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

Optimal. Leaf size=16 \[ \frac{c \log \left (a+b x^2\right )}{2 b} \]

[Out]

(c*Log[a + b*x^2])/(2*b)

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Rubi [A]  time = 0.0045227, antiderivative size = 16, 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, 260} \[ \frac{c \log \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Log[a + b*x^2])/(2*b)

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 260

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

Rubi steps

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

Mathematica [A]  time = 0.002228, size = 16, normalized size = 1. \[ \frac{c \log \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Log[a + b*x^2])/(2*b)

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Maple [A]  time = 0.001, size = 15, normalized size = 0.9 \begin{align*}{\frac{c\ln \left ( b{x}^{2}+a \right ) }{2\,b}} \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)^2,x)

[Out]

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

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Maxima [A]  time = 0.99726, size = 19, normalized size = 1.19 \begin{align*} \frac{c \log \left (b x^{2} + a\right )}{2 \, 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)^2,x, algorithm="maxima")

[Out]

1/2*c*log(b*x^2 + a)/b

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Fricas [A]  time = 1.24215, size = 32, normalized size = 2. \begin{align*} \frac{c \log \left (b x^{2} + a\right )}{2 \, 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)^2,x, algorithm="fricas")

[Out]

1/2*c*log(b*x^2 + a)/b

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Sympy [A]  time = 0.111414, size = 12, normalized size = 0.75 \begin{align*} \frac{c \log{\left (a + b x^{2} \right )}}{2 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)**2,x)

[Out]

c*log(a + b*x**2)/(2*b)

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Giac [B]  time = 1.12881, size = 85, normalized size = 5.31 \begin{align*} -\frac{1}{2} \, c{\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 )} - \frac{a 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)^2,x, algorithm="giac")

[Out]

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