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

Optimal result

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

[Out]

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

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.95 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^4} \, dx=-\frac {A b}{x}+A c x+b C x+\frac {1}{2} B c x^2+\frac {1}{3} c C x^3-\frac {a (2 A+3 x (B+2 C x))}{6 x^3}+b B \log (x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {c C \,x^{3}}{3}+\frac {B \,x^{2} c}{2}+A c x +C b x +b B \ln \left (x \right )-\frac {a B}{2 x^{2}}-\frac {A b +C a}{x}-\frac {a A}{3 x^{3}}\)	\(55\)
risch	\(\frac {c C \,x^{3}}{3}+\frac {B \,x^{2} c}{2}+A c x +C b x +\frac {\left (-A b -C a \right ) x^{2}-\frac {B a x}{2}-\frac {A a}{3}}{x^{3}}+b B \ln \left (x \right )\)	\(56\)
norman	\(\frac {\left (-A b -C a \right ) x^{2}+\left (A c +C b \right ) x^{4}-\frac {A a}{3}-\frac {B a x}{2}+\frac {B c \,x^{5}}{2}+\frac {c C \,x^{6}}{3}}{x^{3}}+b B \ln \left (x \right )\)	\(59\)
parallelrisch	\(\frac {2 c C \,x^{6}+3 B c \,x^{5}+6 A c \,x^{4}+6 B b \ln \left (x \right ) x^{3}+6 C b \,x^{4}-6 A b \,x^{2}-6 C a \,x^{2}-3 B a x -2 A a}{6 x^{3}}\)	\(67\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^4} \, dx=\frac {2 \, C c x^{6} + 3 \, B c x^{5} + 6 \, B b x^{3} \log \left (x\right ) + 6 \, {\left (C b + A c\right )} x^{4} - 3 \, B a x - 6 \, {\left (C a + A b\right )} x^{2} - 2 \, A a}{6 \, x^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^4} \, dx=B b \log {\left (x \right )} + \frac {B c x^{2}}{2} + \frac {C c x^{3}}{3} + x \left (A c + C b\right ) + \frac {- 2 A a - 3 B a x + x^{2} \left (- 6 A b - 6 C a\right )}{6 x^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^4} \, dx=\frac {1}{3} \, C c x^{3} + \frac {1}{2} \, B c x^{2} + B b \log \left (x\right ) + {\left (C b + A c\right )} x - \frac {3 \, B a x + 6 \, {\left (C a + A b\right )} x^{2} + 2 \, A a}{6 \, x^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^4} \, dx=\frac {1}{3} \, C c x^{3} + \frac {1}{2} \, B c x^{2} + C b x + A c x + B b \log \left ({\left | x \right |}\right ) - \frac {3 \, B a x + 6 \, {\left (C a + A b\right )} x^{2} + 2 \, A a}{6 \, x^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int \frac {\left (A+B x+C x^2\right ) \left (a+b x^2+c x^4\right )}{x^4} \, dx=x\,\left (A\,c+C\,b\right )-\frac {\left (A\,b+C\,a\right )\,x^2+\frac {B\,a\,x}{2}+\frac {A\,a}{3}}{x^3}+\frac {B\,c\,x^2}{2}+\frac {C\,c\,x^3}{3}+B\,b\,\ln \left (x\right ) \]

[In]

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

[Out]

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