\(\int \frac {(A+B x+C x^2) (a+b x^2+c x^4)^2}{x^5} \, dx\) [18]

Optimal result

Integrand size = 28, antiderivative size = 148 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^5} \, dx=-\frac {a^2 A}{4 x^4}-\frac {a^2 B}{3 x^3}-\frac {a (2 A b+a C)}{2 x^2}-\frac {2 a b B}{x}+B \left (b^2+2 a c\right ) x+\frac {1}{2} \left (2 A b c+\left (b^2+2 a c\right ) C\right ) x^2+\frac {2}{3} b B c x^3+\frac {1}{4} c (A c+2 b C) x^4+\frac {1}{5} B c^2 x^5+\frac {1}{6} c^2 C x^6+\left (A \left (b^2+2 a c\right )+2 a b C\right ) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {1642} \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^5} \, dx=-\frac {a^2 A}{4 x^4}-\frac {a^2 B}{3 x^3}+\frac {1}{2} x^2 \left (C \left (2 a c+b^2\right )+2 A b c\right )+\log (x) \left (A \left (2 a c+b^2\right )+2 a b C\right )-\frac {a (a C+2 A b)}{2 x^2}+B x \left (2 a c+b^2\right )-\frac {2 a b B}{x}+\frac {1}{4} c x^4 (A c+2 b C)+\frac {2}{3} b B c x^3+\frac {1}{5} B c^2 x^5+\frac {1}{6} c^2 C x^6 \]

[In]

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

[Out]

-1/4*(a^2*A)/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 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*} \text {integral}& = \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] (verified)

Time = 0.06 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.88 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^5} \, dx=-\frac {a^2 \left (3 A+4 B x+6 C x^2\right )}{12 x^4}+\frac {a \left (-A b-2 b B x+c x^3 (2 B+C x)\right )}{x^2}+\frac {1}{60} x \left (30 b^2 (2 B+C x)+10 b c x (6 A+x (4 B+3 C x))+c^2 x^3 (15 A+2 x (6 B+5 C x))\right )+\left (A \left (b^2+2 a c\right )+2 a b C\right ) \log (x) \]

[In]

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

[Out]

-1/12*(a^2*(3*A + 4*B*x + 6*C*x^2))/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]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {c^{2} C \,x^{6}}{6}+\frac {B \,c^{2} x^{5}}{5}+\frac {A \,c^{2} x^{4}}{4}+\frac {C b c \,x^{4}}{2}+\frac {2 b B c \,x^{3}}{3}+A b c \,x^{2}+C a c \,x^{2}+\frac {C \,b^{2} x^{2}}{2}+2 B a c x +B \,b^{2} x +\left (2 A a c +A \,b^{2}+2 a b C \right ) \ln \left (x \right )-\frac {a \left (2 A b +C a \right )}{2 x^{2}}-\frac {2 a b B}{x}-\frac {a^{2} A}{4 x^{4}}-\frac {a^{2} B}{3 x^{3}}\)	\(139\)
norman	\(\frac {\left (\frac {1}{4} A \,c^{2}+\frac {1}{2} C b c \right ) x^{8}+\left (-A a b -\frac {1}{2} C \,a^{2}\right ) x^{2}+\left (A b c +a c C +\frac {1}{2} b^{2} C \right ) x^{6}+\left (2 B a c +B \,b^{2}\right ) x^{5}-\frac {A \,a^{2}}{4}-\frac {B \,a^{2} x}{3}+\frac {B \,c^{2} x^{9}}{5}+\frac {c^{2} C \,x^{10}}{6}-2 B a b \,x^{3}+\frac {2 b B c \,x^{7}}{3}}{x^{4}}+\left (2 A a c +A \,b^{2}+2 a b C \right ) \ln \left (x \right )\)	\(139\)
risch	\(\frac {c^{2} C \,x^{6}}{6}+\frac {B \,c^{2} x^{5}}{5}+\frac {A \,c^{2} x^{4}}{4}+\frac {C b c \,x^{4}}{2}+\frac {2 b B c \,x^{3}}{3}+A b c \,x^{2}+C a c \,x^{2}+\frac {C \,b^{2} x^{2}}{2}+2 B a c x +B \,b^{2} x +\frac {-2 B a b \,x^{3}+\left (-A a b -\frac {1}{2} C \,a^{2}\right ) x^{2}-\frac {B \,a^{2} x}{3}-\frac {A \,a^{2}}{4}}{x^{4}}+2 A \ln \left (x \right ) a c +A \ln \left (x \right ) b^{2}+2 C \ln \left (x \right ) a b\)	\(143\)
parallelrisch	\(\frac {10 c^{2} C \,x^{10}+12 B \,c^{2} x^{9}+15 A \,c^{2} x^{8}+30 C b c \,x^{8}+40 b B c \,x^{7}+60 A b c \,x^{6}+60 C a c \,x^{6}+30 C \,b^{2} x^{6}+120 A \ln \left (x \right ) x^{4} a c +60 A \ln \left (x \right ) x^{4} b^{2}+120 B a c \,x^{5}+60 B \,b^{2} x^{5}+120 C \ln \left (x \right ) x^{4} a b -120 B a b \,x^{3}-60 A a b \,x^{2}-30 C \,a^{2} x^{2}-20 B \,a^{2} x -15 A \,a^{2}}{60 x^{4}}\)	\(161\)

[In]

int((C*x^2+B*x+A)*(c*x^4+b*x^2+a)^2/x^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^5} \, dx=\frac {10 \, C c^{2} x^{10} + 12 \, B c^{2} x^{9} + 40 \, B b c x^{7} + 15 \, {\left (2 \, C b c + A c^{2}\right )} x^{8} + 30 \, {\left (C b^{2} + 2 \, {\left (C a + A b\right )} c\right )} x^{6} - 120 \, B a b x^{3} + 60 \, {\left (B b^{2} + 2 \, B a c\right )} x^{5} + 60 \, {\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} \log \left (x\right ) - 20 \, B a^{2} x - 15 \, A a^{2} - 30 \, {\left (C a^{2} + 2 \, A a b\right )} x^{2}}{60 \, x^{4}} \]

[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

Sympy [A] (verification not implemented)

Time = 1.32 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.03 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^5} \, dx=\frac {2 B b c x^{3}}{3} + \frac {B c^{2} x^{5}}{5} + \frac {C c^{2} x^{6}}{6} + x^{4} \left (\frac {A c^{2}}{4} + \frac {C b c}{2}\right ) + x^{2} \left (A b c + C a c + \frac {C b^{2}}{2}\right ) + x \left (2 B a c + B b^{2}\right ) + \left (2 A a c + A b^{2} + 2 C a b\right ) \log {\left (x \right )} + \frac {- 3 A a^{2} - 4 B a^{2} x - 24 B a b x^{3} + x^{2} \left (- 12 A a b - 6 C a^{2}\right )}{12 x^{4}} \]

[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*(
-12*A*a*b - 6*C*a**2))/(12*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.94 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^5} \, dx=\frac {1}{6} \, C c^{2} x^{6} + \frac {1}{5} \, B c^{2} x^{5} + \frac {2}{3} \, B b c x^{3} + \frac {1}{4} \, {\left (2 \, C b c + A c^{2}\right )} x^{4} + \frac {1}{2} \, {\left (C b^{2} + 2 \, {\left (C a + A b\right )} c\right )} x^{2} + {\left (B b^{2} + 2 \, B a c\right )} x + {\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} \log \left (x\right ) - \frac {24 \, B a b x^{3} + 4 \, B a^{2} x + 3 \, A a^{2} + 6 \, {\left (C a^{2} + 2 \, A a b\right )} x^{2}}{12 \, x^{4}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.96 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^5} \, dx=\frac {1}{6} \, C c^{2} x^{6} + \frac {1}{5} \, B c^{2} x^{5} + \frac {1}{2} \, C b c x^{4} + \frac {1}{4} \, A c^{2} x^{4} + \frac {2}{3} \, B b c x^{3} + \frac {1}{2} \, C b^{2} x^{2} + C a c x^{2} + A b c x^{2} + B b^{2} x + 2 \, B a c x + {\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} \log \left ({\left | x \right |}\right ) - \frac {24 \, B a b x^{3} + 4 \, B a^{2} x + 3 \, A a^{2} + 6 \, {\left (C a^{2} + 2 \, A a b\right )} x^{2}}{12 \, x^{4}} \]

[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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^5} \, dx=x^4\,\left (\frac {A\,c^2}{4}+\frac {C\,b\,c}{2}\right )-\frac {x^2\,\left (\frac {C\,a^2}{2}+A\,b\,a\right )+\frac {A\,a^2}{4}+\frac {B\,a^2\,x}{3}+2\,B\,a\,b\,x^3}{x^4}+x^2\,\left (\frac {C\,b^2}{2}+A\,c\,b+C\,a\,c\right )+\ln \left (x\right )\,\left (A\,b^2+2\,C\,a\,b+2\,A\,a\,c\right )+\frac {B\,c^2\,x^5}{5}+\frac {C\,c^2\,x^6}{6}+B\,x\,\left (b^2+2\,a\,c\right )+\frac {2\,B\,b\,c\,x^3}{3} \]

[In]

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

[Out]

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