\(\int \frac {(a+b x^2) (A+B x+C x^2+D x^3)}{x^4} \, dx\) [69]

Optimal result

Integrand size = 26, antiderivative size = 54 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=-\frac {a A}{3 x^3}-\frac {a B}{2 x^2}-\frac {A b+a C}{x}+b C x+\frac {1}{2} b D x^2+(b B+a D) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 54, 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 = {1816} \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=-\frac {a C+A b}{x}-\frac {a A}{3 x^3}+\log (x) (a D+b B)-\frac {a B}{2 x^2}+b C x+\frac {1}{2} b D x^2 \]

[In]

Int[((a + b*x^2)*(A + B*x + C*x^2 + D*x^3))/x^4,x]

[Out]

-1/3*(a*A)/x^3 - (a*B)/(2*x^2) - (A*b + a*C)/x + b*C*x + (b*D*x^2)/2 + (b*B + a*D)*Log[x]

Rule 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \left (b C+\frac {a A}{x^4}+\frac {a B}{x^3}+\frac {A b+a C}{x^2}+\frac {b B+a D}{x}+b D x\right ) \, dx \\ & = -\frac {a A}{3 x^3}-\frac {a B}{2 x^2}-\frac {A b+a C}{x}+b C x+\frac {1}{2} b D x^2+(b B+a D) \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=-\frac {a A}{3 x^3}-\frac {a B}{2 x^2}+\frac {-A b-a C}{x}+b C x+\frac {1}{2} b D x^2+(b B+a D) \log (x) \]

[In]

Integrate[((a + b*x^2)*(A + B*x + C*x^2 + D*x^3))/x^4,x]

[Out]

-1/3*(a*A)/x^3 - (a*B)/(2*x^2) + (-(A*b) - a*C)/x + b*C*x + (b*D*x^2)/2 + (b*B + a*D)*Log[x]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {D b \,x^{2}}{2}+b C x +\left (B b +D a \right ) \ln \left (x \right )-\frac {a A}{3 x^{3}}-\frac {A b +C a}{x}-\frac {a B}{2 x^{2}}\)	\(49\)
norman	\(\frac {\left (-A b -C a \right ) x^{2}+b C \,x^{4}-\frac {A a}{3}-\frac {B a x}{2}+\frac {b D x^{5}}{2}}{x^{3}}+\left (B b +D a \right ) \ln \left (x \right )\)	\(52\)
parallelrisch	\(-\frac {-3 b D x^{5}-6 B \ln \left (x \right ) x^{3} b -6 b C \,x^{4}-6 D \ln \left (x \right ) x^{3} a +6 A b \,x^{2}+6 C a \,x^{2}+3 B a x +2 A a}{6 x^{3}}\)	\(62\)

[In]

int((b*x^2+a)*(D*x^3+C*x^2+B*x+A)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {3 \, D b x^{5} + 6 \, C b x^{4} + 6 \, {\left (D a + B b\right )} x^{3} \log \left (x\right ) - 3 \, B a x - 6 \, {\left (C a + A b\right )} x^{2} - 2 \, A a}{6 \, x^{3}} \]

[In]

integrate((b*x^2+a)*(D*x^3+C*x^2+B*x+A)/x^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=C b x + \frac {D b x^{2}}{2} + \left (B b + D a\right ) \log {\left (x \right )} + \frac {- 2 A a - 3 B a x + x^{2} \left (- 6 A b - 6 C a\right )}{6 x^{3}} \]

[In]

integrate((b*x**2+a)*(D*x**3+C*x**2+B*x+A)/x**4,x)

[Out]

C*b*x + D*b*x**2/2 + (B*b + D*a)*log(x) + (-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 = 0.20 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {1}{2} \, D b x^{2} + C b x + {\left (D a + B b\right )} \log \left (x\right ) - \frac {3 \, B a x + 6 \, {\left (C a + A b\right )} x^{2} + 2 \, A a}{6 \, x^{3}} \]

[In]

integrate((b*x^2+a)*(D*x^3+C*x^2+B*x+A)/x^4,x, algorithm="maxima")

[Out]

1/2*D*b*x^2 + C*b*x + (D*a + B*b)*log(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 = 0.32 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {1}{2} \, D b x^{2} + C b x + {\left (D a + B b\right )} \log \left ({\left | x \right |}\right ) - \frac {3 \, B a x + 6 \, {\left (C a + A b\right )} x^{2} + 2 \, A a}{6 \, x^{3}} \]

[In]

integrate((b*x^2+a)*(D*x^3+C*x^2+B*x+A)/x^4,x, algorithm="giac")

[Out]

1/2*D*b*x^2 + C*b*x + (D*a + 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 = 6.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {b\,x^2\,D}{2}+a\,\ln \left (x\right )\,D+C\,b\,x-\frac {A\,a}{3\,x^3}-\frac {A\,b}{x}-\frac {B\,a}{2\,x^2}-\frac {C\,a}{x}+B\,b\,\ln \left (x\right ) \]

[In]

int(((a + b*x^2)*(A + B*x + C*x^2 + x^3*D))/x^4,x)

[Out]

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