\(\int \frac {(A+B x+C x^2) (a+b x^2+c x^4)^2}{x^7} \, dx\) [20]

Optimal result

Integrand size = 28, antiderivative size = 149 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^7} \, dx=-\frac {a^2 A}{6 x^6}-\frac {a^2 B}{5 x^5}-\frac {a (2 A b+a C)}{4 x^4}-\frac {2 a b B}{3 x^3}-\frac {A \left (b^2+2 a c\right )+2 a b C}{2 x^2}-\frac {B \left (b^2+2 a c\right )}{x}+2 b B c x+\frac {1}{2} c (A c+2 b C) x^2+\frac {1}{3} B c^2 x^3+\frac {1}{4} c^2 C x^4+\left (2 A b c+\left (b^2+2 a c\right ) C\right ) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 149, 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} \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^7} \, dx=-\frac {a^2 A}{6 x^6}-\frac {a^2 B}{5 x^5}-\frac {A \left (2 a c+b^2\right )+2 a b C}{2 x^2}+\log (x) \left (C \left (2 a c+b^2\right )+2 A b c\right )-\frac {a (a C+2 A b)}{4 x^4}-\frac {B \left (2 a c+b^2\right )}{x}-\frac {2 a b B}{3 x^3}+\frac {1}{2} c x^2 (A c+2 b C)+2 b B c x+\frac {1}{3} B c^2 x^3+\frac {1}{4} c^2 C x^4 \]

[In]

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

[Out]

-1/6*(a^2*A)/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 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 (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] (verified)

Time = 0.07 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.97 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^7} \, dx=-\frac {b^2 B}{x}+b c x (2 B+C x)+\frac {1}{12} c^2 x^3 (4 B+3 C x)+\frac {A \left (-b^2+c^2 x^4\right )}{2 x^2}-\frac {a^2 (10 A+3 x (4 B+5 C x))}{60 x^6}-\frac {a \left (3 A \left (b+2 c x^2\right )+2 x \left (2 b B+3 b C x+6 B c x^2\right )\right )}{6 x^4}+\left (2 A b c+\left (b^2+2 a c\right ) C\right ) \log (x) \]

[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]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {c^{2} C \,x^{4}}{4}+\frac {B \,c^{2} x^{3}}{3}+\frac {A \,c^{2} x^{2}}{2}+C b c \,x^{2}+2 B b c x +\left (2 A b c +2 a c C +b^{2} C \right ) \ln \left (x \right )-\frac {2 A a c +A \,b^{2}+2 a b C}{2 x^{2}}-\frac {a^{2} A}{6 x^{6}}-\frac {a^{2} B}{5 x^{5}}-\frac {B \left (2 a c +b^{2}\right )}{x}-\frac {a \left (2 A b +C a \right )}{4 x^{4}}-\frac {2 a b B}{3 x^{3}}\)	\(136\)
norman	\(\frac {\left (\frac {1}{2} A \,c^{2}+C b c \right ) x^{8}+\left (-\frac {1}{2} A a b -\frac {1}{4} C \,a^{2}\right ) x^{2}+\left (-A a c -\frac {1}{2} A \,b^{2}-a b C \right ) x^{4}+\left (-2 B a c -B \,b^{2}\right ) x^{5}-\frac {A \,a^{2}}{6}-\frac {B \,a^{2} x}{5}+\frac {B \,c^{2} x^{9}}{3}+\frac {c^{2} C \,x^{10}}{4}-\frac {2 B a b \,x^{3}}{3}+2 b B c \,x^{7}}{x^{6}}+\left (2 A b c +2 a c C +b^{2} C \right ) \ln \left (x \right )\)	\(141\)
risch	\(\frac {c^{2} C \,x^{4}}{4}+\frac {B \,c^{2} x^{3}}{3}+\frac {A \,c^{2} x^{2}}{2}+C b c \,x^{2}+2 B b c x +\frac {\left (-2 B a c -B \,b^{2}\right ) x^{5}+\left (-A a c -\frac {1}{2} A \,b^{2}-a b C \right ) x^{4}-\frac {2 B a b \,x^{3}}{3}+\left (-\frac {1}{2} A a b -\frac {1}{4} C \,a^{2}\right ) x^{2}-\frac {B \,a^{2} x}{5}-\frac {A \,a^{2}}{6}}{x^{6}}+2 A \ln \left (x \right ) b c +2 C \ln \left (x \right ) a c +C \ln \left (x \right ) b^{2}\)	\(142\)
parallelrisch	\(\frac {15 c^{2} C \,x^{10}+20 B \,c^{2} x^{9}+30 A \,c^{2} x^{8}+60 C b c \,x^{8}+120 A \ln \left (x \right ) x^{6} b c +120 b B c \,x^{7}+120 C \ln \left (x \right ) x^{6} a c +60 C \ln \left (x \right ) x^{6} b^{2}-120 B a c \,x^{5}-60 B \,b^{2} x^{5}-60 A a c \,x^{4}-30 A \,b^{2} x^{4}-60 C a b \,x^{4}-40 B a b \,x^{3}-30 A a b \,x^{2}-15 C \,a^{2} x^{2}-12 B \,a^{2} x -10 A \,a^{2}}{60 x^{6}}\)	\(161\)

[In]

int((C*x^2+B*x+A)*(c*x^4+b*x^2+a)^2/x^7,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.97 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^7} \, dx=\frac {15 \, C c^{2} x^{10} + 20 \, B c^{2} x^{9} + 120 \, B b c x^{7} + 30 \, {\left (2 \, C b c + A c^{2}\right )} x^{8} + 60 \, {\left (C b^{2} + 2 \, {\left (C a + A b\right )} c\right )} x^{6} \log \left (x\right ) - 40 \, B a b x^{3} - 60 \, {\left (B b^{2} + 2 \, B a c\right )} x^{5} - 30 \, {\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} - 12 \, B a^{2} x - 10 \, A a^{2} - 15 \, {\left (C a^{2} + 2 \, A a b\right )} x^{2}}{60 \, x^{6}} \]

[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

Sympy [A] (verification not implemented)

Time = 28.25 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.06 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^7} \, dx=2 B b c x + \frac {B c^{2} x^{3}}{3} + \frac {C c^{2} x^{4}}{4} + x^{2} \left (\frac {A c^{2}}{2} + C b c\right ) + \left (2 A b c + 2 C a c + C b^{2}\right ) \log {\left (x \right )} + \frac {- 10 A a^{2} - 12 B a^{2} x - 40 B a b x^{3} + x^{5} \left (- 120 B a c - 60 B b^{2}\right ) + x^{4} \left (- 60 A a c - 30 A b^{2} - 60 C a b\right ) + x^{2} \left (- 30 A a b - 15 C a^{2}\right )}{60 x^{6}} \]

[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) + (-
10*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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.94 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^7} \, dx=\frac {1}{4} \, C c^{2} x^{4} + \frac {1}{3} \, B c^{2} x^{3} + 2 \, B b c x + \frac {1}{2} \, {\left (2 \, C b c + A c^{2}\right )} x^{2} + {\left (C b^{2} + 2 \, {\left (C a + A b\right )} c\right )} \log \left (x\right ) - \frac {40 \, B a b x^{3} + 60 \, {\left (B b^{2} + 2 \, B a c\right )} x^{5} + 30 \, {\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} + 12 \, B a^{2} x + 10 \, A a^{2} + 15 \, {\left (C a^{2} + 2 \, A a b\right )} x^{2}}{60 \, x^{6}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.95 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^7} \, dx=\frac {1}{4} \, C c^{2} x^{4} + \frac {1}{3} \, B c^{2} x^{3} + C b c x^{2} + \frac {1}{2} \, A c^{2} x^{2} + 2 \, B b c x + {\left (C b^{2} + 2 \, C a c + 2 \, A b c\right )} \log \left ({\left | x \right |}\right ) - \frac {40 \, B a b x^{3} + 60 \, {\left (B b^{2} + 2 \, B a c\right )} x^{5} + 30 \, {\left (2 \, C a b + A b^{2} + 2 \, A a c\right )} x^{4} + 12 \, B a^{2} x + 10 \, A a^{2} + 15 \, {\left (C a^{2} + 2 \, A a b\right )} x^{2}}{60 \, x^{6}} \]

[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

Mupad [B] (verification not implemented)

Time = 7.87 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.91 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )^2}{x^7} \, dx=x^2\,\left (\frac {A\,c^2}{2}+C\,b\,c\right )-\frac {x^2\,\left (\frac {C\,a^2}{4}+\frac {A\,b\,a}{2}\right )+x^5\,\left (B\,b^2+2\,B\,a\,c\right )+\frac {A\,a^2}{6}+x^4\,\left (\frac {A\,b^2}{2}+C\,a\,b+A\,a\,c\right )+\frac {B\,a^2\,x}{5}+\frac {2\,B\,a\,b\,x^3}{3}}{x^6}+\ln \left (x\right )\,\left (C\,b^2+2\,A\,c\,b+2\,C\,a\,c\right )+\frac {B\,c^2\,x^3}{3}+\frac {C\,c^2\,x^4}{4}+2\,B\,b\,c\,x \]

[In]

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

[Out]

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