\(\int \frac {(a+b x^2)^2 (A+B x^2)}{x^8} \, dx\) [21]

Optimal result

Integrand size = 20, antiderivative size = 53 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^8} \, dx=-\frac {a^2 A}{7 x^7}-\frac {a (2 A b+a B)}{5 x^5}-\frac {b (A b+2 a B)}{3 x^3}-\frac {b^2 B}{x} \]

[Out]

-1/7*a^2*A/x^7-1/5*a*(2*A*b+B*a)/x^5-1/3*b*(A*b+2*B*a)/x^3-b^2*B/x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, 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} \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^8} \, dx=-\frac {a^2 A}{7 x^7}-\frac {a (a B+2 A b)}{5 x^5}-\frac {b (2 a B+A b)}{3 x^3}-\frac {b^2 B}{x} \]

[In]

Int[((a + b*x^2)^2*(A + B*x^2))/x^8,x]

[Out]

-1/7*(a^2*A)/x^7 - (a*(2*A*b + a*B))/(5*x^5) - (b*(A*b + 2*a*B))/(3*x^3) - (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 (\frac {a^2 A}{x^8}+\frac {a (2 A b+a B)}{x^6}+\frac {b (A b+2 a B)}{x^4}+\frac {b^2 B}{x^2}\right ) \, dx \\ & = -\frac {a^2 A}{7 x^7}-\frac {a (2 A b+a B)}{5 x^5}-\frac {b (A b+2 a B)}{3 x^3}-\frac {b^2 B}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^8} \, dx=-\frac {35 b^2 x^4 \left (A+3 B x^2\right )+14 a b x^2 \left (3 A+5 B x^2\right )+3 a^2 \left (5 A+7 B x^2\right )}{105 x^7} \]

[In]

Integrate[((a + b*x^2)^2*(A + B*x^2))/x^8,x]

[Out]

-1/105*(35*b^2*x^4*(A + 3*B*x^2) + 14*a*b*x^2*(3*A + 5*B*x^2) + 3*a^2*(5*A + 7*B*x^2))/x^7

Maple [A] (verified)

Time = 2.49 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {a^{2} A}{7 x^{7}}-\frac {a \left (2 A b +B a \right )}{5 x^{5}}-\frac {b \left (A b +2 B a \right )}{3 x^{3}}-\frac {b^{2} B}{x}\)	\(48\)
norman	\(\frac {-b^{2} B \,x^{6}+\left (-\frac {1}{3} b^{2} A -\frac {2}{3} a b B \right ) x^{4}+\left (-\frac {2}{5} a b A -\frac {1}{5} a^{2} B \right ) x^{2}-\frac {a^{2} A}{7}}{x^{7}}\)	\(53\)
risch	\(\frac {-b^{2} B \,x^{6}+\left (-\frac {1}{3} b^{2} A -\frac {2}{3} a b B \right ) x^{4}+\left (-\frac {2}{5} a b A -\frac {1}{5} a^{2} B \right ) x^{2}-\frac {a^{2} A}{7}}{x^{7}}\)	\(53\)
gosper	\(-\frac {105 b^{2} B \,x^{6}+35 A \,b^{2} x^{4}+70 B a b \,x^{4}+42 a A b \,x^{2}+21 a^{2} B \,x^{2}+15 a^{2} A}{105 x^{7}}\)	\(56\)
parallelrisch	\(-\frac {105 b^{2} B \,x^{6}+35 A \,b^{2} x^{4}+70 B a b \,x^{4}+42 a A b \,x^{2}+21 a^{2} B \,x^{2}+15 a^{2} A}{105 x^{7}}\)	\(56\)

[In]

int((b*x^2+a)^2*(B*x^2+A)/x^8,x,method=_RETURNVERBOSE)

[Out]

-1/7*a^2*A/x^7-1/5*a*(2*A*b+B*a)/x^5-1/3*b*(A*b+2*B*a)/x^3-b^2*B/x

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^8} \, dx=-\frac {105 \, B b^{2} x^{6} + 35 \, {\left (2 \, B a b + A b^{2}\right )} x^{4} + 15 \, A a^{2} + 21 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{105 \, x^{7}} \]

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^8,x, algorithm="fricas")

[Out]

-1/105*(105*B*b^2*x^6 + 35*(2*B*a*b + A*b^2)*x^4 + 15*A*a^2 + 21*(B*a^2 + 2*A*a*b)*x^2)/x^7

Sympy [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^8} \, dx=\frac {- 15 A a^{2} - 105 B b^{2} x^{6} + x^{4} \left (- 35 A b^{2} - 70 B a b\right ) + x^{2} \left (- 42 A a b - 21 B a^{2}\right )}{105 x^{7}} \]

[In]

integrate((b*x**2+a)**2*(B*x**2+A)/x**8,x)

[Out]

(-15*A*a**2 - 105*B*b**2*x**6 + x**4*(-35*A*b**2 - 70*B*a*b) + x**2*(-42*A*a*b - 21*B*a**2))/(105*x**7)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^8} \, dx=-\frac {105 \, B b^{2} x^{6} + 35 \, {\left (2 \, B a b + A b^{2}\right )} x^{4} + 15 \, A a^{2} + 21 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{105 \, x^{7}} \]

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^8,x, algorithm="maxima")

[Out]

-1/105*(105*B*b^2*x^6 + 35*(2*B*a*b + A*b^2)*x^4 + 15*A*a^2 + 21*(B*a^2 + 2*A*a*b)*x^2)/x^7

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^8} \, dx=-\frac {105 \, B b^{2} x^{6} + 70 \, B a b x^{4} + 35 \, A b^{2} x^{4} + 21 \, B a^{2} x^{2} + 42 \, A a b x^{2} + 15 \, A a^{2}}{105 \, x^{7}} \]

[In]

integrate((b*x^2+a)^2*(B*x^2+A)/x^8,x, algorithm="giac")

[Out]

-1/105*(105*B*b^2*x^6 + 70*B*a*b*x^4 + 35*A*b^2*x^4 + 21*B*a^2*x^2 + 42*A*a*b*x^2 + 15*A*a^2)/x^7

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^8} \, dx=-\frac {x^2\,\left (\frac {B\,a^2}{5}+\frac {2\,A\,b\,a}{5}\right )+x^4\,\left (\frac {A\,b^2}{3}+\frac {2\,B\,a\,b}{3}\right )+\frac {A\,a^2}{7}+B\,b^2\,x^6}{x^7} \]

[In]

int(((A + B*x^2)*(a + b*x^2)^2)/x^8,x)

[Out]

-(x^2*((B*a^2)/5 + (2*A*a*b)/5) + x^4*((A*b^2)/3 + (2*B*a*b)/3) + (A*a^2)/7 + B*b^2*x^6)/x^7