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

Optimal. Leaf size=106 \[ -\frac{a^2}{2 c^3 x^2}+\frac{a (b c-a d)}{c^3 \left (c+d x^2\right )}-\frac{(b c-a d)^2}{4 c^2 d \left (c+d x^2\right )^2}-\frac{a (2 b c-3 a d) \log \left (c+d x^2\right )}{2 c^4}+\frac{a \log (x) (2 b c-3 a d)}{c^4} \]

[Out]

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

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Rubi [A]  time = 0.114206, antiderivative size = 106, 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} \[ -\frac{a^2}{2 c^3 x^2}+\frac{a (b c-a d)}{c^3 \left (c+d x^2\right )}-\frac{(b c-a d)^2}{4 c^2 d \left (c+d x^2\right )^2}-\frac{a (2 b c-3 a d) \log \left (c+d x^2\right )}{2 c^4}+\frac{a \log (x) (2 b c-3 a d)}{c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.089042, size = 99, normalized size = 0.93 \[ \frac{-\frac{2 a^2 c}{x^2}-\frac{c^2 (b c-a d)^2}{d \left (c+d x^2\right )^2}+\frac{4 a c (b c-a d)}{c+d x^2}+2 a (3 a d-2 b c) \log \left (c+d x^2\right )+4 a \log (x) (2 b c-3 a d)}{4 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 149, normalized size = 1.4 \begin{align*}{\frac{3\,\ln \left ( d{x}^{2}+c \right ){a}^{2}d}{2\,{c}^{4}}}-{\frac{\ln \left ( d{x}^{2}+c \right ) ab}{{c}^{3}}}-{\frac{{a}^{2}d}{4\,{c}^{2} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{ab}{2\,c \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{{b}^{2}}{4\,d \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{{a}^{2}d}{{c}^{3} \left ( d{x}^{2}+c \right ) }}+{\frac{ab}{{c}^{2} \left ( d{x}^{2}+c \right ) }}-{\frac{{a}^{2}}{2\,{c}^{3}{x}^{2}}}-3\,{\frac{\ln \left ( x \right ){a}^{2}d}{{c}^{4}}}+2\,{\frac{a\ln \left ( x \right ) b}{{c}^{3}}} \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)^3,x)

[Out]

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

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Maxima [A]  time = 1.00577, size = 192, normalized size = 1.81 \begin{align*} -\frac{2 \, a^{2} c^{2} d - 2 \,{\left (2 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{4} +{\left (b^{2} c^{3} - 6 \, a b c^{2} d + 9 \, a^{2} c d^{2}\right )} x^{2}}{4 \,{\left (c^{3} d^{3} x^{6} + 2 \, c^{4} d^{2} x^{4} + c^{5} d x^{2}\right )}} - \frac{{\left (2 \, a b c - 3 \, a^{2} d\right )} \log \left (d x^{2} + c\right )}{2 \, c^{4}} + \frac{{\left (2 \, a b c - 3 \, a^{2} d\right )} \log \left (x^{2}\right )}{2 \, c^{4}} \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)^3,x, algorithm="maxima")

[Out]

-1/4*(2*a^2*c^2*d - 2*(2*a*b*c*d^2 - 3*a^2*d^3)*x^4 + (b^2*c^3 - 6*a*b*c^2*d + 9*a^2*c*d^2)*x^2)/(c^3*d^3*x^6
+ 2*c^4*d^2*x^4 + c^5*d*x^2) - 1/2*(2*a*b*c - 3*a^2*d)*log(d*x^2 + c)/c^4 + 1/2*(2*a*b*c - 3*a^2*d)*log(x^2)/c
^4

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Fricas [B]  time = 1.48117, size = 524, normalized size = 4.94 \begin{align*} -\frac{2 \, a^{2} c^{3} d - 2 \,{\left (2 \, a b c^{2} d^{2} - 3 \, a^{2} c d^{3}\right )} x^{4} +{\left (b^{2} c^{4} - 6 \, a b c^{3} d + 9 \, a^{2} c^{2} d^{2}\right )} x^{2} + 2 \,{\left ({\left (2 \, a b c d^{3} - 3 \, a^{2} d^{4}\right )} x^{6} + 2 \,{\left (2 \, a b c^{2} d^{2} - 3 \, a^{2} c d^{3}\right )} x^{4} +{\left (2 \, a b c^{3} d - 3 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \log \left (d x^{2} + c\right ) - 4 \,{\left ({\left (2 \, a b c d^{3} - 3 \, a^{2} d^{4}\right )} x^{6} + 2 \,{\left (2 \, a b c^{2} d^{2} - 3 \, a^{2} c d^{3}\right )} x^{4} +{\left (2 \, a b c^{3} d - 3 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \log \left (x\right )}{4 \,{\left (c^{4} d^{3} x^{6} + 2 \, c^{5} d^{2} x^{4} + c^{6} d x^{2}\right )}} \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)^3,x, algorithm="fricas")

[Out]

-1/4*(2*a^2*c^3*d - 2*(2*a*b*c^2*d^2 - 3*a^2*c*d^3)*x^4 + (b^2*c^4 - 6*a*b*c^3*d + 9*a^2*c^2*d^2)*x^2 + 2*((2*
a*b*c*d^3 - 3*a^2*d^4)*x^6 + 2*(2*a*b*c^2*d^2 - 3*a^2*c*d^3)*x^4 + (2*a*b*c^3*d - 3*a^2*c^2*d^2)*x^2)*log(d*x^
2 + c) - 4*((2*a*b*c*d^3 - 3*a^2*d^4)*x^6 + 2*(2*a*b*c^2*d^2 - 3*a^2*c*d^3)*x^4 + (2*a*b*c^3*d - 3*a^2*c^2*d^2
)*x^2)*log(x))/(c^4*d^3*x^6 + 2*c^5*d^2*x^4 + c^6*d*x^2)

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Sympy [A]  time = 2.1941, size = 139, normalized size = 1.31 \begin{align*} - \frac{a \left (3 a d - 2 b c\right ) \log{\left (x \right )}}{c^{4}} + \frac{a \left (3 a d - 2 b c\right ) \log{\left (\frac{c}{d} + x^{2} \right )}}{2 c^{4}} - \frac{2 a^{2} c^{2} d + x^{4} \left (6 a^{2} d^{3} - 4 a b c d^{2}\right ) + x^{2} \left (9 a^{2} c d^{2} - 6 a b c^{2} d + b^{2} c^{3}\right )}{4 c^{5} d x^{2} + 8 c^{4} d^{2} x^{4} + 4 c^{3} d^{3} x^{6}} \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)**3,x)

[Out]

-a*(3*a*d - 2*b*c)*log(x)/c**4 + a*(3*a*d - 2*b*c)*log(c/d + x**2)/(2*c**4) - (2*a**2*c**2*d + x**4*(6*a**2*d*
*3 - 4*a*b*c*d**2) + x**2*(9*a**2*c*d**2 - 6*a*b*c**2*d + b**2*c**3))/(4*c**5*d*x**2 + 8*c**4*d**2*x**4 + 4*c*
*3*d**3*x**6)

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Giac [A]  time = 1.18844, size = 239, normalized size = 2.25 \begin{align*} \frac{{\left (2 \, a b c - 3 \, a^{2} d\right )} \log \left (x^{2}\right )}{2 \, c^{4}} - \frac{{\left (2 \, a b c d - 3 \, a^{2} d^{2}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, c^{4} d} - \frac{2 \, a b c x^{2} - 3 \, a^{2} d x^{2} + a^{2} c}{2 \, c^{4} x^{2}} + \frac{6 \, a b c d^{3} x^{4} - 9 \, a^{2} d^{4} x^{4} + 16 \, a b c^{2} d^{2} x^{2} - 22 \, a^{2} c d^{3} x^{2} - b^{2} c^{4} + 12 \, a b c^{3} d - 14 \, a^{2} c^{2} d^{2}}{4 \,{\left (d x^{2} + c\right )}^{2} c^{4} 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)^3,x, algorithm="giac")

[Out]

1/2*(2*a*b*c - 3*a^2*d)*log(x^2)/c^4 - 1/2*(2*a*b*c*d - 3*a^2*d^2)*log(abs(d*x^2 + c))/(c^4*d) - 1/2*(2*a*b*c*
x^2 - 3*a^2*d*x^2 + a^2*c)/(c^4*x^2) + 1/4*(6*a*b*c*d^3*x^4 - 9*a^2*d^4*x^4 + 16*a*b*c^2*d^2*x^2 - 22*a^2*c*d^
3*x^2 - b^2*c^4 + 12*a*b*c^3*d - 14*a^2*c^2*d^2)/((d*x^2 + c)^2*c^4*d)