(C*a^2 + 2*A*a*b)*x^2)/x^6

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Sympy [A]  time = 29.7179, size = 153, normalized size = 1.03 \begin{align*} 2 B b c x + \frac{B c^{2} x^{3}}{3} + \frac{C c^{2} x^{4}}{4} + x^{2} \left (\frac{A c^{2}}{2} + C b c\right ) + \left (2 A b c + 2 C a c + C b^{2}\right ) \log{\left (x \right )} - \frac{10 A a^{2} + 12 B a^{2} x + 40 B a b x^{3} + x^{5} \left (120 B a c + 60 B b^{2}\right ) + x^{4} \left (60 A a c + 30 A b^{2} + 60 C a b\right ) + x^{2} \left (30 A a b + 15 C a^{2}\right )}{60 x^{6}} \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**7,x)

[Out]

2*B*b*c*x + B*c**2*x**3/3 + C*c**2*x**4/4 + x**2*(A*c**2/2 + C*b*c) + (2*A*b*c + 2*C*a*c + C*b**2)*log(x) - (1
0*A*a**2 + 12*B*a**2*x + 40*B*a*b*x**3 + x**5*(120*B*a*c + 60*B*b**2) + x**4*(60*A*a*c + 30*A*b**2 + 60*C*a*b)
+ x**2*(30*A*a*b + 15*C*a**2))/(60*x**6)

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Giac [A]  time = 1.12944, size = 190, normalized size = 1.28 \begin{align*} \frac{1}{4} \, C c^{2} x^{4} + \frac{1}{3} \, B c^{2} x^{3} + C b c x^{2} + \frac{1}{2} \, A c^{2} x^{2} + 2 \, B b c x +{\left (C b^{2} + 2 \, C a c + 2 \, A b c\right )} \log \left ({\left | x \right |}\right ) - \frac{40 \, B a b x^{3} + 60 \,{\left (B b^{2} + 2 \, B a c\right )} x^{5} + 30 \,{\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} + 12 \, B a^{2} x + 10 \, A a^{2} + 15 \,{\left (C a^{2} + 2 \, A a b\right )} x^{2}}{60 \, x^{6}} \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^7,x, algorithm="giac")

[Out]

1/4*C*c^2*x^4 + 1/3*B*c^2*x^3 + C*b*c*x^2 + 1/2*A*c^2*x^2 + 2*B*b*c*x + (C*b^2 + 2*C*a*c + 2*A*b*c)*log(abs(x)
) - 1/60*(40*B*a*b*x^3 + 60*(B*b^2 + 2*B*a*c)*x^5 + 30*(2*C*a*b + A*b^2 + 2*A*a*c)*x^4 + 12*B*a^2*x + 10*A*a^2
+ 15*(C*a^2 + 2*A*a*b)*x^2)/x^6