\(\int \frac {x (A+B x^2)}{(a+b x^2)^2} \, dx\) [79]

Optimal result

Integrand size = 18, antiderivative size = 41 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=-\frac {A b-a B}{2 b^2 \left (a+b x^2\right )}+\frac {B \log \left (a+b x^2\right )}{2 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, 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 = {455, 45} \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B \log \left (a+b x^2\right )}{2 b^2}-\frac {A b-a B}{2 b^2 \left (a+b x^2\right )} \]

[In]

Int[(x*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {-A b+a B}{2 b^2 \left (a+b x^2\right )}+\frac {B \log \left (a+b x^2\right )}{2 b^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.53 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {B \ln \left (b \,x^{2}+a \right )}{2 b^{2}}-\frac {A b -B a}{2 b^{2} \left (b \,x^{2}+a \right )}\)	\(38\)
norman	\(\frac {B \ln \left (b \,x^{2}+a \right )}{2 b^{2}}-\frac {A b -B a}{2 b^{2} \left (b \,x^{2}+a \right )}\)	\(38\)
risch	\(\frac {B \ln \left (b \,x^{2}+a \right )}{2 b^{2}}-\frac {A}{2 b \left (b \,x^{2}+a \right )}+\frac {B a}{2 b^{2} \left (b \,x^{2}+a \right )}\)	\(47\)
parallelrisch	\(-\frac {-B \ln \left (b \,x^{2}+a \right ) x^{2} b -B \ln \left (b \,x^{2}+a \right ) a +A b -B a}{2 b^{2} \left (b \,x^{2}+a \right )}\)	\(50\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.07 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B a - A b + {\left (B b x^{2} + B a\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B \log {\left (a + b x^{2} \right )}}{2 b^{2}} + \frac {- A b + B a}{2 a b^{2} + 2 b^{3} x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.98 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B a - A b}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}} + \frac {B \log \left (b x^{2} + a\right )}{2 \, b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.59 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=-\frac {B {\left (\frac {\log \left (\frac {{\left | b x^{2} + a \right |}}{{\left (b x^{2} + a\right )}^{2} {\left | b \right |}}\right )}{b} - \frac {a}{{\left (b x^{2} + a\right )} b}\right )}}{2 \, b} - \frac {A}{2 \, {\left (b x^{2} + a\right )} b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.90 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B\,\ln \left (b\,x^2+a\right )}{2\,b^2}-\frac {A\,b-B\,a}{2\,b^2\,\left (b\,x^2+a\right )} \]

[In]

int((x*(A + B*x^2))/(a + b*x^2)^2,x)

[Out]

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