\(\int \frac {(a+b x^2) (A+B x^2)}{x^6} \, dx\) [9]

Optimal result

Integrand size = 18, antiderivative size = 31 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^6} \, dx=-\frac {a A}{5 x^5}-\frac {A b+a B}{3 x^3}-\frac {b B}{x} \]

[Out]

-1/5*a*A/x^5+1/3*(-A*b-B*a)/x^3-b*B/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, 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} \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^6} \, dx=-\frac {a B+A b}{3 x^3}-\frac {a A}{5 x^5}-\frac {b B}{x} \]

[In]

Int[((a + b*x^2)*(A + B*x^2))/x^6,x]

[Out]

-1/5*(a*A)/x^5 - (A*b + a*B)/(3*x^3) - (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 (\frac {a A}{x^6}+\frac {A b+a B}{x^4}+\frac {b B}{x^2}\right ) \, dx \\ & = -\frac {a A}{5 x^5}-\frac {A b+a B}{3 x^3}-\frac {b B}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^6} \, dx=-\frac {a A}{5 x^5}+\frac {-A b-a B}{3 x^3}-\frac {b B}{x} \]

[In]

Integrate[((a + b*x^2)*(A + B*x^2))/x^6,x]

[Out]

-1/5*(a*A)/x^5 + (-(A*b) - a*B)/(3*x^3) - (b*B)/x

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90

method	result	size
default	\(-\frac {A b +B a}{3 x^{3}}-\frac {b B}{x}-\frac {a A}{5 x^{5}}\)	\(28\)
norman	\(\frac {-b B \,x^{4}+\left (-\frac {A b}{3}-\frac {B a}{3}\right ) x^{2}-\frac {A a}{5}}{x^{5}}\)	\(30\)
risch	\(\frac {-b B \,x^{4}+\left (-\frac {A b}{3}-\frac {B a}{3}\right ) x^{2}-\frac {A a}{5}}{x^{5}}\)	\(30\)
gosper	\(-\frac {15 b B \,x^{4}+5 A b \,x^{2}+5 B a \,x^{2}+3 A a}{15 x^{5}}\)	\(32\)
parallelrisch	\(-\frac {15 b B \,x^{4}+5 A b \,x^{2}+5 B a \,x^{2}+3 A a}{15 x^{5}}\)	\(32\)

[In]

int((b*x^2+a)*(B*x^2+A)/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/3*(A*b+B*a)/x^3-b*B/x-1/5*a*A/x^5

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^6} \, dx=-\frac {15 \, B b x^{4} + 5 \, {\left (B a + A b\right )} x^{2} + 3 \, A a}{15 \, x^{5}} \]

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^6,x, algorithm="fricas")

[Out]

-1/15*(15*B*b*x^4 + 5*(B*a + A*b)*x^2 + 3*A*a)/x^5

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^6} \, dx=\frac {- 3 A a - 15 B b x^{4} + x^{2} \left (- 5 A b - 5 B a\right )}{15 x^{5}} \]

[In]

integrate((b*x**2+a)*(B*x**2+A)/x**6,x)

[Out]

(-3*A*a - 15*B*b*x**4 + x**2*(-5*A*b - 5*B*a))/(15*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^6} \, dx=-\frac {15 \, B b x^{4} + 5 \, {\left (B a + A b\right )} x^{2} + 3 \, A a}{15 \, x^{5}} \]

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^6,x, algorithm="maxima")

[Out]

-1/15*(15*B*b*x^4 + 5*(B*a + A*b)*x^2 + 3*A*a)/x^5

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^6} \, dx=-\frac {15 \, B b x^{4} + 5 \, B a x^{2} + 5 \, A b x^{2} + 3 \, A a}{15 \, x^{5}} \]

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^6,x, algorithm="giac")

[Out]

-1/15*(15*B*b*x^4 + 5*B*a*x^2 + 5*A*b*x^2 + 3*A*a)/x^5

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^6} \, dx=-\frac {B\,b\,x^4+\left (\frac {A\,b}{3}+\frac {B\,a}{3}\right )\,x^2+\frac {A\,a}{5}}{x^5} \]

[In]

int(((A + B*x^2)*(a + b*x^2))/x^6,x)

[Out]

-((A*a)/5 + x^2*((A*b)/3 + (B*a)/3) + B*b*x^4)/x^5