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

Optimal. Leaf size=24 \[ \frac{c \log (x)}{a}-\frac{c \log \left (a+b x^2\right )}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.0130802, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {21, 266, 36, 29, 31} \[ \frac{c \log (x)}{a}-\frac{c \log \left (a+b x^2\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 266

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rubi steps

\begin{align*} \int \frac{a c+b c x^2}{x \left (a+b x^2\right )^2} \, dx &=c \int \frac{1}{x \left (a+b x^2\right )} \, dx\\ &=\frac{1}{2} c \operatorname{Subst}\left (\int \frac{1}{x (a+b x)} \, dx,x,x^2\right )\\ &=\frac{c \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )}{2 a}-\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{a+b x} \, dx,x,x^2\right )}{2 a}\\ &=\frac{c \log (x)}{a}-\frac{c \log \left (a+b x^2\right )}{2 a}\\ \end{align*}

Mathematica [A]  time = 0.0050465, size = 24, normalized size = 1. \[ c \left (\frac{\log (x)}{a}-\frac{\log \left (a+b x^2\right )}{2 a}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

c*(Log[x]/a - Log[a + b*x^2]/(2*a))

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Maple [A]  time = 0.003, size = 23, normalized size = 1. \begin{align*}{\frac{c\ln \left ( x \right ) }{a}}-{\frac{c\ln \left ( b{x}^{2}+a \right ) }{2\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.999539, size = 34, normalized size = 1.42 \begin{align*} -\frac{c \log \left (b x^{2} + a\right )}{2 \, a} + \frac{c \log \left (x^{2}\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.22329, size = 54, normalized size = 2.25 \begin{align*} -\frac{c \log \left (b x^{2} + a\right ) - 2 \, c \log \left (x\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.190086, size = 17, normalized size = 0.71 \begin{align*} c \left (\frac{\log{\left (x \right )}}{a} - \frac{\log{\left (\frac{a}{b} + x^{2} \right )}}{2 a}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(log(x)/a - log(a/b + x**2)/(2*a))

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Giac [A]  time = 1.15705, size = 35, normalized size = 1.46 \begin{align*} \frac{c \log \left (x^{2}\right )}{2 \, a} - \frac{c \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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