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

Optimal. Leaf size=76 \[ -\frac{b c-a d}{2 a^2 \left (a+b x^2\right )}+\frac{(2 b c-a d) \log \left (a+b x^2\right )}{2 a^3}-\frac{\log (x) (2 b c-a d)}{a^3}-\frac{c}{2 a^2 x^2} \]

[Out]

-c/(2*a^2*x^2) - (b*c - a*d)/(2*a^2*(a + b*x^2)) - ((2*b*c - a*d)*Log[x])/a^3 + ((2*b*c - a*d)*Log[a + b*x^2])
/(2*a^3)

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Rubi [A]  time = 0.0740186, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {446, 77} \[ -\frac{b c-a d}{2 a^2 \left (a+b x^2\right )}+\frac{(2 b c-a d) \log \left (a+b x^2\right )}{2 a^3}-\frac{\log (x) (2 b c-a d)}{a^3}-\frac{c}{2 a^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-c/(2*a^2*x^2) - (b*c - a*d)/(2*a^2*(a + b*x^2)) - ((2*b*c - a*d)*Log[x])/a^3 + ((2*b*c - a*d)*Log[a + b*x^2])
/(2*a^3)

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0458944, size = 64, normalized size = 0.84 \[ \frac{\frac{a (a d-b c)}{a+b x^2}+(2 b c-a d) \log \left (a+b x^2\right )+2 \log (x) (a d-2 b c)-\frac{a c}{x^2}}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a*c)/x^2) + (a*(-(b*c) + a*d))/(a + b*x^2) + 2*(-2*b*c + a*d)*Log[x] + (2*b*c - a*d)*Log[a + b*x^2])/(2*a^
3)

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Maple [A]  time = 0.013, size = 86, normalized size = 1.1 \begin{align*} -{\frac{c}{2\,{a}^{2}{x}^{2}}}+{\frac{\ln \left ( x \right ) d}{{a}^{2}}}-2\,{\frac{bc\ln \left ( x \right ) }{{a}^{3}}}-{\frac{\ln \left ( b{x}^{2}+a \right ) d}{2\,{a}^{2}}}+{\frac{bc\ln \left ( b{x}^{2}+a \right ) }{{a}^{3}}}+{\frac{d}{2\,a \left ( b{x}^{2}+a \right ) }}-{\frac{bc}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.09233, size = 105, normalized size = 1.38 \begin{align*} -\frac{{\left (2 \, b c - a d\right )} x^{2} + a c}{2 \,{\left (a^{2} b x^{4} + a^{3} x^{2}\right )}} + \frac{{\left (2 \, b c - a d\right )} \log \left (b x^{2} + a\right )}{2 \, a^{3}} - \frac{{\left (2 \, b c - a d\right )} \log \left (x^{2}\right )}{2 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((2*b*c - a*d)*x^2 + a*c)/(a^2*b*x^4 + a^3*x^2) + 1/2*(2*b*c - a*d)*log(b*x^2 + a)/a^3 - 1/2*(2*b*c - a*d
)*log(x^2)/a^3

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Fricas [A]  time = 1.55478, size = 248, normalized size = 3.26 \begin{align*} -\frac{a^{2} c +{\left (2 \, a b c - a^{2} d\right )} x^{2} -{\left ({\left (2 \, b^{2} c - a b d\right )} x^{4} +{\left (2 \, a b c - a^{2} d\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) + 2 \,{\left ({\left (2 \, b^{2} c - a b d\right )} x^{4} +{\left (2 \, a b c - a^{2} d\right )} x^{2}\right )} \log \left (x\right )}{2 \,{\left (a^{3} b x^{4} + a^{4} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a^2*c + (2*a*b*c - a^2*d)*x^2 - ((2*b^2*c - a*b*d)*x^4 + (2*a*b*c - a^2*d)*x^2)*log(b*x^2 + a) + 2*((2*b
^2*c - a*b*d)*x^4 + (2*a*b*c - a^2*d)*x^2)*log(x))/(a^3*b*x^4 + a^4*x^2)

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Sympy [A]  time = 1.00116, size = 70, normalized size = 0.92 \begin{align*} \frac{- a c + x^{2} \left (a d - 2 b c\right )}{2 a^{3} x^{2} + 2 a^{2} b x^{4}} + \frac{\left (a d - 2 b c\right ) \log{\left (x \right )}}{a^{3}} - \frac{\left (a d - 2 b c\right ) \log{\left (\frac{a}{b} + x^{2} \right )}}{2 a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a*c + x**2*(a*d - 2*b*c))/(2*a**3*x**2 + 2*a**2*b*x**4) + (a*d - 2*b*c)*log(x)/a**3 - (a*d - 2*b*c)*log(a/b
+ x**2)/(2*a**3)

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Giac [A]  time = 1.16426, size = 113, normalized size = 1.49 \begin{align*} -\frac{{\left (2 \, b c - a d\right )} \log \left (x^{2}\right )}{2 \, a^{3}} - \frac{2 \, b c x^{2} - a d x^{2} + a c}{2 \,{\left (b x^{4} + a x^{2}\right )} a^{2}} + \frac{{\left (2 \, b^{2} c - a b d\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{3} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*c - a*d)*log(x^2)/a^3 - 1/2*(2*b*c*x^2 - a*d*x^2 + a*c)/((b*x^4 + a*x^2)*a^2) + 1/2*(2*b^2*c - a*b*d
)*log(abs(b*x^2 + a))/(a^3*b)