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

Optimal. Leaf size=58 \[ -\frac{a^2}{2 c x^2}+\frac{(b c-a d)^2 \log \left (c+d x^2\right )}{2 c^2 d}+\frac{a \log (x) (2 b c-a d)}{c^2} \]

[Out]

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

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Rubi [A]  time = 0.0576856, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {446, 88} \[ -\frac{a^2}{2 c x^2}+\frac{(b c-a d)^2 \log \left (c+d x^2\right )}{2 c^2 d}+\frac{a \log (x) (2 b c-a d)}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

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

Mathematica [A]  time = 0.027774, size = 60, normalized size = 1.03 \[ \frac{a^2 (-c) d-2 a d x^2 \log (x) (a d-2 b c)+x^2 (b c-a d)^2 \log \left (c+d x^2\right )}{2 c^2 d x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 81, normalized size = 1.4 \begin{align*}{\frac{d\ln \left ( d{x}^{2}+c \right ){a}^{2}}{2\,{c}^{2}}}-{\frac{\ln \left ( d{x}^{2}+c \right ) ab}{c}}+{\frac{\ln \left ( d{x}^{2}+c \right ){b}^{2}}{2\,d}}-{\frac{{a}^{2}}{2\,c{x}^{2}}}-{\frac{\ln \left ( x \right ){a}^{2}d}{{c}^{2}}}+2\,{\frac{a\ln \left ( x \right ) b}{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.985696, size = 95, normalized size = 1.64 \begin{align*} \frac{{\left (2 \, a b c - a^{2} d\right )} \log \left (x^{2}\right )}{2 \, c^{2}} - \frac{a^{2}}{2 \, c x^{2}} + \frac{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (d x^{2} + c\right )}{2 \, c^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.33753, size = 159, normalized size = 2.74 \begin{align*} -\frac{a^{2} c d -{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \log \left (d x^{2} + c\right ) - 2 \,{\left (2 \, a b c d - a^{2} d^{2}\right )} x^{2} \log \left (x\right )}{2 \, c^{2} d x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.42595, size = 49, normalized size = 0.84 \begin{align*} - \frac{a^{2}}{2 c x^{2}} - \frac{a \left (a d - 2 b c\right ) \log{\left (x \right )}}{c^{2}} + \frac{\left (a d - b c\right )^{2} \log{\left (\frac{c}{d} + x^{2} \right )}}{2 c^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14002, size = 123, normalized size = 2.12 \begin{align*} \frac{{\left (2 \, a b c - a^{2} d\right )} \log \left (x^{2}\right )}{2 \, c^{2}} + \frac{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, c^{2} d} - \frac{2 \, a b c x^{2} - a^{2} d x^{2} + a^{2} c}{2 \, c^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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