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

Optimal result

Integrand size = 18, antiderivative size = 29 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^3} \, dx=-\frac {a A}{2 x^2}+\frac {1}{2} b B x^2+(A b+a B) \log (x) \]

[Out]

-1/2*a*A/x^2+1/2*b*B*x^2+(A*b+B*a)*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} \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^3} \, dx=\log (x) (a B+A b)-\frac {a A}{2 x^2}+\frac {1}{2} b B x^2 \]

[In]

Int[((a + b*x^2)*(A + B*x^2))/x^3,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^3} \, dx=-\frac {a A}{2 x^2}+\frac {1}{2} b B x^2+(A b+a B) \log (x) \]

[In]

Integrate[((a + b*x^2)*(A + B*x^2))/x^3,x]

[Out]

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

Maple [A] (verified)

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

method	result	size
default	\(-\frac {a A}{2 x^{2}}+\frac {b B \,x^{2}}{2}+\left (A b +B a \right ) \ln \left (x \right )\)	\(26\)
risch	\(-\frac {a A}{2 x^{2}}+\frac {b B \,x^{2}}{2}+A \ln \left (x \right ) b +B \ln \left (x \right ) a\)	\(26\)
norman	\(\frac {-\frac {A a}{2}+\frac {b B \,x^{4}}{2}}{x^{2}}+\left (A b +B a \right ) \ln \left (x \right )\)	\(28\)
parallelrisch	\(\frac {b B \,x^{4}+2 A \ln \left (x \right ) x^{2} b +2 B \ln \left (x \right ) x^{2} a -A a}{2 x^{2}}\)	\(35\)

[In]

int((b*x^2+a)*(B*x^2+A)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^3} \, dx=\frac {B b x^{4} + 2 \, {\left (B a + A b\right )} x^{2} \log \left (x\right ) - A a}{2 \, x^{2}} \]

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^3} \, dx=- \frac {A a}{2 x^{2}} + \frac {B b x^{2}}{2} + \left (A b + B a\right ) \log {\left (x \right )} \]

[In]

integrate((b*x**2+a)*(B*x**2+A)/x**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^3} \, dx=\frac {1}{2} \, B b x^{2} + \frac {1}{2} \, {\left (B a + A b\right )} \log \left (x^{2}\right ) - \frac {A a}{2 \, x^{2}} \]

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^3} \, dx=\frac {1}{2} \, B b x^{2} + \frac {1}{2} \, {\left (B a + A b\right )} \log \left (x^{2}\right ) - \frac {B a x^{2} + A b x^{2} + A a}{2 \, x^{2}} \]

[In]

integrate((b*x^2+a)*(B*x^2+A)/x^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x^3} \, dx=\ln \left (x\right )\,\left (A\,b+B\,a\right )-\frac {A\,a}{2\,x^2}+\frac {B\,b\,x^2}{2} \]

[In]

int(((A + B*x^2)*(a + b*x^2))/x^3,x)

[Out]

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