3.177 \(\int \frac{(a+b x^2)^2}{x^5 (c+d x^2)} \, dx\)

Optimal. Leaf size=75 \[ -\frac{a^2}{4 c x^4}-\frac{a (2 b c-a d)}{2 c^2 x^2}-\frac{(b c-a d)^2 \log \left (c+d x^2\right )}{2 c^3}+\frac{\log (x) (b c-a d)^2}{c^3} \]

[Out]

-a^2/(4*c*x^4) - (a*(2*b*c - a*d))/(2*c^2*x^2) + ((b*c - a*d)^2*Log[x])/c^3 - ((b*c - a*d)^2*Log[c + d*x^2])/(
2*c^3)

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Rubi [A]  time = 0.0672466, antiderivative size = 75, 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} \[ -\frac{a^2}{4 c x^4}-\frac{a (2 b c-a d)}{2 c^2 x^2}-\frac{(b c-a d)^2 \log \left (c+d x^2\right )}{2 c^3}+\frac{\log (x) (b c-a d)^2}{c^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^5*(c + d*x^2)),x]

[Out]

-a^2/(4*c*x^4) - (a*(2*b*c - a*d))/(2*c^2*x^2) + ((b*c - a*d)^2*Log[x])/c^3 - ((b*c - a*d)^2*Log[c + d*x^2])/(
2*c^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 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^5 \left (c+d x^2\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^2}{x^3 (c+d x)} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{a^2}{c x^3}-\frac{a (-2 b c+a d)}{c^2 x^2}+\frac{(b c-a d)^2}{c^3 x}-\frac{d (b c-a d)^2}{c^3 (c+d x)}\right ) \, dx,x,x^2\right )\\ &=-\frac{a^2}{4 c x^4}-\frac{a (2 b c-a d)}{2 c^2 x^2}+\frac{(b c-a d)^2 \log (x)}{c^3}-\frac{(b c-a d)^2 \log \left (c+d x^2\right )}{2 c^3}\\ \end{align*}

Mathematica [A]  time = 0.0437776, size = 72, normalized size = 0.96 \[ -\frac{a c \left (a c-2 a d x^2+4 b c x^2\right )-4 x^4 \log (x) (b c-a d)^2+2 x^4 (b c-a d)^2 \log \left (c+d x^2\right )}{4 c^3 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^5*(c + d*x^2)),x]

[Out]

-(a*c*(a*c + 4*b*c*x^2 - 2*a*d*x^2) - 4*(b*c - a*d)^2*x^4*Log[x] + 2*(b*c - a*d)^2*x^4*Log[c + d*x^2])/(4*c^3*
x^4)

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Maple [A]  time = 0.006, size = 116, normalized size = 1.6 \begin{align*} -{\frac{\ln \left ( d{x}^{2}+c \right ){a}^{2}{d}^{2}}{2\,{c}^{3}}}+{\frac{\ln \left ( d{x}^{2}+c \right ) abd}{{c}^{2}}}-{\frac{\ln \left ( d{x}^{2}+c \right ){b}^{2}}{2\,c}}-{\frac{{a}^{2}}{4\,c{x}^{4}}}+{\frac{\ln \left ( x \right ){a}^{2}{d}^{2}}{{c}^{3}}}-2\,{\frac{a\ln \left ( x \right ) bd}{{c}^{2}}}+{\frac{\ln \left ( x \right ){b}^{2}}{c}}+{\frac{{a}^{2}d}{2\,{c}^{2}{x}^{2}}}-{\frac{ab}{c{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^5/(d*x^2+c),x)

[Out]

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

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Maxima [A]  time = 1.00865, size = 130, normalized size = 1.73 \begin{align*} -\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^{3}} + \frac{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (x^{2}\right )}{2 \, c^{3}} - \frac{a^{2} c + 2 \,{\left (2 \, a b c - a^{2} d\right )} x^{2}}{4 \, c^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^5/(d*x^2+c),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.26399, size = 213, normalized size = 2.84 \begin{align*} -\frac{2 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{4} \log \left (d x^{2} + c\right ) - 4 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{4} \log \left (x\right ) + a^{2} c^{2} + 2 \,{\left (2 \, a b c^{2} - a^{2} c d\right )} x^{2}}{4 \, c^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^5/(d*x^2+c),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.3518, size = 66, normalized size = 0.88 \begin{align*} \frac{- a^{2} c + x^{2} \left (2 a^{2} d - 4 a b c\right )}{4 c^{2} x^{4}} + \frac{\left (a d - b c\right )^{2} \log{\left (x \right )}}{c^{3}} - \frac{\left (a d - b c\right )^{2} \log{\left (\frac{c}{d} + x^{2} \right )}}{2 c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/x**5/(d*x**2+c),x)

[Out]

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

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Giac [B]  time = 1.15413, size = 188, normalized size = 2.51 \begin{align*} \frac{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (x^{2}\right )}{2 \, c^{3}} - \frac{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, c^{3} d} - \frac{3 \, b^{2} c^{2} x^{4} - 6 \, a b c d x^{4} + 3 \, a^{2} d^{2} x^{4} + 4 \, a b c^{2} x^{2} - 2 \, a^{2} c d x^{2} + a^{2} c^{2}}{4 \, c^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^5/(d*x^2+c),x, algorithm="giac")

[Out]

1/2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(x^2)/c^3 - 1/2*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*log(abs(d*x^2 + c))
/(c^3*d) - 1/4*(3*b^2*c^2*x^4 - 6*a*b*c*d*x^4 + 3*a^2*d^2*x^4 + 4*a*b*c^2*x^2 - 2*a^2*c*d*x^2 + a^2*c^2)/(c^3*
x^4)