\(\int \frac {(a+b x^2)^2 (A+B x^2)}{x^4} \, dx\) [17]

Optimal result

Integrand size = 20, antiderivative size = 48 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^4} \, dx=-\frac {a^2 A}{3 x^3}-\frac {a (2 A b+a B)}{x}+b (A b+2 a B) x+\frac {1}{3} b^2 B x^3 \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {459} \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^4} \, dx=-\frac {a^2 A}{3 x^3}+b x (2 a B+A b)-\frac {a (a B+2 A b)}{x}+\frac {1}{3} b^2 B x^3 \]

[In]

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

[Out]

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

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^4} \, dx=-\frac {a^2 A}{3 x^3}+\frac {-2 a A b-a^2 B}{x}+b (A b+2 a B) x+\frac {1}{3} b^2 B x^3 \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.48 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96

method	result	size
default	\(\frac {b^{2} B \,x^{3}}{3}+A \,b^{2} x +2 B a b x -\frac {a^{2} A}{3 x^{3}}-\frac {a \left (2 A b +B a \right )}{x}\)	\(46\)
risch	\(\frac {b^{2} B \,x^{3}}{3}+A \,b^{2} x +2 B a b x +\frac {\left (-2 a b A -a^{2} B \right ) x^{2}-\frac {a^{2} A}{3}}{x^{3}}\)	\(50\)
norman	\(\frac {\frac {b^{2} B \,x^{6}}{3}+\left (b^{2} A +2 a b B \right ) x^{4}+\left (-2 a b A -a^{2} B \right ) x^{2}-\frac {a^{2} A}{3}}{x^{3}}\)	\(52\)
gosper	\(-\frac {-b^{2} B \,x^{6}-3 A \,b^{2} x^{4}-6 B a b \,x^{4}+6 a A b \,x^{2}+3 a^{2} B \,x^{2}+a^{2} A}{3 x^{3}}\)	\(55\)
parallelrisch	\(\frac {b^{2} B \,x^{6}+3 A \,b^{2} x^{4}+6 B a b \,x^{4}-6 a A b \,x^{2}-3 a^{2} B \,x^{2}-a^{2} A}{3 x^{3}}\)	\(55\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^4} \, dx=\frac {B b^{2} x^{6} + 3 \, {\left (2 \, B a b + A b^{2}\right )} x^{4} - A a^{2} - 3 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{3 \, x^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^4} \, dx=\frac {B b^{2} x^{3}}{3} + x \left (A b^{2} + 2 B a b\right ) + \frac {- A a^{2} + x^{2} \left (- 6 A a b - 3 B a^{2}\right )}{3 x^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^4} \, dx=\frac {1}{3} \, B b^{2} x^{3} + {\left (2 \, B a b + A b^{2}\right )} x - \frac {A a^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{3 \, x^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^4} \, dx=\frac {1}{3} \, B b^{2} x^{3} + 2 \, B a b x + A b^{2} x - \frac {3 \, B a^{2} x^{2} + 6 \, A a b x^{2} + A a^{2}}{3 \, x^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^4} \, dx=x\,\left (A\,b^2+2\,B\,a\,b\right )-\frac {x^2\,\left (B\,a^2+2\,A\,b\,a\right )+\frac {A\,a^2}{3}}{x^3}+\frac {B\,b^2\,x^3}{3} \]

[In]

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

[Out]

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