3.136 \(\int \frac{x (a c+b c x^2)}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=17 \[ -\frac{c}{2 b \left (a+b x^2\right )} \]

[Out]

-c/(2*b*(a + b*x^2))

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Rubi [A]  time = 0.0044847, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {21, 261} \[ -\frac{c}{2 b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-c/(2*b*(a + b*x^2))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{x \left (a c+b c x^2\right )}{\left (a+b x^2\right )^3} \, dx &=c \int \frac{x}{\left (a+b x^2\right )^2} \, dx\\ &=-\frac{c}{2 b \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0020816, size = 17, normalized size = 1. \[ -\frac{c}{2 b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-c/(2*b*(a + b*x^2))

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Maple [A]  time = 0.001, size = 16, normalized size = 0.9 \begin{align*} -{\frac{c}{2\,b \left ( b{x}^{2}+a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c/b/(b*x^2+a)

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Maxima [A]  time = 0.982895, size = 22, normalized size = 1.29 \begin{align*} -\frac{c}{2 \,{\left (b^{2} x^{2} + a b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c/(b^2*x^2 + a*b)

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Fricas [A]  time = 1.22489, size = 32, normalized size = 1.88 \begin{align*} -\frac{c}{2 \,{\left (b^{2} x^{2} + a b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c/(b^2*x^2 + a*b)

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Sympy [A]  time = 0.305678, size = 15, normalized size = 0.88 \begin{align*} - \frac{c}{2 a b + 2 b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c/(2*a*b + 2*b**2*x**2)

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Giac [A]  time = 1.67979, size = 20, normalized size = 1.18 \begin{align*} -\frac{c}{2 \,{\left (b x^{2} + a\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c/((b*x^2 + a)*b)