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

Optimal. Leaf size=141 \[ -\frac {b^2 (b c-3 a d) \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^3}+\frac {\log (x)}{a^2 c^2}+\frac {b^2}{2 a \left (a+b x^2\right ) (b c-a d)^2}-\frac {d^2 (3 b c-a d) \log \left (c+d x^2\right )}{2 c^2 (b c-a d)^3}+\frac {d^2}{2 c \left (c+d x^2\right ) (b c-a d)^2} \]

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Rubi [A]  time = 0.17, antiderivative size = 141, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\frac {b^2 (b c-3 a d) \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^3}+\frac {\log (x)}{a^2 c^2}+\frac {b^2}{2 a \left (a+b x^2\right ) (b c-a d)^2}-\frac {d^2 (3 b c-a d) \log \left (c+d x^2\right )}{2 c^2 (b c-a d)^3}+\frac {d^2}{2 c \left (c+d x^2\right ) (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

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]]

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 133, normalized size = 0.94 \begin {gather*} \frac {1}{2} \left (\frac {b^2 (b c-3 a d) \log \left (a+b x^2\right )}{a^2 (a d-b c)^3}+\frac {2 \log (x)}{a^2 c^2}+\frac {b^2}{a \left (a+b x^2\right ) (b c-a d)^2}+\frac {d^2 (a d-3 b c) \log \left (c+d x^2\right )}{c^2 (b c-a d)^3}+\frac {d^2}{c \left (c+d x^2\right ) (b c-a d)^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 11.97, size = 540, normalized size = 3.83 \begin {gather*} \frac {a b^{3} c^{4} - a^{2} b^{2} c^{3} d + a^{3} b c^{2} d^{2} - a^{4} c d^{3} + {\left (a b^{3} c^{3} d - a^{3} b c d^{3}\right )} x^{2} - {\left (a b^{3} c^{4} - 3 \, a^{2} b^{2} c^{3} d + {\left (b^{4} c^{3} d - 3 \, a b^{3} c^{2} d^{2}\right )} x^{4} + {\left (b^{4} c^{4} - 2 \, a b^{3} c^{3} d - 3 \, a^{2} b^{2} c^{2} d^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) - {\left (3 \, a^{3} b c^{2} d^{2} - a^{4} c d^{3} + {\left (3 \, a^{2} b^{2} c d^{3} - a^{3} b d^{4}\right )} x^{4} + {\left (3 \, a^{2} b^{2} c^{2} d^{2} + 2 \, a^{3} b c d^{3} - a^{4} d^{4}\right )} x^{2}\right )} \log \left (d x^{2} + c\right ) + 2 \, {\left (a b^{3} c^{4} - 3 \, a^{2} b^{2} c^{3} d + 3 \, a^{3} b c^{2} d^{2} - a^{4} c d^{3} + {\left (b^{4} c^{3} d - 3 \, a b^{3} c^{2} d^{2} + 3 \, a^{2} b^{2} c d^{3} - a^{3} b d^{4}\right )} x^{4} + {\left (b^{4} c^{4} - 2 \, a b^{3} c^{3} d + 2 \, a^{3} b c d^{3} - a^{4} d^{4}\right )} x^{2}\right )} \log \relax (x)}{2 \, {\left (a^{3} b^{3} c^{6} - 3 \, a^{4} b^{2} c^{5} d + 3 \, a^{5} b c^{4} d^{2} - a^{6} c^{3} d^{3} + {\left (a^{2} b^{4} c^{5} d - 3 \, a^{3} b^{3} c^{4} d^{2} + 3 \, a^{4} b^{2} c^{3} d^{3} - a^{5} b c^{2} d^{4}\right )} x^{4} + {\left (a^{2} b^{4} c^{6} - 2 \, a^{3} b^{3} c^{5} d + 2 \, a^{5} b c^{3} d^{3} - a^{6} c^{2} d^{4}\right )} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.39, size = 321, normalized size = 2.28 \begin {gather*} -\frac {{\left (b^{4} c - 3 \, a b^{3} d\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )}} - \frac {{\left (3 \, b c d^{3} - a d^{4}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, {\left (b^{3} c^{5} d - 3 \, a b^{2} c^{4} d^{2} + 3 \, a^{2} b c^{3} d^{3} - a^{3} c^{2} d^{4}\right )}} + \frac {b^{3} c^{2} d x^{4} - 2 \, a b^{2} c d^{2} x^{4} + a^{2} b d^{3} x^{4} + b^{3} c^{3} x^{2} + a b^{2} c^{2} d x^{2} + a^{2} b c d^{2} x^{2} + a^{3} d^{3} x^{2} + 3 \, a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + 3 \, 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 (b d x^{4} + b c x^{2} + a d x^{2} + a c\right )}} + \frac {\log \left (x^{2}\right )}{2 \, a^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 225, normalized size = 1.60 \begin {gather*} \frac {a \,d^{3}}{2 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right ) c}-\frac {a \,d^{3} \ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right )^{3} c^{2}}-\frac {b^{3} c}{2 \left (a d -b c \right )^{3} \left (b \,x^{2}+a \right ) a}-\frac {3 b^{2} d \ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right )^{3} a}+\frac {b^{3} c \ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right )^{3} a^{2}}+\frac {b^{2} d}{2 \left (a d -b c \right )^{3} \left (b \,x^{2}+a \right )}+\frac {3 b \,d^{2} \ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right )^{3} c}-\frac {b \,d^{2}}{2 \left (a d -b c \right )^{3} \left (d \,x^{2}+c \right )}+\frac {\ln \relax (x )}{a^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.20, size = 295, normalized size = 2.09 \begin {gather*} -\frac {{\left (b^{3} c - 3 \, a b^{2} d\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (a^{2} b^{3} c^{3} - 3 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} - a^{5} d^{3}\right )}} - \frac {{\left (3 \, b c d^{2} - a d^{3}\right )} \log \left (d x^{2} + c\right )}{2 \, {\left (b^{3} c^{5} - 3 \, a b^{2} c^{4} d + 3 \, a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3}\right )}} + \frac {b^{2} c^{2} + a^{2} d^{2} + {\left (b^{2} c d + a b d^{2}\right )} x^{2}}{2 \, {\left (a^{2} b^{2} c^{4} - 2 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2} + {\left (a b^{3} c^{3} d - 2 \, a^{2} b^{2} c^{2} d^{2} + a^{3} b c d^{3}\right )} x^{4} + {\left (a b^{3} c^{4} - a^{2} b^{2} c^{3} d - a^{3} b c^{2} d^{2} + a^{4} c d^{3}\right )} x^{2}\right )}} + \frac {\log \left (x^{2}\right )}{2 \, a^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.44, size = 193, normalized size = 1.37 \begin {gather*} \frac {\frac {a^2\,d^2+b^2\,c^2}{2\,a\,c\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}+\frac {b\,d\,x^2\,\left (a\,d+b\,c\right )}{2\,a\,c\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}}{b\,d\,x^4+\left (a\,d+b\,c\right )\,x^2+a\,c}+\frac {\ln \relax (x)}{a^2\,c^2}-\frac {b^2\,\ln \left (b\,x^2+a\right )\,\left (3\,a\,d-b\,c\right )}{2\,a^2\,{\left (a\,d-b\,c\right )}^3}-\frac {d^2\,\ln \left (d\,x^2+c\right )\,\left (a\,d-3\,b\,c\right )}{2\,c^2\,{\left (a\,d-b\,c\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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