3.249 \(\int \frac{1}{x^3 (a+b x^2) (c+d x^2)^2} \, dx\)

Optimal. Leaf size=126 \[ \frac{b^3 \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^2}-\frac{\log (x) (2 a d+b c)}{a^2 c^3}+\frac{d^2}{2 c^2 \left (c+d x^2\right ) (b c-a d)}-\frac{d^2 (3 b c-2 a d) \log \left (c+d x^2\right )}{2 c^3 (b c-a d)^2}-\frac{1}{2 a c^2 x^2} \]

[Out]

-1/(2*a*c^2*x^2) + d^2/(2*c^2*(b*c - a*d)*(c + d*x^2)) - ((b*c + 2*a*d)*Log[x])/(a^2*c^3) + (b^3*Log[a + b*x^2
])/(2*a^2*(b*c - a*d)^2) - (d^2*(3*b*c - 2*a*d)*Log[c + d*x^2])/(2*c^3*(b*c - a*d)^2)

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Rubi [A]  time = 0.144465, antiderivative size = 126, 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} \[ \frac{b^3 \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^2}-\frac{\log (x) (2 a d+b c)}{a^2 c^3}+\frac{d^2}{2 c^2 \left (c+d x^2\right ) (b c-a d)}-\frac{d^2 (3 b c-2 a d) \log \left (c+d x^2\right )}{2 c^3 (b c-a d)^2}-\frac{1}{2 a c^2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^2)*(c + d*x^2)^2),x]

[Out]

-1/(2*a*c^2*x^2) + d^2/(2*c^2*(b*c - a*d)*(c + d*x^2)) - ((b*c + 2*a*d)*Log[x])/(a^2*c^3) + (b^3*Log[a + b*x^2
])/(2*a^2*(b*c - a*d)^2) - (d^2*(3*b*c - 2*a*d)*Log[c + d*x^2])/(2*c^3*(b*c - a*d)^2)

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

Mathematica [A]  time = 0.236739, size = 117, normalized size = 0.93 \[ \frac{1}{2} \left (\frac{b^3 \log \left (a+b x^2\right )}{a^2 (b c-a d)^2}-\frac{2 \log (x) (2 a d+b c)}{a^2 c^3}+\frac{\frac{c d^2}{\left (c+d x^2\right ) (b c-a d)}+\frac{d^2 (2 a d-3 b c) \log \left (c+d x^2\right )}{(b c-a d)^2}-\frac{c}{a x^2}}{c^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x^2)*(c + d*x^2)^2),x]

[Out]

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

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Maple [A]  time = 0.018, size = 170, normalized size = 1.4 \begin{align*}{\frac{{d}^{3}\ln \left ( d{x}^{2}+c \right ) a}{{c}^{3} \left ( ad-bc \right ) ^{2}}}-{\frac{3\,{d}^{2}\ln \left ( d{x}^{2}+c \right ) b}{2\,{c}^{2} \left ( ad-bc \right ) ^{2}}}-{\frac{{d}^{3}a}{2\,{c}^{2} \left ( ad-bc \right ) ^{2} \left ( d{x}^{2}+c \right ) }}+{\frac{b{d}^{2}}{2\,c \left ( ad-bc \right ) ^{2} \left ( d{x}^{2}+c \right ) }}-{\frac{1}{2\,a{c}^{2}{x}^{2}}}-2\,{\frac{\ln \left ( x \right ) d}{a{c}^{3}}}-{\frac{\ln \left ( x \right ) b}{{a}^{2}{c}^{2}}}+{\frac{{b}^{3}\ln \left ( b{x}^{2}+a \right ) }{2\,{a}^{2} \left ( ad-bc \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^2+a)/(d*x^2+c)^2,x)

[Out]

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

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Maxima [A]  time = 1.17474, size = 254, normalized size = 2.02 \begin{align*} \frac{b^{3} \log \left (b x^{2} + a\right )}{2 \,{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )}} - \frac{{\left (3 \, b c d^{2} - 2 \, a d^{3}\right )} \log \left (d x^{2} + c\right )}{2 \,{\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )}} - \frac{b c^{2} - a c d +{\left (b c d - 2 \, a d^{2}\right )} x^{2}}{2 \,{\left ({\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x^{4} +{\left (a b c^{4} - a^{2} c^{3} d\right )} x^{2}\right )}} - \frac{{\left (b c + 2 \, a d\right )} \log \left (x^{2}\right )}{2 \, a^{2} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)/(d*x^2+c)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 14.2487, size = 601, normalized size = 4.77 \begin{align*} -\frac{a b^{2} c^{4} - 2 \, a^{2} b c^{3} d + a^{3} c^{2} d^{2} +{\left (a b^{2} c^{3} d - 3 \, a^{2} b c^{2} d^{2} + 2 \, a^{3} c d^{3}\right )} x^{2} -{\left (b^{3} c^{3} d x^{4} + b^{3} c^{4} x^{2}\right )} \log \left (b x^{2} + a\right ) +{\left ({\left (3 \, a^{2} b c d^{3} - 2 \, a^{3} d^{4}\right )} x^{4} +{\left (3 \, a^{2} b c^{2} d^{2} - 2 \, a^{3} c d^{3}\right )} x^{2}\right )} \log \left (d x^{2} + c\right ) + 2 \,{\left ({\left (b^{3} c^{3} d - 3 \, a^{2} b c d^{3} + 2 \, a^{3} d^{4}\right )} x^{4} +{\left (b^{3} c^{4} - 3 \, a^{2} b c^{2} d^{2} + 2 \, a^{3} c d^{3}\right )} x^{2}\right )} \log \left (x\right )}{2 \,{\left ({\left (a^{2} b^{2} c^{5} d - 2 \, a^{3} b c^{4} d^{2} + a^{4} c^{3} d^{3}\right )} x^{4} +{\left (a^{2} b^{2} c^{6} - 2 \, a^{3} b c^{5} d + a^{4} c^{4} d^{2}\right )} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)/(d*x^2+c)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x**2+a)/(d*x**2+c)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.15844, size = 347, normalized size = 2.75 \begin{align*} \frac{b^{4} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \,{\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )}} - \frac{{\left (3 \, b c d^{3} - 2 \, a d^{4}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \,{\left (b^{2} c^{5} d - 2 \, a b c^{4} d^{2} + a^{2} c^{3} d^{3}\right )}} + \frac{b^{3} c^{2} d x^{4} + b^{3} c^{3} x^{2} - 2 \, a b^{2} c^{2} d x^{2} + 6 \, a^{2} b c d^{2} x^{2} - 4 \, a^{3} d^{3} x^{2} - 2 \, a b^{2} c^{3} + 4 \, a^{2} b c^{2} d - 2 \, a^{3} c d^{2}}{4 \,{\left (a^{2} b^{2} c^{4} - 2 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2}\right )}{\left (d x^{4} + c x^{2}\right )}} - \frac{{\left (b c + 2 \, a d\right )} \log \left (x^{2}\right )}{2 \, a^{2} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)/(d*x^2+c)^2,x, algorithm="giac")

[Out]

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