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

Optimal. Leaf size=99 \[ -\frac{b (b c-2 a d) \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^2}+\frac{\log (x)}{a^2 c}-\frac{d^2 \log \left (c+d x^2\right )}{2 c (b c-a d)^2}+\frac{b}{2 a \left (a+b x^2\right ) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.105449, antiderivative size = 99, 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, 72} \[ -\frac{b (b c-2 a d) \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^2}+\frac{\log (x)}{a^2 c}-\frac{d^2 \log \left (c+d x^2\right )}{2 c (b c-a d)^2}+\frac{b}{2 a \left (a+b x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

b/(2*a*(b*c - a*d)*(a + b*x^2)) + Log[x]/(a^2*c) - (b*(b*c - 2*a*d)*Log[a + b*x^2])/(2*a^2*(b*c - a*d)^2) - (d
^2*Log[c + d*x^2])/(2*c*(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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.107655, size = 97, normalized size = 0.98 \[ \frac{2 \log (x)-\frac{a \left (a d^2 \left (a+b x^2\right ) \log \left (c+d x^2\right )+b c (a d-b c)\right )+b c \left (a+b x^2\right ) (b c-2 a d) \log \left (a+b x^2\right )}{\left (a+b x^2\right ) (b c-a d)^2}}{2 a^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.016, size = 139, normalized size = 1.4 \begin{align*} -{\frac{{d}^{2}\ln \left ( d{x}^{2}+c \right ) }{2\,c \left ( ad-bc \right ) ^{2}}}+{\frac{\ln \left ( x \right ) }{{a}^{2}c}}+{\frac{b\ln \left ( b{x}^{2}+a \right ) d}{a \left ( ad-bc \right ) ^{2}}}-{\frac{{b}^{2}\ln \left ( b{x}^{2}+a \right ) c}{2\,{a}^{2} \left ( ad-bc \right ) ^{2}}}-{\frac{bd}{2\, \left ( ad-bc \right ) ^{2} \left ( b{x}^{2}+a \right ) }}+{\frac{{b}^{2}c}{2\,a \left ( ad-bc \right ) ^{2} \left ( b{x}^{2}+a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01521, size = 185, normalized size = 1.87 \begin{align*} -\frac{d^{2} \log \left (d x^{2} + c\right )}{2 \,{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )}} - \frac{{\left (b^{2} c - 2 \, a b d\right )} \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{b}{2 \,{\left (a^{2} b c - a^{3} d +{\left (a b^{2} c - a^{2} b d\right )} x^{2}\right )}} + \frac{\log \left (x^{2}\right )}{2 \, a^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 6.41115, size = 443, normalized size = 4.47 \begin{align*} \frac{a b^{2} c^{2} - a^{2} b c d -{\left (a b^{2} c^{2} - 2 \, a^{2} b c d +{\left (b^{3} c^{2} - 2 \, a b^{2} c d\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) -{\left (a^{2} b d^{2} x^{2} + a^{3} d^{2}\right )} \log \left (d x^{2} + c\right ) + 2 \,{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2} +{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{2}\right )} \log \left (x\right )}{2 \,{\left (a^{3} b^{2} c^{3} - 2 \, a^{4} b c^{2} d + a^{5} c d^{2} +{\left (a^{2} b^{3} c^{3} - 2 \, a^{3} b^{2} c^{2} d + a^{4} b c d^{2}\right )} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.19523, size = 247, normalized size = 2.49 \begin{align*} -\frac{d^{3} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \,{\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )}} - \frac{{\left (b^{3} c - 2 \, a b^{2} d\right )} \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{b^{3} c x^{2} - 2 \, a b^{2} d x^{2} + 2 \, a b^{2} c - 3 \, a^{2} b d}{2 \,{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )}{\left (b x^{2} + a\right )}} + \frac{\log \left (x^{2}\right )}{2 \, a^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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