\(\int \frac {(a+b x^2)^2 (A+B x^2)}{x^6} \, dx\) [19]

Optimal result

Integrand size = 20, antiderivative size = 48 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^6} \, dx=-\frac {a^2 A}{5 x^5}-\frac {a (2 A b+a B)}{3 x^3}-\frac {b (A b+2 a B)}{x}+b^2 B x \]

[Out]

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

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

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^6} \, dx=-\frac {a^2 A}{5 x^5}-\frac {a (2 A b+a B)}{3 x^3}-\frac {b (A b+2 a B)}{x}+b^2 B x \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.50 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {a^{2} A}{5 x^{5}}-\frac {a \left (2 A b +B a \right )}{3 x^{3}}-\frac {b \left (A b +2 B a \right )}{x}+b^{2} B x\)	\(45\)
risch	\(b^{2} B x +\frac {\left (-b^{2} A -2 a b B \right ) x^{4}+\left (-\frac {2}{3} a b A -\frac {1}{3} a^{2} B \right ) x^{2}-\frac {a^{2} A}{5}}{x^{5}}\)	\(51\)
norman	\(\frac {b^{2} B \,x^{6}+\left (-b^{2} A -2 a b B \right ) x^{4}+\left (-\frac {2}{3} a b A -\frac {1}{3} a^{2} B \right ) x^{2}-\frac {a^{2} A}{5}}{x^{5}}\)	\(52\)
gosper	\(-\frac {-15 b^{2} B \,x^{6}+15 A \,b^{2} x^{4}+30 B a b \,x^{4}+10 a A b \,x^{2}+5 a^{2} B \,x^{2}+3 a^{2} A}{15 x^{5}}\)	\(56\)
parallelrisch	\(-\frac {-15 b^{2} B \,x^{6}+15 A \,b^{2} x^{4}+30 B a b \,x^{4}+10 a A b \,x^{2}+5 a^{2} B \,x^{2}+3 a^{2} A}{15 x^{5}}\)	\(56\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^6} \, dx=\frac {15 \, B b^{2} x^{6} - 15 \, {\left (2 \, B a b + A b^{2}\right )} x^{4} - 3 \, A a^{2} - 5 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{15 \, x^{5}} \]

[In]

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

[Out]

1/15*(15*B*b^2*x^6 - 15*(2*B*a*b + A*b^2)*x^4 - 3*A*a^2 - 5*(B*a^2 + 2*A*a*b)*x^2)/x^5

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^6} \, dx=B b^{2} x + \frac {- 3 A a^{2} + x^{4} \left (- 15 A b^{2} - 30 B a b\right ) + x^{2} \left (- 10 A a b - 5 B a^{2}\right )}{15 x^{5}} \]

[In]

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

[Out]

B*b**2*x + (-3*A*a**2 + x**4*(-15*A*b**2 - 30*B*a*b) + x**2*(-10*A*a*b - 5*B*a**2))/(15*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^6} \, dx=B b^{2} x - \frac {15 \, {\left (2 \, B a b + A b^{2}\right )} x^{4} + 3 \, A a^{2} + 5 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{15 \, x^{5}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^6} \, dx=B b^{2} x - \frac {30 \, B a b x^{4} + 15 \, A b^{2} x^{4} + 5 \, B a^{2} x^{2} + 10 \, A a b x^{2} + 3 \, A a^{2}}{15 \, x^{5}} \]

[In]

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

[Out]

B*b^2*x - 1/15*(30*B*a*b*x^4 + 15*A*b^2*x^4 + 5*B*a^2*x^2 + 10*A*a*b*x^2 + 3*A*a^2)/x^5

Mupad [B] (verification not implemented)

Time = 4.85 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^6} \, dx=B\,b^2\,x-\frac {x^2\,\left (\frac {B\,a^2}{3}+\frac {2\,A\,b\,a}{3}\right )+x^4\,\left (A\,b^2+2\,B\,a\,b\right )+\frac {A\,a^2}{5}}{x^5} \]

[In]

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

[Out]

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