3.18 \(\int \frac{(A+B x+C x^2) (a+b x^2+c x^4)^2}{x^5} \, dx\)

Optimal. Leaf size=148 \[ -\frac{a^2 A}{4 x^4}-\frac{a^2 B}{3 x^3}+\frac{1}{2} x^2 \left (C \left (2 a c+b^2\right )+2 A b c\right )+\log (x) \left (A \left (2 a c+b^2\right )+2 a b C\right )-\frac{a (a C+2 A b)}{2 x^2}+B x \left (2 a c+b^2\right )-\frac{2 a b B}{x}+\frac{1}{4} c x^4 (A c+2 b C)+\frac{2}{3} b B c x^3+\frac{1}{5} B c^2 x^5+\frac{1}{6} c^2 C x^6 \]

[Out]

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

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Rubi [A]  time = 0.141568, antiderivative size = 148, 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} \[ -\frac{a^2 A}{4 x^4}-\frac{a^2 B}{3 x^3}+\frac{1}{2} x^2 \left (C \left (2 a c+b^2\right )+2 A b c\right )+\log (x) \left (A \left (2 a c+b^2\right )+2 a b C\right )-\frac{a (a C+2 A b)}{2 x^2}+B x \left (2 a c+b^2\right )-\frac{2 a b B}{x}+\frac{1}{4} c x^4 (A c+2 b C)+\frac{2}{3} b B c x^3+\frac{1}{5} B c^2 x^5+\frac{1}{6} c^2 C x^6 \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x + C*x^2)*(a + b*x^2 + c*x^4)^2)/x^5,x]

[Out]

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

Mathematica [A]  time = 0.0826919, size = 130, normalized size = 0.88 \[ -\frac{a^2 \left (3 A+4 B x+6 C x^2\right )}{12 x^4}+\log (x) \left (A \left (2 a c+b^2\right )+2 a b C\right )+\frac{a \left (-A b-2 b B x+c x^3 (2 B+C x)\right )}{x^2}+\frac{1}{60} x \left (10 b c x (6 A+x (4 B+3 C x))+c^2 x^3 (15 A+2 x (6 B+5 C x))+30 b^2 (2 B+C x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x + C*x^2)*(a + b*x^2 + c*x^4)^2)/x^5,x]

[Out]

-(a^2*(3*A + 4*B*x + 6*C*x^2))/(12*x^4) + (a*(-(A*b) - 2*b*B*x + c*x^3*(2*B + C*x)))/x^2 + (x*(30*b^2*(2*B + C
*x) + 10*b*c*x*(6*A + x*(4*B + 3*C*x)) + c^2*x^3*(15*A + 2*x*(6*B + 5*C*x))))/60 + (A*(b^2 + 2*a*c) + 2*a*b*C)
*Log[x]

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Maple [A]  time = 0.008, size = 144, normalized size = 1. \begin{align*}{\frac{{c}^{2}C{x}^{6}}{6}}+{\frac{B{c}^{2}{x}^{5}}{5}}+{\frac{A{x}^{4}{c}^{2}}{4}}+{\frac{C{x}^{4}bc}{2}}+{\frac{2\,bBc{x}^{3}}{3}}+A{x}^{2}bc+C{x}^{2}ac+{\frac{C{x}^{2}{b}^{2}}{2}}+2\,Bacx+B{b}^{2}x-2\,{\frac{Bab}{x}}-{\frac{Aab}{{x}^{2}}}-{\frac{C{a}^{2}}{2\,{x}^{2}}}-{\frac{A{a}^{2}}{4\,{x}^{4}}}-{\frac{B{a}^{2}}{3\,{x}^{3}}}+2\,A\ln \left ( x \right ) ac+A\ln \left ( x \right ){b}^{2}+2\,C\ln \left ( x \right ) ab \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^5,x)

[Out]

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

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Maxima [A]  time = 0.983737, size = 188, normalized size = 1.27 \begin{align*} \frac{1}{6} \, C c^{2} x^{6} + \frac{1}{5} \, B c^{2} x^{5} + \frac{2}{3} \, B b c x^{3} + \frac{1}{4} \,{\left (2 \, C b c + A c^{2}\right )} x^{4} + \frac{1}{2} \,{\left (C b^{2} + 2 \,{\left (C a + A b\right )} c\right )} x^{2} +{\left (B b^{2} + 2 \, B a c\right )} x +{\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} \log \left (x\right ) - \frac{24 \, B a b x^{3} + 4 \, B a^{2} x + 3 \, A a^{2} + 6 \,{\left (C a^{2} + 2 \, A a b\right )} x^{2}}{12 \, x^{4}} \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^5,x, algorithm="maxima")

[Out]

1/6*C*c^2*x^6 + 1/5*B*c^2*x^5 + 2/3*B*b*c*x^3 + 1/4*(2*C*b*c + A*c^2)*x^4 + 1/2*(C*b^2 + 2*(C*a + A*b)*c)*x^2
+ (B*b^2 + 2*B*a*c)*x + (2*C*a*b + A*b^2 + 2*A*a*c)*log(x) - 1/12*(24*B*a*b*x^3 + 4*B*a^2*x + 3*A*a^2 + 6*(C*a
^2 + 2*A*a*b)*x^2)/x^4

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Fricas [A]  time = 1.23705, size = 346, normalized size = 2.34 \begin{align*} \frac{10 \, C c^{2} x^{10} + 12 \, B c^{2} x^{9} + 40 \, B b c x^{7} + 15 \,{\left (2 \, C b c + A c^{2}\right )} x^{8} + 30 \,{\left (C b^{2} + 2 \,{\left (C a + A b\right )} c\right )} x^{6} - 120 \, B a b x^{3} + 60 \,{\left (B b^{2} + 2 \, B a c\right )} x^{5} + 60 \,{\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} \log \left (x\right ) - 20 \, B a^{2} x - 15 \, A a^{2} - 30 \,{\left (C a^{2} + 2 \, A a b\right )} x^{2}}{60 \, x^{4}} \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^5,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.62166, size = 151, normalized size = 1.02 \begin{align*} \frac{2 B b c x^{3}}{3} + \frac{B c^{2} x^{5}}{5} + \frac{C c^{2} x^{6}}{6} + x^{4} \left (\frac{A c^{2}}{4} + \frac{C b c}{2}\right ) + x^{2} \left (A b c + C a c + \frac{C b^{2}}{2}\right ) + x \left (2 B a c + B b^{2}\right ) + \left (2 A a c + A b^{2} + 2 C a b\right ) \log{\left (x \right )} - \frac{3 A a^{2} + 4 B a^{2} x + 24 B a b x^{3} + x^{2} \left (12 A a b + 6 C a^{2}\right )}{12 x^{4}} \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**5,x)

[Out]

2*B*b*c*x**3/3 + B*c**2*x**5/5 + C*c**2*x**6/6 + x**4*(A*c**2/4 + C*b*c/2) + x**2*(A*b*c + C*a*c + C*b**2/2) +
x*(2*B*a*c + B*b**2) + (2*A*a*c + A*b**2 + 2*C*a*b)*log(x) - (3*A*a**2 + 4*B*a**2*x + 24*B*a*b*x**3 + x**2*(1
2*A*a*b + 6*C*a**2))/(12*x**4)

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Giac [A]  time = 1.11442, size = 192, normalized size = 1.3 \begin{align*} \frac{1}{6} \, C c^{2} x^{6} + \frac{1}{5} \, B c^{2} x^{5} + \frac{1}{2} \, C b c x^{4} + \frac{1}{4} \, A c^{2} x^{4} + \frac{2}{3} \, B b c x^{3} + \frac{1}{2} \, C b^{2} x^{2} + C a c x^{2} + A b c x^{2} + B b^{2} x + 2 \, B a c x +{\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} \log \left ({\left | x \right |}\right ) - \frac{24 \, B a b x^{3} + 4 \, B a^{2} x + 3 \, A a^{2} + 6 \,{\left (C a^{2} + 2 \, A a b\right )} x^{2}}{12 \, x^{4}} \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^5,x, algorithm="giac")

[Out]

1/6*C*c^2*x^6 + 1/5*B*c^2*x^5 + 1/2*C*b*c*x^4 + 1/4*A*c^2*x^4 + 2/3*B*b*c*x^3 + 1/2*C*b^2*x^2 + C*a*c*x^2 + A*
b*c*x^2 + B*b^2*x + 2*B*a*c*x + (2*C*a*b + A*b^2 + 2*A*a*c)*log(abs(x)) - 1/12*(24*B*a*b*x^3 + 4*B*a^2*x + 3*A
*a^2 + 6*(C*a^2 + 2*A*a*b)*x^2)/x^4