\(\int \frac {(A+B x+C x^2) (a+b x^2+c x^4)}{x^5} \, dx\) [8]

Optimal result

Integrand size = 26, antiderivative size = 63 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^5} \, dx=-\frac {a A}{4 x^4}-\frac {a B}{3 x^3}-\frac {A b+a C}{2 x^2}-\frac {b B}{x}+B c x+\frac {1}{2} c C x^2+(A c+b C) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {1642} \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^5} \, dx=-\frac {a C+A b}{2 x^2}-\frac {a A}{4 x^4}-\frac {a B}{3 x^3}+\log (x) (A c+b C)-\frac {b B}{x}+B c x+\frac {1}{2} c C x^2 \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^5} \, dx=-\frac {a \left (3 A+4 B x+6 C x^2\right )}{12 x^4}+\frac {-A b-2 b B x+c x^3 (2 B+C x)}{2 x^2}+(A c+b C) \log (x) \]

[In]

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

[Out]

-1/12*(a*(3*A + 4*B*x + 6*C*x^2))/x^4 + (-(A*b) - 2*b*B*x + c*x^3*(2*B + C*x))/(2*x^2) + (A*c + b*C)*Log[x]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {c C \,x^{2}}{2}+B c x +\left (A c +C b \right ) \ln \left (x \right )-\frac {A b +C a}{2 x^{2}}-\frac {b B}{x}-\frac {a A}{4 x^{4}}-\frac {a B}{3 x^{3}}\)	\(56\)
risch	\(\frac {c C \,x^{2}}{2}+B c x +\frac {-B b \,x^{3}+\left (-\frac {A b}{2}-\frac {C a}{2}\right ) x^{2}-\frac {B a x}{3}-\frac {A a}{4}}{x^{4}}+A \ln \left (x \right ) c +C \ln \left (x \right ) b\)	\(57\)
norman	\(\frac {\left (-\frac {A b}{2}-\frac {C a}{2}\right ) x^{2}+B c \,x^{5}-\frac {A a}{4}-\frac {B a x}{3}-B b \,x^{3}+\frac {c C \,x^{6}}{2}}{x^{4}}+\left (A c +C b \right ) \ln \left (x \right )\)	\(59\)
parallelrisch	\(\frac {6 c C \,x^{6}+12 A \ln \left (x \right ) x^{4} c +12 B c \,x^{5}+12 C \ln \left (x \right ) x^{4} b -12 B b \,x^{3}-6 A b \,x^{2}-6 C a \,x^{2}-4 B a x -3 A a}{12 x^{4}}\)	\(69\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^5} \, dx=\frac {6 \, C c x^{6} + 12 \, B c x^{5} + 12 \, {\left (C b + A c\right )} x^{4} \log \left (x\right ) - 12 \, B b x^{3} - 4 \, B a x - 6 \, {\left (C a + A b\right )} x^{2} - 3 \, A a}{12 \, x^{4}} \]

[In]

integrate((C*x^2+B*x+A)*(c*x^4+b*x^2+a)/x^5,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.96 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^5} \, dx=B c x + \frac {C c x^{2}}{2} + \left (A c + C b\right ) \log {\left (x \right )} + \frac {- 3 A a - 4 B a x - 12 B b x^{3} + x^{2} \left (- 6 A b - 6 C a\right )}{12 x^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^5} \, dx=\frac {1}{2} \, C c x^{2} + B c x + {\left (C b + A c\right )} \log \left (x\right ) - \frac {12 \, B b x^{3} + 4 \, B a x + 6 \, {\left (C a + A b\right )} x^{2} + 3 \, A a}{12 \, x^{4}} \]

[In]

integrate((C*x^2+B*x+A)*(c*x^4+b*x^2+a)/x^5,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^5} \, dx=\frac {1}{2} \, C c x^{2} + B c x + {\left (C b + A c\right )} \log \left ({\left | x \right |}\right ) - \frac {12 \, B b x^{3} + 4 \, B a x + 6 \, {\left (C a + A b\right )} x^{2} + 3 \, A a}{12 \, x^{4}} \]

[In]

integrate((C*x^2+B*x+A)*(c*x^4+b*x^2+a)/x^5,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^5} \, dx=\ln \left (x\right )\,\left (A\,c+C\,b\right )-\frac {B\,b\,x^3+\left (\frac {A\,b}{2}+\frac {C\,a}{2}\right )\,x^2+\frac {B\,a\,x}{3}+\frac {A\,a}{4}}{x^4}+B\,c\,x+\frac {C\,c\,x^2}{2} \]

[In]

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

[Out]

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