\(\int \frac {(a+b x^2) (A+B x^2)}{x^7} \, dx\) [10]

Optimal result

Integrand size = 18, antiderivative size = 33 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^7} \, dx=-\frac {a A}{6 x^6}-\frac {A b+a B}{4 x^4}-\frac {b B}{2 x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {457, 77} \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^7} \, dx=-\frac {a B+A b}{4 x^4}-\frac {a A}{6 x^6}-\frac {b B}{2 x^2} \]

[In]

Int[((a + b*x^2)*(A + B*x^2))/x^7,x]

[Out]

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

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x) (A+B x)}{x^4} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {a A}{x^4}+\frac {A b+a B}{x^3}+\frac {b B}{x^2}\right ) \, dx,x,x^2\right ) \\ & = -\frac {a A}{6 x^6}-\frac {A b+a B}{4 x^4}-\frac {b B}{2 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^7} \, dx=-\frac {a A}{6 x^6}+\frac {-A b-a B}{4 x^4}-\frac {b B}{2 x^2} \]

[In]

Integrate[((a + b*x^2)*(A + B*x^2))/x^7,x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85

method	result	size
default	\(-\frac {a A}{6 x^{6}}-\frac {b B}{2 x^{2}}-\frac {A b +B a}{4 x^{4}}\)	\(28\)
norman	\(\frac {-\frac {b B \,x^{4}}{2}+\left (-\frac {A b}{4}-\frac {B a}{4}\right ) x^{2}-\frac {A a}{6}}{x^{6}}\)	\(30\)
risch	\(\frac {-\frac {b B \,x^{4}}{2}+\left (-\frac {A b}{4}-\frac {B a}{4}\right ) x^{2}-\frac {A a}{6}}{x^{6}}\)	\(30\)
gosper	\(-\frac {6 b B \,x^{4}+3 A b \,x^{2}+3 B a \,x^{2}+2 A a}{12 x^{6}}\)	\(32\)
parallelrisch	\(-\frac {6 b B \,x^{4}+3 A b \,x^{2}+3 B a \,x^{2}+2 A a}{12 x^{6}}\)	\(32\)

[In]

int((b*x^2+a)*(B*x^2+A)/x^7,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^7} \, dx=-\frac {6 \, B b x^{4} + 3 \, {\left (B a + A b\right )} x^{2} + 2 \, A a}{12 \, x^{6}} \]

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^7,x, algorithm="fricas")

[Out]

-1/12*(6*B*b*x^4 + 3*(B*a + A*b)*x^2 + 2*A*a)/x^6

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^7} \, dx=\frac {- 2 A a - 6 B b x^{4} + x^{2} \left (- 3 A b - 3 B a\right )}{12 x^{6}} \]

[In]

integrate((b*x**2+a)*(B*x**2+A)/x**7,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^7} \, dx=-\frac {6 \, B b x^{4} + 3 \, {\left (B a + A b\right )} x^{2} + 2 \, A a}{12 \, x^{6}} \]

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^7,x, algorithm="maxima")

[Out]

-1/12*(6*B*b*x^4 + 3*(B*a + A*b)*x^2 + 2*A*a)/x^6

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^7} \, dx=-\frac {6 \, B b x^{4} + 3 \, B a x^{2} + 3 \, A b x^{2} + 2 \, A a}{12 \, x^{6}} \]

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^7,x, algorithm="giac")

[Out]

-1/12*(6*B*b*x^4 + 3*B*a*x^2 + 3*A*b*x^2 + 2*A*a)/x^6

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^7} \, dx=-\frac {\frac {B\,b\,x^4}{2}+\left (\frac {A\,b}{4}+\frac {B\,a}{4}\right )\,x^2+\frac {A\,a}{6}}{x^6} \]

[In]

int(((A + B*x^2)*(a + b*x^2))/x^7,x)

[Out]

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