3.1.19 \(\int \frac {(A+B x+C x^2) (a+b x^2+c x^4)^2}{x^6} \, dx\) [19]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 143, normalized size of antiderivative = 1.00, 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 = {1642} \begin {gather*} -\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 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(a^2*A)/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 1642

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*}

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Mathematica [A]
time = 0.05, size = 142, normalized size = 0.99 \begin {gather*} -\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 b^2+2 a A c+2 a b C}{x}+2 A b c x+\left (b^2+2 a c\right ) C 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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(a^2*A)/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 = 0.02, size = 135, normalized size = 0.94

method	result	size
default	\(\frac {c^{2} C \,x^{5}}{5}+\frac {B \,c^{2} x^{4}}{4}+\frac {A \,c^{2} x^{3}}{3}+\frac {2 C b c \,x^{3}}{3}+b B c \,x^{2}+2 b c A x +2 a c C x +C \,b^{2} x -\frac {a^{2} A}{5 x^{5}}-\frac {a^{2} B}{4 x^{4}}-\frac {a b B}{x^{2}}-\frac {2 a c A +A \,b^{2}+2 a b C}{x}-\frac {a \left (2 A b +a C \right )}{3 x^{3}}+B \left (2 a c +b^{2}\right ) \ln \left (x \right )\)	\(135\)
risch	\(\frac {c^{2} C \,x^{5}}{5}+\frac {B \,c^{2} x^{4}}{4}+\frac {A \,c^{2} x^{3}}{3}+\frac {2 C b c \,x^{3}}{3}+b B c \,x^{2}+2 b c A x +2 a c C x +C \,b^{2} x +\frac {\left (-2 a c A -A \,b^{2}-2 a b C \right ) x^{4}-a b B \,x^{3}+\left (-\frac {2}{3} a b A -\frac {1}{3} a^{2} C \right ) x^{2}-\frac {a^{2} B x}{4}-\frac {a^{2} A}{5}}{x^{5}}+2 B \ln \left (x \right ) a c +B \ln \left (x \right ) b^{2}\)	\(139\)
norman	\(\frac {\left (\frac {1}{3} A \,c^{2}+\frac {2}{3} b c C \right ) x^{8}+\left (-\frac {2}{3} a b A -\frac {1}{3} a^{2} C \right ) x^{2}+\left (2 b c A +2 a c C +C \,b^{2}\right ) x^{6}+\left (-2 a c A -A \,b^{2}-2 a b C \right ) x^{4}+b B c \,x^{7}-\frac {a^{2} A}{5}+\frac {B \,c^{2} x^{9}}{4}-\frac {a^{2} B x}{4}+\frac {c^{2} C \,x^{10}}{5}-a b B \,x^{3}}{x^{5}}+\left (2 a c B +b^{2} B \right ) \ln \left (x \right )\)	\(140\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 138, normalized size = 0.97 \begin {gather*} \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 {gather*}

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 = 0.36, size = 145, normalized size = 1.01 \begin {gather*} \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 {gather*}

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 = 4.47, size = 155, normalized size = 1.08 \begin {gather*} 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 {gather*}

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 = 5.55, size = 140, normalized size = 0.98 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.05, size = 136, normalized size = 0.95 \begin {gather*} x^3\,\left (\frac {A\,c^2}{3}+\frac {2\,C\,b\,c}{3}\right )-\frac {x^2\,\left (\frac {C\,a^2}{3}+\frac {2\,A\,b\,a}{3}\right )+\frac {A\,a^2}{5}+x^4\,\left (A\,b^2+2\,C\,a\,b+2\,A\,a\,c\right )+\frac {B\,a^2\,x}{4}+B\,a\,b\,x^3}{x^5}+x\,\left (C\,b^2+2\,A\,c\,b+2\,C\,a\,c\right )+\ln \left (x\right )\,\left (B\,b^2+2\,B\,a\,c\right )+\frac {B\,c^2\,x^4}{4}+\frac {C\,c^2\,x^5}{5}+B\,b\,c\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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