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

Optimal result

Integrand size = 18, antiderivative size = 26 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^4} \, dx=-\frac {a A}{3 x^3}-\frac {A b+a B}{x}+b B x \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {459} \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^4} \, dx=-\frac {a B+A b}{x}-\frac {a A}{3 x^3}+b B x \]

[In]

Int[((a + b*x^2)*(A + B*x^2))/x^4,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^4} \, dx=-\frac {a A}{3 x^3}+\frac {-A b-a B}{x}+b B x \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96

method	result	size
default	\(b B x -\frac {a A}{3 x^{3}}-\frac {A b +B a}{x}\)	\(25\)
risch	\(b B x +\frac {\left (-A b -B a \right ) x^{2}-\frac {A a}{3}}{x^{3}}\)	\(28\)
norman	\(\frac {b B \,x^{4}+\left (-A b -B a \right ) x^{2}-\frac {A a}{3}}{x^{3}}\)	\(29\)
gosper	\(-\frac {-3 b B \,x^{4}+3 A b \,x^{2}+3 B a \,x^{2}+A a}{3 x^{3}}\)	\(31\)
parallelrisch	\(-\frac {-3 b B \,x^{4}+3 A b \,x^{2}+3 B a \,x^{2}+A a}{3 x^{3}}\)	\(31\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^4} \, dx=\frac {3 \, B b x^{4} - 3 \, {\left (B a + A b\right )} x^{2} - A a}{3 \, x^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^4} \, dx=B b x + \frac {- A a + x^{2} \left (- 3 A b - 3 B a\right )}{3 x^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^4} \, dx=B b x - \frac {3 \, {\left (B a + A b\right )} x^{2} + A a}{3 \, x^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^4} \, dx=B b x - \frac {3 \, B a x^{2} + 3 \, A b x^{2} + A a}{3 \, x^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.88 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^4} \, dx=B\,b\,x-\frac {\left (A\,b+B\,a\right )\,x^2+\frac {A\,a}{3}}{x^3} \]

[In]

int(((A + B*x^2)*(a + b*x^2))/x^4,x)

[Out]

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