\(\int \frac {(a+b x^2)^2 (A+B x+C x^2+D x^3)}{x^4} \, dx\) [77]

Optimal result

Integrand size = 28, antiderivative size = 98 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=-\frac {a^2 A}{3 x^3}-\frac {a^2 B}{2 x^2}-\frac {a (2 A b+a C)}{x}+b (A b+2 a C) x+\frac {1}{2} b (b B+2 a D) x^2+\frac {1}{3} b^2 C x^3+\frac {1}{4} b^2 D x^4+a (2 b B+a D) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 98, 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 = {1816} \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=-\frac {a^2 A}{3 x^3}-\frac {a^2 B}{2 x^2}+b x (2 a C+A b)-\frac {a (a C+2 A b)}{x}+\frac {1}{2} b x^2 (2 a D+b B)+a \log (x) (a D+2 b B)+\frac {1}{3} b^2 C x^3+\frac {1}{4} b^2 D x^4 \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=-\frac {2 a A b}{x}+a b x (2 C+D x)-\frac {a^2 (2 A+3 x (B+2 C x))}{6 x^3}+\frac {1}{12} b^2 x \left (12 A+x \left (6 B+4 C x+3 D x^2\right )\right )+a (2 b B+a D) \log (x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {b^{2} D x^{4}}{4}+\frac {C \,b^{2} x^{3}}{3}+\frac {b^{2} B \,x^{2}}{2}+D a b \,x^{2}+A \,b^{2} x +2 C a b x +a \left (2 B b +D a \right ) \ln \left (x \right )-\frac {a^{2} A}{3 x^{3}}-\frac {a \left (2 A b +C a \right )}{x}-\frac {a^{2} B}{2 x^{2}}\)	\(92\)
norman	\(\frac {\left (\frac {1}{2} B \,b^{2}+D a b \right ) x^{5}+\left (b^{2} A +2 C a b \right ) x^{4}+\left (-2 a b A -C \,a^{2}\right ) x^{2}-\frac {a^{2} A}{3}+\frac {C \,b^{2} x^{6}}{3}-\frac {a^{2} B x}{2}+\frac {b^{2} D x^{7}}{4}}{x^{3}}+\left (2 a b B +D a^{2}\right ) \ln \left (x \right )\)	\(98\)
parallelrisch	\(\frac {3 b^{2} D x^{7}+4 C \,b^{2} x^{6}+6 b^{2} B \,x^{5}+12 D a b \,x^{5}+12 A \,b^{2} x^{4}+24 B \ln \left (x \right ) x^{3} a b +24 C a b \,x^{4}+12 D \ln \left (x \right ) x^{3} a^{2}-24 a A b \,x^{2}-12 C \,a^{2} x^{2}-6 a^{2} B x -4 a^{2} A}{12 x^{3}}\)	\(110\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {3 \, D b^{2} x^{7} + 4 \, C b^{2} x^{6} + 6 \, {\left (2 \, D a b + B b^{2}\right )} x^{5} + 12 \, {\left (2 \, C a b + A b^{2}\right )} x^{4} + 12 \, {\left (D a^{2} + 2 \, B a b\right )} x^{3} \log \left (x\right ) - 6 \, B a^{2} x - 4 \, A a^{2} - 12 \, {\left (C a^{2} + 2 \, A a b\right )} x^{2}}{12 \, x^{3}} \]

[In]

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

[Out]

1/12*(3*D*b^2*x^7 + 4*C*b^2*x^6 + 6*(2*D*a*b + B*b^2)*x^5 + 12*(2*C*a*b + A*b^2)*x^4 + 12*(D*a^2 + 2*B*a*b)*x^
3*log(x) - 6*B*a^2*x - 4*A*a^2 - 12*(C*a^2 + 2*A*a*b)*x^2)/x^3

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {C b^{2} x^{3}}{3} + \frac {D b^{2} x^{4}}{4} + a \left (2 B b + D a\right ) \log {\left (x \right )} + x^{2} \left (\frac {B b^{2}}{2} + D a b\right ) + x \left (A b^{2} + 2 C a b\right ) + \frac {- 2 A a^{2} - 3 B a^{2} x + x^{2} \left (- 12 A a b - 6 C a^{2}\right )}{6 x^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {1}{4} \, D b^{2} x^{4} + \frac {1}{3} \, C b^{2} x^{3} + \frac {1}{2} \, {\left (2 \, D a b + B b^{2}\right )} x^{2} + {\left (2 \, C a b + A b^{2}\right )} x + {\left (D a^{2} + 2 \, B a b\right )} \log \left (x\right ) - \frac {3 \, B a^{2} x + 2 \, A a^{2} + 6 \, {\left (C a^{2} + 2 \, A a b\right )} x^{2}}{6 \, x^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {1}{4} \, D b^{2} x^{4} + \frac {1}{3} \, C b^{2} x^{3} + D a b x^{2} + \frac {1}{2} \, B b^{2} x^{2} + 2 \, C a b x + A b^{2} x + {\left (D a^{2} + 2 \, B a b\right )} \log \left ({\left | x \right |}\right ) - \frac {3 \, B a^{2} x + 2 \, A a^{2} + 6 \, {\left (C a^{2} + 2 \, A a b\right )} x^{2}}{6 \, x^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.83 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {b^2\,x^4\,D}{4}+\frac {a^2\,\ln \left (x^2\right )\,D}{2}-\frac {A\,\left (a^2+6\,a\,b\,x^2-3\,b^2\,x^4\right )}{3\,x^3}+\frac {B\,\left (b^2\,x^4-a^2+4\,a\,b\,x^2\,\ln \left (x\right )\right )}{2\,x^2}+\frac {C\,\left (-3\,a^2+6\,a\,b\,x^2+b^2\,x^4\right )}{3\,x}+a\,b\,x^2\,D \]

[In]

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

[Out]

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