\(\int \frac {(a+b x^2) (A+B x^2)}{x} \, dx\) [4]

Optimal result

Integrand size = 18, antiderivative size = 29 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x} \, dx=\frac {1}{2} (A b+a B) x^2+\frac {1}{4} b B x^4+a A \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (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} \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x} \, dx=\frac {1}{2} x^2 (a B+A b)+a A \log (x)+\frac {1}{4} b B x^4 \]

[In]

Int[((a + b*x^2)*(A + B*x^2))/x,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x} \, dx=\frac {1}{2} (A b+a B) x^2+\frac {1}{4} b B x^4+a A \log (x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93

method	result	size
norman	\(\left (\frac {A b}{2}+\frac {B a}{2}\right ) x^{2}+\frac {b B \,x^{4}}{4}+a A \ln \left (x \right )\)	\(27\)
default	\(\frac {b B \,x^{4}}{4}+\frac {A b \,x^{2}}{2}+\frac {B a \,x^{2}}{2}+a A \ln \left (x \right )\)	\(28\)
parallelrisch	\(\frac {b B \,x^{4}}{4}+\frac {A b \,x^{2}}{2}+\frac {B a \,x^{2}}{2}+a A \ln \left (x \right )\)	\(28\)
risch	\(\frac {b B \,x^{4}}{4}+\frac {A b \,x^{2}}{2}+\frac {B a \,x^{2}}{2}+\frac {b \,A^{2}}{4 B}+\frac {A a}{2}+\frac {B \,a^{2}}{4 b}+a A \ln \left (x \right )\)	\(50\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x} \, dx=\frac {1}{4} \, B b x^{4} + \frac {1}{2} \, {\left (B a + A b\right )} x^{2} + A a \log \left (x\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x} \, dx=A a \log {\left (x \right )} + \frac {B b x^{4}}{4} + x^{2} \left (\frac {A b}{2} + \frac {B a}{2}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x} \, dx=\frac {1}{4} \, B b x^{4} + \frac {1}{2} \, {\left (B a + A b\right )} x^{2} + \frac {1}{2} \, A a \log \left (x^{2}\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x} \, dx=\frac {1}{4} \, B b x^{4} + \frac {1}{2} \, B a x^{2} + \frac {1}{2} \, A b x^{2} + \frac {1}{2} \, A a \log \left (x^{2}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right )}{x} \, dx=x^2\,\left (\frac {A\,b}{2}+\frac {B\,a}{2}\right )+\frac {B\,b\,x^4}{4}+A\,a\,\ln \left (x\right ) \]

[In]

int(((A + B*x^2)*(a + b*x^2))/x,x)

[Out]

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