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

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

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 126, 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} \[ \frac {b^3 \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^2}-\frac {\log (x) (2 a d+b c)}{a^2 c^3}+\frac {d^2}{2 c^2 \left (c+d x^2\right ) (b c-a d)}-\frac {d^2 (3 b c-2 a d) \log \left (c+d x^2\right )}{2 c^3 (b c-a d)^2}-\frac {1}{2 a c^2 x^2} \]

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 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^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*}

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Mathematica [A]  time = 0.24, 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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fricas [B]  time = 2.76, size = 302, normalized size = 2.40 \[ -\frac {a b^{2} c^{4} - 2 \, a^{2} b c^{3} d + a^{3} c^{2} d^{2} + {\left (a b^{2} c^{3} d - 3 \, a^{2} b c^{2} d^{2} + 2 \, a^{3} c d^{3}\right )} x^{2} - {\left (b^{3} c^{3} d x^{4} + b^{3} c^{4} x^{2}\right )} \log \left (b x^{2} + a\right ) + {\left ({\left (3 \, a^{2} b c d^{3} - 2 \, a^{3} d^{4}\right )} x^{4} + {\left (3 \, a^{2} b c^{2} d^{2} - 2 \, a^{3} c d^{3}\right )} x^{2}\right )} \log \left (d x^{2} + c\right ) + 2 \, {\left ({\left (b^{3} c^{3} d - 3 \, a^{2} b c d^{3} + 2 \, a^{3} d^{4}\right )} x^{4} + {\left (b^{3} c^{4} - 3 \, a^{2} b c^{2} d^{2} + 2 \, a^{3} c d^{3}\right )} x^{2}\right )} \log \relax (x)}{2 \, {\left ({\left (a^{2} b^{2} c^{5} d - 2 \, a^{3} b c^{4} d^{2} + a^{4} c^{3} d^{3}\right )} x^{4} + {\left (a^{2} b^{2} c^{6} - 2 \, a^{3} b c^{5} d + a^{4} c^{4} d^{2}\right )} x^{2}\right )}} \]

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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giac [B]  time = 0.35, size = 257, normalized size = 2.04 \[ \frac {b^{4} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )}} - \frac {{\left (3 \, b c d^{3} - 2 \, a d^{4}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, {\left (b^{2} c^{5} d - 2 \, a b c^{4} d^{2} + a^{2} c^{3} d^{3}\right )}} + \frac {b^{3} c^{2} d x^{4} + b^{3} c^{3} x^{2} - 2 \, a b^{2} c^{2} d x^{2} + 6 \, a^{2} b c d^{2} x^{2} - 4 \, a^{3} d^{3} x^{2} - 2 \, a b^{2} c^{3} + 4 \, a^{2} b c^{2} d - 2 \, a^{3} c d^{2}}{4 \, {\left (a^{2} b^{2} c^{4} - 2 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2}\right )} {\left (d x^{4} + c x^{2}\right )}} - \frac {{\left (b c + 2 \, a d\right )} \log \left (x^{2}\right )}{2 \, a^{2} c^{3}} \]

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)

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maple [A]  time = 0.02, size = 170, normalized size = 1.35 \[ -\frac {a \,d^{3}}{2 \left (a d -b c \right )^{2} \left (d \,x^{2}+c \right ) c^{2}}+\frac {a \,d^{3} \ln \left (d \,x^{2}+c \right )}{\left (a d -b c \right )^{2} c^{3}}+\frac {b^{3} \ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right )^{2} a^{2}}+\frac {b \,d^{2}}{2 \left (a d -b c \right )^{2} \left (d \,x^{2}+c \right ) c}-\frac {3 b \,d^{2} \ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right )^{2} c^{2}}-\frac {2 d \ln \relax (x )}{a \,c^{3}}-\frac {b \ln \relax (x )}{a^{2} c^{2}}-\frac {1}{2 a \,c^{2} x^{2}} \]

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]

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

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maxima [A]  time = 1.13, size = 188, normalized size = 1.49 \[ \frac {b^{3} \log \left (b x^{2} + a\right )}{2 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )}} - \frac {{\left (3 \, b c d^{2} - 2 \, a d^{3}\right )} \log \left (d x^{2} + c\right )}{2 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )}} - \frac {b c^{2} - a c d + {\left (b c d - 2 \, a d^{2}\right )} x^{2}}{2 \, {\left ({\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x^{4} + {\left (a b c^{4} - a^{2} c^{3} d\right )} x^{2}\right )}} - \frac {{\left (b c + 2 \, a d\right )} \log \left (x^{2}\right )}{2 \, a^{2} c^{3}} \]

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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mupad [B]  time = 0.93, size = 171, normalized size = 1.36 \[ \frac {b^3\,\ln \left (b\,x^2+a\right )}{2\,\left (a^4\,d^2-2\,a^3\,b\,c\,d+a^2\,b^2\,c^2\right )}-\frac {\frac {1}{2\,a\,c}+\frac {x^2\,\left (2\,a\,d^2-b\,c\,d\right )}{2\,a\,c^2\,\left (a\,d-b\,c\right )}}{d\,x^4+c\,x^2}+\frac {\ln \left (d\,x^2+c\right )\,\left (2\,a\,d^3-3\,b\,c\,d^2\right )}{2\,a^2\,c^3\,d^2-4\,a\,b\,c^4\,d+2\,b^2\,c^5}-\frac {\ln \relax (x)\,\left (2\,a\,d+b\,c\right )}{a^2\,c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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