\(\int \frac {(a+b x^2)^2 (A+B x^2)}{x^5} \, dx\) [18]

Optimal result

Integrand size = 20, antiderivative size = 51 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^5} \, dx=-\frac {a^2 A}{4 x^4}-\frac {a (2 A b+a B)}{2 x^2}+\frac {1}{2} b^2 B x^2+b (A b+2 a B) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {457, 77} \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^5} \, dx=-\frac {a^2 A}{4 x^4}-\frac {a (a B+2 A b)}{2 x^2}+b \log (x) (2 a B+A b)+\frac {1}{2} b^2 B x^2 \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^5} \, dx=-\frac {4 a A b x^2-2 b^2 B x^6+a^2 \left (A+2 B x^2\right )}{4 x^4}+b (A b+2 a B) \log (x) \]

[In]

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

[Out]

-1/4*(4*a*A*b*x^2 - 2*b^2*B*x^6 + a^2*(A + 2*B*x^2))/x^4 + b*(A*b + 2*a*B)*Log[x]

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90

method	result	size
default	\(-\frac {a^{2} A}{4 x^{4}}-\frac {a \left (2 A b +B a \right )}{2 x^{2}}+\frac {b^{2} B \,x^{2}}{2}+b \left (A b +2 B a \right ) \ln \left (x \right )\)	\(46\)
norman	\(\frac {\left (-a b A -\frac {1}{2} a^{2} B \right ) x^{2}-\frac {a^{2} A}{4}+\frac {b^{2} B \,x^{6}}{2}}{x^{4}}+\left (b^{2} A +2 a b B \right ) \ln \left (x \right )\)	\(52\)
risch	\(\frac {b^{2} B \,x^{2}}{2}+\frac {\left (-a b A -\frac {1}{2} a^{2} B \right ) x^{2}-\frac {a^{2} A}{4}}{x^{4}}+A \ln \left (x \right ) b^{2}+2 B \ln \left (x \right ) a b\)	\(52\)
parallelrisch	\(\frac {2 b^{2} B \,x^{6}+4 A \ln \left (x \right ) x^{4} b^{2}+8 B \ln \left (x \right ) x^{4} a b -4 a A b \,x^{2}-2 a^{2} B \,x^{2}-a^{2} A}{4 x^{4}}\)	\(60\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^5} \, dx=\frac {2 \, B b^{2} x^{6} + 4 \, {\left (2 \, B a b + A b^{2}\right )} x^{4} \log \left (x\right ) - A a^{2} - 2 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{4 \, x^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^5} \, dx=\frac {B b^{2} x^{2}}{2} + b \left (A b + 2 B a\right ) \log {\left (x \right )} + \frac {- A a^{2} + x^{2} \left (- 4 A a b - 2 B a^{2}\right )}{4 x^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^5} \, dx=\frac {1}{2} \, B b^{2} x^{2} + \frac {1}{2} \, {\left (2 \, B a b + A b^{2}\right )} \log \left (x^{2}\right ) - \frac {A a^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{4 \, x^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.41 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^5} \, dx=\frac {1}{2} \, B b^{2} x^{2} + \frac {1}{2} \, {\left (2 \, B a b + A b^{2}\right )} \log \left (x^{2}\right ) - \frac {6 \, B a b x^{4} + 3 \, A b^{2} x^{4} + 2 \, B a^{2} x^{2} + 4 \, A a b x^{2} + A a^{2}}{4 \, x^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.91 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2\right )}{x^5} \, dx=\ln \left (x\right )\,\left (A\,b^2+2\,B\,a\,b\right )-\frac {x^2\,\left (\frac {B\,a^2}{2}+A\,b\,a\right )+\frac {A\,a^2}{4}}{x^4}+\frac {B\,b^2\,x^2}{2} \]

[In]

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

[Out]

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