\(\int \frac {A+B x^2}{x^3 (a+b x^2)^2} \, dx\) [83]

Optimal result

Integrand size = 20, antiderivative size = 76 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^2} \, dx=-\frac {A}{2 a^2 x^2}-\frac {A b-a B}{2 a^2 \left (a+b x^2\right )}-\frac {(2 A b-a B) \log (x)}{a^3}+\frac {(2 A b-a B) \log \left (a+b x^2\right )}{2 a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 76, 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, 78} \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {(2 A b-a B) \log \left (a+b x^2\right )}{2 a^3}-\frac {\log (x) (2 A b-a B)}{a^3}-\frac {A b-a B}{2 a^2 \left (a+b x^2\right )}-\frac {A}{2 a^2 x^2} \]

[In]

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

[Out]

-1/2*A/(a^2*x^2) - (A*b - a*B)/(2*a^2*(a + b*x^2)) - ((2*A*b - a*B)*Log[x])/a^3 + ((2*A*b - a*B)*Log[a + b*x^2
])/(2*a^3)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.84 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {-\frac {a A}{x^2}+\frac {a (-A b+a B)}{a+b x^2}+2 (-2 A b+a B) \log (x)+(2 A b-a B) \log \left (a+b x^2\right )}{2 a^3} \]

[In]

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

[Out]

(-((a*A)/x^2) + (a*(-(A*b) + a*B))/(a + b*x^2) + 2*(-2*A*b + a*B)*Log[x] + (2*A*b - a*B)*Log[a + b*x^2])/(2*a^
3)

Maple [A] (verified)

Time = 2.52 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {A}{2 a^{2} x^{2}}+\frac {\left (-2 A b +B a \right ) \ln \left (x \right )}{a^{3}}+\frac {b \left (\frac {\left (2 A b -B a \right ) \ln \left (b \,x^{2}+a \right )}{b}-\frac {a \left (A b -B a \right )}{b \left (b \,x^{2}+a \right )}\right )}{2 a^{3}}\)	\(76\)
norman	\(\frac {-\frac {A}{2 a}+\frac {b \left (2 A b -B a \right ) x^{4}}{2 a^{3}}}{x^{2} \left (b \,x^{2}+a \right )}-\frac {\left (2 A b -B a \right ) \ln \left (x \right )}{a^{3}}+\frac {\left (2 A b -B a \right ) \ln \left (b \,x^{2}+a \right )}{2 a^{3}}\)	\(78\)
risch	\(\frac {-\frac {\left (2 A b -B a \right ) x^{2}}{2 a^{2}}-\frac {A}{2 a}}{x^{2} \left (b \,x^{2}+a \right )}-\frac {2 \ln \left (x \right ) A b}{a^{3}}+\frac {\ln \left (x \right ) B}{a^{2}}+\frac {\ln \left (-b \,x^{2}-a \right ) A b}{a^{3}}-\frac {\ln \left (-b \,x^{2}-a \right ) B}{2 a^{2}}\)	\(89\)
parallelrisch	\(-\frac {4 A \ln \left (x \right ) x^{4} b^{2}-2 A \ln \left (b \,x^{2}+a \right ) x^{4} b^{2}-2 B \ln \left (x \right ) x^{4} a b +B \ln \left (b \,x^{2}+a \right ) x^{4} a b -2 A \,b^{2} x^{4}+B a b \,x^{4}+4 A \ln \left (x \right ) x^{2} a b -2 A \ln \left (b \,x^{2}+a \right ) x^{2} a b -2 B \ln \left (x \right ) x^{2} a^{2}+B \ln \left (b \,x^{2}+a \right ) x^{2} a^{2}+a^{2} A}{2 a^{3} x^{2} \left (b \,x^{2}+a \right )}\)	\(146\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.54 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^2} \, dx=-\frac {A a^{2} - {\left (B a^{2} - 2 \, A a b\right )} x^{2} + {\left ({\left (B a b - 2 \, A b^{2}\right )} x^{4} + {\left (B a^{2} - 2 \, A a b\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left ({\left (B a b - 2 \, A b^{2}\right )} x^{4} + {\left (B a^{2} - 2 \, A a b\right )} x^{2}\right )} \log \left (x\right )}{2 \, {\left (a^{3} b x^{4} + a^{4} x^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {- A a + x^{2} \left (- 2 A b + B a\right )}{2 a^{3} x^{2} + 2 a^{2} b x^{4}} + \frac {\left (- 2 A b + B a\right ) \log {\left (x \right )}}{a^{3}} - \frac {\left (- 2 A b + B a\right ) \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {{\left (B a - 2 \, A b\right )} x^{2} - A a}{2 \, {\left (a^{2} b x^{4} + a^{3} x^{2}\right )}} - \frac {{\left (B a - 2 \, A b\right )} \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {{\left (B a - 2 \, A b\right )} \log \left (x^{2}\right )}{2 \, a^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {{\left (B a - 2 \, A b\right )} \log \left (x^{2}\right )}{2 \, a^{3}} + \frac {B a x^{2} - 2 \, A b x^{2} - A a}{2 \, {\left (b x^{4} + a x^{2}\right )} a^{2}} - \frac {{\left (B a b - 2 \, A b^{2}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{3} b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.90 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x^2}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {\ln \left (b\,x^2+a\right )\,\left (2\,A\,b-B\,a\right )}{2\,a^3}-\frac {\frac {A}{2\,a}+\frac {x^2\,\left (2\,A\,b-B\,a\right )}{2\,a^2}}{b\,x^4+a\,x^2}-\frac {\ln \left (x\right )\,\left (2\,A\,b-B\,a\right )}{a^3} \]

[In]

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

[Out]

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