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

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

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Rubi [A]  time = 0.15, 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} \begin {gather*} -\frac {b^2}{2 a^2 \left (a+b x^2\right ) (b c-a d)}+\frac {b^2 (2 b c-3 a d) \log \left (a+b x^2\right )}{2 a^3 (b c-a d)^2}-\frac {\log (x) (a d+2 b c)}{a^3 c^2}-\frac {1}{2 a^2 c x^2}+\frac {d^3 \log \left (c+d x^2\right )}{2 c^2 (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.16, size = 119, normalized size = 0.94 \begin {gather*} \frac {1}{2} \left (\frac {b^2 (2 b c-3 a d) \log \left (a+b x^2\right )}{a^3 (b c-a d)^2}-\frac {2 \log (x) (a d+2 b c)}{a^3 c^2}+\frac {b^2}{a^2 \left (a+b x^2\right ) (a d-b c)}-\frac {1}{a^2 c x^2}+\frac {d^3 \log \left (c+d x^2\right )}{c^2 (b c-a d)^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 5.58, size = 303, normalized size = 2.40 \begin {gather*} -\frac {a^{2} b^{2} c^{3} - 2 \, a^{3} b c^{2} d + a^{4} c d^{2} + {\left (2 \, a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + a^{3} b c d^{2}\right )} x^{2} - {\left ({\left (2 \, b^{4} c^{3} - 3 \, a b^{3} c^{2} d\right )} x^{4} + {\left (2 \, a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) - {\left (a^{3} b d^{3} x^{4} + a^{4} d^{3} x^{2}\right )} \log \left (d x^{2} + c\right ) + 2 \, {\left ({\left (2 \, b^{4} c^{3} - 3 \, a b^{3} c^{2} d + a^{3} b d^{3}\right )} x^{4} + {\left (2 \, a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + a^{4} d^{3}\right )} x^{2}\right )} \log \relax (x)}{2 \, {\left ({\left (a^{3} b^{3} c^{4} - 2 \, a^{4} b^{2} c^{3} d + a^{5} b c^{2} d^{2}\right )} x^{4} + {\left (a^{4} b^{2} c^{4} - 2 \, a^{5} b c^{3} d + a^{6} c^{2} d^{2}\right )} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.13, size = 189, normalized size = 1.50 \begin {gather*} \frac {d^{3} \log \left (d x^{2} + c\right )}{2 \, {\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )}} + \frac {{\left (2 \, b^{3} c - 3 \, a b^{2} d\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}\right )}} - \frac {a b c - a^{2} d + {\left (2 \, b^{2} c - a b d\right )} x^{2}}{2 \, {\left ({\left (a^{2} b^{2} c^{2} - a^{3} b c d\right )} x^{4} + {\left (a^{3} b c^{2} - a^{4} c d\right )} x^{2}\right )}} - \frac {{\left (2 \, b c + a d\right )} \log \left (x^{2}\right )}{2 \, a^{3} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.02, size = 171, normalized size = 1.36 \begin {gather*} \frac {\ln \left (b\,x^2+a\right )\,\left (2\,b^3\,c-3\,a\,b^2\,d\right )}{2\,a^5\,d^2-4\,a^4\,b\,c\,d+2\,a^3\,b^2\,c^2}-\frac {\frac {1}{2\,a\,c}-\frac {x^2\,\left (2\,b^2\,c-a\,b\,d\right )}{2\,a^2\,c\,\left (a\,d-b\,c\right )}}{b\,x^4+a\,x^2}+\frac {d^3\,\ln \left (d\,x^2+c\right )}{2\,\left (a^2\,c^2\,d^2-2\,a\,b\,c^3\,d+b^2\,c^4\right )}-\frac {\ln \relax (x)\,\left (a\,d+2\,b\,c\right )}{a^3\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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