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

Optimal. Leaf size=41 \[ \frac{B \log \left (a+b x^2\right )}{2 b^2}-\frac{A b-a B}{2 b^2 \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0363941, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {444, 43} \[ \frac{B \log \left (a+b x^2\right )}{2 b^2}-\frac{A b-a B}{2 b^2 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(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)

Rule 444

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]

Rule 43

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])

Rubi steps

\begin{align*} \int \frac{x \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+B x}{(a+b x)^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{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]  time = 0.0122215, size = 41, normalized size = 1. \[ \frac{a B-A b}{2 b^2 \left (a+b x^2\right )}+\frac{B \log \left (a+b x^2\right )}{2 b^2} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.008, size = 47, normalized size = 1.2 \begin{align*}{\frac{B\ln \left ( b{x}^{2}+a \right ) }{2\,{b}^{2}}}-{\frac{A}{2\,b \left ( b{x}^{2}+a \right ) }}+{\frac{Ba}{2\,{b}^{2} \left ( b{x}^{2}+a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03535, size = 54, normalized size = 1.32 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.22481, size = 92, normalized size = 2.24 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.507564, size = 36, normalized size = 0.88 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.23471, size = 88, normalized size = 2.15 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)