\(\int \frac {(a+b x^2) (A+B x^2)}{x^5} \, dx\) [8]

Optimal result

Integrand size = 18, antiderivative size = 29 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^5} \, dx=-\frac {a A}{4 x^4}-\frac {A b+a B}{2 x^2}+b B \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, 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} \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^5} \, dx=-\frac {a B+A b}{2 x^2}-\frac {a A}{4 x^4}+b B \log (x) \]

[In]

Int[((a + b*x^2)*(A + B*x^2))/x^5,x]

[Out]

-1/4*(a*A)/x^4 - (A*b + a*B)/(2*x^2) + b*B*Log[x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^5} \, dx=-\frac {a A}{4 x^4}+\frac {-A b-a B}{2 x^2}+b B \log (x) \]

[In]

Integrate[((a + b*x^2)*(A + B*x^2))/x^5,x]

[Out]

-1/4*(a*A)/x^4 + (-(A*b) - a*B)/(2*x^2) + b*B*Log[x]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90

method	result	size
default	\(b B \ln \left (x \right )-\frac {A b +B a}{2 x^{2}}-\frac {a A}{4 x^{4}}\)	\(26\)
norman	\(\frac {\left (-\frac {A b}{2}-\frac {B a}{2}\right ) x^{2}-\frac {A a}{4}}{x^{4}}+b B \ln \left (x \right )\)	\(29\)
risch	\(\frac {\left (-\frac {A b}{2}-\frac {B a}{2}\right ) x^{2}-\frac {A a}{4}}{x^{4}}+b B \ln \left (x \right )\)	\(29\)
parallelrisch	\(-\frac {-4 B b \ln \left (x \right ) x^{4}+2 A b \,x^{2}+2 B a \,x^{2}+A a}{4 x^{4}}\)	\(33\)

[In]

int((b*x^2+a)*(B*x^2+A)/x^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^5} \, dx=\frac {4 \, B b x^{4} \log \left (x\right ) - 2 \, {\left (B a + A b\right )} x^{2} - A a}{4 \, x^{4}} \]

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^5,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^5} \, dx=B b \log {\left (x \right )} + \frac {- A a + x^{2} \left (- 2 A b - 2 B a\right )}{4 x^{4}} \]

[In]

integrate((b*x**2+a)*(B*x**2+A)/x**5,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^5} \, dx=\frac {1}{2} \, B b \log \left (x^{2}\right ) - \frac {2 \, {\left (B a + A b\right )} x^{2} + A a}{4 \, x^{4}} \]

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^5,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^5} \, dx=\frac {1}{2} \, B b \log \left (x^{2}\right ) - \frac {3 \, B b x^{4} + 2 \, B a x^{2} + 2 \, A b x^{2} + A a}{4 \, x^{4}} \]

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^5,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^5} \, dx=B\,b\,\ln \left (x\right )-\frac {\left (\frac {A\,b}{2}+\frac {B\,a}{2}\right )\,x^2+\frac {A\,a}{4}}{x^4} \]

[In]

int(((A + B*x^2)*(a + b*x^2))/x^5,x)

[Out]

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