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

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

[Out]

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

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Rubi [A]  time = 0.0301275, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {21, 266, 44} \[ -\frac{c \log \left (a+b x^2\right )}{2 a^2}+\frac{c \log (x)}{a^2}+\frac{c}{2 a \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

c/(2*a*(a + b*x^2)) + (c*Log[x])/a^2 - (c*Log[a + b*x^2])/(2*a^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 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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{a c+b c x^2}{x \left (a+b x^2\right )^3} \, dx &=c \int \frac{1}{x \left (a+b x^2\right )^2} \, dx\\ &=\frac{1}{2} c \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} c \operatorname{Subst}\left (\int \left (\frac{1}{a^2 x}-\frac{b}{a (a+b x)^2}-\frac{b}{a^2 (a+b x)}\right ) \, dx,x,x^2\right )\\ &=\frac{c}{2 a \left (a+b x^2\right )}+\frac{c \log (x)}{a^2}-\frac{c \log \left (a+b x^2\right )}{2 a^2}\\ \end{align*}

Mathematica [A]  time = 0.0138165, size = 34, normalized size = 0.83 \[ \frac{c \left (\frac{a}{a+b x^2}-\log \left (a+b x^2\right )+2 \log (x)\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 38, normalized size = 0.9 \begin{align*}{\frac{c}{2\,a \left ( b{x}^{2}+a \right ) }}+{\frac{c\ln \left ( x \right ) }{{a}^{2}}}-{\frac{c\ln \left ( b{x}^{2}+a \right ) }{2\,{a}^{2}}} \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)^3,x)

[Out]

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

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Maxima [A]  time = 0.978968, size = 54, normalized size = 1.32 \begin{align*} \frac{c}{2 \,{\left (a b x^{2} + a^{2}\right )}} - \frac{c \log \left (b x^{2} + a\right )}{2 \, a^{2}} + \frac{c \log \left (x^{2}\right )}{2 \, a^{2}} \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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.2695, size = 120, normalized size = 2.93 \begin{align*} \frac{a c -{\left (b c x^{2} + a c\right )} \log \left (b x^{2} + a\right ) + 2 \,{\left (b c x^{2} + a c\right )} \log \left (x\right )}{2 \,{\left (a^{2} b x^{2} + a^{3}\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)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.432526, size = 36, normalized size = 0.88 \begin{align*} c \left (\frac{1}{2 a^{2} + 2 a b x^{2}} + \frac{\log{\left (x \right )}}{a^{2}} - \frac{\log{\left (\frac{a}{b} + x^{2} \right )}}{2 a^{2}}\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)**3,x)

[Out]

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

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Giac [A]  time = 1.12721, size = 69, normalized size = 1.68 \begin{align*} \frac{c \log \left (x^{2}\right )}{2 \, a^{2}} - \frac{c \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{2}} + \frac{b c x^{2} + 2 \, a c}{2 \,{\left (b x^{2} + a\right )} a^{2}} \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)^3,x, algorithm="giac")

[Out]

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